An interesting comment has been placed on the 2012DA14 flyby thread by talkshop regular ‘Scute’ (Andrew Cooper) which investigates the possibility that the Russian Meteor was indeed related to the asteroid. This was dismissed at the time but Scute’s investigation of the orbital dynamics seems to raise doubt about this:
The Chelyabinsk Meteor and a possible link with 2012DA14
I think the idea of the Russian meteor being related to 2012DA14 should be resurrected. I say resurrected because the idea was so roundly slapped down by NASA within hours of the impact and never discussed again. Most of the information below was gleaned from NASA’s own JPL Horizons ephemeris for 2012DA14.
Let me begin by addressing a few myths that seemed to sew it up regarding the lack of any link between the two
Firstly, the direction of approach was not on the night side of the earth but on the day side (2012DA14 flipped under and up round the back only in the last 5 hours) and the radiant was not, as variously described, “the South Pole” or -81 degrees (implied by the above as being -81 to the night side), but at -69 degrees on the sunward side.
Secondly, the radiant had a right ascension of almost exactly 00 hours ,that is, 30 degrees east of the sun (which was at 21 hours 54 min of RA on the day) in the equatorial plane. The Russian (Chebarkul) meteor came in at 13 degrees east of the sun in local horizontal coordinates.
Thirdly, the incoming trajectory of the meteor was not north-south but on an azimuth of 99 degrees i.e. 9 degrees south of east. Since it was sunrise this meant that the meteor came from a direction close to the sun (13 degrees east of it), in other words, coming in over a great circle running down the globe to the south, although a better approximation would be south east, This was possible because the Earth’s axis was tilted back by 12.5 on that date, making a late sunrise for Chebarkul, so watching the sunrise on a somewhat tighter, northern latitude line meant looking along a straight line that soon scribed south eastwards in lower latitudes (rather than curving round the 55 degree North line).
Fourthly, 2012DA14 was not going “too slow” for a related fragment to arrive at 17km/second: its radiant, relative velocity to the Earth before being accelerated was 12600mph. That is 5.6 km/ sec. If you add to that the freefall velocity of 11.2 km/sec (the corollary of escape velocity) you get 16.8 km/sec. Add to that the eastward rotation of the earth at 55degrees north at an Azimuth of 9 degrees south of east (0.2 km/sec) you arrive at precisely 17km/sec. This is the same calculation that Zuluaga and Ferrin (and now, NASA) must have done in reverse for their version of the reconstruction of the trajectory: I calculated the radial speed of their hypothesised orbits at the Earth’s position (r value/ radius from sun=1AU) on the day of impact (but without the Earth’s gravitational influence added) and ended up with 34.8 and 35.2 km/sec for the 2 posited orbits. That amounts to 5 and 5.4 km/sec relative to the Earth, respectively. Adding the freefall velocity and the eastward rotation you get 16.4 and 16.8km/sec. The difference between these posited orbits and the posited 2012DA14 fragment is that they invoke the head-on trajectory solution with little or no curvature as they are pulled into the gravity well. If it’s a bulls-eye hit the curvature is zero. The Zuluaga and Ferrin video shows the meteor coming in from about 3 degrees above the solar plane. The NASA video now shows the same.
I believe there are multiple solutions between head on (3 degree inclination) and an 11.6 degree inclination (11.6 deg being the solar plane analogue to the -69 degree radiant from a geocentric view). These various solutions involve increasing degrees of curvature as the meteor is pulled into the Earth’s gravitational well but curvature in freefall doesn’t make the final velocity any slower. By the way, none of these high curvature scenarios would involve capture in orbit- it’s either impact or escape along a hyperbolic path back out of the well. And high curvature means a 50 degree ‘q angle’ (half way through the turn). It’s the q angle that determines how far round to the north the meteor can curve and still impact rather than miss and escape.
The last myth is that at 55 degrees latitude Cherbarkul is too far North for any fragments to hit. This is not true because from a -69 degree radiant, the ‘equator line’, or tangent line, that 2012DA14 could see from below was shunted upward (from the real equator) on the sunward side by 21 degrees. Due to the vagaries of the trajectory of any purported fragment (see below), it would still require around 55 degrees of hyperbolic turn from that raised equator line onwards, although that describes a track all the way to Chebarkul- atmospheric entry would begin at 50 degrees round. For an idea of the sort of curvature needed for the fragment, I plotted a curve from first principles (subtracting the freefall component of velocity from the baseline, straight velocity of 17km/sec) and came up with a q angle of 50 degrees. For some perspective, the Apollo missions came in on a hyperbolic orbit at 11.2 km/sec and still got 69 degrees round the back of the globe, regardless of rotation and a sedate 1080 mile reentry. Even 2012DA14 curved 30 degrees or more for a q angle of 15 deg and that was under the influence of one tenth the gravity. 50 degrees is probably an upper limit but that is exactly what is required for a 2012DA14 fragment riding along on the radiant angle to turn in hyperbolically and hit Chebarkul at a low trajectory.
One other point, though not classifiable as a myth is that 2012DA14 is being characterised by NASA as a CO or CV type chondrite (carbonaceous with calcium and aluminium inclusions) based on spectral observations whereas they say the Chebarkul meteor is a stony chondrite because the few pieces found so far “are reported to be silicate rich”. This, they say, rules out any link. This may be true but the whole tenor of their delivery is one of running scared and has to be looked at in the light of the following quotes, all within a minute on one video:
[re 2012DA14] “there was no danger of a collisions, NASA assured people”
“… In a one in a million chance that still has NASA scientists shaking their heads”.
“These are rare events and it’s incredible to see them happen on the same day.”
As things stand as of 5th March 2013 , I feel that the evidence presented here is more convincing than some spectral measurements not chiming with a few reported silicate bearing fragments.
The radiant is on the right at 00hours 30M. The purported fragment didn’t have to be on this line , just slightly displaced and parallel to it a few hours back up range at say 80-100,000 miles.
So we now have a completely different, sunward radiant of -69 degrees of declination. In fact, for a fragment travelling 50,000 miles sunward from the track line of 2012DA14 and 16 hours ahead, it would probably come in from nearer to a -66 to -67 degree trajectory before being really hiked round on its hyperbolic orbit (its close encounter-to-impact trajectory). 2012DA14 itself turned in by 2-3 degrees from its side of the track in the last day before starting its 30 degree hyperbolic turn. A -66 or -67 incoming trajectory for the hypothetical fragment means the raised ‘equator line’ was up to 24 degrees higher on the sunward side, reducing the required curvature even further, possibly to 47 degrees before atmospheric entry. Indeed, that sunward fragment track, displaced as it is seemingly arbitrarily, at 50,000 miles over and parallel to 2012DA14 actually allows for 10-12,000 miles or 2 degrees of inward curvature from 03:00 on 14th February, 24 hours up range. This is before being really hiked in from a -66 to -67 degree point at about 22:00 UTC on February 14th, some 5 hours from impact.
You can download the ephemeris for 2012DA14 from JPL horizons website (click ‘web interface’ and enter ‘observer’ ‘geocentric’ and dates from 1st Feb 2013 to 28th Feb 2013 at hourly intervals). I suggest the entire month so as to give a better feel of what’s going on. It’s easy to scroll up and down quickly. It doesn’t show speeds but I got the figure of 12600 mph from a news item and checked it against the orbital speed and inclination of 2012DA14 for the vertical component, then derived the horizontal geocentric component from that. Those vectors do pretty well add up to a 12600mph, 69 degree slope until one or two days out (300k to 600k miles). The JPL video has 2012DA14 at about 13,500 mph at 4 hours out. I apologise for the lack of links, they are currently playing havoc with my formatting. I might do another separate comment with some links.
I have described the trajectory of the hypothetical fragment. Now I need to describe where to look for it in a forensic sense- using astrodynamics software to rerun different scenarios with the meteor exiting from a narrow window around the proposed path. If anyone here has astrodynamics software, please feel free to join in and prove it one way or the other. I have no software so feedback would be welcome.
The best place to look for the fragment is emerging from a window bounded by a geocentric declination of between -59 and -65 degrees and a right ascension of between 22H and 2H 30 minutes. The declination angles of the window aperture are less than the radiant and trajectory angle because the fragment is now cutting up through the angle lines, past the radiant, trying to get level with the earth in the same way that 2012DA14 cut the other way through the angles to come up the back: the radiant becomes irrelevant at this point and that’s why it was totally misleading to talk about a -81 degree radiant.
Also, because the fragment is cutting up through the declination angles, its own trajectory angle cannot be described with declination angles any more- except for a bulls eye geocentric hit. It has to be aiming over the top of the Earth to get a chance of being pulled in hyperbolically over the ‘new’ equator line for a hit. This means that whatever declination angle is chosen for the instantaneous position of the emergence of the fragment through the window, its actual trajectory angle would need to be greater by 3 to 6 degrees or so in order to aim it away from the geocentre to two or three thousand miles out from the equator line, that is, two or three thousand miles from the disc it sees above it and with the same right ascension as the RA of the fragment emerging at the window. This angle cannot be geocentric because it has to look as if it will miss the Earth. Looked at another way, it is a line running parallel to a declination angle that starts 3 to 6 degrees in from the fragment emergence point, measured radially inward. This jiggling of inputs for the fragments own inclination is nevertheless bounded by the upper limit of 69 deg, the true radiant angle. There is also potentially a small amount of leftward (solar plane y axis) component as they emerge round towards the 2H 30 mark because these are trying to pass by the side of the Earth but get caught. I think these, around the 0 to 1H mark (0 to 30 degrees of right ascension) are the best candidates.
That arrival window describes a curved slit sitting somewhere roughly below where the southern tip of New Zealand was at the time of impact. The fragment would be emerging at a trajectory angle of between -65 and -69 degrees of declination (that is, its own path angle as opposed to the box perimeter angles). This needs to be set at 10 hours 20 mins (clock time) up range (about 150k to 170k miles?) so that the fragment would be coming through it at 17:00 UTC on February 14th 2013. Some time not long after that the trajectory angle would start to curve in noticeably on its hyperbolic path.
I extended the window round to 2H 30M because of the angle discrepancy between the radiant and the final trajectory. This was up to 17 degrees when playing with equatorial-to-horizontal coords for Chebarkul (but one-way calcs made this inexact) and only 8.5 degrees using old fashioned cotton stuck to a globe.
Incidentally, using the globe method resulted in a view, looking downrange along the trajectory, identical to the several Youtube videos of 2012DA14 when freeze-framed at 3:20 UTC on 15th February. It was the view from DA14 as the Chebarkul meteor hit- you can see the beginning of the meteor’s ground track over Taiwan before it disappears over the ‘equator line’ at 22 deg north. You can see that its hyperbolic trajectory extends round and down past and almost parallel to you to your right, in other words, the same trajectory but displaced 50k miles to the right. This ‘almost parallel’ track would correspond to the sunward track of the fragment not being quite sunward but with a slight right ascension from that track of a few thousand miles so that the fragment came from further round towards 2H 30 but up and in on the correct line. This allows for the 8.5 degree disparity.
It should be said here that 2012DA14 seems to have corkscrewed itself by between 5 and 12.5 degrees depending on where you make the cutoff between true radiant and local approach. It is apparent in the ephemeris (you can’t see its subtlety in any video or diagram). It may seem to defy physics (not orbiting on a great circle) but I would ascribe it to the slowing down 2012DA14 as its orbit goes from inside track to outside track and its relative prograde motion with respect to the Earth went nearly to zero. It was almost stationary in the prograde vector, sitting at -86 declination and ready to be plucked up along the most convenient longitude line. Because there was some forward motion still, it did get plucked up pretty well opposite but not 180 degrees- the entire pass was 4 or 5 degrees inside one hemisphere, resulting in a 5 to 12.5 degree twist. If this happened to the proposed Chebarkul fragment, it would have skewed in exactly the necessary way to bring it round to this new ‘wrong’ 8.5 degree-off trajectory and heading over Taiwan, China, and into Chebarkul. This is why the search box extends round to 2H 30M. It’s because the fragment oversteps ever so slightly before skewing. For that reason, once you get round to 00H to 2H, the trajectory emerging from the box will have a sideways component to the left (positive RA) ie not radially inward. These candidates will be skewing 8-10 degrees anticlockwise as you look down on them as they rise and will probably do so over the last 80,000 miles. This is how their local ‘radiant’ is skewed round. That’s if DA14 is anything to go by.
The fragment would be travelling at about 13,000 mph through the window but it would be best to plug into the software ephemeris details for 2012DA14 for the speeds from far out so as not to start at an arbitrarily high speed. However, beware of piggy-backing on 2012DA14 data to extrapolate hypothesised fragment data. I have seen, among other things, a rather amusing graph that relied on like-for-like parallel trajectories with earth skimming fragments refusing to bend round under the influence of 10 x the acceleration 2012DA14 was experiencing. That is their ‘proof’ that the fragments can’t make it north of the equator. That was on a respected astronomy blog.
If all goes to plan, you should see all manner of near misses, flying over Asia and Russia, a few direct hits on the Southern Hemisphere in a wide band from New Zealand northwards and a few Northern Hemisphere hits, one of them right on Chebarkul. You’ll have to play with it though, maybe venture a little way outside the window if needs be. The worst that can happen is some very interesting near misses- but I really do think there will be hits. This need for playing around is reflected in the fact that some solutions could imply the need to add 5 or 10,000 miles to the track line displacement further uprange (back down the slope) for a possible 60,000mile or so displacement. You can’t use the 2012DA14 trajectory, displaced and pasted to the other track- it’s bending the wrong way from 24 hours out. Even a mirror image wouldn’t be faithful due to its greater radial displacement.
THE POSITION OF THE FRAGMENT IN ORBIT- THE BIGGER PICTURE:
I will acknowledge that when NASA said that the Chebarkul meteor was not following along the same path as 2012DA14 they were right but only in a highly technical sense. Any fragment passing within a radius of a few thousand miles of the trajectory would not have impacted the other side of the Earth. But when considering the possibility meteor showers, you have to think in terms of millions of miles, even for asteroidal showers such as the Geminids and the Quarantids, because the Earth takes days to travel through them. I think NASA was clutching at straws. You have to look at the bigger picture and besides, the proposition as put forward in this comment isn’t played out on a vast scale in solar system terms.
I propose a fragment riding just 200,000 miles above 2012DA14 (north with respect to the solar plane). Just stating that baldly might understandably invite querying as to why they should be related. However, visualising it scaled down, it would be the equivalent of two tiny pieces of rock on an orbit 54 metres round, mirroring each other’s every move in speed, inclination and eccentricity, all the while staying exactly 2 centimetres apart, one directly above the other. I would consider those two pieces as related, one broken off from the other.
When the 65,000 mph prograde element of the Earth’s orbit is removed we get the geocentric element of relative movement between 2012DA14 (along with its hypothetical fragment) and Earth. Gone is the gentle 10 degree inclination with respect to the Earth’s orbit, with the asteroid climbing gently up a slope and sedately past the night side. It’s turned into a precipitous 69 degree climb, skewing round to vertical as 2012DA14 was apparently dragged up from under the South Pole and slung shot vertically above. That does serve a purpose for geocentric calculations and visualisations but it is as well to remember that it helps to plug mentally into the elongated, gentle slope version from time to time so as to get a good feel for what is really happening to the Asteroid and the purported fragment as they pass Earth.
Once the frame of reference snaps to geocentric you see the fragment rising up ahead of 2012DA14 and slightly to one side and they start to look a little disjointed. But when you snap back to solar system view with them both sailing along, rising up alongside Earth, you see that one is directly above the other: the 50,000 mile displacement is really just a 200,000 mile vertical displacement which, when looking down the -69 degree radiant makes them appear to be 50,000 miles apart. That said, there may be a fraction of further displacement to the outside of the orbit track too to allow for the skewing effect. They only ride on different tracks because of their vertical displacement and those two tracks went either side of the Earth. One was too close and the fragment hit (hypothetically). There’s something telling about that vertical nature of the relationship: the proposed fragment is following the exact same track but directly above. If it had been shifted a few thousand miles long ago, perhaps by a small collision, then previous close encounters, passing beneath the Earth would have widened that gap quite dramatically: 0.01mm/sec^2 differentials in gravitational acceleration add up to 4m/sec over a four day encounter within a million miles. That’s tens of thousands of miles per year. 200,000 miles isn’t as far as it seems.
I doubt if its possible but old sky scans might show up the culprit: at 12 million miles it would be 15 minutes of arc displaced from 2012DA14 when looking straight down the 69 degree track and would be offset at around the ’9 o’clock’ mark. For the 2012 pass at 6 million miles it would have been around one degree offset above at about the 12 o’clock mark.
Happy hunting!
Scute
Tallbloke's Talkshop 
Andrew,
I think we can call that a success
Well, we hit the planet. But did it land where you expected it to?
I’ve been wondering how you advanced 7 hours up the radiant. Because the way you do it seems different from the way I do it. You just asked for AC4’s vectors at two different times. But I would have wanted the Earth’s vectors on those two times as well – because when AC4 has moved forward 7 hours, it has moved up the radiant towards the Earth, But it’s on the radiant from the Earth that is 7 hours ahead, and what we want is for it to be on the radiant of the Earth now – i.e. 8 Feb 0:00.
So what I would do is to find the relative x,y,z distance of AC4 from the Earth on 8 Feb 7:44, and then use that relative distance to find the advanced position along the radiant towards the Earth on 8 Feb 0:00. And the same has to be done with the relative velocity of AC4 with respect of the Earth.
I’ll work this out tomorrow, after I’ve got hold of Earth’s position on 8 Feb 7:44. It’ll be interesting to see how the result compares with your figures.
I’m interested because today I’ve started creating a line of rocks up the radiant from DA14, each one 12 hours along from the last one. I’m doing this so that if I want to advance, say, 3.2 days, I can go up the line of rocks and interpolate between them to get a fairly good DA14 advanced location, from which I can construct new impacting rocks that are advanced or retarded by several days (e.g. the Cuban fireball was about 3 days advanced).
Frank,
I agree with all you say. I got the relative Earth position on 00:00:00 8th Feb by calculating the approach vector in xy and multiplying by the advance time. It should be accurate to about 50 km I think because the approach vector in xy hardly changes (radiant angle is rock solid at that time) but I’ll grant you, getting Earth’s position at 7:44:33 on the 8th would be a bit more accurate so if you want to supply that, I can use it when I calculate it later.
It’s the velocity vector inaccuracies that will make more of a difference in the final calculation I think.
Andrew
…oh and I should explain, the main reason for getting AC4 vectors for 7:44:33 on 8th Feb was to get the increase in speed in each axis which was around 2 metres per second from 00:00:00.
Andrew,
I took your AC4 advanced (now called AC4-a744) and moved its start z position upwards until it was just trimming the Earth. This body (which I’ve called AC4-b744) impacts at 54.5 degrees north at 3:08 am on 15 feb 2013, as you can see here. The red portion of the trail is inside the Earth.
I had a shot at advancing AC4 by 7 hrs and 44 minutes but screwed up the numbers somewhere, and so just used your figures.
I also now have a line of rocks, each separated by 12 hours, with DA14 in the middle of them, which proceed in line ahead formation and each of which make a closest approach of about 30,000 km at 12 hour intervals. I should be able to use this line of milestone rocks to position start locations for Cuban and other fireballs.
Andrew,
Here are Earth locations (km-s) from JPL for 7:44 and 7:45
2456331.822222222, A.D. 2013-Feb-08 07:44:00.0000,
-1.124265131791330E+08, 9.542108858143042E+07, -1.020506229326847E+04,
-1.982121334476033E+01, -2.277816795316002E+01, -7.922912810874160E-05,
2456331.822916667, A.D. 2013-Feb-08 07:45:00.0000,
-1.124277024435649E+08, 9.541972188417424E+07, -1.020506704173208E+04,
-1.982093438629824E+01, -2.277840725221657E+01, -7.905283311568714E-05,
Frank,
Here are some new state vectors for AC4- (advanced). I did it the other way, comparing distance to Earth at 00:00 and 7:44 on 8th Feb. the Y position vector was only 15km off but the x was 1000km or so different, so one of those pesky JPL numbers may have jumped a point or two. However, a larger x may bring it in for all I know.
When I was out I suddenly thought I must have left the 4m/sec z speed increase off the advanced fragment’s vectors because it would account for the shortfall. But I had put it on so I have no real differences here except for the x position vector and 5-6 centimetres per second in speeds for x and y (20-30km over a week).
Adjusted state vectors for AC4- advanced on 00:00:00 8th Feb 2013: km and km/sec.
x : -1.10546484 E8
y: 9.4713718 E7
z: -2.636450 E6
vx: -2.2071067 E1
vy: -2.0531067 E1
vz: 4.174 E0
Your rock train idea sounds interesting. It reminds me that you said you might be able to fire rocks backwards with your new computer. What about firing AC4 backwards and seeing where it passes Earth in 2012. Even if it were out by 5 million km it would require little adjustment to make it pass as it ‘should’ in both years.
Andrew
Andrew,
I pasted in your new values, and your latest rock also smacked into the South China Sea almost exactly like the last one. I then lifted it up the z axis so that its initial z position became -2.632250E+06 (all other vectors unchanged) so that it just grazed the Earth.
It seemed almost indistinguishable from the previous 54.5 degree N grazer, but I took the two images and superimposed them on each other, and this showed that they were coming down slightly different lines at a slightly different angle (in the x-y dimension anyway). The newest one is the left hand line, with the short impact trail.
Andrew,
Since your latest suggestion was coming in at a more acute angle than the last one, after you’d changes x,, y, vx, vy very slightly, I went and did the opposite to a greater extent.
The result was rather pleasing after I’d managed to get it to just graze the Earth, in that it was now coming in at a much shallower angle and grazing Finland (very far north!), but 6 hours later than intended.
Here are the vectors I used, if you’re interested:
x, y, z (km) -1.105041E+08, 9.473875E+07, -2.724550E+06,
vx, vy, vz (km/s) -2.2061067E+01, -2.0631067E+01, 4.174E+00,
Frank
Yes, that is interesting. You are starting to circularise the orbit and so it is going to come in further east. It is also likely to steal a degree or two more latitude because as it is pushed further to the outside track it sees further up the latitudes because of the tilt of the Earth. I noticed you dropped the z significantly and yet still grazed at 61N which you did before but only with AC4’s much higher z.
If it is to be truly circularised the velocities need to be adjusted so that they are coming in from a degree or so different in xy. And the xy position on the 8th for such a fragment would need to be about, and I’m half guessing, 350,000km further out. If I use my Earth position to AC4, I get 247,605 km further out than AC4 in -x and 249,918 in +y.
Do you want me to do it properly with the state vectors and velocity change to get the one degree? In the meantime, if you added the above values to AC4 744 (a or b, the one that hit) keeping z the same I would hope it would pass behind the Earth. Then a tweak less in minus vx (less negative) and more in minus vy (more negative) would swing it round on a truly circularised orbit. I say “less” and “more” because that’s what it is in absolute scalar speed.
Andrew
Frank
I note that, looking again at your Finland hit you did indeed decrease x and increase y which is very interesting. It also makes sense that they were both adjusted by the same degree (it’s a near 45 deg angle to xy axes on 8th Feb- see below). I was thinking somewhere along the lines of a 40m/sec adjustment for x and y for the 350,000 km displacement but haven’t worked out quite how to do it yet. Your numbers, scaled up, suggest 60 or 70 m/sec. The two x and y adjustments would be almost equal in magnitude, as were yours, because DA14 and AC fragments were at 47.1 deg to the y axis on 00:00:00 on 8th Feb 2013. For a one deg change, it is changing from a 47.1 deg angle to the y axis to a 46.1 deg angle.
Andrew
Andrew,
Do you want me to do it properly with the state vectors and velocity change to get the one degree?
I would prefer that, given my propensity to misinterpret what you want.
I’m not entirely sure what you are trying to do at the moment. You seem to be focused on the Chelyabinsk rock. Are you trying to circularise the orbit of a candidate rock? Are you hoping that this will bring the rock along a path that comes in just south of east at Chelyabinsk? Do you think you will be able to do so?
I fooled around a bit today, adjusting orbits, but in the end decided that I didn’t know what I was trying to do, or how to do it. So I’m inclined to leave it to you to circularise orbits and so on, and I’ll just test your suggestions.
I now have a chain of 17 milestone rocks which represent DA14, and 8 rocks advanced by 12 hour intervals, and 8 rocks retarded by 12 hour intervals. Up until now, we’ve based everything on DA14 8 feb 2013 0:00. What I’m hoping is that I can use these 17 milestones to find fairly accurately where DA14 will be at any time between 4 days ahead of it, and 4 days behind, relative to the Earth, by interpolating between positions. This now seems to be working, and I can also add offsets to these positions. So that if I want to consider a rock that’s 39.7 hours ahead of DA14, I can find where DA14 will be, relative to the Earth, and offset by x,y,z values from there. I’m hoping that this will make it much easier to explore the other fireballs (and perhaps Chelyabinsk as well), by sliding up and down the milestones to get the right time and date..
Frank
Yes, I am more focused on the Chelyabinsk rock. For me it’s like a talisman rock. The reason for this is that it is so well documented and the others so poorly documented that it emerged as the only one for which I can do any meaningful calculations. I suppose my ultimate goal is to have a candidate rock which lands in Chelyabinsk at the right azimuth and atmospheric altitude and can be traced back down an orbit which passes Earth a year earlier, most likely across its bow, to circularise it.
One other reason for focusing on circularisation is that a few comments back you mentioned the approach angle for all these rocks and that it was a little improved but nowhere near enough to be at the right azimuth for Chelyabinsk. The inclination change was not going to help on that front because it was all in z. That would only improve the reach northwards. You alluded to my not being too worried about the 20 or so degrees needed to pull these hits round east of the sun. I can’t say I’m totally sanguine about clawing back that much without losing the DA14 narrative but I thought maybe now is the time to strike out more boldly across xy and find that one degree or so of crossover angle which I hope will go a long way towards it. If the crossover angle is reduced, the orbit is more circular and the fragment must approach from further round towards the evening terminator.
“Are you trying to circularise the orbit of a candidate rock?”
Yes. It would be a candidate rock based on AC4-b744. That hit at 54.5 N at around the right time. It also has the 7.83 deg inclination albeit with a tiny z value added to the previous AC.
“Are you hoping that this will bring the rock along a path that comes in just south of east at Chelyabinsk?”
Yes, but with these milestones as priorities in chronological order:
1) Send a rock on a close pass at 6-12 degrees east of sun (proof of concept)
2) Land a rock anywhere on Earth at 6-12 degrees east of sun
3) Land a rock anywhere along 55 N at 6-12 deg east of sun.
4) Land a rock near Chelyabinsk at 6-12 deg east of sun.
5) Achieve number 4 with correct atmospheric altitude and time of entry.
We may achieve two of these in one go. We may not achieve number 4 easily or perfectly but it is number 3 which proves the process entirely because we know it is reproducible for Chelyabinsk. Number 5 would be a bonus. By far the biggest signature in the whole excercise is number 1 because if there is no angle change at all it is pointless continuing.
“Do you think you will be able to do so?”
My gut says it is definitely possible for such a rock/orbit to exist coming in from that area. A slight concern is aiming it with the right vx and vy and getting really close on the first attempt. But when all is said and done we now have a track record of my supplying vectors which contain the narrative, getting close to target and you refining them without losing the narrative. For example AC4-b744 is probably only 0.02 deg off its original intended inclination of 7.83 and definitely no more than 0.1 deg. So your adjustments kept the narrative completely intact.
I like your chain of rocks approach. It speeds up the process like factory templates, almost. Why didn’t you do what you did for the 2012 rock train (48 either side, was it?) where they were an hour apart? Or is that pointless if you can offset the time and position well enough with 12 hour gaps?
Andrew
Andrew,
Yes, I am more focused on the Chelyabinsk rock. For me it’s like a talisman rock. The reason for this is that it is so well documented and the others so poorly documented that it emerged as the only one for which I can do any meaningful calculations.
I agree that it’s well documented. For most of the others we have single reports, more or less. But there are nevertheless places and times and dates, so I will be interested in seeing if I can get rocks to come down there and then. I’m thinking of doing it very roughly, simply to show how a DA14 companion might have approached.
I can’t say I’m totally sanguine about clawing back that much without losing the DA14 narrative
Another narrative may emerge. If it proves to be possible to land rocks in all the places reported, we’ll end up with a partial map of the rock cloud. That might tell us what shape it has, and what the distribution of rocks in it might be like. Such clouds might well have their own dynamic behaviour, with outlying rocks orbiting around the cloud (I was very interested when I modelled the little rock cloud that got stretched as it made a close approach to the earth that the rocks in it were bunching together before that happened).
It may be best to forget about the ‘DA14 narrative’ until you have something that lands on Chelyabiinsk, and then see how it might relate to DA14. i.e. not feel too constrained by the narrative at the outset.
Why didn’t you do what you did for the 2012 rock train (48 either side, was it?) where they were an hour apart? Or is that pointless if you can offset the time and position well enough with 12 hour gaps?
With the 2012 rock train I was simply putting rocks along DA14’s orbit, ahead of it and behind it, and wasn’t concerned with the Earth’s relative position. With the new rock train I’m advancing up the radiant towards the Earth, and calculating their positions relative to the Earth in the manner we discussed a day or two back -.i.e. finding the positions and speeds of rocks relative to the Earth in several hours time, and using these numbers to offset them from the Earth on 8 feb 2013 0:00. It’s a more laborious set of calculations.
Also, I wanted the 2012 rock train to be fairly densely populated – one every hour – so that I could watch the behaviour of the train as the Earth passed near it. The new train is just intended to be used to roughly position rocks (like Halo1093 or AC4) along the radiant.so that they’re arriving at about the right time. So if, say, the Ufa fireball was 2 days before DA14’s closest approach, I will be able to go 2 days along the new train to position test rocks. With a bit of an effort, I might also be able to say how I want it to be approaching the Earth (e.g. above or below, left or right or centre). But I haven’t worked out that bit yet. Anyway, it should speed up testing.
Anyway, I look forward to your suggestions for new rocks. I may be a bit busy over the next week or so, but not so busy that I won’t be able to try out a few things whenever you have some figures..
Frank
I’ve made good progress on the circularised orbit. I’ve worked it all out as usual, both position and adjusted xy velocities for the new angle. I’m holding back though because I’m working out the absolute distances of the Earth and the new circularised fragment from the sun at 00:00 on the 8th. From that I get their absolute velocities so as to give the new fragment a bit more cred by adjusting its state vectors a bit more.
I should have something for you tomorrow, Sunday, evening.
Andrew
Frank
I’ve done two circularised fragments. The first is heavily based on AC4b744. For the second I went back to the drawing board, working with a slightly altered eccentricity to reflect the true qualities of the whole orbit rather than a snapshot from the 8th-15th Feb. Goodness knows if it will work but it looks right and seems to be pointing the right way. I did notice an issue with the orbit angle at the end (based on 365.25 days per year and not 325.2563…) which possibly affects it coming in 9 minutes early. That could mean 3000km but it’s easy to slide up and down the orbit to adjust.
First circularised fragment (km and km/sec):
x: -1.10871621 E08
y: + 9.5043037 E07
z: – 2.632250 E06
vx: -2.158286 E 01
vy: -2.104350 E 01
vz: +4.174 E 00
Second circularised fragment (km and km/sec):
x: -1.10899997 E 08
y: +9.5070692 E 07
z: -2.637727 E 06
vx: -2.162482716 E 01
vy: -2.107547255 E 01
vz: +4.174 E 00
I think the most important thing is to see what the incoming angle is, whether it hits or not. So, if you could measure that or do a snapshot from a little way out where it’s not curving it would be good. I’ve calculated for 12 degrees east of sun in both cases.
Good luck!
Andrew
Frank,
I misread a number that shoved that second one 8000km too far back down the orbit. I can redo the position vectors tomorrow. In the meantime you’re welcome to try it and see if it crosses the ecliptic 8,000km behind! The velocities remain unchanged. But it does highlight how easy it is to miss something like decimal points etc.
Andrew
Andrew,
I named the first one AC6a. It doesn’t seem to come in on a noticeably different angle (in the x-y plane anyway) than other rocks. This snapshot shows the approach from distance. My new DA14 milestone rock train is off to the left, sweeping round the Earth. AC6a is on the right, and performs a fairly sharp turn after impact, with relative-z briefly turning positive.
AC6a impacts, but in the southern hemisphere.
Impact: 15 Feb 2013 01:25:36 48.4 degrees South between Antactica and South Africa, coming in approximately W – E. Not a graze.
impact relative rx, ry, rz (m) -4213920.682937622 -2199047.624847412 -4228950.9727625605
I’ll try the second one, AC6b a bit later.
AC6b, 1000, 0.01, 2456331.500000000, A.D. 2013-Feb-08 07:44:00.0000, -1.10899997E+08, +9.5070692E+07, -2.637727E+06, -2.162482716E+01, -2.107547255E+01, 4.174E+00,
AC6b closest approach: 4.8998556E7 m on 15 Feb 2013 01:57:20 No impact.
Approach path shows AC6b initially to the right of the DA14 milestone train, but left of it at closest approach.
Frank
Thanks, that was very informative if a little disappointing in some ways. I’ll do a longer comment later but in the meantime here are the revised state vectors for the second fragment, AC6b.
One important point is that it needs to be fired at 00:00:00 on the 8th Feb. Your details show you fired it at 07:44 so what I have below should come in closer. I should have stated the time along with the vectors. Apologies for that.
Revised state vectors for AC6b km and km/sec:
x: -1.10848921 E08
y: +9.5065161 E07
z: -2.636632 E06
vx: -2.162482716 E01
vy: -2.107547255 E01
vz: +4.174 E00
To be fired at 00:00:00 on 8th Feb 2013.
I realise there’s a question mark over the first attempt not coming in way round east like I’d hoped but it would be good to see what these revised vectors do. This is because it was a revamped eccentricity, absolute velocity and approach angle and it’s gratifying it came as close as it did. So with the 00:00 firing time and the position vectors where they should have been in the first place, it may get even closer. Any residuals would be good feedback for highlighting errors in initial conditions assumptions and for the consequent adjustments I would need to make in any future calculations.
Thanks
Andrew
Andrew,
The time and date in that field aren’t used in my model. It’s info that JPL provide which I don’t use (yet). They all start at 8 Feb 2013 0:00.
Here are your latest vectors read back:
AC6b-rev, 1000, 0.01, 2456331.500000000, A.D. 2013-Feb-08 00:00:00.0000, -1.10848921E+08, +9.5065161E+07, -2.636632E+06, -2.162482716E+01, -2.107547255E+01, 4.174E+00,
Here’s the approach of AC6b-rev, at a fairly distinct angle to the DA14 milestones to its left.
AC6b-rev impacts on 15 Feb 2013 01:44:11. at 0.7 degrees N (just above the equator) in a grazing impact.
AC6b-rev rel dist x,y,z -6303599.5950164795, 855228.2938079834, -285030.5937670488 m
AC6b-rev relative velocity vx, vy, vz 3299.6067337613076, 3073.22080892944, 11289.002498559727 m/s
This is a very interesting one. I’ve included both the x,y,z relative distances from the centre of the Earth, and the relative x,y,z velocities to the Earth, because it wasn’t immediately clear what this particular brute was doing. The blue approach line says that relative z is negative, and it lands more or less exactly on the equator, in a grazing impact (red line), which turns green as it exits the Earth with relative z positive.
Judging by the relative velocities, which are all positive, it would seem that this one arrived on the far side of the Earth, and impacted on the side facing us, and the red line is running under Europe approx SW-NE. The red impact line actually runs from the equator to about 55 degrees N. So it’s a bit deeper in than a graze.
If it was a bit further out, and didn’t impact, it would be a good candidate for explaining the night-time fireballs seen in Britain and Europe around 15 Feb 2013.
I subtracted 680 km from AC6b-rev’s start x location, to push it out from the Earth slightly, and now have a true grazing impact at 28.6 degrees N in Algeria. It’s actually going almost due north.
I think I’m going to have to get into the trigonometry I wrote 4-5 years back, which is almost incomprehensible to me now, so as to find both latitude and longitude, and impact approach angle and altitude and azimuth.
This will take deep concentration on my part.
But I also want to use this model to develop a simple climate simulation. And also start using 3D instead of my simple view down the z axis. onto the x-y plane. And I want to add stars.
Frank
Thank you for redoing that. It certainly is an interesting one and explains a lot of things.
One is that these west-to-east evening and night time ones could be a signature. It makes sense because of the tilt of the earth makes it possible.
I was going to say this anyway, after your last comment in which you said:
“AC6a impacts, but in the southern hemisphere.
Impact: 15 Feb 2013 01:25:36 48.4 degrees South between Antactica and South Africa, coming in approximately W – E. Not a graze.”
AC6a is betraying the same behaviour.
The other thing I’ve learnt is that, as long as the initial vx and vy are similar, it is always going to come in at almost 90 deg to the prograde orbital path. The reason AC6-rev is at about a one deg angle (probably 1.35 deg) to the DA14 milestones is that I swung the crossover angle round by 1.34 deg. So it’s coming in at almost 90 degrees to that new angle.
However, I’m wondering if all these rocks are coming in from west of the sun because DA14 was a fraction ahead in its orbit. It is telling that on 8th Feb 00:00 DA14 is doing 30,144 m/s in xy and Earth is doing 30,195 m/s essentially in xy of course as it’s on the ecliptic. Earth is catching up from behind at 51 m/s. This means an impact from slightly behind which is further exaggerated by the rocks falling backwards into the gravity well from ahead and accelerating in that direction to boot.
In contrast, AC6-rev is doing 30,196 m/s (almost identical to Earth) in xy at that time and because of the higher speed it is set back along its orbit by 32,000 km at which point it is close to being directly in line along a radius from the sun. So that shunting back along the orbit and speeding it up is at least partially responsible for the angle change.
That xy speed and position change was itself a fortuitous consequence of changing the eccentricity to reflect the half million km romp across xy. It was dictated by the new eccentricity and the eccentricity, in turn, reflects the fattening of the orbit by about 1%. Without that change it would be somewhat ersatz, doing a speed that its radius shouldn’t allow. The eccentricity was reduced from 0.1081 to 0.0866 and the new speed for that eccentricity at that point was calculated. Hence the 51 m/s change in xy speed at the same radius from the sun. If I tweaked the perihelion a couple of degrees round as well, it would add yet more speed. All these changes are slight and would retain a DA14 signature.
The natural extension of the above argument is to fatten the orbit more, slide further back down the orbit and approach the Earth from behind. I did already do some calclulations a few weeks back when I thought DA14 really was catching up with the Earth on the 8th. To get rocks coming in at 9-12 deg east of the sun would require a genuine catch-up speed from behind of between 0.5 and 0.7 km/sec. For NASA’s azimuth of 105 degrees, it would probably be only 0.3 km/sec. That’s precisely the sort of catch up speeds that DA14 or its circularised companions have if they come from behind and have had their eccentricity slightly changed.
Andrew
Impact: 15 Feb 2013 01:25:36 48.4 degrees South between Antactica and South Africa, coming in approximately W – E. Not a graze.”
AC6a is betraying the same behaviour.
Actually, I was wrong about this, because I was looking through the Earth to the far side, and neglected to flip the image. AC6a was coming in roughly E-W, not W-E.
Earth is catching up from behind at 51 m/s. This means an impact from slightly behind which is further exaggerated by the rocks falling backwards into the gravity well from ahead and accelerating in that direction to boot.
I’ve been vaguely wondering too whether rocks that are ahead of the Earth could fall back onto it. And with very low relative speeds, these can land anywhere. I’ve managed to get low speed rocks to go three-quarters of the way round the Earth before impacting.
The way I see it is that, if there’s a rock train lying along DA14’s orbit, then as the Earth gets closer and closer to it, before passing through it after DA14 has passed, it will be slowing the rocks in the rock train ahead of it. And they’re not moving much faster than the Earth anyway.
There could be a lot of very complex things going on.
Frank
We crossed comments. I just saw your 680km x adjustment shot which looks good. I think that’s quite far north seeing as its coming up from the night side.
If you adjusted about 8500 back in x and y you might get a hit near Chelyabinsk. However, that might not tell us much because we know it’s possible and we know the angle is a degree or so round.
Knowing the atmospheric entry angle would be good. It is difficult to work it out from the vectors but by the same token it’s a challenge, I’m sure, to programme it.
I couldn’t follow the Earth-relative distances for AC6-rev by the way. Were they for closest approach inside the Earth? They looked too small (close to centre) to be grazing just underneath and the z value was negative. Is that because you are subtracting from positive z? I’m sure there is logic in it.
I could follow the velocities but I wondered if that was v’s at impact or at closest approach inside the Earth. They sum to 12.1 km/sec so I presume they are the impact velocity. It’s interesting that the x and y relative velocity ratios are largely retained despite the gravity well. It suggests it went right underneath and not up the side. And if it goes right underneath (67 south, not the Pole) it will cut SW to NE across the longitudes.
Andrew
Impact velocities.
But they’re all estimates. I only discover impact a second or two after it’s happened, by which time the rock could have travelled a fair distance/
Frank
I’ve been sliding back along the orbit of AC6-rev to a particular point behind the Earth in its orbit. Then I worked out geocentric xy relative velocities that would bring a new rock in 7 degrees east of the sun. I made some good progress but then I thought I’d made a mistake converting the xy radiant into absolute vx and vy. But on closer scrutiny, I think it’s OK. It bears similar characteristics to AC6-rev and therefore DA14.
I should be able to give you something late tomorrow. I just need to convert it to state vectors.
Andrew
Andrew,
Unfortunately, I won’t be able to run any simulation models for a while. My new PC stopped working this morning. Stone dead. It’s going to be shipped back to its maker tomorrow. It has the Netbeans development environment and all the files on it.
I may be able to fall back on earlier models, but the computer they’re on is half dead too.
Frank
Sorry to hear you have computer problems. I shall continue to post any vector suggestions for you to try when it comes back or if you get the older one going again. Seeing as you can’t run anything now, I’ll hold off on the AC6-rev (7 deg east version) because I thought of another way of doing it which may be more accurate and at least a check against what I have done already. So I’ll probably post them up over the weekend.
Are you able to check in still? There’s always other stuff coming up such as:
http://www.bbc.co.uk/news/uk-22460642
I nearly didn’t link this because it is thought to be linked to Halley’s comet. But then I thought how you always have a good handle on all the different meteors that fell in Feb and might conjecture something interesting.
Another reason is that it was on a mid evening hour line and glowing green. There were several at this sort of time in Feb, including SF and Manchester so they were coming in on the same hour angle although that doesn’t necessarily mean the same great circle.
But, and its a big but, whatever great circle it was, it would have swung round by about 82 degrees in the 83 days since then so an evening one would have been in the early hours in February.
Andrew
Andrew,
Yes, I’m still able to check in. I just don’t have the Netbeans development environment and the files with which to model the stuff we’ve been doing, although I might be able to fall back on another computer. I’m hoping that my dead computer comes back soon, revitalised.
Yes, I noticed the UK fireball report, and saw the video. But it happened in May, not February. All the same, I thought it was interesting that it was going S-N, just after I’d modelled something that did pretty much the same over Europe (the one that came in low over Algeria) in February.
They said it was glowing green because it probably had copper in it. I’ve wondered if the green might have been something to do with the temperature of it. The California fireball was also said to glow green. And they thought it was something to do with Halley’s comet (which I have modeled, and actually does pass quite near the Earth around this time of year).
Frank
I hope your computer problems are not too bad and get resolved soon.
Here are the state vectors for a fragment designed to come in 6 degrees east of the sun, when you do get the computer back.
It is based on AC6b-rev. I slid 438,000 km back down its orbit to a point that was behind the Earth in its orbit. That point was 6 deg east of of the sun as seen from Earth. Then I worked out the x and y relative velocities required for the xy distance it had to travel in the requisite time whilst maintaining that 6 degrees all the way to impact.
I then subsumed those values into the absolute vx and vy velocities. vz stays the same.
The position vectors might be a little bit off because there was a lot of fine tuning due to relying on AC6b-rev’s xy crossover angle and then accounting for this and the curve in the orbit over the course of the slide-back. I was going to try another method to check it but haven’t had time and I think it’s best to get something out there rather than endless fine-tuning without even firing it off.
Besides, by far the most important thing to bear in mind here is the proof of concept which is that it should come in at a substantially different angle, preferably at 6 deg east of the sun. If it is approaching the Earth at that angle just outside the gravity well then that is a success which can easily be adjusted in the position vectors.
The new fragment is still closely related to AC6b-rev and therefore DA14. Its absolute speed might be 0.2 km/sec faster than ideal but that can be addressed either by reducing the inclination a tad more or invoking a rotation eastwards of its perihelion by a few degrees.
State vectors for new fragment to come in 6 deg east of sun (km and km/sec):
To be fired at 00:00:00 on 8th Feb 2013.
x: -1.10562356 E8
y: +9.5389182 E7
z: -2.696627 E6
vx: -2.1860303 E1
vy: -2.1360743 E1
vz: +4.174 E0
Good luck,
Andrew
Hi, guys. Long time no talk to.
Andrew, you said . . .
“Here are the state [sic] vectors for a fragment designed to come in 6 degrees east of the sun, when you do get the computer back.
It is based on AC6b-rev. I slid 438,000 km back down its orbit to a point that was behind the Earth in its orbit. That point was 6 deg east of of the sun as seen from Earth. Then I worked out the x and y relative velocities required for the xy distance it had to travel in the requisite time whilst maintaining that 6 degrees all the way to impact.”
Unless I read this wrong, this is not quite correct. The radiant point was 6° east of the Sun (which rose at 111°), and the heading was 279° (my calculations). The 6° E meant that the radiant was at 105°°. The 6° was not maintained as a path.
My 279° was based on ONE video in Yekult with the person on foot and who panned overhead from one end of the path to the other, and which had clear alignments on buildings. Based on the condition of the trail it was within 2-5 minutes of the passage of the object.
You might have been talking about some other object. In which case good on ya for getting the 6° east of the Sun. If so, ignore the rest of my comment here.
Steve
Andrew,
My computer should be returned tomorrow (with a new power pack), and so I hope to be able to try out your new vectors on Wednesday or so.
Steve,
My 279° was based on ONE video in Yekult with the person on foot and who panned overhead from one end of the path to the other, and which had clear alignments on buildings.
Which video was that? Do you have a link? And I can’t find anywhere called “Yekult”.
Frank
Thanks for the update.
Andrew
Hi Steve
There were several estimates for the azimuth and heading. They seemed to centre on 99 degrees and 105 degrees, i.e. 12 or 6 degrees east of the sun because it just happened to be rising. I would plump for the 99 degrees which would send it along a 279 deg heading as you calculated. 99 deg looks right for various reasons, not least of all that it’s easier to get a fit over Yekult and it’s the best fit in the middle of the two extremes on the Wiki page.
However, since NASA seems so sure that their ‘government sensors’ were right in detecting a 105 degree azimuth, I thought I would go for that for the time being. I’m taking baby steps towards the 99 degrees and it’s easier to stay close to DA14 DNA with a 105 deg azimuth. When it comes to it, I doubt if 99 degrees is all that much more difficult but I thought why not try and land one on NASA’s azimuth on the way to 99 degrees.
Andrew
Frank
It’s Yetkul.
This is from the Wiki Chelyabinsk Meteor page (which is why I now call it Chelyabinsk meteor as I promised). It’s a map of the trajectory. Although Yetkul has its Russian name here, you can see it to the SE of Korkino tucked in the angle between the darker, central trajectory line and a road. It’s by a sort of kidney-shaped lake and there is ample evidence the path went right over the lake.
It’s best to click on full resolution.
I tried to find Steve’s vid from Yetkul but with no luck. I know it’s in the comments at Stefan Geen’s site. I’ll keep trying.
Is this Hohmann Calculator of any use to anyone? I found it here (last comment).
Frank
The Hohmann transfer is what you might call short and surgical because it involves just a few seconds to a few minutes of burn. That last commenter who linked it alluded to this.
He also asked whether they had taken Earth’s gravity in to account. The Kaira people who were probably conversant with the NASA narrative when they posted this, would have been taking Earth’s gravity into account, as did NASA and the Colombians too.
This is a knotty area though. I suspect this commenter just added the 11.2 km/sec to DA14’s speed which you can’t do. But there is a confounding issue in that the Colombians’ orbit is doing 5.6 km/sec faster than the Earth which is almost exactly DA14’s approach speed. That’s what led me to make the same mistake because I mistook their absolute speed surplus as being suspiciously identical to DA14. That made me rush it through without realising the Colombians had a 15 deg or so crossover which added a bigger xy vector- and at the time it was just a comment and not destined to be an article so I was less careful.
I think this commenter is doing the same thing by using all the Earth’s gravity well acceleration when he can’t really do that. I still think the fact that the Colombian fragment having an absolute velocity surplus identical to DA14 approach speed is fishy.
Anyway, I still bookmarked the link. It might be fun to play with so thanks.
As for us, we can’t rely on instantaneous punches. We need to keep bearing in mind the slow action of the gravity well which we have been doing assiduously up till now with our radiant-sliding, checking speeds and sliding back. Those seemingly innocuous 2 or 3 metre per second increases turn into a rather ugly 2000km between the 8th and the 15th.
Andrew
Andrew,
I got some code running on an old computer, and tried out your latest vectors as AC7. It missed the Earth by quite a long way.
Then, while retaining your velocity vectors, I moved its start position until it just grazed the Earth. The image shows a close-up of the impact at 59.5 degrees N, and also in bottom left the relative paths of AC7a, DA14, and Halo1093 as they approach the Earth.
AC7a is the same as AC7 except x = x – 6.65E+07, y = y – 1.74E+08, z = z – 2.60E+07 metres
It’s quite promising, although it comes in about 5 hours late. Halo1093 impacts at almost the right time as the Chelyabinsk fireball, and so I transposed the path of AC7a onto the Halo1093 impact screen capture. It’s coming in at a very promising angle. It just needs to land a bit further on than it does.
I haven’t been thinking much about these various rocks though. What I think I’m going to try to do is to convert my Orbit.java code to 3D. If I can do that then we should be able to look at the Earth from anywhere we like, including riding along on one of these rocks, or from the surface of the Earth. I’ll retain my vector map of the transparent Earth, so it shouldn’t look too different. It does of course require me to fire up my trigonometry a bit, but I’ve done 3D stuff several times before.
Tinkered a bit more to produce AC7b, also transposed.
Frank
It looks as though AC7b went through the same hole in Lake Chebarkul.
I was impressed by the other two as well, despite the initial hitches so well done for tweaking them. I was in the process of doing a longer comment regarding those when I saw your AC7b comment so I may not finish that till tomorrow. It’s just a post mortem on why AC7 was as far out as it was in xy.
The main thing is that it is coming in at a substantially different angle (exactly in line with the sun) though why it’s not further round, I couldn’t say but it might become apparent.
At first, I thought your adjustments for AC7b might take us a bit far away from DA14 but these adjustments of 50-150,000 km are small in the context: even on your geocentric snapshots it’s all happening within a small square 2 million km off towards the sun but when you turn to heliocentric, that small square is sitting between two tram lines that are about 10 screen widths long which suggests any angle changes are even smaller than you might think.
That said, it would be highly informative (if this is possible) to switch off the Earth/moon and allow all the rocks fly round to February 2014. Looking down the z at them and comparing their ellipses would speak volumes. By ‘the rocks’ I mean AC7, AC7a, AC7b, along with DA14, Halo1093 and possibly AC6b-rev (because it had its eccentricity changed and is therefore a template fragment)….but I did hear you when you said you were concentrating on converting to 3D. So maybe that’s a wish list for now.
But I will add another thing to that list. I’m itching to know AC7b’s latitude and angle of entry. Is it just the relative vx/vy/vz to Earth that you have? I could work out the angle to the ecliptic from these and then slide the Earth in at the relevant latitude and subtract.
Andrew
Carried on tinkering. I advanced AC7b about 5 hours forward to create AC7V (which may also be going fractionally more slowly) to arrive at the right time. This gets quite near Chelyabinsk a few minutes early. State vectors at for 8 feb 0:00 are included in the image.
I haven’t actually done anything towards 3D today. Maybe tomorrow.
Frank
Well, I must say that’s all very encouraging. Notice how, even with the 11 degree swing round it is still curving up to the top of the page away from the (slightly more horizontal) angle we want. This is a good thing though! Since we already know we have to swing round by at least 6 degrees, we are going to swing past the point where it starts curving back the other way to a more favourable angle. That means the radiant will have swung east of the atmospheric trajectory line and the ‘bird cage wire’ effect will curve it in the last couple of degrees. So if we do go for 6 degrees more so as to achieve a 105 azimuth we may get a 103 azimuth.
Your speed reduction doesn’t look drastic and it changes the orbit angle/crossover by only 1/500th of a degree. Even the change from 11 deg east of sun to where we are was only a 0.09 degree change. I was quite surprised by that but I triple checked it- it’s just the cotangent of the vx and vy and taking into account the 438,000km shunt back in the orbit.
Andrew
Andrew,
I realised this morning that I could very easily “switch off” the Earth and Moon’s gravity in my simulation model, simply by reducing their mass to a few kilograms, and leaving everything else unchanged.
I then ran the simulation on from 8 Feb 2013, and changed the screen view to barycentric, and plotted out the orbits of DA14, AC7, and AC7V over the next year or so. Here’s a screen capture that overlays AC7V (red) on DA14 (blue/green). I’d guess that AC7V’s orbit is more circular. DA14 was almost back at its start position on 7 Feb 2014. AC7V was at more or less the same place on 19 Feb 2014. I’ve also got your original AC7 orbit.
I haven’t plotted AC7a or AC7b. But AC7V is simply AC7b advanced 5 hours, so should have the same orbit.
I’ve been thinking a bit about this circularisation business, and it seems to me that if an orbit is fully circularised, the barycentre will lie at the centre of that circle. But it also seems to me that if an orbit is to become circularised it must retain the same kinetic energy as the non-circularised orbit. So I’ve been thinking that if I wanted to circularise the orbit of DA14, I should find its mean kinetic energy, and use this to find its mean speed, and hence the radius of its circular orbit. I haven’t actually tried doing this, however. Does this seem like the right idea?
Frank
“I realised this morning that I could very easily “switch off” the Earth and Moon’s gravity in my simulation model, simply by reducing their mass to a few kilograms, and leaving everything else unchanged.”
That was a cool idea. Yes, it looks as if AC7b is more circularised as it should be. Remember, a tiny portion of that extra xy radius compared with DA14 is due to the 2.5 degree reduced inclination which allows it to present more of its ellipse shape to us than DA14. However it is an extremely small component. Even DA14 itself is presenting 98.35% of its semimajor axis at 10.33 degrees.
Could you send the Earth round with them so I can see how far inside the orbit they are? They keep snapping to circles in my mind even though I’ve measured diameters this way and that to prove they have the eccentricity they should have.
I would say that being 11 days out (8th to 19th) isn’t a problem right now because we are learning an awful lot from these fragments, but it would be good to get any polished-up candidates down to 3 days or less. That keeps them in the cloud. This may be a long way off- don’t want to throw the baby out with the bath water. You said AC7b was essentially the same as AC7 but it was 24 m/s slower in x and 20 m/sec slower in y. That’s 31.24 m/sec which might account for some of that. Also, in my post mortem for AC7, which is now getting pushed out by the exigencies of today’s tour around the sun, I mention how I didn’t bother to adjust the eccentricity to fit the new xy angle. That’s why I said it might be 0.2km/sec too fast, which on a less circularised orbit would most likely increase its period (think of the Colombian fragment with an absolute speed of 35+ km/sec and a four year period).
As for the kinetic energy conservation, you are right, but the total energy of each rock is the sum of its kinetic energy and potential energy. You may recall my mentioning fragments ‘trading’ KE gained for PE many comments ago. This is intimately linked to conservation of momentum and it is that which is implied in Kepler’s laws and the vis viva equation.
But I think what you need to look at is the mean anomaly which allows you to compute the position of a body anywhere along its orbit at any time. This equation is related to the Keplerian law of sweeping out equal areas in equal times. I always have this sweeping concept in mind when dealing with the approach of DA14 fragments, slowing down from perihelion, but I haven’t got round to using the mean anomaly equation. I used a laborious method, employing three different equations (eccentricity, vis viva and v squared equation) in sequence and jumbled around to establish various key values which were substituted into the next equation. This gave me the correct absolute speed for a particular radius and eccentricity. It was 10 jumbo pad pages of calcs and I did it only for AC6-rev. It is telling that this was the only off-piste fragment that has had a direct hit. I think it was because it wasn’t being given a speed that really belonged to another radius and subtely different crossover angle/eccentricity.
My round-the-houses method was giving me a single set of values for a single spot in the orbit. The mean anomaly allows you to do this more flexibly. I did it the other way because a) the equations were more familiar to me and b) I only needed values for one point in the orbit, the 8th Feb firing point. I probably was shadowing aspects of the mean anomaly equation. All these equations are derived from each other and look similar. They all have Kepler DNA.
There are other related anomaly equations but I think mean anomaly is the one you need. I shall need to start looking at it too.
Andrew
Frank
A couple of other points:
I shall start applying myself soon to getting round to 105 deg.
Do you remember that we conjectured DA14 as sitting at the back of the front half your split rock train when it returned in 2013? With AC7, we have established that the Chelyabinsk meteor, if DA14-related, came in from the front of the back half of the split train. That’s a very neat fit to the model.
It follows that perhaps all the fragments came in from two narrow angle bandwidths centred on 10 degrees either side of the sun. If that is right, there was a dearth of rocks in the middle so none came from there. That’s somewhat speculative but as we are trying to narrow down where they all came from, those two radiants would be a good place to start. The times, positions and atmospheric trajectories might be categorisable into two sets each containing a signature from its respective radiant.
Andrew
Andrew,
Here’s a screenshot showing the Earth’s orbit, with Moon, DA14, and all other nearby rocks removed.
And here’s a screenshot of the above overlaid onto the DA14-AC7V screenshot. The Earth’s orbit seems to exactly overlay AC7V’s orbit, suggesting that they’re very similar in the x-y plane.
Do you remember that we conjectured DA14 as sitting at the back of the front half your split rock train when it returned in 2013? With AC7, we have established that the Chelyabinsk meteor, if DA14-related, came in from the front of the back half of the split train. That’s a very neat fit to the model.
What I remember finding was that, a bit to my surprise, the Earth blew a hole in my conjectural fore-and-aft rock train lying along DA14’s orbit back during the 2012 close approach. DA14 was lying fairly near the end of the front half of that rock train. The effect of the Earth’s passage was to slow down rocks near the end of the front half of the train. And speed up rocks near the front of the back half of the train. These slowed and speeded rocks then moved away in different directions from the train, and the hole in the rock train gradually got bigger. I don’t recall doing any further investigation. But I do remember you conjecturing (before I’d looked at the rock train, I believe) that the Chelyabinsk rock may well have been one that had been disturbed in 2012.
At the time you suggested this, I thought it was a bit of a long shot, because it supposed that a) there was a rock train along DA14’s orbit lying fore and aft of it, b) this rock train was disturbed in 2012 to push a few rocks into different orbits, and c) one of these disturbed rocks landed on Chelyabinsk. But I’m now beginning to think that it’s maybe not such a long shot after all. Supposition a) is probably pretty factual, even if there’s no evidence for it that I know of. Supposition b) is less suppositional, because I’ve seen that the Earth does indeed disturb the rock train in 2012, and in a very particular way. In fact, it creates a completely new gradually lengthen rock train at an angle to the original DA14 rock train.
I haven’t really looked at where this rock train goes. But it wouldn’t be too difficult to do so. And we might find that it passes the Earth a year later, with rocks coming in from a new angle, and perhaps quite a long way from the DA14 rock train. One of these might be our Chelyabinsk rock. The close approach of this rock train would of course spawn yet another rock train….
A further iteration might be to make the DA14 rock train into a cylinder of rocks rather than a line of rocks.
So that might be a good future project, perhaps when (and if) I’ve got 3D working.
I think the good thing about such a line of investigation is that we could have rocks that are a long way from DA14 but are nevertheless (quite literally) its offspring. But if we did this, we’d have to stick to trying out rocks that were in daughter trains.
Frank
Thanks for superimposing the Earth. It’s heartening to see that what has been ‘modelled’ on paper is reproduced in the computer model. The fact that AC7b sits perfectly under the circle of the Earth in xy means that its slight elliptical shape is squashed into a circle by its 7.83 degree inclination.
Your account of the story of the split rock train in response to my comment is as I remember it. However, in addition, I remember that there was a 2012 snapshot and a 2013 return snapshot showing the gap- I then wrote a comment that worked on the supposition that the 2012 accelerated rocks ended up overlying the front of the rear half in 2013; and the decelerated rocks ended up overlying the rear of the front train. This supposition was based on the known circularisation of DA14 by the same process over several years (as acknowledged by NASA).
It is true that at the time we only had the beginning and end snapshots and I was speculating on the behaviour of the particles in the intervening year of their orbit. So, yes, it would be great if you could run them round the whole orbit between the two snapshots so we can see. Whatever they do I’m sure it will look very beautiful and teach us a lot in the process. The worst that can happen is that the process of redistribution and overlaying is the mirror image of what I’m suggesting- that’s still very encouraging.
I should make a small clarification here. You did show 4 superimposed shots for 8th Feb through to April (22nd ). I tried to divine what was happening to the affected rocks but it was still too early in the orbit.
The sections of the train either side of the gap have overlaid rocks, hoovered from the gap. Those short sections appear to be about twice the density of the rest of the train. This is my assumption because the darker, denser red bits are both about half the length of the gap (from memory). Moreover, they are centred on the DA14 radiant and the AC7b radiant which is very promising indeed. If a particular radiant has twice the number of rocks, the chances of a hit from that angle is doubled.
You said:
“I think the good thing about such a line of investigation is that we could have rocks that are a long way from DA14 but are nevertheless (quite literally) its offspring.”
Exactly. And this concept is what NASA was studiously avoiding despite observing DA14’s relentless perturbation over several years. The notion that they never imagined this same process would be stretching any companion rocks away by hundreds of thousands of km is hard to countenance.
Andrew
Frank
I found the relevant images, linked below although I realise they are probably quite accessible for you. It was the second comment down in this, the second page of comments. It was actually April 2nd, that last composite. The 2013 gap and the radiants I referred to isn’t quite as neat as I remember in that the trailing half is set back and not aimed at be Earth. However, that has a lot to do with initial conditions. If you did do a cylinder I suspect we would see all sorts of weird stuff happening, with closer rocks and more favourable radiants on that trailing edge.
I also said AC7b when I meant AC7V. And to be more precise, the proposed trailing sweet spot that contains the Chelyabinsk rock is 7 degrees further round than the AC7V radiant.
I was trying to keep it simple because it’s sometimes good not to muddy a neat concept with all sorts of caveats and qualifications. AC7V is in line with the sun and needs to swing round a bit. That clarification is more for future referring back to avoid head scratching. I re-read the thread quite a lot.
Andrew,
I dug out my 1 feb 2012 rock train, and ran it for about 2 years, taking snapshots at irregular intervals of a geocentric view with one fixed scale, with the Earth at centre (green blob), and combining them. Times and dates displayed were screwy, so I’ve crossed them out. The first one shows the initial rock train nearing the Earth at 1, with DA14 near the front of it, and then 2, 3, 4 , 5 showing the growing gap that appears after Earth’s close approach, and daughter sidestreams. When after about a year the train returns to the Earth, there are 2 distinct streams with a gap between them, and DA14 is towards the end of the the leading chain.
The second one shows the next 12 months, as the broken rock stream gets re-broken in the 2013 close approach. By 2014, the gap in the rock train has got much larger, and Earth passes through the gap in 2014.
I hope these make sense. Essentially, the rock train travels anticlockwise round the Earth, and as it does so, its hands point a bit like a clock.
When I originally did these last month, I was mostly interested that the rock train got broken in 2012. I didn’t spend too much time thinking about what was happening.
Very roughly what’s happening is that when the Earth passes through the rock train, it pulls rocks inward from either end of the rock train, and these gradually peel off in opposite directions over the next year, to form two new rock trains.
But I’m beginning to think that, by the time it comes back round to the Earth, the original rock train has completely peeled off. If so, my initial rock train may be too short, and I ought to extend it forward from DA14, so that some of the original rock train remains in place.
Andrew,
It rather looks as if I was using a longer rock train back in April than I was today, judging by the images you’ve just posted. I may have truncated it subsequently because I thought I didn’t need the front half (and may have been working on a slow computer and felt I needed to).
Frank
That’s very interesting. Does it take you long to knock out these labelled overlays? I should think it’s quite a lot of work so thanks for doing it.
It looks as though my suspicions are confirmed. The slowed rocks on the inward track (bottom of the right hand bunch in ‘3’, fanning towards sun) fold back on the rock train by ‘4’ and overlay the back part of the forward train from that point onwards. The reverse happens to the other train.
This is still a slight assumption but I can’t see how the left hand bunch in ‘3’ can get ahead because it is flung further from the sun than it would have been without the Earth’s influence. Moreover, the back of that stream constitutes the back of the original pre-pass rock train so that has to be at the back of the whole configuration at ‘4’.
There’s one small confounding issue here. I was assuming that circularised rocks would still be behind the Earth on approach in 2013, providing us with the 7 deg east of sun radiant. It seems that this is not the case, at least with these initial conditions but it should be remembered that these rocks have been whizzing round close to us probably for centuries or millenia with close encounters every 85 to 170 years (recently). Chelyabinsk could have been perturbed to a forward train in 1928 and slowly dropped back by 500,000 km over 85 years (0.186 metres per second).
I can see from the angles of the trains in 3 and 4 that they are probably early March and early September and 5 must be about late October. Although the distances in z are larger, it shows how all these rocks, on a 360-370 day period hug together within a few million km of each other and the Earth. They are pulsating out and coming back together and into line at the node where they were first disturbed. Very aesthetic.
I do wonder whether a cylinder/train if wide enough will have a less disturbed area that populates the gap in 2013 and provides us with the 7 degree radiant. I think there is some mileage in that idea (of yours!).
Andrew
Does it take you long to knock out these labelled overlays? I should think it’s quite a lot of work so thanks for doing it.
It didn’t take long. An hour or two. Some things are harder than others. For example, to push a rock forward 5 hours down its track (like I did with AC7V) is quite a lot of work for me. And of course I don’t like to do it. But jigging a rock left or right , or up and down, is easy.
Tomorrow I plan to extend the 1 Feb 2012 rock train forwards and backwards. i.e. make it longer. Because I think that maybe when DA14 passed the Earth on 15 Feb 2013 19:25 UTC, a lot of rocks hit the Earth at almost precisely that time because that’s where the bend in the rock train was. And the Earth went straight through the bend.
Andrew,
I dug out a long 2012 rock train, and added it to yesterday’s short one, and ran it for a year. I got the dates about right this time. Here is a composite geocentric view of 7 different dates running from 4 feb 2012 to 28 jan 2013, progressing anticlockwise around the Earth. The dark lines are yesterday’s short train, and the dotted lines curving out either end of the short train are the long train. The Earth is the green dot at the centre. I hope you’re able to disentangle the images.
I think that one thing that is clear about this is that the Earth passed through a very densely populated part of the broken rock train at more or less exactly the time of DA14’s closest approach, and this would explain the peak of rocks at that time. But the rock train at this point in time consists of two rock trains which are travelling almost parallel to each other.
I forgot to save the image with the Earth in green, with its name next to it. It’s the double dot at the centre of the image.
Also notable is the deformation of the long train after the close approach in Aug 2012.
Frank
That’s another bit of the puzzle sorted then. I always look forward to seeing what you’ll come up with. It was the July snapshot that did it for me. You can easily see the sequence now with July resembling both May and October. October is particularly majestic now that you have the long train in, and it looks somewhat more foreboding than January because it is still whipping back on itself, betraying the denser populated bend that will eventually lash the Earth (to strain the metaphor a tad).
I was able to tease out all the rock streams and could see the two dots at the very centre . However, didn’t you mean the two dash-like dots near the centre, just off it though, around the two o’clock mark? I assume that you did mean those but I wrote the long drawn out paragraph below querying the two tiny central dots. When I got to the end of the paragraph I thought you must have meant those other two offset ones. But I’ve left the paragraph in to show my reasoning. If you agree it’s the two offset ones then you can just say ‘yes it’s them’ and avoid reading the turgid prose.
In the second of the two snapshots I linked a couple of comments back it shows the long chain (though not as long as this one) in red. The four images are from just before close encounter 2012 to April 2012 and includes March 18th. You have the Earth and moon labelled and said it was a composite so I assume the Earth is fixed and the successive snapshots of the train dance around and away from the Earth as they move through their orbit. However, in this latest composite, which includes a similar snapshot at March 15th, you say the Earth is the two dots at the centre. That puts it straight down the orbit of the train from the opening/fanning breach and off the train orbit by only a smidgin. It doesn’t look configured the same as the linked one and in that, the March snapshot was only 3 days different. What’s more, I can see a couple of tiny dots off to the right of the March 15th line in the new version and in about the same place as in my linked version. Those couple of dots seem to be in the right place for the Feb 2012 and Jan 2013 pics in this new version too if they were to mimic the the linked version too… as well as mimicking the outsider track slowing and inside track quickening of DA14 at all the right times and distances.
Anyway, It doesn’t detract from the orbiting patterns which look intuitively absolutely right too me.
That bend and subsequent denser patch really does speak volumes.
Andrew
P.S. I’m quite busy this weekend. I shall have time to relax, ponder new snapshots and make brief comments so please do keep slinging stuff round. It’s very entertaining as well as educational. Being busy, my brain will only be geared for fun and games, not calculating the 7 degree extra radiant. So I’ll start that on Monday, I think.
Yes, it’s them 🙂
It’s occurred to me that if the Earth has such an impact on the rock train in 2012, what happened in previous years?
Accordingly, I’ve set up a long rock train in 2005, and I’m currently watching what happens to it over the next 8 years. And quite a lot is happening, although the rocks aren’t being thrown around quite as spectacularly as you’ve just been seeing.
I doubt if my model is accurate enough to do long periods of time. But I might be able to correct it by repositioning DA14 (and the rest of the rock train) each year to NASA’s location.
Frank
I think that’s a good idea to go further back especially as NASA did the same thing to see how the Earth was affecting DA14. They said its period was around 368 a few years ago and was shortened to 366.2.
Was stuff happening even in 2005? I suppose I’ll see in due course. I can see how the tail of a long train would be affected at that time but DA 14 would have been crossing well ahead of us, about 10-20 million km ahead?
I was interested to hear that the train was affected in the August crossover. If the train takes several days to chug through that point on the Earth’s orbit the stragglers certainly would be affected when the Earth arrived at that point. It would be a possible explanation for the the many northern hemisphere punches of 1927 and 1928. Several weeks back, I was looking at a Tunguska link. It was difficult to prove anything due to the nodes changing from June to August/Feb (as Steve had said) but in the process of calculating I used the 368-365.25= circa 2.7 days period difference up to 2013 and divided that into the number of days in the year from June 30th to Feb 15 (the proportion of Earth’s orbit). It was close to 105 years. I then divided 2.7 into the number of days between June 30th to and the August node date and came up with a date of 1927 and 1928 for the two closest approach years for DA14 (ie, it was always going through that August point but before we got there. Each year we would be closer when it passed through until, in 1927 and 8 we were right on top of it.) I then double-checked this by dividing the 2.7 into the number of days between the August 12th and February 15th nodes and came up with 85, which brings us from 1928 to 2013. So if the train was a few days long it wouldn’t be crossing our path in most of the intervening 85 years. It would be crossing ahead of us at both nodes.
So then I looked for candidate hits in 1927 or 1928. Being the descending node, any northern hemisphere hit would be a high inclination punch and the sweetest spot would be on the twighlight side of the Earth. So I did a search and came up with this, 3 days away from DA14’s pass in 1928:
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1929PA…..37..275O&db_key=AST
page_ind=0&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES
An
Andrew,
It took a couple of hours to run the simulation last night. What I did was to get DA14’s vectors at 12 hour intervals from 1 Jan 2005 to 1 Mar 2005 and use these to create a chain of rocks with the real DA14 at the 1 Feb 2005 location. All other bodies including Earth and Moon were placed at their 1 Feb positions. So I started out with a nice long arc of rocks.
I then took snapshots of this arc on 28 Dec of each year subsequently. This was so that the rock train had moved on one year, and also so that the arc of rocks lay more or less horizontally on the screen so that I could stack them one after another into an image showing the evolution of the rock train.
Eventually, DA14 actually did make its closest approach to Earth in mid-Feb 2013, but about 20 hours early (assuming the calendar date is accurate), and 83,000 km from the Earth, and on the wrong side of it. I’m not too surprised at this. I don’t expect my model to be accurate over long periods of time.
I’ve combined all this information into one image that shows 8 snapshots of the rock train, with DA14’s closest approach details shown in a separate box.
There’s quite a lot to be said about this. Firstly, as you can see, even by the end of 2005, the train has been disturbed both by the Feb and Aug Earth intercepts. Two large gaps open up and widen. DA14 is in a relatively unaffected part of train. But (and I have no explanation for this) it starts out at one end of this intermediate train, and travels all the way to the other end.It can be seen as a red pixel at some points.
However, rocks don’t seem to being torn off the rock train and hurled into deep space. Instead, many of them seem to be moving to and fro along the rock train, and slightly above it or below it. I think I can understand why this happening, and it’s because during the Feb and Aug Earth intercepts, rocks are being giving a fairly strong impulse along the rock train in one direction or other.
There’s a lot more that can be said, but the conclusion that I draw is that starting a rock train in 2012 isn’t good enough. There’s a lot going on in the years before then.
For some unknown reason, DA14’s closest approach box is truncated in this image. But it doesn’t really matter, because that was really about the model’s accuracy. DA14 was, in this case, the very last rock in the train approaching Earth.
I was a bit puzzled at DA14’s motion. I wondered if the Earth’s proximity was the cause of this motion. To find out, I reduced the Earth’s mass to near zero, and ran the simulation again. This time, as expected, the rock train was hardly disturbed at all. I captured images of the rock train as each year passed, and found that the entire rock train was slowly moving along the orbit. This image shows screen captures on the same date over several years, superimposed one above the other (with earth as green blob).
This would seem to mean that DA14 was gradually moving back down its orbit (retrograde) towards the point where the Earth was biting lumps out of the rock train lying along that orbit. Or maybe I’m misinterpreting it.
Andrew,
I might be making some progress at last.
After looking at the rock chain earlier, I began to wonder how such a chain might form. I supposed that some body must have broken up at some point. What happened when a large body broke up? A while back, I created a cubical cloud of rocks, and watched what happened to it as it passed low over the Earth. I thought I’d create a similar cloud of rocks around DA14, to see what happened to it over time.. So, starting in 2005, I created a 48,000 km side length cubical cloud of 343 rocks ( 7 cubed) around it, and sat back to watch. I didn’t really expect anything much to happen at all.
In fact, what happened was quite extraordinary.
The first thing that happened was that the cubical rock cloud gradually got longer (and perhaps thinner), extending out along the orbit of DA14. Over the next 7 years it became many millions of km long, and also retained its shape.
Then, after the close approach of 2012, the entire rock cloud gradually folded in half, so that by the time it came around to the Earth in Feb 2013, it presented an extraordinary appearance, almost as if the entire collection of rocks looked like a giant fireball. The Earth then proceeded to fly right through it, hurling rocks in all directions.
It was all completely unexpected. I was quite stunned.
There are some problems, however. After its 8 year journey, DA14 (which should be a single red dot somewhere in that cloud) isn’t quite where it should be. The Earth ought to be going straight through the wide head of the ‘giant fireball’ in which rocks seem to be going in all sorts of different directions.
This looks like a very promising way of generating rocks falling onto the Earth around 15 Feb.
Frank
Wow! That’s a quite a coup. Sometimes people say of this sort of thing, ‘oh, but it’s just a theory’. Well, it is but when it explains specific events at specific times without any fudging, I would call it a sound working hypothesis- at the very least. You could have done that experiment back in 2005 and walked away thinking it’s just a computer game. But if it really went on to crash through the Earth 8 years later, as it did, the computer ‘game’ takes on an eerie reality and people have to give the initial inputs (cloud circa 50,000 km across) a fair degree of credibility. I’m impressed.
One point about the ‘fireball’ shaped stream. It shows it going between the Earth and the moon. You said, “then” the Earth went right through it hurling rocks everywhere. Does that mean the ‘fireball’ of rocks is a few thousand km under the Earth in the snapshot just before hitting? Or was this close approach between Earth and moon enough to hurl rocks everywhere? I do acknowledge that, from your recent experiments, this close is close enough to hurl rocks into significantly different orbits and I also acknowledge that after 8 years, this close to Earth is as good as saying right through. I only ask because you said ‘then’ the Earth went through and the snapshot is showing what appears to be the ‘fireball’ as you described it before being broken up.
Andrew
Frank
An answer to your question:
Your snapshot of DA14 in successive years, getting closer to Earth is to do with its slower orbital period. So you are right. If you imagine the Feb 15th spot in our orbit, we go through that point on Feb 15th every year but DA14 went through it only on the 15th in 2013. But in 2012 it crossed our orbit one and a half days ahead (your experiment showed that because the Earth went through the train, down range of its orbit, one and a half days later). That’s in keeping with the 366.2 day orbit. So DA14 was going through the same point every 366.2 days and we were going through that point every 365.25. So we were catching up every year which is the same as saying DA14 was drifting back towards Earth each year. The exact amount is confounded by the increasing circularisation and period reduction in each successive year and is as much as 2.7 days but the principle still holds.
This same principle is what I was referring to when describing my calcs for the year in which the DA14 train/cloud passed through the August node at the same time as the Earth. One could say it simply passes through that point every year and be right. But between 1908 and 1926 it was passing through that point long before the Earth got there. By applying the 2.7 day rule as the orbital period difference purportedly was (NASA) and dividing 2.7 days (2.66 degrees) into the arc between June 30th and August 12th I came up with 19 or so years. Adding that to 1908 gave me the date, 1927 or 28, the train would be passing through at the same time as we did.
Although, I must reiterate my scepticism of a Tungunska link. By far the more convincing evidence was working the 2.7 day, 2.66 degree calculation back from 2013 to 1928 (85 lots of 2.66 fit into the arc between the nodes). And I seem to remember that there was a tiny fudge of a degree or so too but that just meant the Texas fireball was an extra day off DA14 which is no big deal in the science of trains and clouds.
Andrew
One point about the ‘fireball’ shaped stream. It shows it going between the Earth and the moon. You said, “then” the Earth went right through it hurling rocks everywhere. Does that mean the ‘fireball’ of rocks is a few thousand km under the Earth in the snapshot just before hitting? Or was this close approach between Earth and moon enough to hurl rocks everywhere?
It’s rather remarkable that it really does look like a gigantic fireball, even though it isn’t. In fact it’s just a collection of rocks approaching the Earth along the usual radiant with which we’re very familiar. It’s just that for the purpose of this simulation run, the view is DA14centric rather than geocentric, and so the Earth and Moon arrive into this picture from top left. The rocks get hurled around as the Earth passes through the ‘fireball’.
The trouble is that, after 8 years, DA14 isn’t where it should be. It should pass the Earth 30,000 km left of it, not 85,000 km right of it.
What I’m going to do, in order to improve accuracy, is get hold of historical DA14 vectors for every year, and use these to correct the vectors of my DA14 (and its accompanying rock cloud) every year, so that when it finally approaches the Earth on 15 Feb 2013, everything is pretty much exactly where it should be. It may also mean that I can run the simulation very fast, making automatic corrections as I go along. At the moment it’s taking about 15 minutes to process a year, so 8 years 2005-2013 takes 2 hours. If I can go four times faster, a year will take 3 minutes. And a century will take 300 minutes, or 5 hours (I’m thinking of starting the rock cloud in 1913 to allow it to grow to a respectable size).
I’ve already got the annual DA14 vectors from NASA, and I’ve written the code that reads the vectors. I now just need to write the code that adjusts all the rock cloud vectors each year. But I should have that working some time tomorrow.
The result (unless I’ve got something badly wrong) should be that the ‘gigantic fireball’ passes straight through the Earth, or vice versa, with multiple impacts on and around 15 Feb 2013. It should be really spectacular.
If I manage to get it all working tomorrow (I’m too tired to continue tonight), I’ll produce a set of images showing the gradual extension of the cubical rock cloud into a very long rock train over the first 7 years, and then the process by which it bends back round after the 2012 encounter, and then finally its explosive impact on the Earth.
As a taster, here’s what happened when the Earth passed through the “gigantic fireball” train today, just after the image I posted earlier. Tomorrow it should be going straight through the front end, ahead of DA14.
One last thought. The “gigantic fireball” is actually a single rock train that has folded in half. And it appears to be some 40,000 km across. Which is about what my original cube size was. Can you remember how far ahead of DA14 Halo1093 or AC7 were? That should determine how big to make my original cube 8 years earlier.
There’s going to be zero problems linking these rocks to DA14. They’ll all be lumps that split off and slowly floated away over a few centuries..They’ll all have the exact same DNA.
Frank
I’ve described, below, a method for finding the size of the cube using Halo 1093 and AC7. However there are a couple of caveats. Firstly, the resultant cube has Halo 1093 in one corner and AC7 diagonally opposite in all three dimensions. This means the cube is sitting there on 8th Feb neatly orientated to be in line with the xyz coordinate system. My little brain finds that untidy and wants to swivel the cube 45 degrees to match the orbit. It’s a minor point though and I think you are transporting the cube back in time to 2005 and centring it on DA14 aren’t you? So maybe I’m being pedantic and you just need the dimensions.
Secondly, AC7 has its radiant coming directly from the sun and not the required 7 to 12 degrees further round east for Chelyabinsk . This means you really only have half the cube needed after calculating with AC7. Our future 7 to 12 deg fragment will be further over in plus x, plus y, same z but I don’t think it will be double the distance. Perhaps two-thirds as much. However, why not just double the x and y sides for the results you get using the method below to be safe? If too large, the outliers won’t affect the core cloud if spaced out (assuming their mass is very small). In fact their relative lack of disturbance may allow us to distill out the core cloud.
Method as follows:
You can determine the side of the cube in each dimension by getting the position vectors for Halo 1093 and AC7 on 00:00:00 8th Feb 2013 and do the following:
Subtract Halo x from AC7 x to get side length of cube in x.
Subtract Halo y from AC7 y to get y side length
Subtract Halo z from AC7 z to get z length.
I have to be very careful subtracting minus x from minus x to get plus x and getting plus x. It’s a plus x value of a line which is nevertheless still in the minus x part of the coordinate system. Both y values are positive values so the result of the subtraction is a positive value in a positive quadrant so no problem there. z is rather like x though. You should end up with a ‘cube’ about 500,000km on sides x and y but, I believe, somewhat different (shorter?) in z.
I think this will get you near to what you want.
Andrew
Andrew,
A long day programming. In order to make my model more accurate, I got hold of the positions of the planets, and DA14, every year on 1 Feb for the past century, and used this to correct my simulation annually. This makes it much more accurate. And also much faster.
Starting on 1 Feb 2005, I constructed a cube of 343 rocks with a side length of 48,000 km, and watched as the cube gradually extended into a long train. This train had a hole punched in it in Feb 2012, which resulted in one end folding back round creating the ‘fireball’ appearance.
The Earth then punched through the head of the ‘fireball’ on 15 Feb 2013. In this combination view, the Earth is seen approaching the head of the ‘fireball’ on 15 Feb, and passing through it a bit later, and dragging out a spray of rocks behind it after it has passed through. I’d hoped for some Earth impacts, but the nearest any rock got to it was 12,000 km.
The good thing about this is that we now have a very plausible rock cloud around DA14, through which the Earth passes. It is not extensive enough however to include something like Halo1093 or AC7V. So I’ll probably be trying to make the cloud bigger, and also trying to get an idea of what sort of velocity vectors rocks in it have got.
All in all, I think things have got a lot more more promising.
Frank
As your amanuensis for AC7V/Halo cube calcs, I used the position info for AC7V and Halo1093 in the below comments of yours.
http://cfrankdavis.files.wordpress.com/2013/05/ac7v.pnghttps://tallbloke.wordpress.com/2013/03/05/andrew-cooper-were-the-recent-asteroid-flyby-and-russian-meteor-strike-events-linked/comment-page-1/#comment-45656
Their x difference is 131,902 km
their y difference is 492,095 km
Their z difference is 703,763 km
I realised that It was too simplistic just to use these values to construct a cube. This shape is more like a cereal packet and when I considered pushing the cloud back up the orbit so as to encompass a potential 7 degree east radiant, it involved reducing the x value of AC7V even further to be the same as Halo1093, giving us a ‘cube’ that had only two dimensions, a sort of A4 page shape! Clearly if they are in a cube, they are at the same point on opposite sides and the plane between them (the A4 sheet) slices the cube in half or thereabouts. This is all hypothetical anyway but we have to start somewhere. So…
Their distance apart in xy is 509,466km
Their distance apart in xyz is 868,814
Although I think AC7 shows great promise, it is 703,000 km higher due to its 7.83 inclination and you are currently centring on DA14 at a 10.33 deg inclination. So why not reduce the large z value to be in keeping with the x and y values?
If they are 509,466 km apart in xy, that could be your cube side with z reduced to that value for neatness.
If you extended it to encompass a 7 degree east fragment, I would guess that with the reduced x and increased y to shove it back by the required 300,000 km, it would put it a further 200,000km from Halo 1093. So that would mean a cube of 700,000 km on a side. That might give fragments at 7 degrees east but they may not have a 7 degree radiant.
Andrew
Frank
The two links to your two comments containing position vectors fused so it didn’t work. Here they are again, separated:
Andrew,
Thanks for your calculations. I suppose that over the past couple of days I’ve seen how rock clouds can grow into very long trains. What they don’t seem to do is get any wider laterally. If anything, they shrink slightly. But what we need are rocks that lie quite a long way out laterally from any rock train if we are to reproduce Halo1093 or AC7.
I ended up yesterday more or less reproducing my earlier rock trains. The longer they become, the more they get knocked about, it seems. Although for the most part, in the absence of a close encounter, rocks seem to just get shuffled up and down the rock train.
Anyway, now that I have 100 years of data for planets and DA14, I may try running the model for 100 years, if only to find out whether DA14 got anywhere near anything over that time, and see what an accompanying rock train might look like. It should take two or three hours on my fastest computer, using the correction/adjustment feature I got working yesterday.
I might also give some of these rocks a slight push to move them gradually further from the sun, and perhaps circularise them slightly.
Frank
I’m looking forward to your 100-year run. Could you look out for any anomalous behaviour in August 1927 and 1928?
You said,
“I suppose that over the past couple of days I’ve seen how rock clouds can grow into very long trains. What they don’t seem to do is get any wider laterally. If anything, they shrink slightly. But what we need are rocks that lie quite a long way out laterally from any rock train if we are to reproduce Halo1093 or AC7.”
Your cube is 49,000 km on a side and, from what you say above, I gather that this cube extrudes whilst keeping roughly the same or slightly diminished cross sectional area. Is that right?
If so, a 500,000 km-per-side cube should extrude whilst keeping roughly the same cross sectional area. That would mean that Halo 1093 and AC7V type candidates should arrive in 2013 because the ‘fireball’ head and tail would have a 500,000 by 500,000 km cross-sectional area. Would you agree with that reasoning? I might be missing something.
Of course, I have to qualify this again. By ‘candidates’ I mean position only and not vector/angle to the sun. In other words, I mean that rocks from the original 500,000km cube would pass through space in Feb 2013 near to the position vectors of Halo1093 and AC7V. That same reasoning doesn’t allow for a definitive inference of any substantial change in the radiant angle of these rocks. However, I can guess that with a radius difference of 500,000km across the width of the ‘fireball’ stream, they would by definition have different absolute velocity vectors and therefore different geocentric radiant angles.
In fact, I believe that it could be the above phenomenon which is causing DA14 to drift back within the stream. We had already established that it arrived nearer each year (Earth catching up) due to its period being 366.2 days but, as you said, it was difficult to explain how it drifted back within the rock train, because the whole train should behave as DA14 does and arrive each year stepped back by the same amount. But if they are orbiting at subtly different radii, up to 49,000 km apart, you would see some relative drift of rocks within the stream over the years and, perhaps, by a few million km.
Can you pick rocks within the stream, other than DA14, and examine their long-term drift while noting whether they are on the inside track or outside track? Even if they move from inside to outside at different points in their orbits, it means they must have different absolute xy speeds as they approach Earth even if it gives a measly one degree or so of radiant shift.
If the above paragraph holds true, then the 500,000km cross-section rock train would exhibit even larger absolute velocity variations leading to larger geocentric xy radiant angles.
This might not be enough to give the required radiants for HALO1093 and AC7V but I think it has a lot of promise because rocks orbiting at different radii cannot help but have different absolute speeds and are therefore catching up with the Earth at different rates. I liken it to a motorway in rush hour where everything is whizzing along close-packed at 70mph but a passenger looking sideways out of his window sees an elegant drifting back and forth of the different lanes. To someone on a bridge overlooking the three lanes, that subtlety is lost.
And AC7V’s absolute velocity x and y values were not so very different from AC5 and its DA14-radiant progenitors. In other words, it was drifting forward on a slightly faster track, a miniscule angle difference but a whole 11 degrees, geocentric, further round east.
Andrew
Your cube is 49,000 km on a side and, from what you say above, I gather that this cube extrudes whilst keeping roughly the same or slightly diminished cross sectional area. Is that right?
Yes, that’s about right, although I have the impression that the lateral width of the train narrows slightly.
Also, the shape changes as the train proceeds around the orbit. It seems to lengthen at perihelion, and to shorten at aphelion. I say ‘seems’ because I have just been looking at the view down onto the x-y plane.
If so, a 500,000 km-per-side cube should extrude whilst keeping roughly the same cross sectional area. That would mean that Halo 1093 and AC7V type candidates should arrive in 2013 because the ‘fireball’ head and tail would have a 500,000 by 500,000 km cross-sectional area. Would you agree with that reasoning?
I tried one almost that size yesterday, and when it arrived up at the Earth, it had retained roughly the original spacing. However there were only 3 or 4 rocks visible passing through the solar system in 2013, the other 340 lying along an enormously long rock train.
I’m looking forward to your 100-year run. Could you look out for any anomalous behaviour in August 1927 and 1928?
I started one today, but abandoned it after about 40 years. From a very small rock cloud with 100 km spacing between rocks, it steadily lengthened over the next 20 years to the point that its nose had caught up with its tail. However, it wasn’t producing an orbital path as I had expected, but a spiral, with the tail spiralling slowly outwards.
I’m not sure at the moment whether this is a real effect or not. There is a possibility that my annual adjustment of the rock cloud’s locations and velocities may be generating this effect, because I use the difference between DA14’s computed position and NASA’s position to adjust all rocks in the rock cloud. This is OK when the rock cloud is quite short, but doesn’t make sense if it extends in a circle right round the Sun.
However, it may equally be the result of encounters with the Earth. In mid-Feb 1918, it seems that the rock cloud passes within about 440,000 km of the Earth, and this has quite a big effect, and may be generating the spiral.
Equally, it may be a consequence of using long time steps.
I’ll get to the bottom of it in the end.
Anyway, I didn’t notice anything in 1927/28. But I wasn’t looking for anything.
In fact, I believe that it could be the above phenomenon which is causing DA14 to drift back within the stream. We had already established that it arrived nearer each year (Earth catching up) due to its period being 366.2 days but, as you said, it was difficult to explain how it drifted back within the rock train, because the whole train should behave as DA14 does and arrive each year stepped back by the same amount. But if they are orbiting at subtly different radii, up to 49,000 km apart, you would see some relative drift of rocks within the stream over the years and, perhaps, by a few million km.
Rock clouds are proving to be rather complex things. Rocks are being tugged to and fro the whole time, and they don’t all behave like DA14. Even when there isn’t a close encounter, some rocks are being slightly accelerated, and others decelerated, and fractional changes in relative speeds cause rocks to bunch up in some places, spread apart in others.
Can you pick rocks within the stream, other than DA14, and examine their long-term drift while noting whether they are on the inside track or outside track? Even if they move from inside to outside at different points in their orbits, it means they must have different absolute xy speeds as they approach Earth even if it gives a measly one degree or so of radiant shift.
I want to do this. But for the moment all I’ve done is to change my naming conventions for rocks to include their x,y,z offset from DA14 at origin. So when I see closest approach figures, I know where they started life.
Anyway, at the moment, I’m trying to find out whether the spirals I’ve been seeing are illusory or not.
Frank
Thanks for the feedback
You are right about the stretching out at perihelion. Even the Earth varies by 1km per second between perihelion and aphelion, DA14 more so. So the stretching out is like cars pulling away from traffic lights or, more realistically, speeding up after rubber-necking. The traffic stream stretches.
“I tried one almost that size yesterday, and when it arrived up at the Earth, it had retained roughly the original spacing. However there were only 3 or 4 rocks visible passing through the solar system in 2013, the other 340 lying along an enormously long rock train.”
I don’t fully understand this. You say that it had retained roughly the original spacing but that it was an enormously long rock train that seems to have straddled more than the solar system. Doubtless this makes sense at a deeper resolution than I am giving it. Can you explain it a bit more?
Andrew
Andrew,
I started with a cubic rock cloud of 343 rocks with an 8,000 km spacing between rocks, and that was the one I used to generate the images over the last couple of days.
I then increased the spacing to 100,000 km, and found that the resulting rock train was not only thicker but also longer. Since both clouds had 343 rocks in them, that meant that the second train was very dispersed.
I then decreased the spacing to 1,000 km, and then 100 km. These produced thin and relatively short trains.
However, it’s beginning to look as if there’s some sort of problem with my new (as of yesterday) annual adjustment to put planets in the right place. If I take out this adjustment, the rock trains are much shorter. It looks as if there’s an error which gets multiplied.
Thanks Frank
I’ll let you get on with your suspected glitch before asking any more questions. With luck the big cloud may still have a future by the sound of it.
Andrew
Andrew,
It looks like there is indeed something happening in 1927 and 1928. I’m working on the problem of the length of my rock trains, but I noticed that in 1928 my latest rock train had folded in half. And in August a hole appeared in it. I’ll try running it again, and get a screen capture or two.
I’m also pretty sure that there’s a big event in 1918, which seems to have the effect of throwing the little rock train I have at that time into a quite new orbit.
It’s beginning to look as if every 10 years or so something fairly major happens to DA14. In which case, since these events result in trains of rocks detaching themselves, there are probably rocks all over the shop by the time we get to 2013..
I’m currently running my model for a year, and then on 1 Feb of each year I’m correcting the vectors of all known bodies. I can’t correct my rock train bodies in this way, because they’re not known bodies. So the best I can do is adjust them so that they are in the same relative position to the corrected DA!4, and the same relative speeds. But these are estimates. And at the moment my estimates seem to result in a rock train that lengthens a lot more rapidly than one which is just let run without any corrections or adjustments. Whatever I end up doing, I think my rock trains are going to be educated guesses about the positions and speeds of rocks.
Andrew,
I’ve got a couple of composite images showing Earth encounters with the DA14 rock train in 1927. In the first of them, the rock train is shown intact in red in Jan 1927. Thereafter, it begins to bend as a continues in its orbit throughout the rest of 1927. And in the second image the sequence continues through 1928 with a hole gradually appearing in the rock train due to a second encounter with the Earth. Offset to the bottom right are images of the rock train in 1929 and 1930. Quite a lot was happening.
The rock train grew from a small cube of rocks in 1913. But most likely there was a rock train in place at that time.
I’ve had the thought that DA14 might be spawning new rocks as it goes along. So another approach to making rock clouds would be to have DA14 emit rocks (perhaps as a result of impacts upon it) at some small relative velocity. That way, there would always be a rock cloud around it. It might also emit rocks every time it had a fairly close encounter with the Earth
Frank
That’s very intriguing. Was this phenomenon in 1927 and 1928 the only time this happened last century?
I thought DA14 was going to be nearer the Earth instead of trailing several weeks behind.
What I find intriguing is that, despite DA14 being nowhere near the Earth, it is only one week ahead of the bend. The Earth is so far ahead, at all times if the year, that it can’t have caused the bend. Would you agree or am I not seeing the full picture?
Andrew
Whoops, I forgot about your last comment saying something happens to DA14 every ten years and there must be rocks all over the shop. So forget my first question because that answers it.
It’s still a nice coincidence that I focused specifically on 1927 and 1928 and now you have some rock antics in those two years. But it seems to be happening for a different reason.
Andrew,
I’m a bit puzzled myself at what’s happening in 1927 and 1928. It’s fairly clear that the Earth passes through the rock train in Feb 1928, and blows a hole in it. It would appear that it passed close to the end of it the year before, but I can’t see where.
Also, it’s the rock train that is being affected, and not DA14,
I haven’t looked again at what happened in 1918, but I believe that DA14 and its shorter rock train were thrown into a new orbit.
I haven’t run the thing for the full 100 years, but it rather looks to me that any rock train lying along DA14’s orbit simply gets shredded over the subsequent century. But our interest should really be centred upon DA14 and the rocks in its near vicinity. It doesn’t really matter what happened to the rest of the rocks.
Frank
“But our interest should really be centred upon DA14 and the rocks in its near vicinity. It doesn’t really matter what happened to the rest of the rocks.”
Agreed. If the rock train was really long we wouldn’t have had all the rocks come in one week in one year. Although I find the extrusion principle acceptable, there must be a concentration around DA14 for that short burst to have happened. Of course, that makes me wonder whether, for that to be the case, DA14 must have emitted those rocks in recent years for them not all to have decayed from a cloud into a train. But with a Roche limit of much less than 28,000km, the Earth couldn’t be the culprit. It was millions of km away. A direct hit on DA14 is a vanishingly small chance. So I have no answer to that conundrum.
Andrew
Gents: In a couple of days I will move this post forward in time so comments remain open. The URL will change to https://tallbloke.wordpress.com/2013/05/05/andrew-cooper-were-the-recent-asteroid-flyby-and-russian-meteor-strike-events-linked/
Thanks for your ongoing interesting debate – Rog
Rog,
Thank you for allowing us to keep on scribbling on the margins of your blog. 🙂
Andrew,
The broad lesson I’ve drawn from my historical excursion over the pat few days is that any rock train that developed around DA14 would not have turned into a nice elliptical train of rocks circling the Sun, but would have been fairly rapidly dismembered, and the rocks sprayed throughout the solar system.
However, in all likelihood those which were in very close proximity to DA14 may well have continued accompanying it throughout its many adventures, although most likely gradually dispersing.
Accordingly, it seems to me that it would probably be more productive to return to the ‘fireball’ rock cloud that is generated by DA14’s rock train close approach to the Earth in February 2012, and to your circularisation ideas. During 2012 the rock train fore and aft of DA14 is turned back on itself, becoming two merged rock trains that have folded together to form the ‘fireball’, through the head of which the Earth passes on 15 Feb 2013. I’m thinking now of concentrating on the head of that ‘fireball’, and gradually and judiciously enlarging it to encompass the Earth.
So what I’m thinking of doing is to find out which rocks in my current rock clouds are nearest to the Earth in Feb 2012, and use these to extrapolate/interpolate a set of rocks around them, so that instead of my 343 rocks being dispersed along a very long train, they are all concentrated around DA14 at the head of the fireball. This should make for a lot of close approaches/impacts on 15 feb 2013. It’s my old ‘shotgun blast’ technique, slightly reworked. What we should have is a fairly plausible set of rocks in different places in the rock cloud ‘fireball’, and travelling at slightly different velocities.
In the meanwhile I’m thinking that I may run my solar system for 100 years simply to find out when DA14 comes closest to the Earth each year. I’ll be correcting all bodies using NASA’s Horizons vectors as I go along. This might be quite instructive. Incidentally, NASA’s vectors now go back to 1600 AD. Last I heard (from you?) they only went back to about 1900 AD.
Andrew,
But with a Roche limit of much less than 28,000km, the Earth couldn’t be the culprit. It was millions of km away. A direct hit on DA14 is a vanishingly small chance. So I have no answer to that conundrum.
The Earth is regularly nudging the rocks in DA14’s rock cloud to and fro. But I don’t think that tidal forces need be the only explanation for the break-up of such an asteroid.
For a start, there are other things flying around, and it’s possible that DA14 is the remnant of some collision. After all, it seems that whenever asteroids are found, they usually have impact craters on them. I saw a documentary about the Chelyabinsk meteor in which it was said that fragments of it showed that it was one which had collided with something at some point in its history.
But I’ve also supposed that, when bits are knocked off something like DA14, some of these may go into orbit around it (I have come across at least one instance of this), and may fall back upon it. There might be small rocks continually bouncing off the surface of DA14. These would not be slowed like terrestrial rocks, because there would be no atmosphere to do so. So bodies like DA14 might be being eroded by the continual impact of tiny hammers.
In addition there’s heating and cooling. If DA14 were not spinning (or even if it was) its sun-facing side would be much hotter than its shaded side. The regular expansion and contraction of rocks might result in rock ‘fatigue’ not dissimilar to metal fatigue. Once a body had fractured internally so that it had become a pile of separate rocks in contact with each other, it seems quite plausible that these might drift apart.
Also, DA14 and its accompanying rock cloud might capture other bodies. It was said of the Chelyabinsk meteor that it was made of different stuff than DA14, so couldn’t have been a companion. But I was not convinced.
Incidentally, as part of an investigation of global warming, a year or two back I constructed a little simulation model of solar terrestrial heating and cooling as the Earth revolved But I didn’t know how to model the atmosphere very well. It might be quite fun to construct a similar model of DA14,using known rock thermal conductivity and capacitance,and braking stresses and so on.
Rog,
Thanks for bringing the thread forward. Appreciated.
Frank,
I have had several thoughts on emitting rocks and also on the idea of focusing on the rocks in DA14’s vicinity. I haven’t typed them all up yet so I’ll post them as I go. The following ideas were typed up last night before seeing your two comments this afternoon. I have noted your views on collisions etc and agree with all of them. My only concern is the regularity with which they happen. So anyway, my comment from last night hasn’t been edited in the light of your answers to my ‘conundrum’ on the emitting/ collisions/Roche limit. It is mostly concerned with an idea for a method of emitting while keeping a DA14 provenance. I shall post two more ideas on the related subject of returning to DA14’s vicinity and circularisation. One will be about the rocks that are 500,000km to 700,000km back from the head of the fireball and should, I believe, be circularised, crossing DA14’s orbit to the inside track. The second is focusing on the several outliers (not outriders) that I see in various snapshots. These are slippery but I think they are ultra close encounters and hold much promise.
Last nights comment:
You said:
“I’ve had the thought that DA14 might be spawning new rocks as it goes along. So another approach to making rock clouds would be to have DA14 emit rocks (perhaps as a result of impacts upon it) at some small relative velocity. That way, there would always be a rock cloud around it. It might also emit rocks every time it had a fairly close encounter with the Earth.”
The Roche limit for DA14 and Earth is less than 28,000km. How much less I don’t know but they knew that, with the 2013 close encounter, it would jiggle a bit but wouldn’t shred. Having said that, the bigger DA14 of old, if it existed could have been looser, more elongated and spinning faster, all factors that would increase the Roche limit and permit a greater chance of shredding.
Apropos emitting rocks, I had wondered about a one-off generation of cloud rocks. The reason was that I was wondering if your rocks on the periphery of the cube were qualified to exhibit DA14 velocity vectors. I presume all 343 rocks have the same v as DA14? If so they get constrained to follow a different orbit. But if there is a sideways explosion of 343 rocks from DA14, they would exhibit a truer vxyz provenance from DA14. Like the more ersatz ones mentioned above, they would have different orbits from DA14, but they would be truer to reality.
By sideways explosion, I mean sending rocks outwards radially in a star that is perpendicular to the orbit. I think a conventional explosion with rocks sent forward and aft could confound things a little because the increased/decreased v would be confounded with the increased/decreased v of being sent to the inside or outside track. Then again, if you’re not concerned about that, a proper explosion like a slow-motion bomb would be pretty cool.
I would suggest making DA14 temporarily 1kg and the exploded rocks the same or less so that there would be a vanishingly small escape velocity. That way all the delta v goes into the required movement as opposed to overcoming DA14 gravity (which I admit is miniscule anyway). The delta v of the explosion would probably need to be in the order of 1m/s I would have thought, ie each rock leaving DA14 at that speed in its designated direction. DA14 could be put back to its old mass once they had got thousands of km away.
I’m not sure I would agree with emitting rocks on a regular basis unless it’s every few centuries. I keep having to remind myself that we could have been waltzing with DA14 and its companions for millenia and so a long and continually evolving train is likely- even with only occasional emissions/explosions. I also keep in mind that DA14 could be a sibling of many similar sized or bigger rocks. In other words, it could be an anonymous cloud rock albeit an above average one, with the parent body happily orbiting 5 million km .
Andrew
Frank
I am often seeing outliers (as opposed to outriders) in your snapshots. These are rocks that are either showing behaviour that is an extreme version of the gentler rock train perturbation or are so extreme that they lose any sort of provenance from one snapshot overlay to the next. Far from being tiresome anomalies- and I admit, I do blot them out subconsciously for the most part- I think they could be the secret for substantially different radiants and inclinations. I think they are all sub 10,000 km close encounters and as such the great circle around which they pass has a vastly greater influence on the angle of their inclination and their eccentricity, quite apart from the obvious greater slingshot effect. That is, where they are flung to as opposed to how fast.
This is different from the gentle tug on the rest of the train, a tug which can only be pretty well in one direction due to the distance and radiant which is set for days and weeks at a time. Because these outliers are, as I posit, super-close encounters there are very few of them, only one or two per experimental run and sometimes none.
There was one in the snapshot, linked below, that sat in the middle of the August gap almost unaffected for four years, yet it must have been one of the closest approachers, by definition. I suspect that it was one of my pet polar passes which would affect inclination only and not eccentricity or a different xy track.
Andrew
Andrew,
By sideways explosion, I mean sending rocks outwards radially in a star that is perpendicular to the orbit. I think a conventional explosion with rocks sent forward and aft could confound things a little because the increased/decreased v would be confounded with the increased/decreased v of being sent to the inside or outside track. Then again, if you’re not concerned about that, a proper explosion like a slow-motion bomb would be pretty cool.
I would suggest making DA14 temporarily 1kg and the exploded rocks the same or less so that there would be a vanishingly small escape velocity. That way all the delta v goes into the required movement as opposed to overcoming DA14 gravity (which I admit is miniscule anyway). The delta v of the explosion would probably need to be in the order of 1m/s I would have thought, ie each rock leaving DA14 at that speed in its designated direction. DA14 could be put back to its old mass once they had got thousands of km away.
I think this is a good proposal. The only slight variation I would suggest is that they don’t all have the same delta v, and that they have a very small forward/backward speed. It would depend when they were started. In some ways it would be simplest if the explosion happened just before the Feb 2012 encounter. Or maybe just after.
I spent a while today thinking about asteroids heating and cooling. It seemed to me that small rocks would probably heat up fairly uniformly all over, but large non-rotating rocks would get very hot on one side, and very cold the other. If the hot half expanded relative to the cold half, stresses would be set up that could result in rock failing in tension on its cold, shaded side. Such a sudden failure could, I imagined, result in an asteroid exploding. I already have working code to model the dynamic behaviour of 3D frameworks of nodes connected by struts (it’s actually more or less the same code I use in my orbital simulation model), and I also have code that I wrote to simulate solar heating and cooling of a planet surface. So it shouldn’t be too difficult to bring the two bits of code together to model an asteroid being heated and cooled. Then I might be able to test the inkling I had today that such asteroids might explode. It’s a bit of a long shot, of course, but it might provide a way – other than tidal forces – which could bring disintegration.
There was one in the snapshot, linked below, that sat in the middle of the August gap almost unaffected for four years, yet it must have been one of the closest approachers, by definition.
I noticed that one too. I think the explanation for it is as follows: When the Earth passes over a rock train, it pulls rocks on either side of it towards it, so that the ones on the right of its path are pulled left, and the ones on the left of its path are pulled right. And the nearer these rocks are, the stronger they are pulled, and the faster they accelerate. However, upon occasion, the Earth will pass almost exactly over one of the rocks in the rock train, and give it no left-right pull at all. And that’s what we’re seeing with that lone rock in the middle of the gap. However, that rock has probably been very greatly accelerated upwards or downwards, And if you could see the rock train from the side, it would have a pronounced spike in it, with that rock sitting at the tip of the spike.
Frank
Yep, good comment. I concur on everything.
As for that rock in the middle of the August gap, you are confirming my thoughts on that too. When I said polar, I really meant over or under (north or south in z which is the same as flying over 67 or so north or south).
The reason I plumped for this sort of ‘polar pass’ several weeks ago was that a) it would reduce inclination if passing above circa 67 deg north b) change the radiant on approach the next year if it went only a tiny bit to one side c) would arrive a day or so earlier than its previous year’s orbit d) would land somewhere on that same radiant near the top of the Earth, 67 degrees north e) by arriving earlier would be lower in z and therefore more likely to hit than the last year.
I think you are right that it would look different in z. I believe, in that snapshot, it was a circa 67 deg south pass so that the inclination was increased, not decreased. This would mean that it had difficulty keeping up with the train as seen in xy, looking down in z and so that’s why it drifted backwards. Conversely, if it had been a circa 67 deg north pass it would have been drifting forward and faster in xy.
It’s certainly food for thought. The only trouble is, I suppose it is harder to model because it needs to be an accurate shot over the 67 deg mark….mind you, I just had a thought, you have quite a stable of northern misses in your pursuit of grazers. Could you dig these out and send them on their way to 2014? It would be a similar effect as that from 2012 to 2013.
Andrew
Andrew,
100 years of DA14 closest approaches to Earth ( in metres, times +/- 2.25 hours). My notion that there were regular close approaches seems to have been mistaken. After 1 Feb 1913, there was only one close approach: in Feb 1918. Snapshot shows Earth in green, Moon in blue, DA14 in red.
DA-14 closest approach: 6.9634449E10 on 07 Aug 1913 20:22:40 dt=8192.0 s
DA-14 closest approach: 4.0880796E10 on 16 Aug 1914 06:24:00 dt=8192.0 s
DA-14 closest approach: 1.20976261E10 on 22 Aug 1915 23:57:52 dt=8192.0 s
DA-14 closest approach: 1.59490068E10 on 20 Aug 1916 00:10:40 dt=8192.0 s
DA-14 closest approach: 2.9408684E10 on 07 Feb 1917 03:54:40 dt=8192.0 s
DA-14 closest approach: 4.35704E8 on 17 Feb 1918 05:07:12 dt=8192.0 s
DA-14 closest approach: 4.4532101E9 on 18 Feb 1919 12:58:40 dt=8192.0 s
DA-14 closest approach: 1.2513836E10 on 20 Feb 1920 19:35:28 dt=8192.0 s
DA-14 closest approach: 2.16767713E10 on 22 Feb 1921 17:06:08 dt=8192.0 s
DA-14 closest approach: 3.13907978E10 on 25 Feb 1922 15:38:40 dt=8192.0 s
DA-14 closest approach: 4.1436217E10 on 01 Mar 1923 01:33:52 dt=8192.0 s
DA-14 closest approach: 5.1702182E10 on 03 Mar 1924 06:56:00 dt=8192.0 s
DA-14 closest approach: 6.2074819E10 on 06 Mar 1925 22:38:56 dt=8192.0 s
DA-14 closest approach: 7.2428528E10 on 10 Mar 1926 08:34:08 dt=8192.0 s
DA-14 closest approach: 8.2810249E10 on 14 Mar 1927 01:18:56 dt=8192.0 s
DA-14 closest approach: 9.3121782E10 on 16 Mar 1928 11:14:08 dt=8192.0 s
DA-14 closest approach: 1.03312458E11 on 20 Mar 1929 07:30:09 dt=8192.0 s
DA-14 closest approach: 1.13342734E11 on 23 Mar 1930 17:25:21 dt=8192.0 s
DA-14 closest approach: 1.23255407E11 on 27 Mar 1931 10:10:09 dt=8192.0 s
DA-14 closest approach: 1.32954538E11 on 29 Mar 1932 20:05:21 dt=8192.0 s
DA-14 closest approach: 1.42431912E11 on 02 Apr 1933 09:31:45 dt=8192.0 s
DA-14 closest approach: 1.51667491E11 on 05 Apr 1934 17:10:25 dt=8192.0 s
DA-14 closest approach: 1.60700514E11 on 09 Apr 1935 05:22:09 dt=8192.0 s
DA-14 closest approach: 1.69469608E11 on 11 Apr 1936 13:00:49 dt=8192.0 s
DA-14 closest approach: 1.77985487E11 on 14 Apr 1937 17:21:05 dt=8192.0 s
DA-14 closest approach: 1.86235814E11 on 17 Apr 1938 22:43:13 dt=8192.0 s
DA-14 closest approach: 1.9421877E11 on 20 Apr 1939 21:15:45 dt=8192.0 s
DA-14 closest approach: 2.01856139E11 on 23 Apr 1940 00:21:21 dt=8192.0 s
DA-14 closest approach: 2.09253138E11 on 25 Apr 1941 19:35:29 dt=8192.0 s
DA-14 closest approach: 2.16334746E11 on 28 Apr 1942 20:24:33 dt=8192.0 s
DA-14 closest approach: 2.23098274E11 on 01 May 1943 07:34:25 dt=8192.0 s
DA-14 closest approach: 2.29537006E11 on 03 May 1944 03:50:25 dt=8192.0 s
DA-14 closest approach: 2.35629511E11 on 05 May 1945 07:08:50 dt=8192.0 s
DA-14 closest approach: 2.41389781E11 on 07 May 1946 20:35:14 dt=8192.0 s
DA-14 closest approach: 2.46786277E11 on 09 May 1947 20:22:26 dt=8192.0 s
DA-14 closest approach: 2.51855913E11 on 10 May 1948 20:09:38 dt=8192.0 s
DA-14 closest approach: 2.56562659E11 on 12 May 1949 16:38:26 dt=8192.0 s
DA-14 closest approach: 2.6091461E11 on 14 May 1950 02:46:26 dt=8192.0 s
DA-14 closest approach: 2.64892285E11 on 15 May 1951 17:27:30 dt=8192.0 s
DA-14 closest approach: 2.68512199E11 on 15 May 1952 13:56:18 dt=8192.0 s
DA-14 closest approach: 2.71759032E11 on 16 May 1953 20:45:54 dt=8192.0 s
DA-14 closest approach: 2.74616402E11 on 17 May 1954 17:14:42 dt=8192.0 s
DA-14 closest approach: 2.77128905E11 on 18 May 1955 00:04:18 dt=8192.0 s
DA-14 closest approach: 2.79235199E11 on 17 May 1956 16:00:02 dt=8192.0 s
DA-14 closest approach: 2.80928682E11 on 17 May 1957 14:58:10 dt=8192.0 s
DA-14 closest approach: 2.82231407E11 on 17 May 1958 17:14:42 dt=8192.0 s
DA-14 closest approach: 2.83122434E11 on 18 May 1959 00:04:18 dt=8192.0 s
DA-14 closest approach: 2.83622769E11 on 16 May 1960 14:58:10 dt=8192.0 s
DA-14 closest approach: 2.83721105E11 on 16 May 1961 20:45:55 dt=8192.0 s
DA-14 closest approach: 2.83397652E11 on 16 May 1962 20:45:55 dt=8192.0 s
DA-14 closest approach: 2.8268177E11 on 16 May 1963 16:12:51 dt=8192.0 s
DA-14 closest approach: 2.81571557E11 on 16 May 1964 10:25:07 dt=8192.0 s
DA-14 closest approach: 2.80070226E11 on 16 May 1965 16:12:51 dt=8192.0 s
DA-14 closest approach: 2.78191047E11 on 17 May 1966 08:08:35 dt=8192.0 s
DA-14 closest approach: 2.75923534E11 on 18 May 1967 11:26:59 dt=8192.0 s
DA-14 closest approach: 2.73292362E11 on 18 May 1968 03:22:43 dt=8192.0 s
DA-14 closest approach: 2.70271021E11 on 19 May 1969 17:01:55 dt=8192.0 s
DA-14 closest approach: 2.66869195E11 on 20 May 1970 22:36:51 dt=8192.0 s
DA-14 closest approach: 2.63127925E11 on 22 May 1971 15:34:27 dt=8192.0 s
DA-14 closest approach: 2.59035644E11 on 23 May 1972 10:48:35 dt=8192.0 s
DA-14 closest approach: 2.54600872E11 on 25 May 1973 05:00:51 dt=8192.0 s
DA-14 closest approach: 2.4980077E11 on 27 May 1974 11:37:39 dt=8192.0 s
DA-14 closest approach: 2.44693549E11 on 29 May 1975 09:08:19 dt=8192.0 s
DA-14 closest approach: 2.3925357E11 on 30 May 1976 20:18:11 dt=8192.0 s
DA-14 closest approach: 2.33476637E11 on 02 Jun 1977 01:53:08 dt=8192.0 s
DA-14 closest approach: 2.27410199E11 on 04 Jun 1978 17:36:04 dt=8192.0 s
DA-14 closest approach: 2.21003252E11 on 07 Jun 1979 04:45:56 dt=8192.0 s
DA-14 closest approach: 2.14293496E11 on 08 Jun 1980 20:28:52 dt=8192.0 s
DA-14 closest approach: 2.0726032E11 on 11 Jun 1981 15:43:00 dt=8192.0 s
DA-14 closest approach: 1.99937901E11 on 14 Jun 1982 09:42:28 dt=8192.0 s
DA-14 closest approach: 1.92296993E11 on 17 Jun 1983 12:48:04 dt=8192.0 s
DA-14 closest approach: 1.84396579E11 on 19 Jun 1984 06:47:32 dt=8192.0 s
DA-14 closest approach: 1.762198E11 on 22 Jun 1985 13:24:20 dt=8192.0 s
DA-14 closest approach: 1.67784645E11 on 25 Jun 1986 11:56:52 dt=8192.0 s
DA-14 closest approach: 1.59071388E11 on 28 Jun 1987 21:52:04 dt=8192.0 s
DA-14 closest approach: 1.50142894E11 on 30 Jun 1988 22:41:08 dt=8192.0 s
DA-14 closest approach: 1.4105197E11 on 04 Jul 1989 12:07:32 dt=8192.0 s
DA-14 closest approach: 1.31719537E11 on 07 Jul 1990 15:13:08 dt=8192.0 s
DA-14 closest approach: 1.22219512E11 on 11 Jul 1991 03:24:52 dt=8192.0 s
DA-14 closest approach: 1.12572015E11 on 13 Jul 1992 08:47:00 dt=8192.0 s
DA-14 closest approach: 1.02766215E11 on 17 Jul 1993 00:29:56 dt=8192.0 s
DA-14 closest approach: 9.2812403E10 on 20 Jul 1994 08:08:37 dt=8192.0 s
DA-14 closest approach: 8.2797076E10 on 23 Jul 1995 22:36:53 dt=8192.0 s
DA-14 closest approach: 7.2725815E10 on 26 Jul 1996 08:32:05 dt=8192.0 s
DA-14 closest approach: 6.2599569E10 on 30 Jul 1997 00:15:01 dt=8192.0 s
DA-14 closest approach: 5.2456862E10 on 02 Aug 1998 10:10:13 dt=8192.0 s
DA-14 closest approach: 4.2439291E10 on 05 Aug 1999 22:21:57 dt=8192.0 s
DA-14 closest approach: 3.25578097E10 on 08 Aug 2000 10:33:41 dt=8192.0 s
DA-14 closest approach: 2.29629891E10 on 11 Aug 2001 19:27:01 dt=8192.0 s
DA-14 closest approach: 1.38960159E10 on 15 Aug 2002 03:05:41 dt=8192.0 s
DA-14 closest approach: 5.9115182E9 on 18 Aug 2003 01:38:13 dt=8192.0 s
DA-14 closest approach: 1.26149901E9 on 19 Aug 2004 05:58:29 dt=8192.0 s
DA-14 closest approach: 5.7119887E9 on 21 Aug 2005 11:33:25 dt=8192.0 s
DA-14 closest approach: 1.2823339E10 on 24 Aug 2006 12:22:29 dt=8192.0 s
DA-14 closest approach: 2.05081907E10 on 29 Aug 2007 12:58:45 dt=8192.0 s
DA-14 closest approach: 3.26692332E10 on 11 Apr 2008 19:50:29 dt=8192.0 s
DA-14 closest approach: 2.83483607E10 on 13 Sep 2008 11:16:21 dt=8192.0 s
DA-14 closest approach: 2.52144128E10 on 17 Feb 2009 11:56:53 dt=8192.0 s
DA-14 closest approach: 1.71703654E10 on 14 Feb 2010 13:24:22 dt=8192.0 s
DA-14 closest approach: 9.4506547E9 on 15 Feb 2011 00:47:02 dt=8192.0 s
DA-14 closest approach: 2.60680243E9 on 16 Feb 2012 08:38:30 dt=8192.0 s
DA-14 closest approach: 3.3476168E7 on 15 Feb 2013 19:33:26 dt=512.0 s
It seems that DA14 spent most of the past century a long way from Earth. Which would suggest that it had from Feb 1918 to Feb 2013 to build up a rock train. I just so happened to start my rock train on 1 Feb 1913 just as DA14 was approaching its next rendezvous with Earth, and watched it get shredded. If I’d waited for another 10 years, I suspect I’d have seen a different picture, with a much more stable rock train.
Frank,
That’s great. I’ve gone through the whole lot. What strikes me is that there is a predictable pattern until around the 1980’s and they are at opposite ends of their orbits between 1950 and 1970 (150,000,000km Earth orbit plus 133,000,000km DA14 perihelion = 2.83 E11 metres).
After the 1980’s or so the close approaches flip to the other node (Feb). That is probably because the nodes are/were something like 179 days apart and not exactly half a year apart so instead of getting a close approach every 6 months, DA14 got reined in on the Feb node after the August node had had its turn. But that’s a guess and I realise you are doing whatever is the single closest approach- both nodes would have seen close approaches but August’s were the closest, then February.
I see what you say about starting the rock train at an inauspicious time. Are you going to rerun it, starting post 1913? As you say, it appears it had many years to build up a train. It may be better to go from 1919 but it would also be interesting to see what sort of ripples 1918 would make and then do another new experiment starting from 1919. But you’re the one who has to do it and I just sit back and watch! And I presume this list took some time- it’s clever to programme it for close approaches.
One last point: you had said that something seemed to be happening every 10 years. Was that what might be called a data mirage? A few comments back, it seemed you were concluding that the train was being broken up every ten years and so it wouldn’t have time to build up.
Andrew
Andrew,
Are you going to rerun it, starting post 1913?
I did exactly that, starting in 1924, to be on the very safe side. But my rock train builds up fairly rapidly, and after 30 years or so,it extends around the entire orbit. And once this happens, of course the rock train gets shredded by the Earth.
But I don’t place much faith in my rock trains. I’m correcting planetary vectors every year using NASA data, but my accompanying adjustments of the rock train are just guesses (and not very good guesses). After about 30 years, the Earth is on the far side of the orbit from DA14, and the DA14 rock train which started in 1924 as a tight bunch of rocks now extends around the entire orbit. Also, the rock train is showing signs of spiralling inwards. This is something I’ve seen several times now, but I’m very far from being convinced that it’s not an artifact of my ‘adjustments’, so I’m not placing any credence in it at present. . .
A few comments back, it seemed you were concluding that the train was being broken up every ten years and so it wouldn’t have time to build up.
I think it’s simply that the longer a train gets, the more likely it is to encounter the Earth and be broken up. My current rock train takes 30 years to extend around the entire orbit, but rocks along it would have encountered the Earth long before 30 years were up. Only rocks in the immediate vicinity of DA14 would stay in place. Which is why it seems best to focus on a small cloud around DA14.
Andrew,
I’ve gone back to building rock clouds on 1 Feb 2005. One thing that I thought might be interesting would be to plot the path of Earth relative to DA14 and its accompanying cloud. This image is a DA14centric view down onto the x-y plane, with the DA14 rock cloud a blob in the centre, and Earth(green) and Moon(blue) creating a blue-green path. The curving black lines at the sides are probably Mars and Venus. Earth-Moon start off at top right, and they very roughly circle anticlockwise, before finally moving closer to DA14 in 2012.
What’s interesting to me about this is that the Earth remains in one quadrant relative to DA14 the whole time. And since it seems to be the body with the greatest gravitational effect on DA14, it would seem that from 2005 to 2013 it was pulling DA14 in a +x and +y direction (don’t know about z). The effect would have been small, but fairly consistent. I’ve been wondering what effect that would be likely to have on DA14. It would seem to slow DA14 at one point in its orbit, and speed it up on the opposite side. The Sun and other planets would have little net pull in any direction, I imagine, since they would be circling around DA14.
Frank
“Only rocks in the immediate vicinity of DA14 would stay in place. Which is why it seems best to focus on a small cloud around DA14.”
I agree but I’m glad you did these century-long rock train experiments because it has helped me visualise how the bend behaves and the rocks just up range of the bend when it returns to the Earth. I presume you are much more confident about your 2005-13 run and especially so for the 2012 folding in half and 2013 ‘fireball’ head. This makes perfect sense to me for a close approach, now that I’ve seen so many iterations of rock train approaches and bends.
One of your runs that had a big impression on me was the 1929 one that had the two October snapshots off to the side for the early thirties. The main thing that dawned on me was that the entire folded tail was whipping from inside DA14’s orbit to outside around the August node and back to inside in February. This is fairly obvious when you consider each perturbed rock in isolation because it is the essence of circularisation (those October shots show the folded tail orbiting almost in the wake of the Earth). But when considering the whole tail/bend/Feb-overlap phenomenon, it is clear that in February 2013 the tail was flipping back past the main train to the inside track again. The two halves are swopping places with rocks criss-crossing orbits by the dozen or hundreds or thousands. Nowhere is this behaviour more extreme than right on the bend itself which means right inside the ‘fireball’ head that hits the Earth in 2013. By ‘extreme’, I mean both in terms of rock density and relative vx and vy. That would explain perfectly all those varying trajectories of 11th to 16th of Feb. They are not just ‘overlaying’ and ‘concertinaing’ as we have said before. They are imploding from a tight, dense ‘V-shape’ bend, into the ‘fireball head’ just before the Earth passes through it. It is the xy component of the separate rocks in this implosion (due to different vx and vy’s) that cause the different radiants relative to the sun.
I would guess that if you plotted the summed vx and vy’ of, say, a dozen rocks either side of the bend just before it becomes overlaid, they would show the requisite xy velocity differences to account for the 20 deg of angle variation we need. Or, you could give me the state vectors, including DA14 as a reference rock, for these rocks at that snapshot in time and I could do the sums.
If they didn’t show a full 20 deg difference, then I believe they would show 5 to 10 deg and you might gain the remainder from tweaking the initial conditions in 2012 (most likely the density of rocks at that time, giving rise to closer encounters- this would be in keeping with our thoughts on hanging close on the coat tails of DA14.)
This comment has become my third comment of the three I mentioned I would do concerning the 2012 and 2013 close encounters and staying close to DA14. The other two were the one on outliers and the one on radial emmision of rocks to maintain a close cloud formation. When I outlined what I was going to say in this comment, I mentioned the behaviour of rocks 400-500,000 km down range from the ‘fireball’. I think these rocks are swopping paths too but more slowly. We may still need to invoke the behaviour in this area but the fireball head has the most extreme version of what I’m describing. My only faint concern is, from the snapshots it looked as if the ‘fireball’ encounter would be over in a matter of hours so we might then need to move down range to get the juicy candidates for the east-west Cuban strike.
One more observation that chimes nicely with this analysis is that all the action is in the forward section of the gap which you created originally in the red snapshots in April. I had posited then that the other side of the gap may have played host to all the east-of-sun radiant rocks. That seemed very neat but I was aware that the head of this rear train was rather far back for that to happen. I now realise that the bend (which, after all, defines the position in the train of the 2012 close approach) can’t help but end up in the path of the Earth in 2013. This is because the Earth is relentlessly catching up DA14 in its orbit. The Earth couldn’t help but go crashing through the middle of the mayhem it had caused 2.6 million km ahead of it in 2012. 2.606802 E9 is 2,606,802 million km. That was the 2012 close approach distance. It is exactly the right catch-up rate for a day’s orbital difference: 2,608,848 km per day @ 30.195 km/sec!
We can see all this in your ‘fireball’ snapshots but the above explanation pins a plausible narrative on what, to some, might otherwise seems like too much of a coincidence. Even we were mightily surprised by it but really it was just the higher resolution analysis of that rear end of the concertinaed red train where DA14 was lurking.
I think the ‘fireball’ has much going for it.
Andrew
Frank
We crossed comments. I’ve glanced at your recent one plus the blue/green Earth to DA14 snapshot. Interesting. Will look more closely later.
Andrew
The main thing that dawned on me was that the entire folded tail was whipping from inside DA14′s orbit to outside around the August node and back to inside in February.
Yes, that’s right.
it is clear that in February 2013 the tail was flipping back past the main train to the inside track again. The two halves are swopping places with rocks criss-crossing orbits by the dozen or hundreds or thousands.
Yes, that’s right. Quite what’s happening in the z dimension isn’t clear.
I would guess that if you plotted the summed vx and vy’ of, say, a dozen rocks either side of the bend just before it becomes overlaid, they would show the requisite xy velocity differences to account for the 20 deg of angle variation we need. Or, you could give me the state vectors, including DA14 as a reference rock, for these rocks at that snapshot in time and I could do the sums.
What I’m thinking of doing is to print off the state vectors for the entire rock cloud in a format that I can re-use to start another run. The idea being to interpolate or extrapolate new rocks between rocks in the rock cloud., and also to have a record of a complete ‘fireball’ rock cloud. I could also use this dataset for further separate analysis.
However, since there are 343 rocks in the cloud, this is quite a lot of data. I could fish out a small fragment, but it seems to me easier to just give you the lot. But it would probably be best if I emailed you the data, rather than fill up Rog’s blog with meaningless numbers. If you are agreeable, my email address is cfrankdavis (at) googlemail (dot) com. We could of course continue the discussion here, so that anyone else can join in.
And yes, I am rather more confident about rock clouds starting in 2005, because they’re small enough (I hope) for my annual corrections to be fairly accurate. The problem comes when rock trains start to occupy long curving sections of the orbit.
Frank
I’ve emailed you for the vectors, along with some blather about how they might be presented. You are right about not cluttering up the thread with reams of data, although I think the 100 years of close approaches was interesting and belongs here.
I also agree with you on keeping up the conversation here in case anyone else wants to join in or just read. I’d hate to think that it might become disjointed and incoherent to outside readers due to our conducting discussions in camera. The email will be good for the large data files and I might email you something off topic about modelling the sun’s tachocline when you have a few spare months and terabytes. But not quite yet. I need to do the vectors. They could be a gift for working out the ‘AC7-7 degree-further-east rock’.
Andrew
Frank
I was looking at your blue-green snapshot from earlier today (DA14-centric with earth/moon). You said:
“What’s interesting to me about this is that the Earth remains in one quadrant relative to DA14 the whole time.”
What I see is the Earth going round and round 8 or 9 times which I presume is 8 orbits from 2005-2013. But I’m not sure if I’m seeing it correctly because you say the Earth remains in one quadrant. I can certainly agree that the Earth is pulling in +x, +y every February because it is catching up with DA14 from behind in every one of those years. And (it’s just dawned on me) I think it might be doing the same thing in August due to the nodes not being exactly opposite, Earth therefore being ahead at the node and DA14 coming from outside to inside. Is that what you mean? The rest of the year, wouldn’t it be pulling in many different directions according to what has overtaken what?
As you can see, I’m a bit at sea and yet it looks as though it will be a very useful snapshot when I’m seeing it properly.
Andrew
Frank
Thanks for the vectors. I’m going to ask you a few questions about the initial cube and close approach but could you post up the snapshot of the close approach so it is easy to refer to on the thread? (Or could I have done that? You might do it more professionally unless it’s just a link).
I won’t bombard you with too many questions about the vectors at first because I know you need to answer the last comment about the DA14-centric spiral. I would like to be able to grasp that one.
Andrew
You can email me and transfer data various ways if you want it on WP servers. We (readers who donate) pay a small amount to allow .zip archives so anything can be bundled.
Too tired to write more tonight.
Andrew,
‘To try to clarify the blue-green snapshot, let’s first look at the same view of the apparent motion of Jupiter around DA14 over about 12 years. It goes in a circle around it, regularly coming nearer and turning briefly retrograde. What we are looking at is an almost geocentric view of Jupiter, since DA14 is in a similar orbit. And it’s also a Ptolemaic view of the solar system, in which Jupiter goes round the Earth, but has an epicyclic motion around the circle.
The same applies with Mars. over a shorter period. It also makes an circular epicyclic journey round the Earth/DA14. The light grey lines are the barycentre’s apparent motion, which can be ignored.
Both Jupiter and Mars go round and round in a circle, so the net gravitational pull they exert on DA14 roughly cancels out over a period of about 10 years. However, in the case of the Earth over the past 8 years, the gravitational pull on DA14 has been consistently in one direction: +x and +y.
Now this is just the last 8 years, and I would expect to see the Earth’s apparent motion around DA14 to correspond to something rather more circular (I could plot it, but these 100 year runs are quite time-consuming).
However, assuming that the Earth is the principal gravitational influence on DA14, I think it’s quite interesting that it has been consistently exerting a slight pull on DA14 in the +x and +y direction for the past 8 years. Whether it is at all meaningful is another question. If the Earth was removed, DA14 may well have continued almost exactly as before.
tchannon,
Thank you for your offer, How do I find your email address should I want to create a zip file?
Andrew,
A couple of snapshots of the ‘fireball’ around the time of closest approach to Earth, using my old trick of colouring bodies with z greater than Earth z green, and ones less than blue.
As DA14 and its attendant cloud come up the z axis, the first snapshot shows that the green arm on which DA14 belongs is very slightly higher up the z axis than the other blue arm. But in the second snapshot 4 minutes later, the other arm has mostly turned green too. So the non-DA14 arm would seem to be inclined slightly downwards relative to the other arm.
The ‘ fireball’ seems to turn to follow the Earth. Here’s a view of the ‘fireball’ in April 2013. It would appear to be tracking along behind the Earth. If DA14’s rock cloud was highly reflective, it might look a bit like a comet.
We should perhaps think of some name for this object with its two arms that fold together. ‘Fireball’ isn’t really right.
Frank
All fun and games here. Electricity sub station blew up spontaneously and so I’ve been without power, light, Internet, telephone, hot food…
So, I went with the flow, sat supping red wine and reading Lord Chesterfield’s Letters (1774 edition) in the sunset then by candlelight (for all of 5 mins) to complete the Romantic tableau. Now catapulted rudely into 21st Century by the switch-on and pondering the implications of integrating the area under a spiral to derive the nett force and renaming ‘fireballs’. Sorry I’m a bit behind.
I get the blue/green DA14 centric snapshot now so thanks for the extra help. It certainly is pulling in +x,+y most of the time but I sort of duped myself into thinking there were no other quadrants involved at all. At a casual glance, it looks that way but if I draw an x and y axis line through DA14 I can see that there is a chunk of the spiral to the left that’s pulling in -x, +y and another chunk to bottom right that’s pulling in +x, -y. I knew this had to be happening somewhere in the orbits as one raced ahead of the other and back.
So sorry to involve you in another explanation but it did make me see what I should have seen anyway. I still look at it and think it’s entirely in +x,+y, but it’s an illusion, a kind of subconscious bias.
I think this relentless 8 year bias definitely did have an effect on DA14 and is the primary factor in its circularisation from a period of circa 368 days to 366.2. I suspect the main influence was in each February pass. NASA said the period had been shortened thus in a series of close passes in the ‘last few years’. They are aware of everything we are discovering, I’m sure.
I am guessing (using a ruler though) that the spiral is about 65 million km across in the xy direction. If I follow the 2012 encounter back round, it is the biggest circle and therefore constitutes the greatest distance between them about 6 months before. I would have thought it was the other way round but I do remember the NASA video which showed DA14 a long way back round and catching up fast for the close encounter. It seems it wasn’t epicyclic in the last few years but was so in the first few? Otherwise, I have it back to front.
This snapshot is a very useful little tool for referring back to. I wish I had a gallery of best images on the thread and a clip board of best comments instead of scrolling back. I’ve been back to the red gap snapshot numerous times and also your first set of 8th Feb data from the first comments in March. I suppose I could do desktop links but it’s not be same.
I put the protractor on DA14 and measured the angle of the line of c of g across the spiral (if you cut that shape out and balanced it on a knife edge). It looks like 60 deg from the x-axis. However, that is biased to the large circles that don’t have much influence. The c of g of the inner circles and nested epicycles is more like 56 deg. So, 56 degrees, and on 15th Feb the Earth is at 56 degrees. So the average pull over 8 years is at exactly the right angle to pull DA14 into line for the Feb 15th pass.
I shall carry on with vector observations tomorrow but, to sign off in the same vein as that in which I began…that is the sum of my lucubrations for tonight. 🙂
Andrew
Very appropriate to read Lord Chesterfield’s letters by candlelight, since he probably wrote them that way.
Anyway, back to DA14, as you know, my usual technique has been to fire off shotgun blasts of rocks in all directions, and watch where they go.
Yesterday I constructed an exploding cube of rocks around DA14, and soon came to realise that it was very easily adaptable to fire rocks in any direction I cared to. So I spent much of yesterday trying to move the ‘fireball’ nearer the Earth, firing off shotgun blasts of rocks from DA14 on 1 Feb 2005. In fact the relative speeds I’ve been using have been very low,in the range 0.5 to 5 m/s. And the effect is to move rocks around in the rock train that develops. I soon found that sending rocks off in a -x and +y direction tended to result in them getting closer to the Earth.
This snapshot shows the close approach of the tip of the resulting shotgun ‘fireball’ to Earth. The closest approach here is about 1000 km above the the surface of the far side of the Earth somewhere around the equator. DA14 can be seen in the background moving along a slightly different path to the other strand of rocks which lie between it and the Earth, and make their closest approach at about the same time, or a few minutes later.
However, what we want are rocks that are a day or two ahead of DA14, in a more circular orbit. So I’m wondering what direction and what speed to fire off new shotgun blasts of rocks to achieve this. Perhaps you might have some suggestions? Since 0.5 to 5 m/s movements result in rocks ending up only 20,000 km from DA14 after 8 years, it would seem that much higher speeds will be required.
I’ve been wondering a bit how DA14 might have been the origin of such rocks travelling at hundreds or even thousands of metres per second. My suggestion a few days ago was that DA14 had exploded under thermal stress. But I had the thought yesterday that, as DA14 proceeded along its orbit, it would have had a ‘bow wave’ of smaller rocks bouncing off it (and also slowing it). It seemed entirely plausible that some of these might have flown off at high speeds in one direction or other, mostly from the front end of DA14. One of these might have been the Chelyabinsk meteor (which could have been made of a quite different material than DA14) Equally rocks making a close approach to DA14 might have been deflected into a new orbit.
Further to the ‘bow wave’ idea, most of the planets in the solar system wouldn’t have ‘bow waves’ of rocks bouncing off them, because they have atmospheres that slow incoming rocks, and frequently capture them. The Moon, lacking an atmosphere, probably would have a ‘bow wave’, except that it is a pretty massive object with a relatively high escape velocity, and also it seems to be a pile of loose rocks, at least on its surface, which would absorb an impact much like an atmosphere..
But DA14 seems to be a big lump of fused iron, and would make a perfect surface for rocks to bounce off. It might be seen as a sort of giant ‘tennis racket’. Such rocks would have an attendant cloud of stones that had hit them and bounced off, rather than ones which had been prised loose by tidal forces.
Frank
I wouldn’t worry too much about the velocity changes needed to come in on the Chelyabinsk radiant. Although it’s true they would have a slightly faster speed in -x, +y, there is give and take between the vx and vy. It doesn’t quite cancel out (cancelling would imply a pure angle change with no change to absolute speed) but it is fairly close. All this tugging over the last 8 years is changing angles of the vectors more than the scalar speeds. It’s true you do see a speed change but it doesn’t imply an input of energy- it all comes out in the wash with circularised fragments giving up less KE to PE and therefore keeping up their average speed despite being slowed initially. Moreover, checking my calculations for the 2012 pass, there was a velocity change (slowing/circularisation) of 5m/sec per day. So that’s at least 20-25 m/sec over the 5 day close encounter.
I also rechecked your very useful 100 year close encounter list, above, and found that there was a very close encounter in August 2004 of 1.26 million km. I worked out that would affect DA14 by 100m/sec over 5 days. If you add in another 8 years worth of less-close encounters, the changes mount up to 100’s of m/sec in absolute changes for DA14. As for relative changes for fragments that are (somehow) already 100,000km distant it should be at least approaching 100m/sec over 8 years or 9 years- especially with that relentless xy pull from almost every point in the orbit. One other factor to bear in mind that recently dawned on me: if a fragment that is 100,000 km closer is slowed by 5m/sec over 5 days (I’ve checked that the calc for 2.606 million km in 2012), it should arrive way closer (tens of thousands of km, I think, not calculated) the next year. That means its a 6 or 8 m/sec increase the next year, 10 the next etc. It becomes a virtuous cycle. August 2004 was even closer than Feb 2012.
Here is an assesment of how all this chimes with the AC fragments we have slung from 8th Feb 2013. AC6 and AC7 are the most pertinent. For AC7 I used AC6 and did the give and take thing in vx and vy. I added 1.865 km/sec to its minus x velocity but subtracted 1.306 km/sec from its plus y velocity. This was neccesary to change the xy radiant angle. And this made its absolute velocity 30.846 km/sec (ie in x,y, and z) and 30.563 km/sec in x,y.
For guidance on the accuracy of the above procedure, AC6 had a revamped eccentricity and recalculated speed for that eccentricity and radius. That calc gave an absolute velocity of 30.483 km/sec (xyz). AC6 got a hit, first time, which says something for its authenticity as to absolute speed. Now, AC7 was based on AC6 but with the give and take in vx and vy. True, it wasn’t an exact swop and the absolute speed went up to 30.846. This is 1.2% off the more reliable AC6 velocity. It could do with being less but it’s not off by much.
If, as a theoretical exercise, the AC7 inclination went to 0 deg we would be left with the xy velocity as being the absolute v (30.563 km/sec) and that would be only 0.26% out from the more authentic AC6 velocity. I still believe a close ‘polar’ encounter could reduce the inclination to 0 or so deg and transfer all the speed into xy.
So although AC7 might be a bit on the high side for absolute velocity, I think it could just about represent a Chelyabinsk type fragment, especially if it started out 50,000 to 100,000km from DA14. I mentioned at the time that this slight surfeit in absolute v might be accommodated by tweaking the perihelion round by 2 or 3 degrees which would add speed without flouting the conservation of momentum.
As for the August 2004 encounter, here’s an excerpt from your 100 year list:
DA-14 closest approach: 5.9115182E9 on 18 Aug 2003 01:38:13 dt=8192.0 s
DA-14 closest approach: 1.26149901E9 on 19 Aug 2004 05:58:29 dt=8192.0 s
DA-14 closest approach: 5.7119887E9 on 21 Aug 2005 11:33:25 dt=8192.0 s
DA-14 closest approach: 1.2823339E10 on 24 Aug 2006
In August 2004 DA14 is only 1.26 million km away, half the 2012 distance. If you started your run in 2003 or even 2002 I think you might see some fireworks.
There’s one extra bonus. I included the previous year 2003 because its the same day what with the leap year. This implies that the Earth was 5.9 million km behind DA14 in 2003 and moved to 1.26 million km ahead in 2004. This means a pulling in x,y, big time, in 2004. This also means a large relative dispersion of fragments in the cloud. If you also got a 67 degree south close encounter it would reduce the inclination too (67 south because it’s the August node).
So as for suggestions, I would say:
1) start in 2003 or even 2002 if your worries about accumulated inaccuracies don’t impinge too much.
2) do a cube that is 50,000 or 100,000 km across. If you do the diagonal positioning suggested in 3), below, you’d almost get a 100,000 km spacing in a 50,000km cube!
3) maybe put DA14 at the top corner of the cube and our best candidate the opposite corner so that the differential pull is greatest. For a 50,000km cube, that will put them 86,602 km apart. There’s nothing so special a out DA14 that it should occupy the centre. It’s a small rock that got discovered on a whim when an astronomer in Spain decided to scan a non-ecliptic radiant for a change. If DA14 is in the topmost corner in -x,-y,+z relative to the centre of the cube and our candidate is in the opposite corner, +x, y,-z, the line between them would be at an angle of 54.74 deg to the xy plane. This would be nicely in keeping with the August 2004 radiant on approach. Nothing like exactly the same but in keeping with it- and that’s only if I’m right about Earth being ahead. This would put our fragment near to 86,000 km closer as it comes down on the radiant. Incidentally, I say ‘radiant’ but there’s no rock solid radiant like there was in 2013. It’s passing at a distance albeit comparatively close and that means it’s swinging past with a constantly changing angle.
4) i can see that setting up 3), above might necessitate starting in June or early July 2004 so that you can set DA14 in that top-back corner in relation to the Earth and construct the cube with our candidate in the near bottom corner- again, assuming the Earth is ahead but I think it is.
5) Do a run with no relative speed differences in vx vy and vz for DA14 and its cloud companions. So, with all the same initial velocities we can tease out exactly what that differential pull across the cube diagonal is.
It may appear that these suggestions are rather similar to the 50,000km cube runs you did before but there are two differences. Firstly, the 2004 close pass is included. This has 4.3 times the gravitational influence of the 2012 pass (2.6 million squared : 1.26 million squared). Secondly, it involves constructing a cube that specifically places DA14 and our fragment in positions that enhance the differential pull by probably two to three times what it was in the original 49,000 km cubes if DA14 was in the centre of those cubes. I make that 10 to 13 times the differential pull of a conventional cube started in Jan 2012. It could be that August 2004 was what jumbled things up. I know it was included in the 100 year run but it sounds as though by that time the train position was not so reliable as to be riding in close formation like the cube above. It’s only this sort of configuration that will give the requisite +vx and +vy differential between DA14 and the fragment.
It might appear at first glance to be too neatly constructed, but if there was a 50,000 km cloud, with DA14 more towards one end then, by definition, the rocks at the other end will experience a differential pull. This cube just examines the nearer, more accelerated rocks because if anything from the cloud hit Chelyabinsk, it would be these rocks with a greater acceleration in +x,+y.
I hope that’s of some help, Frank. I don’t expect you to follow everything. These are suggestions and help me think too. But I think including August 2004 would be very interesting.
Andrew
Frank
I’ve got an upcoming comment on the rock vectors you emailed me but it’s not quite ready yet.
In the meantime, as an aside, I was always intrigued by the comment in the NASA article linked below. It is comment number 7 and says that people on St Lawrence island, Alaska saw the meteor coming in. It is very matter of fact and holds some credibility for me. I think I mentioned this once before but I don’t think I linked the NASA article and comment.
What makes this sound credible is that St Lawrence island was the only place on Earth, apart from ships at sea, which was in a position to have a chance of seeing the meteor reflecting the sun as it came in at an altitude of maybe 2-4000km. The sun had just set. Further eastwards into Alaska would have meant the meteor being below the horizon; further west into Russia would have meant being dazzled by the sunset; further down the coast of Alaska and Canada would mean tracking back east and deeper past the terminator.
On the other side of the Earth, the terminator was across the Horn of Africa and up through the Caucasus but any observer there would have a 14-24 degree disadvantage with respect to St Lawrence island, depending on the 7-12 deg easterly radiant of Chelyabinsk. That leaves St. Lawrence island as a prime candidate, pretty well the only candidate and lo and behold, up comes a matter of fact comment saying someone saw it, albeit second hand.
I searched to find other references to a sighting from St. Lawrence Island but to no avail. If you get a chance one day, you may have more luck. You seem to find stuff I don’t.
If any witness was able to describe accurately the path of the object, which would have resembled the steady, sharp, pin-point light track the Space station makes (ie not a fireball!), it could settle whether it was DA14 related or coming in on the Colombians’ 3 deg downward trajectory. Seeing as St Lawrence Island is at 63 deg N and was on the terminator at the time, it wasn’t skewed up to the ecliptic pole or down as it would be at night. It was at about the 63 degrees its latitude suggests. The Colombians orbit has a circa 3 deg inclination downwards (it changes as they refine it). This means that if Chelyabinsk was on the Colombian orbit, any witness on St Lawrence would have seen it coming in across the western horizon starting at about a 30 degree altitude due SSW, slanting downwards as it made its way from south to north. It would go over the horizon at around due west. I say they would ‘see’ it but of course, they would only see the sunlit portion of that trajectory which, I would guess would be the early part. And if it were the early part, it would be hanging in the sky for that much longer due to the perspective.
If DA14 related it would be skimming the horizon. I don’t mind admitting it could be just below the horizon if it had the 69 deg radiant (10.33 deg inclination). Looking at my globe from the DA14- related trajectory, St. Lawrence is right on the horizon. But, then again if the inclination was 3-5 degrees instead of 7.83 or 10.33 it would definitely be above but lower than the Colombian trajectory.
Links:
NASA article and comment:
http://blogs.nasa.gov/cm/blog/Watch%20the%20Skies/posts/post_1360947411975.html
Wiki page for St. Lawrence island:
http://en.m.wikipedia.org/wiki/St._Lawrence_Island
Andrew,
Today I constructed a cube of rocks with a 95,000 km side length around DA14 on 1 feb 2005. This snapshot shows its arrival at the Earth on 15 feb 2013. There’s still the same ‘fireball’ as we’ve seen before, but it’s larger, and more dispersed. But it’s not very greatly larger. I tried running it at different speeds (longer or shorter time steps), but got much the same results every time.
It seems that these clouds just get stretched out into long trains, and the bigger the cloud, the longer the train. And these long trains fold in half after feb 2012, and arrive with a concentration of rocks around the folded area, creating the ‘fireball’ effect.
I am however getting increasingly unhappy with the accuracy of my model, particularly the cloud over long periods of time. And so I’ve started looking into improving accuracy by using RK4 (or maybe even RK7) mathematical methods. I’m not sure how much these will help, as they also are approximations. They’re just supposed to be better approximations. Over the next few days I want to make a big effort in this direction.
Re St Lawrence island, I can’t say that I’m much excited. We have no idea what they saw. It may have been DA14, or a plane or a satellite. Many years ago, I was surprised how many satellites could be spotted with the naked eye, if one cared to look directly upwards after sunset on a sunny, cloudless day. One could tell that they were satellites, because if they were tracking west to east, they would at some point enter the Earth’s shadow, and become invisible.
P.S. I have the 13 feb 2013 vectors and 15 feb 2013 closest approaches for the above rock cloud, if you’re interested.
Frank
This is my comment on the rock vectors you sent me. I don’t refer to your recent 95,000km cube but note the comment with two interpolated updates. I interpolated them after I had written this comment in its entirety. Your recent comment includes evidence I acknowledge and agree with but it runs counter to some of what I say here. But not everything so I carry on after the update because most of what I say still has some relevance despite rubbing up against the inconvenient truth of the fold/fireball happening at almost the same place regardless of the train length.
I found myself calling the ‘fireball’ the ‘bend’ or ‘fold’ as we had done before. That was because I was emphasising the density, the overlapping and the difference in vx,vy between the two strands just before they cross over. I think it is useful to remember that the two strands are whipping across each other and only resemble a fireball for a short while. Shall we call it the ‘bend’ or something that emphasises the folding-over process? We could have a derivative term for the period where it resembles a fireball. Maybe something like ‘peak flexion’ or the ‘flexion point’. ‘The flexion point approaches the Earth….’; 10,000km down from the flexion point…’ . Or something like that?
The vector comment is as follows:
I’ve been looking through the rock vectors you emailed me along with the close approach snapshots that you also posted above. The first thing I noted was that it was a very small cube compared with the ones you have been doing. That is informative in itself because it all goes into the mix and your subsequent 0.5 to 5 m/sec adjustments have established the relative vx and vy’s needed to nudge the rocks in +x,+y, by 30,000 km or so. I think the blue/green trajectory trails are especially useful because any angular differences would betray a different vx,vy profile, if large enough, notwithstanding the gravity well curve.
It seems that the general trend is in +vx (less speed in -x direction) with very many rocks possesing a +10 to +12 m/s change from DA14 on 13th Feb. More positive x means a swing to the right, towards Earth).
I believe this original cube had no variations on DA14’s vxyz, just position offsets. Is that right? You then added .5 to 5m/sec in +x,+y to shunt rocks the 30,000 km or so towards the Earth as shown in the second snapshot. I would have thought that the larger vx, vy disparities (shown below) that were brought about by the offsets alone would have produced more Earthward rocks. In the first snapshot, DA14 is near the head of the bend. I have no answer for that but it is a minor point.
Presumably the +x,+y rocks are the ones in the outer strand and the -x,-y rocks are those in the inner strand which looks as though it crossed over the outer strand very recently. I wasn’t able to link specific state vectors to specific rocks in the snapshot because it’s difficult to visualise all relative positions using position vectors. I might have been able to pin them to a particular strand but time constraints wouldn’t allow that.
The maximum vx,vy disparities with DA14 are:
+ vx : 37m/s
-vx : 12 m/s
+vy : circa 5-10m/s ( this is from memory and I was noting the biggest changes only)
-vy : 18 m/s
Since it is a small cube around DA14, it has stayed on the ‘wrong’ side of the sun along with DA14. In other words the Earth went through the gap produced by the folding of the train in 2012.
I always had in mind the red snapshot (from your comment on 22nd April 8:21) which showed DA14 nestling at the back end of the front train but with a significant number of rocks up range, to its right in the snapshot. Judging from the Earth-moon distance (admittedly, grainy pixels), this extra spread back down the orbit was about 640,000 km long. That distance chimes nicely with AC7-V’s position up range, back down the orbit, on the 8th Feb 2013 (500,000 km). That setting-back down the orbit was needed for the large difference in approach angle we achieved.
I’m pretty sure this longer tail-back, beyond DA14 and therefore east of the sun, was achieved because you started with a train that was 48 hours forward and aft of DA14 or thereabouts. If I remember correctly, these rocks were placed hourly up and down the radiant, meaning they were just over 20,000km apart and constituting a train that would therefore run just over 1 million km ahead and 1 million km behind the Earth. That probably corresponds to quite a big initial cube, I admit, but it appears that something between this size and the size of the cube you emailed me, is what is needed to bring some rocks in at 7 deg east of the sun.
///////
UPDATE: I note your recent comment that says the train always folds in half in 2012 and it seems that in your recent experiments this happens in the same place near to DA14 regardless of the length of the train. This somewhat confounds my above speculation. However, you also say in your recent comment that you are somewhat dissatisfied with the accuracy of the model.
I would say that the DA14 accuracy after 8 years would suggest it is very accurate. Then again, these recent runs produced a fold that ended up 20-30,000 km ahead of the Earth in 2013, whereas your red snapshot of April 22nd shows it 600,000km behind. I suppose that reflects some sort of difference in inputs or time stops. If greater accuracy brings about a folding of the rock train just a few hundred thousand km further back i.e. 2 million km from the Earth in 2012 instead of 2.6 million km, it means the fold/end the tail is further back behind the Earth in 2013. This makes an eastward, Chelyabinsk-type radiant readily attainable. The 22nd April snapshot suggests the folding in 2012 really was around 2 million km from the Earth.
Also, maybe the size of the cube, or more precisely, its xy width straddling the orbit really does influence where the fold happens. This would come about due to the outermost fragments coming closer to the Earth even if they have been somewhat extruded into a train.
Anyway, I’ve had some quite positive thoughts about the bend in the light of delving into these rock vectors. I find it intriguing that the Chelyabinsk rock is coming from a point that is in keeping with what is probably the very back of the fold, just before the open space of the swept gap behind. I’m going to treat that as a separate comment because the last few have been a bit long due to being a bit behind.
Andrew
Andrew,
‘Flexion point” is quite nice. But bend or fold or elbow (with two arms) are just as good.
Yes, DA14’s rocks cloud had the same vxyz as it. Except when I built a few exploding rock clouds.
I would say that the DA14 accuracy after 8 years would suggest it is very accurate.
Yes, DA14 is very accurate, but that’s because I correct its position every year on 1 feb. It’s the other rocks in the cloud that are the problem. Ones that never get too far from DA14, are quite accurate too. But after that they seem to be all over the shop,
Anyway, today I’ve started putting together a testbed to look at the simulation’s accuracy, so to the extent that I’ve been focused on anything, it has been that.
Frank
“‘Flexion point” is quite nice. But bend or fold or elbow (with two arms) are just as good.”
That’s good because I use most of those in my comment about the bend. ‘Elbow’ is good too. I won’t post the comment up for now because you are busy. I’ll wait for a quieter moment.
Perhaps, one day, you can explain RK4 and RK7 mathematical methods (potted version:) )
Andrew
Andrew,
I’ve spent the day implementing RK4 in my simulation model. RK4 is a 4th order Runge-Kutta method of estimating solutions for differential equations. Essentially, in my implementation of it, what it does is to compute position and speed of a body after an interval dt, by detecting curvature (which I haven’t been doing) and making adjustments. Basically it looks ahead along the orbit a little. I’ve just adapted some code I found online.
I really didn’t know what to expect.of it. But, a little to my surprise it performed extremely well on my testbed. Using my previous Euler method, a rock near the Earth going in a circular orbit around the Sun would be tens of thousands of km out of position after one orbit (less if I used a smaller dt). But with RK4, even using a large dt, it appeared to be only about 10 metres out of position!
I’ve now integrated the new RK4 code into my main simulation model. If it’s as accurate as it seems to be, not only will DA14 be in the right place, but all the other rocks will be too, more or less. It’ll probably take me a day or two to verify this. And if it turns out to be the case, then I will have laid to rest my concerns about the accuracy of my simulation.
I have, furthermore, seen a way in which I think the greatly enhanced accuracy of RK4 could be improved even more. As I mentioned, the RK4 method looks ahead to find changes in speed and acceleration. It does this at dt/2 and dt, and generates a weighted speed and acceleration. But what it isn’t doing, as it works out new gravitational accelarations at dt/2 and dt, is to take into account the fact that at dt/2 and dt, all the bodies will have moved. My proposal is to assume that these bodies are moving along straight lines during the short period of dt, and using their speed to estimate their future positions, and hence the future acceleration of whatever body I happen to be looking at.
Anyway, it’ll probably take me a day or two to check it all out (and implement my proposed enhancement). After that, with a bit of luck we should be back in business, but this time with a greatly improved simulation model.
Frank
10 metres eh? Yep, not bad, not bad.
These sound like very useful additions and all done with understated expertise. I shall very much look forward to the new runs.
“As I mentioned, the RK4 method looks ahead to find changes in speed and acceleration. It does this at dt/2 and dt, and generates a weighted speed and acceleration.”
That’s what I was going to do with DA14. dt was 6 months dt/2 was 3 months (!!!). My brain isn’t quite as nimble as your programme. The reason was to get an orbital speed maximum and minimum (at the apsides) and one for 11:00 UTC on 15th Feb 2013. That would tell me the the speed DA14 would have had without the Earth’s gravity well and how much it slows per day from perihelion to aphelion (non linear but linear for short stretches).
Did you know that on 15th Feb 2013 DA14 was almost exactly half way (timewise) between perihelion and aphelion? Just another sign of a long and complex ballet between the Earth and DA14. Unlikely to be pure coincidence.
Andrew
Frank
And this is right up our street
http://www.nasa.gov/mission_pages/WISE/news/wise20130529.html
The take-home message is that clouds do spread but appear to be long-lived and identifiable. I wonder if they have any diagrams. I’ll check later.
Andrew,
If my expertise is understated, it is because the expertise is absent.
And I must say that yesterday I was totally stunned by the accuracy of the RK4 upgrade of my code.
Let me describe the little testbed I set up to measure it: I set up a single body in a circular orbit around the motionless Sun, at roughly the distance of the Earth. I did this using G.m1.m2/r^2 = m1.v^2/r to find the tangential velocity v of my test body at radius r. Once I had v, I also had the orbital period length.
I then placed a test body on the orbit, with the required tangential velocity, and measured how far it deviated from a perfect circle at 4 points along the orbit.
Using my Euler method, the deviation after one orbit was several hundred thousand km. Much more than I had imagined.
I then ran the same test using RK4, and the deviation from a true circle after one orbit was just 8 centimetres.
I didn’t believe it at first (and wrote yesterday that it was a few metres). I had expected that RK4 would be an improvement, but would still be a few thousand miles out. But, no, the improvement was one of several orders of magnitude. And it took me most of the day to get my head round that. Furthermore, the error on most of the quadrants of the circle was actually less than 8 centimetres.
In fact, I still don’t really believe it today. And it may well be another week before it really sinks in.
However, once I had upgraded my main simulation model, which is complete with planets and rock clouds and so on,the picture got a lot muddier, and rocks were flying all over the place, much as before. But since I (am slowly beginning to) believe that my simulation model is now very accurate, I’ve started to look around for other sources of inaccuracy.
And the principal culprit seems to be my planet correction/repositioning which periodically kicks in to put rocks where NASA says they’re supposed to be. It’s beginning to look as if my new, far more accurate model is disagreeing with NASA’s figures.
Thinking about this a bit, it seemed that in my new RK4 simulation, the Earth was now arriving about a day earlier at the Earth on 15 feb 2013. And that suggested that its velocity was fractionally too high. And if its velocity was too high, perhaps my Sun was too heavy.
The value of solar mass that I’m using is 1.9891e30 kg. It also happens to be the figure that Google offers when asked. But according to Wikipedia’s Solar Mass page, a more accurate value is 1.98855 +/- 0.00025 kg. Which happens to be very slightly smaller than the value I’m using.
I’m now thinking of dispensing completely with my orbit correction stuff (which anyway plays havoc with the rock cloud), and instead adjust the mass of the Sun (and maybe the Earth, and DA14) until DA14 shows up more or less exactly where it’s supposed to be.
I’m not sure how to do this, but matters are proceeding apace.
Frank
I use the gravitational parameter (Greek letter mu) for my sums. It consists of the mass x G ready-calculated. It’s like buying ready mix concrete because you don’t know the proportions therein. I already knew that the gravitational constant wasn’t as accurately calculated as one might think so I considered mu(sun) as good enough.
I checked just now by dividing mu(sun) through by the gravitational constant and got the sun’s mass as 1.98855…E30 kg. The difference between that and 1.9891 E30 kg is pretty big, a factor of 3616 which in turn implies an inaccuracy of orbital speed on the 8th Feb of around 8 m/sec (because of the linear relationship to the sun’s mass). That seems like a lot of inaccuracy between the two touted figures. My calculation for the Earth’s orbit on the 8th at 00:00:00 was 4.5 m/sec off with your JPL Horizons vectors, summed. I used the gravitational parameter for the sun, as above, and therefore with the more accurate mass baked in. I attributed the shortfall to the other planets (I guessed Jupiter had been pulling on it for several months judging by its position in the winter night sky). A 4.5m/sec inaccuracy was good enough for my purposes when I was just trying to get an idea of where I was in relation to the Earth. It wasn’t used in the final AC vector calculations.
I can see how calculating the curvature yields a better result. It’s encouraging to see it so accurate.
Andrew
Whoops! Small correction: the sun’s mass has a linear relationship with the gravitational force and the square of the orbital speed:
v^2= GM(2/r -1/a)
So the speed anomaly would be much smaller than 8m/sec.
It’s turned out that I’ve been up a bit of a blind alley today, because it seems I didn’t integrate the new RK4 code properly into my main simulation model. The result has been that I’ve been running my old code all today, while imagining it was the new RK4 code! And so have got more or less nowhere at all today. And all my remarks earlier today about the behaviour of the new ‘RK4’ main simulation model are null and void.
However, I now seem to have the RK4 code running. And on the first trial, running with DT=16384 secs, (quite a long time) the uncorrected DA14 made a close approach to the Earth in more or less the right position – which looks very promising indeed. However, I’ve been fiddling with the solar mass, and various other things.
I’ll report further tomorrow.
As for the matter of the solar mass, NASA’s Horizons website gives the following solar mass:
Mass (10^30 kg) = 1.9891
which is the value I’ve mostly been using (but which it seems you haven’t been). I would suppose that this is the mass NASA uses for their calculations. I’m currently using 1.98855, and finding that DA14 makes its close approach a day later than it should. So I will probably return to using the larger value..
Well, it seems that RK4 is at last living up to the promise of the testbed results. I’ve been running with DT = 16384 s, and not correcting any positions of planets or other rocks as I have in the past. And after 8 years DA14 and its accompanying rock cloud show up more or less on schedule, and DA14 makes a close approach to Earth of 28,000 km. However, it’s a day early.
When I used a sun mass of 1.98855e30 kg, it showed up a day late. And now that I’ve gone back to using 1.9891e30 kg, it’s a day early.
But the important thing is that it’s very nearly where it should be, without any corrections being made.. And this means that the rock cloud around DA14 is also more or less exactly where it should be, relative to DA14. And what we’re looking at in the snapshot above started out as a large 300,000 km-side-length cube of rocks on 1 feb 2005.
Apart from that, I’m not seeing very much difference from my previous model results. The rock train gradually lengthens, and gets punctured by the Earth in 2012, and develops a ‘fireball’ bend throughout 2012.
I’m very pleased. It may just be a matter of tweaking the model a bit (reducing sun mass slightly?) to get it right..
Frank
Are you using a gravitational constant of 6.67384 E-11? Theoretically, that ’84’ at the end could be shunted down by 80 to give 6.67304. 0.0008 is the uncertainty for the gravitational constant. I presume the middling figure is preferred though. I don’t think I’ve ever seen it quoted otherwise. But it would be interesting to see if it makes up the difference.
Also, there’s the +/- 0.00025 E 30 kg uncertainty on the sun’s mass so you could shave that off first and see. That’s probably what you were referring to.
Andrew
Frank
It’s promising to see the new run showing the same fold and the ‘fireball’ head.
Another thing that is very useful is seeing the relative vectors of other rocks. What is slightly
troubling for me is that they all appear to be in line meaning no variation in xy vector direction and therefore radiant.
However, those rocks around the Earth seem rigidly straight whereas some of those at bottom left are a bit squiggly or at least have a few offset pixels. Do you reckon there might be some 5 degree angles in there and it’s just that they are short lines?
I would expect the bottom left ones to be angled a bit more due to being the most circularised (depending on which line is the old train and which is the folded part). That is why I mentioned, the other day, the idea of hunting 300,000-400,000 km down the ‘fireball’ tail i.e. ahead of the Earth for these candidates. Then it was a question of hoping the whole fireball could be shunted back (bigger cube or whatever) so that these candidates hit the Earth. But this is a big cube and a much more accurate run so I’m beginning to think that the circularisation can’t be the source of the required radiant alone. It would seem that AC7’s absolute velocity really was made too fast just to achieve that angle change we saw. It served a purpose though in showing how a small vx vy change does indeed change the angle drastically.
I’m not too bothered though. There are other avenues but I’ll let you carry on tweaking for now.
Andrew
Are you using a gravitational constant of 6.67384 E-11?
Very nearly. In my brand new RK4 code the declaration is:
G = .00000000006673;
What mass are you using for the Earth (and Moon) by the way?
Space.com
NASA TV will broadcast several live views from telescopes as the asteroid 1998 QE2 makes its closest approach to Earth.
At 4:59 p.m. EDT, Friday, May 31, 1998 QE2 will pass by Earth at a safe distance of about 3.6 million miles — its closest approach for at least the next two centuries. The asteroid was discovered Aug. 19, 1998, by the Massachusetts Institute of Technology’s Lincoln Near Earth Asteroid Research Program near Socorro, N.M.
10 pm BST, it seems.
Daily Telegraph
Frank
I use the gravitational parameter for Earth, quoted by Wiki as 398,600.4418 km^3 sec^-2. This appears to give an Earth mass of 5.972580131 E24 kg. the Earth page on Wiki gives it as 5.9736 E24 kg. That seems like rather a discrepancy too although just over half of it would be due to my version of G as opposed to the lower value.
I saw the 1998 QE2 story this morning and was going to link it so I’m glad you saw it. I had a vague notion of looking out for fireballs and there was a big one in the US today but there are biggish ones there every few weeks. I didn’t give it much more thought than that at first because it’s passing at 5.8 million km. However, I then remembered your 2012 DA14 run where the Earth went straight through the rock train and therefore in DA14’s wake. I would tentatively suggest doing the same for 1998 QE2 but I realise you are doing a lot right now. I just thought that if we went through in its wake, in real life, we might see some hits that are explicable.
I saw a NASA diagram of the orbit on the BBC Science page. It was a bit crude and if it’s anything to go by, QE2 passes fully on the outside track but I can’t be 100% sure the perspective is reliable. I think they would’ve made more of it if it had crossed our path exactly. But 1 or 2 million km is ok for our purposes and even if it’s just tugging on the train it would be interesting to see the degree of perturbation.
Anyway, in the course of discovering today’s US fireball on the AMS site, linked below, I saw that they had a cluster of 4 fireballs on the 19th-20th May. They were all on the 4AM hour line and all SE to NW. Although this wouldn’t be DA14 related, it is interesting that the AMS find this significant. We keep banging on about azimuth angles matching up; I keep banging on about hour lines. This is the first time I’ve seen anyone at least note that such a cluster type is uncanny.
I infer that they are tacitly suggesting a link which means a train hundreds of thousands of km long. DA14 was doing 476,000 km per day geocentric and it was relatively slow so if the 4 fireballs are linked, the AMS must be thinking in these terms. Such a tacit admission is a first for an astronomer community who refuse to think that Chelyabinsk and DA14 were linked when 100,000 closer than that distance.
Andrew
AMS:
http://www.amsmeteors.org/2013/05/four-fireballs-at-four-in-last-24-hours/
BBC:
http://www.bbc.co.uk/news/science-environment-22736709
Andrew,
I didn’t get to see anything of QE2. If there was a live broadcast, I missed it.
But I did the next best thing tonight, and got hold of (1998 QE2)’s vectors from NASA, and plugged them into my simulation model. Here’s a snapshot of the solar system. DA14 is the red line. QE2 is the elliptical path, and seems to have an orbital period of around 4 years. On I feb 2005 it was a bit outside the orbit of Mars, and on its way in.
Anyway, I also built a large rock cloud round it, and watched that develop. Here’s today’s closest approach. The rock train goes sailing by about 6 million km away.
My simulation was only a few hours out in its estimate of closest approach. It was quite a nice test of it.
It illustrates the difference between QE2 and DA14. On 15 feb, we passed through DA14’s rock train, nicely folded over to create a double-strength train at that point. But today we didn’t pass through QE2’s rock train – unless it happens to be 6 million km wide. And if it was, there would be fireballs every day for weeks or months, because the Earth would be travelling along inside the rock train .
There must be lots of these rock trains steaming round the solar system. But I’m beginning to think that the “February fireballs” of recent years are mostly due to the rock train associated with DA14, and in which DA14 itself might be a relatively junior member..
Andrew,
I woke up this morning thinking that, in many ways, what we’re trying to do with DA14 is construct a wider rock train, with the outer rocks going round slower, and the inner ones faster.And last night, I ended up saying that QE2’s rock train would have had to be 6 million km wide if the Earth was passing through it right now.
Is such a rock train width implausible? Are there rules and regulations about how wide rock trains can be? So today I constructed a 7 x 7 x 7 rock cloud around QE2, with rocks 2 million km apart, and set it running. The cloud gets pretty strung out along the orbit of QE2. But there’s a snapshot showing QE2’s close approach last night, with all the rock names printed alongside them (which is helpful). The names give the original positions of the rocks relative to QE2 at the outset in million kms. QE2 is shown. The nearest rock to Earth is x6y2z6e6km – which started out 6 million km out along the x axis, 2 million km along the y axis, and 6 million km up the z axis -.The Earth and Moon are at the centre, with the arc of about half the Moon (Luna) orbit traced out. Interestingly the rock on the opposite side of the initial cloud – x-6y-2z-6e6km – also appears, just coming into view.
I then used the nearest rock x6y2z6e6km to construct a smaller cloud around it, and ran it again. Here’s the snapshot of its close approach. The naming convention here uses e5 rather than e6, so that 63e5 is 6.3 million km. The closest any rock came was, x63y21z62e5km closest approach: 6.2054861E8 m on 01 Jun 2013 23:54. That’s 620,0000 km, or a bit outside the orbit of the Moon.
So, if there is a 6 million km radius rock train around QE2, we’re in it right now, and getting deeper into it. So maybe we’ll be reading more reports of fireballs.
I’ll try and get some QE2 rocks a bit nearer the Earth, and find their relative postions and speeds. But on the face of it, it would seem that they’re coming from the opposite direction to DA14 rocks like Halo1093 and co. And since QE2 is fairly near its perihelion, the accompanying rocks will be going at top speed.
Yes, they are moving pretty fast relative to the Earth (rock number 3). They wouldn’t ever circle round the Earth a bit like DA14 rocks.
x63y21z62e5km closest approach: 621128.5 km on 01 Jun 2013 23:39:21 dt=2048.0 s
x63y21z62e5km is -225335.67,-538895.7,-211324.25 km relative to 3
x63y21z62e5km is 5.281876,-5.5065136,7.7883863 km/s relative to 3
x66y22z65e5km closest approach: 423373.7 km on 02 Jun 2013 06:28:57 dt=2048.0 s
x66y22z65e5km is -175758.6,-371036.03,-106624.93 km relative to 3
x66y22z65e5km is 5.339302,-5.547598,7.796663 km/s relative to 3
Frank
Thank you for trying out my suggestion in such depth. I wouldn’t mind betting you were making a move for QE2 ephemeris already though! It’s very tempting as a real life test bed rock.
On what date did you start the cube off?
What I find heartening is that although the cube gets strung out, it maintains its width: 7 rocks on a side means 6 gaps @ 2 million km = 12 million km on a side. x6y2z6 is about 5 million km away from QE2. x-6y-2z-5 looks as if it’s about 8 million km away on the opposite side. The appearance of the opposite rock you mention would seem to corroborate this as being the cube width, largely maintained as the train extrudes.
The second snapshot illustrates this further with all those nicely spaced orbits. Interesting how y is always 17e5 km and causes those staggered ranks of 3 or 4, each rank aligned in a straight line front. That’s why I asked when you started the run. I suppose I should ask roughly where it was as well, in relation to its aphelion/ perihelion. But if it was kicked off in the last two years I can eyeball where it was. It’s just that I think it matters in which quadrant the offsets were. It doesn’t affect the experiment, just the post analysis.
Also, does that mean there are another 340 or so rocks out of view? It must’ve strung out a long way.
Andrew
Frank
You mentioned the Fireballs of February being DA14 related. That NASA page on the subject says the following:
“Indeed, a 1990 study by astronomer Ian Holliday suggests that the ‘February Fireballs’ are real. He analyzed photographic records of about a thousand fireballs from the 1970s and 80s and found evidence for a fireball stream intersecting Earth’s orbit in February.”
It then says, “He also found signs of fireball streams in late summer and fall.”
So that means strange behaviour in both February and presumably August and early September. That corresponds to the two nodes for DA14.
Andrew
Andrew,
I started it on 1 feb 2005. It’s my current working start date. Although I know you want to go back to before 2004. As you know, it’s about half way between perihelion and aphelion.
What I find heartening is that although the cube gets strung out, it maintains its width
Yes, I’m glad about that too. I knew the clouds got longer, but I wasn’t entirely sure whether when they got longer, they might get thinner, like an elastic band.
Also, does that mean there are another 340 or so rocks out of view?
Yes. It gets very strung out. But this time I feel I can trust the code. Today I started out Apophis on the same date, to build up a rock train all the way through to its 13 April 2029 close approach – over 24 years -and it arrived on time, without any corrections to the orbit. Not exactly the right time, but the same day.Not exactly the right distance from Earth either, but near enough.
I haven’t fixed on a G or Sun mass yet though. And that makes a difference, I found what NASA maybe use for G here.
Also I’ve been thinking some more about circularising DA14. I know how to work out kinetic energy, but I guess that potential energy will be using acceleration towards the barycentre, and distance from barycentre. I’m thinking that I might find the circular orbit of DA14, and then interpolate between that and the real DA14 to produce a long chain of rocks out from one orbit to the other. All with the same k.e. + p.e. And then advance them and retard them along their orbits. They should approach the Earth at a different angle.
On the February ‘fireballs’ thing, I’ve been watching the Earth punch through the rock train year after year, knocking big holes in it. This probably means that some years the Earth passes through one of these gaps, and there are very few fireballs. Other years, like 2013, it hits a place where the rocks are densely clustered, and there are lots of fireballs.
Frank,
You are certainly along the right lines with your idea of keeping the KE+PE constant.
On a related point, I was thinking about the fact that AC7, though substantially angled round to the east, was too fast and so had too much KE for its PE. AC6 had already suggested that circularisation alone produced a minimal angle change (it was 1 degree or so, commensurate with the half million km shunt in xy that narrowed the crossover by 1.34 degrees). So AC6 had its shortcomings as well as AC7.
That still leaves another avenue, or even points to it. If a fragment is perturbed in 2012 so that it is less circularised as oppose to more circularised, it would come in with a greater xy velocity than a circularised fragment- and do so without necessarily violating the KE+PE rule. This implies more eccentric, slower period, sitting in the rearward train behind the gap. That’s not a favourable spot for a hit. But those rocks are the ones that are slowly tugged from behind in 2012. I’m thinking this rock is more like a close passer like the one that was sitting in the middle of the gap for four years. That particle appeared to have just its inclination altered. This one I’m suggesting would be a close passer around north-west on the night side in 2012. That should make it a little more eccentric and turn its apsides anticlockwise allowing for both a greater incoming angle and a faster speed (closer to perihelion).
I may not be entirely correct with the apsides thing and any eccentricity increase means a longer period when we want a shorter one if anything. But we are always referring back to DA14. If this fragment was made more eccentric or slowed, it could have had its close pass on the 13th Feb 2012, making its return, early on the 15th Feb 2013, represent a slowed rock. (That is, slowed period but slightly faster and more angled at this point from perihelion).
Andrew
Andrew,
Perhaps it would be a good idea if I had a look at what happens to a few of the rocks that get displaced in 2012 through being slowed or speeded. At the moment I have no idea where they go once they’ve broken away from the main rock train.
Frank
Yes, good idea, if you were able to focus on a few specific rocks either side of the 2012 split and do a snapshot of their entire orbits, I think we might see some interesting patterns that look like Venn diagrams. I realise you don’t often do whole orbit snapshots. Is that harder to do than portions? Actually, two rocks, one either side of the split would speak volumes.
Andrew
Andrew,
I set up a 7 x 7 x 7 cube of rocks, with 10,000 km between rocks, centred on DA14 on 1 Feb 2005, and took note of which ones were speeded up during the subsequent 2012 close approach to Earth (by which time the cube had become a rock train). I picked out 3 of these, and in a subsequent run, got this snapshot of their orbits after they broke away from the rock train. DA14 is the red line, and the nearby black line is the Earth. All these speeded rocks were behind DA14 in the rock train. They all moved out into wider orbits with a slower period than that of Earth’s orbit.
It was a bit harder to discern rocks that were slowed during the 2012 encounter, but rock number 62 had started out with x -30,000 km, y +20,000 km, and z -20,000 km relative to DA14. I reason that rock 61, which had the same relative position except y was +10,000 km was probably one which had been slowed in the encounter, rather than speeded. This guess turned out to be correct. This snapshot shows the subsequent orbit of Earth (black), DA14 (red), and rock 61 (blue). Rock 61 drops into a lower orbit with a shorter period than that of Earth’s orbit
I finally put all 343 rocks in a cube between rock 61 and 62’s initial positions, in the hope that the rock train would be thrown every which way.on 2012. And indeed it was. All 343 were thrown into either higher and slower orbits, or lower and faster orbits. So that by Sep 2012,some of them had nearly arrived back at their 15 feb 2012 start positions.
My conclusion is that if you speed rocks, they’ll arrive back at the 15 feb intersect/node later along higher orbits than DA14, and if you slow them, they arrive back earlier along lower orbits.
Since we’re looking for a rock that’s about 18 hours (I forget what it is right now) ahead of DA14, maybe one that has been very slightly slowed on 15 Feb 2012 will arrive a few hours ahead of DA14. So rocks between DA14 and rock 61 and 62 are all likely to be being slowed more than DA14 (which is also being slowed).
Andrew,
I then constructed a cube of rocks between DA14 and rocks 61 & 62, and ran that. This snapshot shows DA14 and these rocks approaching Earth. Most of the rocks are inside DA14’s orbit (and extend down to the bottom left out of view). But a few are slightly further out than DA14, and in a higher orbit. The ones in the lower orbits have closest approaches to Earth earlier than DA14, but they are well inside its orbit.
In this uncorrected run:
DA14 closest approach: 49697.05 km on 15 Feb 2013 18:40:35 dt=16.0 s
DA14 is -47811.18,-13129.134,-3393.2693 km relative to 3
DA14 is -1.2730575,2.8973253,6.6132092 km/s relative to 3
The nearest rock to being 18 hours ahead is this one:
x-25y7z-10e3km closest approach: 1.1081173E7 km on 15 Feb 2013 00:41:15 dt=64.0 s
x-25y7z-10e3km is -6575269.0,-8774608.0,1601400.9 km relative to 3
x-25y7z-10e3km is -0.09988457,1.0529846,5.360232 km/s relative to 3
But it’s 11 million km away.
Missing snapshot from above comment.
Frank
That snapshot of the cube between rocks 61 and 62 being thrown into every which way is what I’d hoped to see. We have seen snippets of this all along in various orbit sections and rock trains but to have this picture is like having a library of possible rock trains.
This helps because although the rock train shots such as the 1928-30 one showed beautiful rock train curves, I had to keep reminding myself that those curves weren’t the actual orbits. They represented a line of rocks that were all in different orbits that were, admittedly, close to that line. I imagined the orbit lines fanning out back behind the rock train like the veins on a very slender leaf fanning out from the central vein. The fact that you have done so many rocks means that I can now do the opposite: see the rock trains cutting across these orbits. This is because this snapshot shows all the orbits once they were completed but on any given date they were all in different places whipping back and forth in two trains as per the standard behaviour first illustrated here by you on April 22nd. I’ve seen so many of those types of shots since then that I can now roughly superimpose the trains on this latest snapshot. It helps me to see what you are doing.
This is all very interesting and has thrown up a number of questions for me. I don’t expect you to answer everything but if you can give me a bit of an idea it would be of help. So in my ramblings below a few questions crop up. There’s loads more I want to say but it’s getting late.
My biggest surprise is that the slowed rocks can cut the corner so much and get back by September 2012. At first, I thought the small size of the orbit must be be an xy illusion, in other words, these smaller orbits were actually largish ones that were drastically inclined down into the z axis. But the speed of their return suggests they are indeed small unless, I suppose, they are inclined as well as running so close to the sun as to bring them back fast. That closer perihelion is obviously happening anyway but I didn’t think that could happen easily while remaining so circular and near to DA14’s inclination.
Even if the smallest orbits are all around 10 deg inclination, I presume there is at least some z action going on here, that is, inclination changes? Are you able to check that or can you only do so by proxy e.g. checking vz of different rocks after the 2012 pass? If there are a few drastically steepened orbits they would nestle into apparently lower orbits in the xy snapshot. That doesn’t change the overall message because at least 335 or so of these rocks have been pulled along the radiant only and can’t have had a near polar pass. Perhaps all 343 have. Only a tiny few could have had drastic inclination changes and it would have been a south polar pass that increased inclination enough to allow a rock to masquerade as a tighter orbit in xy.
Do you know the distance between the rocks in the 61/62 cube train in 2012? I’m trying to get an idea of how many rocks were tugged along the radiant and how many might have passed so close as to behave as outliers with big inclination changes. If the gaps in 2012 are huge and the closest approach is 100,000km it would mean no drastic inclination changes. If they are close there might be a few.
I’m not as obsessed with inclination change as this comment might suggest. It’s more to do with understanding z and its implications for outliers. Then, from there, it would be interesting to find an outlier with high curvature, non polar pass which was flung into a much more eccentric orbit but towards the sun instead of out wide and high. That way it should return early, at high speed and with the correct vx vy.
I shouldn’t worry too much about the 18 hr ahead rock being 11 million miles away. I think initial conditions are everything and we may end up fine-tuning in the hundreds of km rather than the tens of thousands and still finding wild differences a year later.
BTW, I thought it especially resourceful of you to narrow down the field in such a forensic way.
Andrew
When I said about checking any variation in vz’s after 2012, I really meant any variation between the ratio of vz and summed vx,vy. There are a lot of vz increases but it’s the commensurate xy increases that maintain the ratio and therefore the inclination. It’s when the xy increases don’t keep up with the z increases that the inclination steepens. I can crunch those figures if you have them for close approachers as well as a control from about a million km away (all of them analysed when past the gravity well).
Andrew
Do you know the distance between the rocks in the 61/62 cube train in 2012?
No, but I could find out. They started out 1,000 km apart in 2005. In 2012, they got very close to the Earth, with one closest approach of 17,000 km. Which means that the Earth passed right through the rock train, just like in 2013 (were there fireballs reported in feb 2012?). There may even have been impacts, but my RK4 code doesn’t have impact detection re-implemented yet.
Even if the smallest orbits are all around 10 deg inclination, I presume there is at least some z action going on here, that is, inclination changes? Are you able to check that or can you only do so by proxy e.g. checking vz of different rocks after the 2012 pass?
I haven’t looked at z. There could be a lot of z motion, since the Earth is passing very close to these rocks.
Anyway, here’s a snapshot of the 2012 close approach of the small rock train. The Earth is near the top end of the train, which has a length slightly greater than Earth-Moon distance. The Moon is a dot at about 5 o’clock from the Earth.
And here’s a snapshot of the rock orbits after 2012 close approach, but with my usual blue for z Earth z. DA14’s orbit never seems to rise more than 30-40 million km above/below the Earth. And neither does the rocktrain after close approach in 2012. If it had gone further, there would have been further colour changes.
It looks as if most of these rocks are staying in about the same plane as DA14’s orbit. However, some of the innermost and outermost have changed their plane a lot, and are going in -z when DA14 is moving +z.
This makes sense. Most of the rocks in the train are quite far from the Earth. But clearly a few have been close enough to get thrown into a new orbits. But none of these seem to get very high up the z axis in any direction.
Andrew,
I’ll try to respond to your later comment tomorrow. Too sleepy right now. But in the meanttime, I hope my coloured orbits are helpful. Green is above Earth z in +ve direction, blue is below Earth z in -ve direction.
Frank,
Thanks for answering my questions and posting that great snapshot with the z component. It looks as though some close approachers got flipped into a negative inclination: there’s a slender blue band of rocks that takes the inside track, representing the tightest of all the orbits (as seen in xy, at least). This band turns to positive z at the August node when all other rocks turn negative. It then moves to the outside of some of the innermost blue tracks and approaches the Earth in 2013 as a distinct separate band. This distinctive behaviour leads me to believe that it represents the outlier, back door method of taking the inside track i.e. the close passes that are round the back on the night side and towards the radiant antipode. These are then flung back down into negative z and towards the sun instead of flung out and high like all the other rocks that pass behind.
The mirror image of these renegade rocks are the seven weird blue rocks that are flung out wide and low and don’t nest with any of the green outside trackers. I would tentatively posit that these pass just in front of the Earth and don’t behave like their more distant counterparts. I say ‘tentatively’ because I can see issues with that and this is a rushed comment.
I can see other interesting things happening:
1) the negative z inside trackers appear to be angling back in just before feb 2013 as if their perihelion has been precessed by several degrees.
2) it occurred to me that this interesting little band of rocks, if passing close around the back, was passing straight up the 69 radiant that DA14 did in 2013. DA14 passed ahead at 30-45 deg maximum in 2012 as far as I can remember but these rocks, 2.6 million km behind are acting like a dress rehearsal for DA14 in 2013. That means that any close approachers round the back have to pass over the 69 deg north radiant antipode. Some of these would be turned negative and inwards by virtue of passing over that latitude. If they are made more eccentric and with a precessed perihelion, they can hardly help but come back the next year earlier, with faster xy component, eastward radiant and over the day side of the pole with a 55 deg or so antipode.
3) notice that most of the rocks in that snapshot, even the modestly perturbed ones, have their ‘August’ node precessed by about a month. NASA’s fireballs of feb article said that they had found another stream in late summer to early autumn
I’ll have a few more thoughts later I think.
Andrew
Andrew,
In answer to my own question,
(were there fireballs reported in feb 2012?)
the answer is yes.
Feb. 22, 2012 In the middle of the night on February 13th, something disturbed the animal population of rural Portal, Georgia. Cows started mooing anxiously and local dogs howled at the sky. The cause of the commotion was a rock from space.
“At 1:43 AM Eastern, I witnessed an amazing fireball,” reports Portal resident Henry Strickland. “It was very large and lit up half the sky as it fragmented. The event set dogs barking and upset cattle, which began to make excited sounds. I regret I didn’t have a camera; it lasted nearly 6 seconds.”
Strickland witnessed one of the unusual “Fireballs of February.”
“This month, some big space rocks have been hitting Earth’s atmosphere,” says Bill Cooke of NASA’s Meteoroid Environment Office. “There have been five or six notable fireballs that might have dropped meteorites around the United States.”
It’s not the number of fireballs that has researchers puzzled. So far, fireball counts in February 2012 are about normal. Instead, it’s the appearance and trajectory of the fireballs that sets them apart.
“These fireballs are particularly slow and penetrating,” explains meteor expert Peter Brown, a physics professor at the University of Western Ontario. “They hit the top of the atmosphere moving slower than 15 km/s, decelerate rapidly, and make it to within 50 km of Earth’s surface.”
Also Daily Mail:
24 February 2012
Visible across the United States for the whole of this month, the fireballs have become a source of mystery for NASA scientists and other sky-watchers around the world.
Observing the phenomenon for the past 50 years, NASA say that February’s fireballs are different from other months but the reason why is not clear.
‘These fireballs are particularly slow and penetrating,’ said Peter Brown, a physics professor at the University of Western Ontario in Canada.
‘They hit the top of the atmosphere moving slower than 15 kilometers per second (33,500 mph), decelerate rapidly and make it to within 50 kilometers (31 miles) of Earth’s surface.’
….Ranging in size from a basketball to a bus, the annual light-show still has NASA’s best stumped.
‘They all hail from the asteroid belt, but not from a single location in the asteroid belt,’ said Bill Cooke of the Meteoroid Environment Office at NASA’s Marshall Space Flight Center in Huntsville, Alabama.
‘There is no common source for these fireballs, which is puzzling.’
Over the past half-century sky-watchers have launched numerous studies into the February phenomenon.
Their results have been inconclusive, with some detecting more fireballs and others noting no increase.
Here are the closest approach figures for one of the rocks in last night’s run:
x-31y18z-17e3km closest approach: 7681.187 km on 16 Feb 2012 16:54:34 dt=1.0 s
x-31y18z-17e3km is 2602.072,6183.223,3741.3613 km relative to 3
x-31y18z-17e3km is -7.8928633,-1.9164599,8.69193 km/s relative to 3
It’s so close that it has probably accelerated a lot by the time it gets to this point. Here’s one that was a lot further away:
x-29y12z-23e3km closest approach: 296604.03 km on 16 Feb 2012 16:54:36 dt=1.0 s
x-29y12z-23e3km is -177456.72,-235415.03,32601.13 km relative to 3
x-29y12z-23e3km is -1.9970579,2.274871,5.5580683 km/s relative to 3
And this one is an impact:
x-33y14z-23e3km closest approach: 2205.4658 km on 16 Feb 2012 16:55:58 dt=1.0 s
x-33y14z-23e3km is -1359.8522,-363.78058,1697.8534 km relative to 3
x-33y14z-23e3km is 6.0499344,16.99017,8.597491 km/s relative to 3
It would never have flown back out.
I think the second one (which would also have accelerated quite a lot) shows the relative speed of this rock train was much the same as in 2013, and so rocks would have managed to skim the Earth at high latitudes, going relatively slowly, as in 2013.
One thing that has emerged, and that I was slow to see, was that while these rock trains look like trains of rocks all following each other, the ones I’ve been constructing with cubes full of rocks aren’t following each other at all. They swap places the whole time. And it isn’t the same rock that’s at the front of the ‘train’ all the time.
I don’t really think that making cubes of rocks all travelling at the same speed is quite the right thing to do. I think that the outer rocks should be going slower, and the inner ones faster. And rock cubes made that way might not develop into ‘trains’ at all.
So today I was thinking about what happens when a single cubic rock breaks into 343 smaller rocks. Before it breaks up, the rock will have a single P.E. and K.E. And after it has broken up, the 343 rocks will each have a P.E. and K.E. which will sum to equal the original single large value. If the positions of all these rocks is known, the ones that have a greater orbital radius will have a higher P.E, and so will need a lower K.E., by reducing their velocity. And those with a shorter orbital radius will have a lower P.E., and require a larger P.E, by increasing their velocity.
But this assumes that all the rocks have the same P.E. + K.E. Maybe they don’t have to.
I’m just thinking out loud…
Frank
That catchily named x-31y18z-17 e3km is the very candidate I was looking for. I knew it would be hard to model and fire up the 2012 radiant on the correct impact parameter but you’ve managed to bag it as part of your cube.
This rock is passing north east on the evening dark side and half way up the globe. That is perfect for both inclination reduction and eccentricity increase while probably swinging it back to the day side. The close approach is at around 45-50 deg N on around the ten o’clock hour line (over Northern China or Southern Mongolia I think). Looking on the globe, its ground track is nicely on course to hit the 69 degree antipode in Alaska and then get flung back to the day side via Canada and on its way again as one of your super inside trackers. I think its inclination goes to a very low figure, possibly negative. If so, it would be a member of that green track on the inside. We don’t actually want negative inclination, more like 4 to 7.5 deg but this rock looks like a good one. Can you send it on round and show its orbit and close approach credentials in 2013? Maybe superimposed on the other ‘library’ snapshot to see where it sits?
If it is an inside tracker with low inclination it will at least be a good template to work from. As I said in a recent comment, if we get something promising I think we’ll be adjusting in the hundreds of km around it and seeing big swings all over the Earth (but on the right radiant) as a result in 2013. Then adjustments down to 10km might see it swing all over Russia the other end. Then 1 km and see it swing a hundred km either side of Chelyabinsk etc. That’s my hunch.
Other points
1) I summed the vx vy vz of the three rocks you posted up. This close approacher was doing 11.9km/sec, so nicely in keeping with your 12.5 to 12.9 km/sec skimmers in 2013. The impactor was theoretically doing 19 km/sec but only because it was still dropping inside the Earth. The distant one was doing 6.3 km/sec, so that was in keeping with DA14 a couple of days out. So all those speeds make sense.
2) Regarding your June 3rd 9:28 PM comment, I agree that having the same speed for all the rocks is a bit false, especially for large cubes. But if the rocks have different KE+ PE it would only be because their masses are different. They would all still have to have the same speed for a given orbital height.
3) This swapping around of rocks from different sides of the train and drifting up and down the train is something like the pulsating cloud I was talking about- expanding at the apsides and contracting at the nodes.
4) It is interesting to see fireballs in 2012 because it gives some credibility to the idea of even looking at close passes at that time. That’s because what we are doing presupposes a 2.6 million km train.
Andrew
A technical point:
In my previous comment I said ‘half way up the globe on the night side’. I meant half way up the northern part above z=0. I’m trying not to use the e word (equator) which was why I said globe, meaning a sphere with an ecliptic plane running through it delineating +z and -z (and even that is wrong because the Earth isn’t quite on the ecliptic at this time). Still, you use +/- z either side of the geocentre which I find very neat and consequently I find the equator an irrelevant annoyance.
A further confounding issue with this half way up thing is that, by coincidence, the 3741km z component places it in line with the mid-latitudes after the 23.4 deg tilt downwards is added. When all is said and done, it’s around 35 degrees up the sphere from the ‘ecliptic’ on the night side. That’s important because it is imparting a little more xy effect for eccentricity than for inclination which is what we want.
Andrew
Andrew,
That catchily named x-31y18z-17 e3km is the very candidate I was looking for. I knew it would be hard to model and fire up the 2012 radiant on the correct impact parameter but you’ve managed to bag it as part of your cube.
That rock is rock number 322, so I’ll call it that for now. But rather than dropping into an inside track, it’s one of the rocks that is thrown into a wide outside track, as this snapshot of its subsequent orbit shows.
And this shouldn’t be too surprising. Here’s the 2012 close approach again:
x-31y18z-17e3km closest approach: 7681.187 km on 16 Feb 2012 16:54:34 dt=1.0 s
x-31y18z-17e3km is 2602.072,6183.223,3741.3613 km relative to 3
x-31y18z-17e3km is -7.8928633,-1.9164599,8.69193 km/s relative to 3
The Earth at this point will have a negative absolute vx, and a negative absolute vy, and rock 322 has positive relative x and y distance to Earth. So 322 is behind the Earth as it moves along its orbit, and is being strongly accelerated forward along that orbital path. No wonder it moves into a very wide outside orbit, and doesn’t come back for over 2 years.
If you want an inside tracker, I think you’re going to have to look on the other side of the Earth for rocks that are being slowed by the Earth behind them, and that will subsequently drop into lower orbits. This is what happens to DA14 after its close approach in 2013, as the snapshot shows (red line),
It seems to me that what we may be looking for is a rock that isn’t actually slowed in vx and vy, but is given a -vz shunt to reduce its orbital inclination to the ecliptic plane, so that in 2013 it returns above DA14 in the z dimension, and impacts (if it impacts at all) further north on the Earth.
I’ll email you separately all the 343 close approaches of 2012.
Andrew,
For the purposes of finding Chelyabinsk candidate rocks, it seems to me that we don’t really want a rock that has been very strongly disturbed like rock 322 above.. It seems to me that we may want a rock that is much nearer DA14 on 16 feb 2012, but higher up the z axis, and 18 hours ahead of it. Such a rock would hopefully be given a slight -vz shove, and have its orbital inclination reduced. And maybe it would be pushed a bit further out in the x-y plane.
The really great thing about the 16 feb 2012 close approach is that the Earth really is right inside the rock train at this point, and can be used to fire off rocks in almost any direction we like. I created the minicloud above in order to produce maximum disturbance in feb 2012. But, as I say, I don’t think we really want something that has been very strongly deviated as a Chelyabinsk candidate.
Frank
Shame about rock 322. Thanks for trying. I know that the rocks behind are pulled and swing out but this had such a large q angle (at least 45 deg) that is was doing a 90 deg turn, geocentric back towards the inside track. I realise the orbit speeds at circa 30 km/sec dominate over the geocentric speeds of 6 or 8 km/sec but I thought 90 degrees might be enough to take all that increased speed from the pull and direct it back inside the Earth’s orbit albeit less than the geocentric angle. It certainly would have been thrown outwards at less of an angle than the pull alone would suggest. However, I concede that it appears the orbit speed is so dominant as to smudge a 90 degree geocentric turn into just a few degrees prograde, along the orbit.
I still think that snapshot of all the orbits with the strange blue/green mirror image depicting a downward inclination after Feb 2012 is intriguing. This cohort of rocks sitting on the inside track is worth mining if only for info about how the outliers get to where they are. They are interesting in terms of how they turned negative, ie their interaction with the Earth prior to dipping negative. They are also interesting in that they return in a distinct bunch, detached from the rest as if their perihelion has been swung round or their vx,vy changed a lot. However, if they dip after February 2012 it means they come in from above the ecliptic in 2013 and there’s no way I can see of getting a hit at the atmospheric entry angle of Chelyabinsk at 6 km/sec approach and from above. I still think there’s useful info in that mirrored inclination bunch though. Do you remember that nice snapshot of the tight cube that extruded to a train as it passed over the Earth and bunched up again? That went back steeply into negative z so maybe that’s how it happens.
But getting back to the practical matter of a vz kick, what I reckon is best for your reduced inclination rock is one of the inside trackers that is green after Feb 2012 and is at about 6.5 degrees (see below).
Apropos inclination, you mentioned giving rocks a kick in z so as to reduce the inclination. We did already do that, from AC4 (I think) onwards. Certainly AC6 and 7 plus their derivatives had z kicked down to 4.174 km/sec so as to reduce their inclination to around 7.8 degrees from 10.33. That was how you managed to get those high skimmers in Finland and northern Russia. I think 61 degrees was the highest as opposed to 48 deg for Halo 1093. That gain in latitude was commensurate with a 13 degree gain or lift in the radiant from 69 deg to 56. Or presumably 56- I thought it would be 51, giving a 66 skimmer.
Anyway, we have done the z kick thing and it works- but it’s an avenue which is far from exhausted. It seems 7.8 deg inclination isn’t quite enough and it needs to be more like 6.5 to 7 degrees to get the 16-18 deg atmospheric entry angle at 55N. It would be really great if we could do that. A quick calculation (30.610 km/sec x sin 6.5) says it should be 3.465 km/sec in z on 00:00:00 8th Feb 2013. vx and vy have to be adjusted up so as to compensate and keep the absolute speed at 30.610 km/sec. I don’t think I can do that tonight but if you look at AC4 (I think that was the first z 4.174 rock) you can see how vx and vy were adjusted a tiny bit to compensate. Also that rock sat above DA14 in z. A 6.5 deg inclination rock would be sitting that distance higher times a factor of 1.203449 comparing the sine values.
If you could find a rock from those inside trackers whose vectors just outside the gravity well on exit sum to 6.5 deg or so you have a ready-made candidate. I started looking at the rocks you emailed me earlier but the list froze half way through. I think it’s a glitch my end. I’ll look again tomorrow. They are useful for some of this stage of our work but inclination can only be reliable from vectors outside the gravity well but not a long way further because the vz slowly decreases after the node and becomes negative 3 months later.
Sorry not to have concrete state vectors for a 6.5 deg or so rock. I’m a bit behind. If you like, you could email me vectors of inside trackers about a week after Feb 2012 so I can get a close match or use one as a template. They would be green at that time and would have to be 2 million + km away.
Andrew
Frank
Thank you for emailing me rocks earlier which you mentioned were inside trackers which turned blue (negative inclination) after their close encounter on Feb 16th 2012.
There were seven, all told, that turned back down into negative z and, as you say, they were all impactors so we are looking at close approaches between about 1200km and 2200km from geocentre. That is very informative because it shows that the inside trackers can’t go negative in real life. Presumably there were quite a number of other rocks that had close approaches inside the Earth between 2200km and 6371km and which stayed in positive z but probably with very low inclinations. If we want a rock that is at 6.5 deg positive inclination, do you think you might find it skimming just over the surface or 1000, 2000, 3000km above?
I like the idea of inside trackers because they arrive earlier the next year and, I believe, they come in with more -vx and less -vy which is then magnified when you snap to geocentric. It means a more eastward radiant. I know some of them arrive back in September 2012 but it might give us a clue as to whether the xy radiant of an inside tracker is more favourable for an east of sun radiant. However, you don’t want to let that get in the way of finding a lesser inclination. I just mention it because we should keep in mind the vx and vy ratio on its return in 2013, outside the gravity well and compare it with DA14.
I think it is also a more solid approach to get a rock to start before Feb 2012, undergo its close approach, orbit for a year, and land near Chelyabinsk on 15th Feb 2013. That would look so much more credible than firing rocks from 8th Feb, knowing that if you trace them back they don’t quite line up with the train or Earth in 2012.
By the way, whilst it would be nice to get 15th Feb hits, I think it’s less important at first than divining the behaviour, inclination and vx,vy of return rocks even if they are days or weeks out. It’s all info that goes into the pot, rather like these impossible close approachers you sent me that have nevertheless pointed to where the 6.5 or 7 degree inclination rocks might be.
I noticed the close approachers you sent all had the same x and y relative position in the original cube but were stacked on top of each other in z. The close approaches and their corresponding state vectors all seemed to be neatly in proportion as the distances from the geocentre increased. That says a lot about the reliability of the programme, I reckon.
So, my only suggestion for today is to creep up above the 7 rocks you sent me until you are skimming just above 6371 km on your close approach. If you send that/those round, we could look at its vz-to-absolute speed ratio and get the inclination angle. It would need to be on 8th Feb if a similar orbital period to DA 14 or a week from close approach, if much faster, so that the angle was pure.
Andrew
I’ll try shifting those rocks slightly tomorrow. I doubt I’ll be able to get them to exactly 6371 km. It’ll probably be quite a lot higher than that.
I haven’t done much today, but I did try out an idea I had a while back. I’ve noticed that when I build these cubic rock clouds, the ones that end up making close approaches to the Earth tend to be along a line through the cube. Today I wondered what would happen if I extended a long line of rocks out from DA14 on 1 feb 2005 in that direction. I wanted to get one that impacted the Earth on 15 feb 2013 at about 2:30 am UT (same time as the Chelyabinsk rock). So constructed a 500,000 km line with 100 rocks in it, with DA14 at one end.
This snapshot shows the line of rocks approaching Earth on 14 feb 2013, with DA14 the red one on the end at the bottom. In a few hours, the Earth passes through the line, with rocks impacting on it, at around about 2 am on 15 feb 2013. I don’t know why the line is curved. But nevertheless, all the rocks in this cloud are going to pass the Earth at distances slightly less than or slightly more than DA14.
The nearest to 2:30 am 15 feb 2013 closest approach was rock x151y242z280e3km. Those numbers, times 1000 km, are the initial offsets from DA14 on 1 feb 2005.
So it was a great success, at least in terms of finding a rock that would impact the Earth at the required time. What I might try doing next is to take this offset, and try out rocks moving at slightly different speeds (currently all the rocks have the same velocity vectors as DA14) and see where they end up. A small change in the velocity vectors might well add up to quite a large change after 8 years
Incidentally, instead of referring to it as a ‘chelyabinsk candidate rock’, I think it would be easier to just call it ‘che’, with numeric suffixes.
Frank
That sounds like a useful experiment. If I understand correctly, it seems you have identified the line through the original cube which corresponds to the rocks that are strung along the geocentric radiant in 2013. Do reckon that’s one way of describing it?
I like the idea because it focuses on close approachers. The other way, with a train, the Earth punches through and is soon gone. That has rocks strung along the orbit; this has rocks strung almost across the orbit if they are coming up the radiant. However, my interpretation doesn’t quite fit with the curve in the snapshot- unless it is the dense apex of the curve that hits the Earth side on and so resembles a straight line?
If those close encounter rocks you sent me are anything to go by, I would have thought the line through the cube you mention sits vertically in z because those seven rocks sat directly on top of one another. However, I would say intuitively, and from the QE2 train snapshot (rocks keep their xy position), that the line would go diagonally yet steeply in -x,y. So can you tell me where it does run?
The fact it curves suggests it undergoes the same bending, whiplash effect as the rock trains but because it is across the orbit in 2012 but maybe angled a little away from us, there is still enough differential g to slow rocks to subtly different speeds. Because the differential is less than for the conventional trains, which are stretched away along the orbit of DA14, this scenario would result in a fat curve like the one in the snapshot. But that’s only a conjecture.
BTW it wasn’t 2:30 UTC, it was 3:20.
Andrew
Yes, it was the geocentric radiant, more or less.
The line extended from DA14 along a straight line to x +151,000 km ,y +242,000 km, z +280,000 km, and would have extended back in the corresponding negative direction.on the other side.
The line was just a guess. And in fact the intercept wouldn’t have passed through DA14, but near to it. It may be the sum of several errors that results in the curve, because this particular rock train isn’t much affected by close approaches.
Frank
From your comment June 3, 2013 at 9:28 pm:
“One thing that has emerged, and that I was slow to see, was that while these rock trains look like trains of rocks all following each other, the ones I’ve been constructing with cubes full of rocks aren’t following each other at all. They swap places the whole time. And it isn’t the same rock that’s at the front of the ‘train’ all the time.”
It’s occurred to me that these rocks are probably self correcting, by which I mean the outside trackers in the initial cube start too high and compensate by giving up PE on the opposite side of their orbit. The ones on the inside which start to low, compensate by giving up KE on the opposite side. In other words, inside and outside rocks in the cube have the same orbit but they are mirror images of each other. This means they self correct within six months of the cube launch date. It still means the orbits you see probably don’t have quite the right eccentricity and speeds but I think they are probably very close and, the main point for saying this, they don’t gradually get more and more distorted from reality as the years go by. That’s if my hunch is right.
Andrew
Andrew,
I found a rock that might interest you. Rock 321 appears in this snapshot as the innermost orbit. It’s inside the 7 I sent you yesterday, but after feb 2012 it has z positive (z = +4.75 million km on 15 March 2012) while they have z negative.
Here’s its closest approach in 2012
[321] x-31y17z-17e3km closest approach[7]: 10736.227 km on 16 Feb 2012 16:49:34 dt=1.0 s
x-31y17z-17e3km is -6565.8013,-6143.467,5866.41 km relative to 3
x-31y17z-17e3km is -0.4781135,7.5533676,7.3937335 km/s relative to 3
And here’s its state vectors on 13 feb 2013 (km s)
x-31y18z-17e3km,
2.23914622976E8, -2.92260151296E8, 6662918.656,
12.355298828125, 7.49761328125, -0.70517236328125,
In other words, inside and outside rocks in the cube have the same orbit but they are mirror images of each other. This means they self correct
Not sure about the ‘self-correction’ concept. As I see it, they all have different orbits, and the period of these orbits is slightly different. And this means that every time they come back round to the same point in the orbit, they’re all a bit further apart.
Frank
That rock, number 322 appears to be at about 121 degrees (early July position for Earth) and 398 million miles from the sun on Feb 13th. It’s only doing 14km per second which would figure for such a distance. It must be an outside tracker? Correct me if wrong- I wondered if it was in metres but it says ‘km s’ which must mean km and km/sec.
It is low inclination though, about 2.9 degrees. This is reliable only because it is near to the node, if I’m right about the vectors.
Andrew
Hmm… The inside tracker was named as 321 on my orbit display. I wonder if the numbers are screwed up a bit.
I’ll investigate further tomorrow.
And that innermost ellipse, which was what I thought it must be at first, does look interesting. The reason is that it comes in at a much better xy angle. I know it might have a short period but it could be indicative of what to look for in the more subtly changed rocks.
I have become increasingly aware recently that because all these ellipses nest so well, they all end up being instantaneously parallel on their return in Feb 2013. This appears to be somewhat irksome when we want a different xy radiant. However, I think the solution to this is at least partly to be found in the idea of a rock taking a slight inside track and arriving earlier. This would mean that it was crossing the Earth’s orbit before it had nested into that parallel alignment. It would be bumping up against the Earth’s orbit on the 11th or 12th. It would also be faster which is always welcome.
However, we are always thinking about 15th Feb and DA14. But if a train rock had been flipped to the inside track on, say, the 19th or 20th Feb 2012, it would return early but ‘early’ for that rock would mean 15th Feb 2013. Then it would have that greater crossover and higher speed without violating the nested ellipses idea. The only reason our ellipses all nest is that they are all perturbed on the same day. In reality they would be smudged over a 4-7 day period of rock train interaction like a spirograph drawing.
Andrew
I see what happened. In my earlier comment I gave the closest approach of [321] x-31y17z-17e3km, but the state vectors of x-31y18z-17e3km. I copied the wrong values. Correct values should be.
closest approach:(km)
[321] x-31y17z-17e3km closest approach[7]: 10736.227 km on 16 Feb 2012 16:49:34 dt=1.0 s
x-31y17z-17e3km is -6565.8013,-6143.467,5866.41 km relative to 3
x-31y17z-17e3km is -0.4781135,7.5533676,7.3937335 km/s relative to 3
state vectors (km seconds)
[321] x-31y17z-17e3km, 10000.0, 0.025, 2456337.000053561, A.D. 13 Feb 2013 12:00:04 dt=64.0 s,
5.9214286848E7, -4.9552433152E7, 727665.792,
27.776240234375, 38.71473046875, -3.992997314453125,
[321] is quite near the sun, and on the other side of it from DA14, after doing about one and a half orbits since feb 2012, while DA14 has done almost 1 orbit..
Frank
Rock 321 is situated just before the ‘August’ node on Feb 13th 2013. It’s at the equivalent of Earth’s July 31st or August 1st position. It has a speed of about 48km/sec and an inclination of 4.79 degrees. That is, a positive inclination according to the convention but angled down due to approaching the descending node.
So tight ellipses are 50% faster but we suspected it was something like that. It’s still useful info for working out what happens to lesser disturbed rocks.
I did have another thought about this rock though. It would be a coincidence if such a rock went round 3 times to DA14’s 2 times and hit on the same day, a little earlier. But if the train goes right round the orbit and was in an intimate dance with the Earth since 2005, or 1918 or 1600 its not quite as implausible as one might think if you assume many rocks to be in the train. Almost all such similarly disturbed rocks would be careering through the February node on different dates when the Earth was nowhere near and then when the Earth does pass through, it has no choice but to run the gauntlet of crossing a busy motorway. In that scenario this seemingly special rock hits on 15th Feb as simply the next rock travelling down the motorway. There’s no big coincidence, but it only works if one presumes a very long train with very many rocks experiencing close encounters over at least several years. Rock 321 would then be like the lottery winner who thinks ‘why me?’ but if millions are taking part someone has to get the numbers ‘by coincidence’.
This thought occurred to me only because rock 321 happens to have an orbital period that’s almost 50% faster than the Earth’s. I realised that if it’s bearing down on the opposite node a year later it would be bearing down on the February node 2 years later. So if it was perturbed in February 2011 or 2009 or 2007…or 1918 (at 49/51% faster because an even year a century ago) it could have the right characteristics for a February 2013 hit. If the train was a few days wide it increases the chance of such a perturbation further still. For all we know, it could have been perturbed to such an orbit in 1600 and had 206 goes at hitting us so far. Not such a coincidence there.
If all these ellipses of all these different orbits are nesting on February 15th it explains the February fireballs both in terms of the numbers and their coming from different directions. The nesting compresses their orbits down to a narrow tube for that day and a day or two either side. The differences in absolute speed and particularly in vx:vy ratio account for different radiants and the vz difference accounts for inclination.
Andrew
I seem to have lost my broadband connection, so it looks like I’ll be out of action for a while.
Frank
Shame. Hope you get it back soon.
I might as well fill the hiatus with a couple of stored chunks of research which I’ve been holding on to so as not to interrupt the flow. I have been looking into the Colombians’ data and Revolution Square shadows for a while now. This is a good time to share my findings because it follows on fairly smoothly from my last comment on faster rocks in nested ellipses.
One clue that supports faster rocks is that Zuluaga and Ferrin, the Colombians, have +\- 2.6 km/sec error bars for the atmospheric speed of the Chelyabinsk rock. That would give it a lower estimated speed of 14.9km/sec. 15 km/sec is the speed the Russian Academy cited from the beginning but they were drowned out by NASA. I’m sure 14.9 km/sec (14.7 without Earth rotation) is attainable at the Feb node by a faster period rock and a faster period would imply a slightly bigger -x,+y, I believe, giving a more easterly radiant.
NASA are happy to trumpet Zuluaga and Ferrin but the error bars are to be found buried in a supplementary download which appeared on their web site a long time after the speed factor was ‘settled’ on 17.5 km/sec.
I have an idea why such error bars exist. It’s to do with the problem of parallax on the Revolution Square shadows. That 100 deg angle arc they traced only needs to be out by 4 degrees on the eastern measurement to take the speed down from 17.5 km/sec to 14.9 km/sec. This is because the meteor’s azimuth is 105 degrees and the shadow starts at 122 degrees (pointing back to 32 deg south of east). This means the line of the shadow, produced, crosses the trajectory line at a very acute 17 degrees: the shadow hardly moves at first but the rock is travelling a vast distance all the same. At this point, it travels nearly 4km per degree as opposed to 0.5km per degree when the shadow is at 90 degrees to the trajectory. There’s a big potential for angle-to-distance error in the measurement on the easterly margins.
If they haven’t included the camera distortion it could drop even lower. I estimated it at 10% using different maps and methods. That would reduce the speed further to 13.4km/sec. This is due to the base line (road kerb) for tangent calculations being squeezed meaning a perpendicular dropped to that kerb from the shadow tip is too far along it, giving a bigger angle.
Zuluaga and Ferrin have used two other shadow videos in their latest calculations but they only refer to using them for the ascertainment of the azimuth, not the speed. Revolution square is the only video cited for this purpose and the distance-to-time ratio of 77km over 4.4 secs is very much in keeping with Revolution Square.
Getting a faster period rock to hit on the same day as DA14’s pass may still be a long shot, but far less so if there are dozens of nested ellipses. It’s nothing like the coincidence it at first appears and I have to keep reminding myself that we are already dealing with an amazing coincidence- faster rocks in nested ellipses are at least as plausible as an unrelated rock coming from the asteroid belt on the same day.
Andrew
Frank
Here’s another stored comment. It refers to the point I made about pairs of rocks having mirror image orbits and to your reply:
” As I see it, they all have different orbits, and the period of these orbits is slightly different. And this means that every time they come back round to the same point in the orbit, they’re all a bit further apart.”
I would agree with what you say there. My point was more to do with individually paired rocks on opposite sides of the cube (or same distance either side of DA14). My idea is based on the QE2 image where the opposite side rock was running pretty well parallel to the one that was close approaching. You noted that it was just coming into view. That was after 8 years and so the two rocks must have the same type of orbit so as not to get strung apart but they are clearly not on the same actual orbit. So my guess is that at the August node they swap places and so by definition their orbits would be mirror images or near enough notwithstanding other planetary influences.
But even if that is right, it still holds that whilst each pair would stay together, the different pairs would string out along the orbit.
The NASA link I posted a week or so ago mentioned the asteroid families getting strung out so it does happen. I would have thought that if it happens at all then eventually they fill the whole orbit. It would be nicer for us if it stayed as a 2 million km elongated cloud but if the February fireballs are anything to do with DA14, the cloud goes right the way round.
Andrew
Andrew,
No sign of a new hub yet.
But you my be interested to know that I now seem to have my simultion model working in reverse (if I want it to), with the planets and calendar going backwards.
This is something you’ve suggested a couple of times. It should now be possible to reverse a few rocks to see where they came from. It may be of greatest use if started at 3:20 UT on 15 feb 2013 to reverse rocks out from Chelyabinsk (or anywhere else) to find where they come from. This should be of particular interest to you because you’ve spent much more time than I have looking at videos of the fireball. I think you’re even a member of some discussion group about it. I’ve spent very little time on this end of the story, partly because it seems to me very difficult to get the actual radiant along which the rock is coming in, and also because it will have already been slowed by atmospheric drag forces.
Anyway, to launch reversing rocks, it will be necessary to chose a radiant to reverse them back up. And I’m wondering what the best convention for defining this radiant should be. Right ascension and declination are, I beliieve, one notation. But I believe there are others. Perhaps you’d like to suggest one.
I already have code embedded in my model to launch rocks in this manner, but only in the direction of the Earth’s rotation (although the Chelyabinsk rock was coming in very near this). So I need to get my head around the trigonometry I’ve already got embedded, and develop it a bit further to allow different launch radiants.
I’ve also though that it’s probably going to be necessary to take into account the non-spherical nature of the Earth.
Now that reverse motion seems to be working, I’ve been also wondering how to combine the sort of thing I’ve been doing so far with reverse motion. It may be possible to do them at the same time (so to speak). I could run a simulation of two sets of rocks at the same time. One set would come from the vicinity of DA14 and go forward in time. The other set would start at Chelyabinsk, and go backward in time. When the calendar date and time of the two models coincided, I could then see how far apart the rocks all were. The planets should all be in the same places, but test rocks would be a long way apart. But if the launch parameters were varied, they could perhaps be got to marry up at the meeting point. …if you see what I mean.
Frank
Sorry to hear there’s no hub yet. But it’s great that the programme runs backwards now. I’m a bit late so I can’t do the thinking and research to answer everything properly but I’ll have more time tomorrow.
The good thing about the atmospheric trajectory of Chelyabinsk is that Zuluaga and Ferrin, the Colombians, now agree exactly with NASA on almost everything especially the most important thing which is the azimuth. Z+F say 105.1 deg and NASA’s map measures at 105 deg. They are 1.1km/sec different on atmospheric speed but that’s less important than firing direction. Z+F say 17.7 degrees altitude (it was as low as 15.8). Judging by their previous estimates and their error bars, these figures aren’t set in stone but are their best effort.
Also, as I said a couple of comments back the Z+F velocity can be as low as 14.9 km/sec which chimes with the 15km/sec of the Russian Academy. If the camera distortion isn’t factored, it’s even less, around 13km/sec.
You mention the speed loss. It’s not as much as one would think. The speed loss has been calculated by Z+F and it’s close to my calc- very small due due the v^2 effect in 1/2mv^2. It dropped by only 4% in my calc between entry and explosion. For Z+F it was 5%. This is related to the 440kt:90kt ratio of energy figures cited by NASA.
I questioned the NASA energies long ago and had another old comment to post up on it (I had promised Steve I would). Basically, the initial energy figures didn’t match to the velocity they cited of 18.6 km/sec but made it too high (20+ km/sec). This meant that the reduced, revised energies were correct for the stated speed (still cited as 18.6 km/sec). I had pro-ratad from old energies and 18.6 km/sec to get 15km/sec for the new revised energies but when I did it from scratch their revised figures were in accordance with each other and the old figures gave the excessive speed.
One interesting point is that if you extrapolate 18.6 km/sec on the entry angle given, and for the 32.5 secs cited from the infrasound, it’s ‘hitting’ the atmosphere about 149 miles above the Earth (including drop away of curvature)- which is 50 miles above the 99 mile low Earth orbit threshold!
(I’m using miles here because of a tacit preference for miles when sources are citing where the atmosphere ends so all such data is stuck in my head in miles and looks right in miles especially Apollo data, below).
At 13km/sec over 32.5 secs, it would have hit at 89 miles up. This is in keeping with Apollo missions which registered initial drag (5% of g) at 80 miles up and 11km/sec.
It’s true there is atmosphere for hundreds of miles but if if you can sustain an orbit through it I doubt if an infrasound detector can hear that high up.
Andrew
Frank
“I’ve also though that it’s probably going to be necessary to take into account the non-spherical nature of the Earth.”
If you use a horizontal to equatorial conversion I think the oblateness must be taken into account. My web based converter only goes from equatorial to horizontal and I couldn’t find one that went the other way.
However, seeing as your programme is heliocentric it seems bit confusing to go to geocentric equatorial. There is ecliptic coordinates that use the equivalent of RA and declination. At least that keeps everything in line with the ecliptic but your programme is in Cartesian coordinates. Although I’m happy with geocentric RA/declination and also your helio xyz vectors I’m no expert at converting. In theory we could pinpoint the Earth at impact time (56.12? Deg heliocentric) and do a single laborious calculation from 55N out on the correct angle in 3 dimensions. Easy to visualise but prone to mistakes.
Andrew
No time to answer (I’m sat in a pub again, using their wifi).
But BT sent the replacement hub to the wrong address.
Frank
Fingers crossed for speedy delivery.
This is archived comment number 3. I did it a few weeks ago just before your first fireball comment which put the kibosh on it. This was because this comment is mostly related to the apex of the curve, the sharp flexion point, being half a million km further back. Your fireballs are the apex itself, going right through the Earth’s IP zone.
However I think there are some nuggets of truth buried in here which could be thrown out with the bath water if not just noted and put on the back burner. There are some nice, neat circumstantial bits of evidence. Here goes…
As I said, I’ve had some quite positive thoughts about the bend in the light of delving into these rock vectors. I find it intriguing that the Chelyabinsk rock is coming from a point that is in keeping with what is probably the very back of the fold, just before the open space of the swept gap behind. Far from stretching the limits of credulity, it seems that this could be quite the opposite, a signature of where the fold is, where an intensely dense pocket of rocks were coming through and therefore the most likely radiant by far for a hit. If the red snapshot is indicative of what actually happened, it is yet another coincidence that the Chelyabinsk rock is coming right out of a 5-10 degree window that you’ve established completely independently with the computer programme as being the most dense area. Moreover, this area is where the folded tail was whipping back to the inside track so half the rocks had -vx,-vy disparities commensurate with a more easterly x,y approach and, if still lingering at an angle of around 3-4 degrees on the outside track between 8/2 and 15/2, could have been strung along a radiant in line with the Earth like the SL-9 ‘beads on a necklace’. That would explain the early hits.
In fact I think it could well be that the Earth got caught between the two strands like a nut in a nutcracker. DA14 would have been on the inside strand. All those evening hits would have been on the outside strand closing in from the outside track on the backside of the Earth between the twilight terminator and the 11PM hour line. That would possibly explain Cuba and certainly the two Japan hits. Even though one Japan hit was at 1AM, and therefore past the 11PM hour line, it could easily have been lurking in the gravity well just beyond but within the impact parameter, i.e. plucked backwards. There were no hits I know of between 1 and 9 in the morning. That arc of hour lines through the night correseponds to where the two tails would fan out beyond the Earth producing no candidates in that area- a sort of tail shadow.
Although there is nothing in principle to stop rocks from this nightside tail from creeping to the day side, the +z component could have largely swamped the slow xy towards the Earth. Rocks would have been funnelled into the impact parameter like grated cheese and straight into the northern hemisphere. They wouldn’t have a chance to cross beyond the impact parameter. Any escapees would have been unrecorded southern hemisphere hits in the oceans or Antarctica.
Chelyabinsk would come straight from the flexion point in this scenario, a daytime hit from 7-12 deg east of sun. The North American hits might still chime well with the nutcracker idea even if they were a day and a half after Chelyabinsk. This is because of the z component which means that any cloud in 2012, rather than a train, would have been folded like a piece of paper (imagine 10 rock trains stacked closely then being folded). This pleat of this fold wouldn’t have been conveniently straight up and down the z but more likely at an angle tipped away from the Earth due to the circa 30-45 deg downward angle of the DA14 radiant in 2012’s close pass. This would mean Chelyabinsk could hit from the direction of the pleat (bend) on the 15th and the SF fireball could still come a day later from the outside tail or fold. Of course, that particular scenario is very unlikely but I’m covering would-be objections.
All these night time hits are plausible when you consider that the northern hemisphere is tilted downwards on the night side and within much easier reach of either tail radiant. What’s more, the tilt introduces a strong east-west bias to anything coming straight up the z, and more so if plucked a little backwards. That would help explain Cuba and SF.
If the Earth was sandwiched thus, there would, by definition be a period during which the strand was threading its way through the 13,500 mile impact parameter on the backside of the Earth, pelting it with rocks that were in the well and doomed to hit. If the strand was a hundred thousand miles wide it could take days to whip sunward if you map it over roughly from the DA14 xy radiant speed: there’s a 3km/sec geocentric xy radiant for DA14 due to the 5.86 deg crossover but it would be less for rocks in the tail on the dark side due to a relatively circularised orbit- maybe between 1/2 and 2 km/sec. I think this scenario or something broadly similar could explain the whole shower in February 2013.
Even if you can’t model the tail to whip in like a nutcracker, I think there is a lot of mileage to be had out of considering the bend as being 600,000km behind the Earth. So many independently discovered factors fit with Chelyabinsk: density of rocks; whipping tail; half rocks with promising vx,vy; the radiant is the limit of the bend before the gap; the radiant is the limit of plausible absolute velocities. That’s five things we have discovered, one by one, which all point to the Chelyabinsk radiant.
A note on the bend/Chelyabinsk radiant being the limit for vx vy disparities: if you construct a fragment with a 20 degree east radiant it would be firmly ensconced in the gap, way behind the bend and likely have an absolute velocity that is unsustainable in terms of the conservation of momentum when referenced back to DA14’s momentum.
Here are two snapshots which suggest (1st pic) that the fold can occur way back towards Earth from DA14, leaving DA14 on the undisturbed inside track and (2nd pic) that Earth could get sandwiched ( the October shots- although it’s late 20’s that type of nutcracker shape could have crunched the Earth in 2013)
Andrew
Update on thread comments.
The thread lapsed on 14th June 2013 and was reactivated on the 16th. Andrew and Frank carried on communicating via email just as though they were commenting. So I’m going to copy over the emails one by one into the thread so as to retain the narrative and will leave a short comment at the end to signal we are back to commenting directly. There were about 20 emails/comments but a couple were off topic so they will be left out. Everything else copied over will almost completely on topic. I’ll copy over in chunks so we don’t swamp the recent comments list if that’s OK with the mods.
Andrew
From Frank 15th June (from email)
Andrew,
Might be an idea to email Tallbloke. He didn’t seem to want to close down the discussion, which does indeed have a strong narrative.
Today I dug out some info on earth’s oblateness. The polar radius is about 20 km less than the equatorial radius. Elsewhere I found that these oblate spheres have an elliptical cross-section. So I used this to write a bit of code to give a fairly good approximation of the Earth’s oblateness, for use during close approaches. In the process of finding the maths to do this, I came across a new word: “eccentric anomaly” which I’m sure you’re familiar with. Anyway, with the new oblateness code installed, a variant of Halo1093 landed 1.5 degrees further north when the Earth’s oblateness was included than it did using a spherical Earth. Which was quite a lot.
Anyway, seeing that I’ve now coded up, for the very first time, something to do with ellipses, it occurred to me that some of your ellipse-based mathematics might be included in the simulation. Would that be useful?
Cheers,
Frank
From Andrew 15th June 2013 (from email)
Frank
I’ve emailed Rog. That extra 1.5 degrees is a great success. I’m now kicking myself for not thinking of that- it’s like scraping 20km of Earth away at the last moment to give it 20km altitude- and at a highly acute angle as well. I know you mentioned oblateness recently but I think it was to do with converting local azimuth and altitude to RA and declination.
I know about the eccentric anomaly. I think I mentioned it as one of the other two anomalies when I brought up the mean anomaly but I may not have cited it by name. The third one is the true anomaly. All three are intimately related. However, I was struggling to grasp the mean anomaly concept. It reminds me of calculus where I can see the beauty of it and specific scenarios where it applies but am slow to be able to manipulate the terms, churn the algebra. I shall go back to it though. Here are my thoughts on it so far. It may help you understand the explanations on the net which are positively Delphic.
UPDATE: I did go back to it and I think I’ve got it. I’ve described my thoughts below. It’s a bit long-winded and even if it’s not quite right I’m sure it would be good for you to get a feel for the ellipse/circle relationship at the heart of the mean anomaly. Having described it though, I felt uneasy about applying it to circularised fragments. This is because although I knew it was intimately related to the conservation of momentum within a particular orbit it doesn’t quite translate to the conservation of momentum of other fragments thrown into different orbits as far as I can see. I touch on this, below, and dismiss it as a quibble and that is correct if staying within the same ellipse and concerning oneself with conservation of momentum within that ellipse.
For conservation of momentum of DA14 fragments in other orbits I can see that the simple linear momentum x radius works for instantaneous points in the orbit. And since a rock’s mass isn’t changed by being displaced, it is just a simple velocity x radius parity to check on conservation. This may appear to flout the KE conservation but because a higher rock has some extra PE squirrelled away, the momemtum conservation closely resembles KE conservation without regard for PE or even having to have any regard for it. I was going to do this for a higher DA14 fragment out of interest but even though I’m pretty sure of the methodology, I think I would still add up the PE and KE of both just as a sensible check.
Anyway, working out what the heck mean anomaly actually is, mathematically, was interesting and even if you just gloss over this, it will give you a feel for what these rocks are doing when staying within their own ellipse. I would have thought that this couldn’t do anything but enhance your view of the behaviour of perturbed rocks and hopefully allow you to see what equations to bake into the program for those erratics. The behaviour of the rocks in your programme with regard to mean anomaly isn’t an issue as far as I can see because the mean anomaly equation, linked below, is obviously derived from Kepler’s laws as are all your programme equations.
I’m starting off with the eccentric anomaly since you mentioned it and using the diagrams on this Wiki page when visualising it:
http://en.m.wikipedia.org/wiki/Eccentric_anomaly
So here’s my understanding of mean anomaly, complete with unedited concerns.
These are resolved later or in the update above.
I can see how the eccentric anomaly is very useful in keeping tabs on where an orbiting body in an ellipse would be mapped over to its circumscribed circle. That’s not quite the same as circularisation because the focus of the ellipse isn’t the c of g that the circularised ghost would orbit around. But from that basic concept one can see that there is always this catch-up waltz going on between the orbiting body and its circularised ghost.
I think it really could be compared as a truly circularised version if one simply bears in mind that the circle would need to be shunted over to be centralised over the c of g. The only small problem I have with that is that, to be truly circularised there would have to be some give and take with the ellipse being squashed in at the apsides. Indeed, that happened to AC6 and AC7 just by virtue of crunching them through the eccentricity equation and then the semi major axis equation. This is because those equations have ellipse DNA in them. Their semimajor axis was reduced by about a million km.
Still, the whole point of the eccentric anomaly and its derivative, the mean anomaly is to compare the position of a body with its theoretical cousin on the outer circle so my quibbles can’t be valid here. Because the body on the outer circle orbits with a constant speed, one can go forward and back in time along the circle and drop down to the ellipse using the eccentric anomaly to find out where the ellipse body will be at that time.
It’s a neat concept because they can be easily compared catching up and overtaking eachother. They only meet up and overtake at the apsides. The ellipse body overtakes at periapsis and the circularised body at apoapsis.
I bear this in mind when thinking of circularised fragments of DA14 acting like the outer, quasi circle with the ellipse of DA14 circumscribed by that circle. I also think of the Earth as circumscribing DA14 in a very rough sense. However, I say very rough because they don’t share the same perihelion angle. The Earth’s is around Jan 5th and DA14’s is around 18th November. Also, the 10.33 deg inclination throws it off in xy, making DA14 look more circular.
What I don’t get about the mean anomaly is that they say it is an angle, but isn’t really an angle and then say it is an area but really it’s just the length of a line along the orbit. Then they say it’s a ratio. I can see how these statements are related to Kepler’s laws, especially sweeping out equal areas/pie shapes in equal times. What I really don’t get is that it is stated as a single angle: the Earth’s is about 357 deg and DA14’s is 299.9 deg. Those two figures obviously speak of the degree of circularity and diveregence from a 360 deg ideal.
UPDATE: I think what they are trying to say is that, for instance in the case of the Earth, it has a notional eccentric circle that is very slightly bigger and circumscribes the Earth’s ellipse. The 357 degree mean anomaly is the same thing as saying that when the Earth orbits a full 360 degrees of its true orbit it sweeps out an area that is the equivalent of 357 degrees of this notional outer circle. It has actually done 360 degrees but if the area it swept was mapped to the eccentric circle it would only sweep a pie shape of 357 deg. That must be it. That satisfies the concept that it is an area, an angle and a line (along the circumference subtending that angle). It also satisfies the idea of a ratio (to 360 deg). So if I’m right about that, DA14 sweeps out a whole ellipse but that area is the equivalent of a pie slice subtended by 299.9 degrees of its eccentric circle.
These definitions of mean anomaly as angles don’t seem to bear much in common with the Wiki formula linked here:
http://en.m.wikipedia.org/wiki/Mean_anomaly
However, I think they are related when you integrate the formula over a full orbit. I think the 357 deg for Earth actually is the integrated value. That Wiki page formula allows you to find out the position of a particle along its orbit by, in effect, mapping over to its eccentric circle (using the mean motion, a constant speed value that would be maintained on the eccentric circle for the orbital period). This can only be done by using a start point, any start point, and a dt value which is crunched via the mean motion. But since mean motion is related to time and is exactly proportional to equal chunks along the the eccentric circle circumference, this means that any calculation using the mean anomaly equation is in fact using a section of circumference of the eccentric circle to map to. This in turn means that it is using an angle (origin is the sun) that subtends that circumference section and therefore a pie shape and a ratio to define the new particle position- except that it doesn’t need to invoke the pie shape and ratio each time, just the distance to add along the ellipse. The really neat thing here is that although for any given dt the distance to add is going to vary according to where in the ellipse you started (perigee=bigger angle per unit time), the sum of all dt’s around the ellipse add to one definitive pie shape on the eccentric circle. That is defined by an angle which is, in turn, a ratio (to 360 deg). So in the case of the Earth, it sweeps out an area which is 357 360ths of the area of its eccentric circle.
So if I understand it correctly, that seemingly weird angle defining the mean anomaly is the integrated value of all the dt values around the inner ellipse. Those summed dt values give a summed pie shape and angle subtended on the eccentric circle. That angle is 357 degrees for the Earth and 299.9 deg for DA14.
I hope that is of some help.
From Andrew on 15th June 2013 (from email)
Frank
Although I said you didn’t have to worry about the mean anomaly for your programme orbits because they are all derived from Kepler, I think it is of greater importance than I thought. This is because by using the mean anomaly equation you can jump to any time in the orbit a get a single position value without (presumably) having to crunch your way round the orbit doing thousands of cumulative calculations at 1000 or 500 secs dt’s. I realise that once you have done the crunching, the info is stored and you can access a particular time and get the vectors but when you are doing large cubes of 343 rocks or similarly complex runs there is potentially a lot of computer time wasted on the uninteresting rocks. I’m wondering if using the mean anomaly would be quicker and more surgical.
Another reason it might be useful is that, since it’s not using dt steps but piggy-backing on a perfect circle of known geometry and exact time step, it is perfectly accurate, notwithstanding the usual minor unknown perturbations. I know your RK 4 additions made a substantial difference in accuracy so maybe that isn’t an issue anymore but the facility to pick a time in the orbit without crunching the whole orbit and get an instantaneous position, possibly along with its speed would be a boon.
Andrew
From Frank (from email) 15th June 2013
Andrew,
I came across the eccentric anomaly in one of the parametric equations for ellipses on Wikipedia. I wanted a way of generating an ellipse simply using distance and angle. And X(t) = a . cos t, Y(t) = b . sin t was the nearest I could find. But, as the text emphasizes, t is not the angle of the point on the ellipse, but of the eccentric anomaly. But I decided that, for my purposes, I could probably get away with pretending that it was a point on the ellipse, because the Earth is very nearly spherical. But I’d prefer to have an accurate way of finding the point. If you know of one, I’d be very interested. Because all I have to do is to find the latitude of an approaching rock (which I can do), and use this to find the radius of the Earth beneath that point.
And, yes, I thought that including the oblateness of the Earth would probably win a few minutes of arc. And was surprised by 1.5 degrees. A more accurate ellipse-generator would probably refine that number.
But it occurs to me that if I’m going to take account of the oblateness of the Earth (which I am doing), then I really ought to take account of the elevation of Chelyabinsk above sea level as well. And I have no idea what that is. But Chelyabinsk is only about 60 km east of the Ural mountains (highest peak is Mount Narodnaya, approximately 1,895 m), and so may well lie in its foothills. The asteroid approaching from the east on 15 feb 2013 may well have found that ground level was rising in that region (and maybe the atmosphere above such high ground is slightly deeper, a bit like water flowing over a submerged rock).
“This is because by using the mean anomaly equation you can jump to any time in the orbit a get a single position value without (presumably) having to crunch your way round the orbit doing thousands of cumulative calculations at 1000 or 500 secs dt’s.”
This is what I have to do at the moment – crunch my way round the orbit doing millions of calculations. Or at least my computer does. And it would be very handy to be able to set up an elliptical orbit, and move around it. I don’t know how to do that
“It is perfectly accurate, notwithstanding the usual minor unknown perturbations.”
…like the close approach of Earth on 16 Feb 2012. This is why, ultimately, I tend to place rather more faith in numerical simulations than in elliptical geometry that always seems to me to be a bit too ideal (no offence meant to Kepler, of course!).
Cheers,
Frank
From Frank (from email) on 16th June 2013
Andrew,
Chelyabinsk is indeed sited on high ground east of the Urals. It’s 237 m above sea level. Korkino, about 20 km south of it, is 249 m above sea level. Yemanzhelinsk, about 40 km south, is 242 m above sea level. And Chebarkul to the west is 340 m above sea level. And Timiryazevskiy, W-SW of Chelyabinsk, is 100 m above sea level. East of Chelyabinsk, Kurgan is 72 m above sea level. To the north Kamensk-Uralsky is 168 m above sea level. To the northeast Shadrinsk is 83 m above sea level. And to the south Troitsk is 160 m above sea level.
So it rather looks as if Chelyabinsk and Korkino and Yemanzhelinsk are sited on a slightly elevated ridge running N-S. The ridge is about 150 m higher than the surrounding area. And it is highest at Korkino.
If the same ridge is reproduced in the atmosphere above, then the approaching rock would have encountered the ridge. And if it had passed over, it would have hit a higher ridge above the Urals. But this assumed that the atmosphere faithfully reproduces the terrain below.
All of which reminds me that I still don’t know whether a ‘ridge of high pressure’ in the atmosphere entails a physical ridge also.
Frank
From Frank (from email) on 16th June 2013
I got the above info from
http://dateandtime.info/citycoordinates.php?id=1502112
Frank
From Andrew (from email) on 16th June 2013
Frank
No reply from TB yet. I’ll email Tim tomorrow if no answer.
I’ve been busily calculating the exact radius at Chelyabinsk including the altitude so I shall dwell on that and leave your thoughts on elliptical orbits, referring to my green quotes, till tomorrow.
I found the page below which has two answers that are the same. It looks right to me.
http://math.stackexchange.com/questions/22064/calculating-a-point-that-lies-on-an-ellipse-given-an-angle
I hope that link doesn’t mess up the formatting. They often do. Sorry.
So I crunched the two coordinate values given by the two equations given in the second answer on that linked page. I used the Wiki values for the equatorial radius as ‘a’, semi major axis and the polar radius as ‘b’, semi minor axis (6378.1km and 6356.8km).
For theta, I used 55 degrees. Zuluaga and Ferrin’s latest projected impact latitude was 55.0537 degrees. There were some largish numbers elsewhere in the maths so I stuck to a round figure to minimise mistakes. It’s only 5km off in circumferential sense which can only be a few metres or tens of metres at most in an idealised ellipse radius and Z and F have another estimate that’s 0.0421 deg more, so same distance further on the other side, to the north.
I came up with an x coordinate of 3650.12 km to 2 decimal places.
I came up with a y coordinate of 5212.91 km to 2 decimal places.
The x and y are not commensurate with your position vectors. They are just normal exercise book Cartesian coords for a 2D graph representing a 2D slice through an ellipse. That means I could use Pythagoras on a simple triangle to get the radius, the hypotenuse without any accusations of tautology saying, ‘hey but it’s not a true radius if its still an ellipse’.
Using Pythagoras, these coordinates yielded a radius of 6363.79 km at 55N.
Then, I used Z and F’s altitude figure from the March 31st 2013 estimate which is 261.4 metres. You’re right, it is in the foothills. I should note here that the other northward estimate had an altitude of 219 metres. Of course, if they are pinpointing exact spots, almost to the metre in lat and long, their altitudes will swing wildly. So I wouldn’t dwell too much on 261.4 m as being written in stone. Seeing as we need every last arc second, I think it would be reasonable to guess that the average elevation around there is 200-250 metres or take the average of 240 between their two estimates.
But, to be exact, using Z AND F’s 261.4 m (and adding figures to 2 dec places to figures to 4 dec places) I get 6363.79 km plus 0.2614 km= 6364.0514 km for the exact radius at impact.
So…what was your radius, without the altitude? 🙂
Andrew
From Andrew (from email) on 16th June 2013
My last reply crossed with your elevation/ ridges reply. That’s good to get it confirmed that the elevation is between 100 and 300 metres roughly. The area to concentrate on is west of Chebarkul and 13 km west of Miass. The little town of Syrostan was considered much nearer the projected theoretical trajectory end point. Seeing as you came up with figures in the 200’s around that area, with Chebarkul at a 300 m maximum, and Z and F are firmly ensconced in the 200-260 metre range, I would stick with that. The elevation figures to the east and south are useful to know as a guide to the general elevation but the projected end point is now narrowed down to a square of about 10 or 12 km centred on or near Miass. They haven’t actually said that, but I’ve gleaned it from the Z and F estimates and Stefan Geens.
Of course, the altitude error bars are only about 2% of the gain from calculating the oblateness so we are already quite accurate now and just tinkering at the edges. However, I would still agree that it is nice to nail it down to the last few metres.
I don’t think the atmosphere mirrors the terrain below in any significant or permanent manner. In this respect it is like the sea. It finds its own level regardless of what’s below, poking up into it at various altitudes. This is why the air is thin at altitude and pressure, speed of sound, boiling point of water drop away on predictable curves as altitude increases, whether in a plane or up a mountain. See link below. Fig b shows pressure drop off and a superimposed Everest. The prose says that, according to the graph (which is generalised for all points on the Earth), the summit of Everest would experience 30% of the pressure found at sea level. This is notwithstanding the not insignificant effects of temperature variation and humidity, but I don’t think those things would affect the general curve in that graph much.
http://www.nc-climate.ncsu.edu/edu/k12/.AtmStructure
Also, a ridge of high pressure I think is always referring to the front, and so is more to do with the approaching quasi vertical wall of high pressure rather than any bulge on top of it. That said, I know the African warm layer rides up and over the moist lower atmosphere of the Atlantic and I should think it probably varies in height according to the seasons and the temperature in the Sahara. I also heard talk on the Talkshop of the troposphere being higher or lower by a significant amount according to season. But none of these depend on the terrain elevation below as far as I know.
Andrew
From Frank (from email) on 16th June 2013
Andrew,
“So…what was your radius, without the altitude? :)”
The same as yours – because I’ve now swapped over to using the method you’ve just found. 🙂
And the gain in latitude by my Halo1093 variant has increased from 1.5 degrees to 1.55 degrees. So my earlier cruder method of calculating oblate radius wasn’t far out.
Are you calculating the exact radius by simply adding 261 m to the oblate radius. Is the radius of the Earth its radius at sea level?
I think that the answer to my question about the thickness of the atmosphere above a point must be that it does follow the terrain below. If the atmosphere is regarded as being a fluid (like water), then as a stream or river with a constant rectangular cross-section flows at a constant speed downhill, its surface maintains the same height above the bottom at all points. If the cross-section narrows, the stream must flow faster to maintain the same depth, or else get deeper. This will mean that the atmospheric drag forces will be greater over high ground at the same radius. But I’m not entirely sure…
Incidentally, do you have a link for Zuluaga and Ferrin?
Frank
From Frank (from email) on 16th June 2013
“I don’t think the atmosphere mirrors the terrain below in any significant or permanent manner. In this respect it is like the sea. It finds its own level regardless of what’s below.”
Well, I’ve just turned up a NASA article about the Earth Atmosphere Model which says:
The model assumes that the pressure and temperature change only with altitude.
Altitude, note. If they’d said radius, it would have been different. So clearly to calculate air density, you must know the height above ground. And for a body moving at some radius above the Earth, the altitude is going to fall quite a lot as it passes over mountains. The changes in density may be fractional, but they’ll still be present.
Frank
From Andrew (from email) on 16th June 2013
Frank
I’m still intrigued to know the old parametric radius because it must have been very close to this new value, yes? The reason I’m interested is because of getting a feel for the eccentric anomaly at work in the real world.
Yes, I’m adding 261m to the oblate radius. The values of the radius at equator and poles are for sea level and so the oblate sphere radius at 55 north assumes Chelyabinsk to have sunk below the waves of a planet-wide ocean. You then have to build up the terrain above that level.
I wasn’t surprised at the 1.55 degrees for such a small scraping away of radius. When it’s so close to tangential there’s a big difference. Bill Bryson once said of Nebraska, or some such flat state, that you could see for a hundred miles, or 200 if you stood on two telephone directories. Nonsense in the detail but true in essence.
“I think that the answer to my question about the thickness of the atmosphere above a point must be that it does follow the terrain below”.
If you mean thickness to mean density as opposed to vertical thickness or height above the mountains then I understand your point. I don’t think it holds though because it assumes the air wants to flow over these letter-box thin ridges and maybe canyons. But I would say that the mountains simply hold pools of stagnant (as in stationary) air that just won’t budge because of the surrounding mountains. Any surplus flowing over the mountain tops will behave much like the air in lower, flatter terrain notwithstanding the pressure and temperature differences.
But yes, I do agree that if for some reason there is some sort of strong high pressure front that drives large amounts of atmosphere out of the stagnant pools and down and through canyons, I suspect the density would increase with the flow.
Coincidentally, when we get cold easterly winds several times a year, I’ve heard the weathermen say it is cold air from the Urals. It apparently sits in the bowl shaped by the mountains and then suddenly flows out and across Europe. I don’t know what stops it from flowing out before it actually does flow out but whatever it is, it supports your theory and shoots down my objections.
One other thing is that all the major air resistance for the Chelyabinsk rock was experienced in the high stratosphere to mesosphere (23 to 47 km according to Z and F and NASA). The troposphere where all the weather occurs is like a closed-off system of much denser air way down below.
I have two links for Zuluaga and Ferrin. The first is their page that gets updated periodically. It has downloads at the bottom: graphs and extra details. The second is a 10 page paper that I found only a few hours ago and haven’t fully read. It has way more detail and yet it is dated 21st of Feb 2013. I know there are some things in there that are now discounted, like the azimuth. It nevertheless makes interesting reading. One eye opener is their constant reference to the velocity being poorly characterised and have a lowest estimate of 13.4 km/sec. That’s 7.35 km/sec cosmic speed, our sorts of speeds almost. They have three orbits depicted and vaguely mention that two of them show extremes but don’t relate the smaller one to the 13.4 km/sec speed. The orbit is still quite big and I don’t think it does relate to it. The additional details download in the first link, added to the site recently, has velocity error bars that imply a 14.9 km/sec lower limit and I think this is what is being represented in the earlier diagrams.
There’s also no mention of camera distortion which I know exists in the Revolution Square footage and can only reduce the distance covered and therefore the velocity.
Z AND F links:
http://urania.udea.edu.co/sitios/facom/research/chelyabinsk-meteoroid.php?#
Click to access 1302.5377.pdf
Andrew
From Andrew (from email) on 16th June 2013
Frank
I just found this Z and F PDF doc. It appears to be a lot more detailed and they have used another video for the shadows as well. It looks more accurate than I thought they had been, based on their past references to how they used the other videos (only for azimuth) and how much they had relied on Stefan Geens’ Revolution Square calculations for speed.
I haven’t read it yet. Thought I’d send it to you in case you are reading through the older 21st Feb version I sent.
Andrew
Re last comment:
Forgot the link!
Click to access 1303.1796v1.pdf
From Frank (from email) on 16th June 21013
Andrew
“I’m still intrigued to know the old parametric radius.”
I used x(t) = a . cos t and y(t) = b . sin t where a was the equatorial radius of the Earth, and b was the polar radius. In this case t is the eccentric anomaly, but I thought it would probably be very near the latitude, The actual code to find radius using the m.latitude of the incoming Halo1093 rock was:
lx = thisPool.b[3].equatorialRadius * Math.cos( m.latitude );
ly = thisPool.b[3].polarRadius * Math.sin( m.latitude );
rb = Math.sqrt( lx*lx + ly*ly );
But I’m now using the formulae you produced last night to find lx and ly.
On the atmospheric matters, it seems that there’s no definition of where the top of the atmosphere is, because it fades gradually away. But if the ‘top’ of the atmosphere is regarded as being at the point where it falls to some particular arbitrarily-chosen density, then it seems to me that, assuming still air, points of equal density (isosteres) will follow the terrain below. That is the conclusion that I drew from the NASA model that I found last night.
It also seems to me that in areas of high pressure, there must be more air, because atmospheric pressure is the summed mass of air overhead. And if there’s more air overhead, then the Earth’s atmosphere must bulge outwards in high pressure zones, and inwards in low pressure ones. This sort of makes sense to me, and would mean that the ‘top’ (as defined above) of the ‘atmospheric ocean’ would have all sorts of waves and ripples traversing it, and sloshing around all the time.
And if high pressure days are quite often cloudless, then 15 Feb 2013 may well have seen a high pressure area sitting over Chelyabinsk.
But really I’m simply trying to gain a better understanding of the atmosphere, because quite apart from our current enterprise, I’d like to build a simulation model of one to look at its warming and cooling. I am an inveterate modeller of more or less everything I can lay my grubby little hands on, but so far the atmosphere has defeated all my attempts to take hold of it..
Frank
P.S. Thanks for the Z&F links.
From Andrew (from email) on 17th June 2013
Frank
I think it does bulge at high pressure zones and I should think there are small ripples and waves too. I’m not convinced NASA mean atmospheric altitude when they say altitude (and not radius). Whenever I read about altitude it is either explicitly defined as the height above sea level or tacitly assumed to be because it is defined that way in so many other contexts. When this is the case, it applies to a point above sea level whether it’s on land or in mid-air.
Regardless of whether the above is correct, an interesting test of the bulge would be whether planes flying at 35,000 feet go up a bit higher over mountain ranges to stay within their ideal operating pressure. When I fly over the Alps they look closer than the lowlands but that doesn’t mean we haven’t crept up a tiny bit.
Super cells are an example of a big localised pimple, breaking through the tropopause. These are laden with vapour which evaporates and cools the air which then fans out across the top of the tropopause or gets blown to one side causing the familiar supercell anvil. But really, you are talking about the top of the atmosphere bulging. For most practical purposes to do with drag, 80 miles is often cited and that is what I’m sticking with for assumed height at which the Che meteor hit and was detected by the infrasound sensors. That’s why 32.5 secs of noise at 17.5 km/sec doesn’t stack up but it does with 13-13.5 km/sec.
Andrew
End of copied emails.
As per my comment of of 18th June 2013 at 12:18 I have copied over the above 15 emails and now the copying is complete.
From now on, we shall be commenting directly to the thread.
Andrew
By the way, Frank, that Zuluaga and Ferrin paper I linked said that there was some evidence that Chelyabinsk was related to Tunguska. I nearly dropped my cup of tea when I read that. They were citing the more obvious things like latitude and radiant and not explaining how the node changed from June 30th, but even to mention it was an eye opener because it means it’s more than just a vague possibility.
It’s just that they (and also Steve) are linking Chelyabinsk to Tunguska and I have tentatively linked DA14 to Tunguska. It follows that DA14 might be related to Chelyabinsk via Tunguska.
The main reason I questioned the DA14 link to Tunguska was because of the node change but Z and F evidently don’t find that a problem for Chelyabinsk. I trust their expertise. I just think their inputs might be a bit off and their Monte Carlo simulation a bit too reliant on picking the easiest fit, hence the direct hit for Chelyabinsk.
My calcs, based on NASA’s 368 day orbit for DA14 prior to 2011, fitted almost exactly with the Earth’s Tunguska position (June 30th) and Earth’s DA14 close approach position (Feb 15th), that is, if DA14 were to catch up over the 105 years from 1908 to 2013.
The 105 year gap from Tunguska 1908 to 2013 implies a 2.774-day surplus orbital period. DA14’s was 368-365.256= a 2.744-day surplus. And NASA were saying roughly 368 days, not dead on.
If DA14 was related to Tunguska, I think it would have been a very close approacher or a breakaway piece from the actual Tunguska rock (a bounce event) which then got slowed enough to skew it into negative z. That would at least explain the ascending node over that side of the orbit changing to a descending node.
[June 30th to Feb 15 is 291 days. Dividing by 105 gives 2.774. That’s the lag which over 105 years results in meeting up again 291 angle-days further round the Earth’s orbit….notwithstanding the node problem]
Andrew
Andrew,
You’ll like these. The first image shows a now-familiar snapshot of Halo1093 skimming over Mongolia.
But you won’t have seen either this alternative view, of the same event, or this one either
They are respectively the views along the z axis, the y axis, and the x axis.
The other trail is of a rock I launched from Chelyabinsk at 3:20 am on 15 Feb 2013.
Oh yes, that’s pretty cool, Frank. I notice it shows that it’s a tightish impact parameter. I thought it might be 1000km further out although I realise there’s some parallax going on.
So are x and y axis views applicable to all operations now?
Andrew
Frank
Was the launch from Chelyabinsk just playing around or did you get anywhere with it? It looks as if it’s about the same speed as Halo 1093 but in reverse.
Andrew
So are x and y axis views applicable to all operations now?
I hope so. It took a bit of a kludge to get it to work, but fingers crossed, it should work.
Was the launch from Chelyabinsk just playing around
It was just playing around. I wanted to check it would launch from the right place, and it did. I didn’t care where it went.
Andrew,
I decided to reconstruct a few of your AC-series rocks today. It was very easy to reconstruct AC6b and AC6b-rev, because their state vectors are posted upthread. I then spent a while trying to get a variant of these – AC6c – to get somewhere near Chelyabinsk – until it occurred to me that, with a little adjustment, I could maybe use its closest approach velocity vectors in the satellite I launch on 3:20 15 feb 2013 from Chelyabinsk, and then reverse back out to see where it came from.
It worked a treat. And I now have the state vectors of Che1-ac6c on 0:00 8 feb 2013. And it lands smack on Chelyabinsk, of course. That’s Halo1093 coming in at the same time on Mongolia on the right.
I also wound back the orbit a year or two. Che1-ac6c is in a wider, slower orbit than the Earth. It takes a few days longer than the Earth to go round. Which means that it wouldn’t have come out of the ‘fireball’ rock clouds we’ve been looking at.
In the side view along the Y axis, its orbital inclination is also a lot less than DA14’s (also shown). The Earth’s orbit is the horizontal straight line.
Tomorrow, I’ll maybe try and work out what angle the radiant from Chelyabinsk makes east of north. Using my protractor, it looks like it’s around 118 degrees. I seem to remember you saying that the consensus these days is that it came in on 105 degree bearing.
Anyway, I was very pleased. We now have a rock that lands smack on Chelyabinsk at exactly the right time, and which is in a similar orbit as DA14. It’s just not coming in at quite the right angle. And it’s also hard to trace just how it might have got detached from the DA14 rock cloud to wind up in that orbit.
That’s a great success, Frank. Could you do a y axis view of the hit so as to see the atmospheric angle? Also the vxyz on impact would allow me to see the angle too if I can subtract that from the latitude/tilt/azimuth.
The reduced inclination is due to the 4.170 km/sec in z on 8th Feb 00:00 as opposed to 5.339 km/sec.
Yes, the consensus is 105 degrees. Z n F say 105.1 deg.
The fact it is only a few days off in its orbit is very encouraging. How many days longer does it take? Or it may be better just to say the orbital period.
If you can manage that with an old refurbished rock I’m sure we can do even better with various refinements.
Andrew
Andrew,
I thought you might like to see the approach of Che1_ac6c. I haven’t measured the difference of angle between it and DA14 coming up on the left, but it’s quite a lot. Che1 comes in quite a lot faster than DA14.
Here’s the view along the y-axis. And also the view along the x-axis..
The impact velocity vectors were Earth vx – 8781.815 m/s, Earth vy + 8097.159;m/s, Earth vz – 2733.171 m/s; Although I actually launched this rock at 1.01 Earth radius, or 63.71 km above the surface above Chelyabinsk. And I haven’t included for Earth’s oblateness.
Last thing I did last night, to try to get the approach radiant more towards 105 degrees, was to change vx to Earth vx – 8381.815 m/s and vy to Earth vy + 8497.159 m/s. This won me another degree or so. But more interestingly the resulting orbit was only slightly longer than DA14’s, and also less inclined
And I must say I’m not too bothered about 105 degrees. I think getting anywhere in that region would be great. But there’s clearly a whole family of rocks that could land on Chelyabinsk at 3:20 on 15 Feb 2013. It would be interesting to know what Z&F’s approach vx,vy,and vz are. Because that will be one member of the extended family.
Andrew,
Che1 orbit period. Reversing back round from 8 Feb 2013, it looks like it arrives back where it started on about 22 Jan 2013. That’s 16 days longer than the terrestrial year.
22 jan 2012, not 2013.
Lots to talk about there, Frank.
I haven’t checked this over for spelling mistakes etc because I had to shoot out but I think it’s fairly OK. I just wanted to get it out there.
Thanks for the x and y axis shots. They speak volumes because the atmospheric trajectory looks perfect. Has it still got AC6 DNA to get that angle?
The reason I ask is that I had calculated AC6 (or its predecessor) for a 7.83 inclination for an 18 deg atmospheric trajectory which it sort of resembles. However, prior to getting these x and y axis shots you formerly achieved, I believe, 61 deg N skims which would imply some atmospheric altitude angle at 55N but not 18 degrees. Did you play around with the vz when you fired it off or can we deduce that AC6 was indeed quite accurate in its atmospheric altitude angle. It would be good to know if I’m doing something right and I did always puzzle that it didn’t seem enough- but we never measured it or had a sideways view.
As far as I can remember, AC6 was the first if the east-of-sun attempts and prior to today, your best refinement of AC6 was AC7 (based on my AC6b-rev, from memory). AC7 had the start position vectors adjusted so as to get the time right. So, the reason I’m regurgitating all this, partly rhetorically, is that they are all descendants of the AC6 granddaddy and that is why they all exhibit the marked angle from DA14 and the higher speed. I just measured the angle with my protractor and it is indeed 11 degrees which is what AC6 and AC7 were.
Also, all these rocks, since AC4 had 7.83 deg orbital inclinations (or hopefully did- the y axis orbit snapshot looks about right).
That 11 degrees to DA14 was 6-7 degrees short of what I had expected for AC6. It meant it came straight from the sun and needed another 6-7 to get round to the 105 deg azimuth. I believe I needed to include the 7 deg turn in the orbit in the 171.333 hrs of the run from 8th Feb. I swore blind to myself it wasn’t needed but I think I missed the underlying solidity of adding a vx,vy component that operates regardless of the turn. So today I have embarked on the long-promised attempt to get further round. I know you say you’re not too bothered about getting right on 105 deg but I’ll mention two things. One is I don’t think it’s going to be difficult to get near and secondly, the main thing in this new calculation isn’t so much the radiant but the speed. What I’m attempting to do is reduce its absolute speed to comply with DA14 (PE/KE) then convert that speed into the proportions of vxyz for the correct vx,vy radiant and vz/7.83 deg inclination. That way it should be only a few days longer in its orbit, or even a tad faster because the vx,vy angle isn’t dependent on absolute speed.
Having mentioned its coming in faster, I just summed those impact velocity vectors and it comes to 12.254 km/sec which is the same or slower than Halo, isn’t it? I always thought AC6/7 was faster because it started off with 300m/sec or more added in absolute velocity. But I realise you said you did play around with the x and y velocities for the angle so maybe that’s why.
You asked about Z & F’s vxyz. I think it’s in the ‘other details’ document on their web page under the ‘downloads section’. Theirs will sum to 17.5 km/sec but you have a point because the proportions give the angle. We just have to reduce them in line with an absolute velocity of 12.2- 13km/sec. Also the NASA page has the cosmic geocentric vxyz which, summed, come to about 13 km/sec. That results in 18.6 impact velocity though and corresponds to the Apollo class orbit similar to Z & F presumably.
http://urania.udea.edu.co/sitios/facom/pages/chelyabinsk-meteoroid.rs/files/chelyabinsk-meteoroid7t6ln/ChEI-ZFG2013-summary.txt
http://neo.jpl.nasa.gov/fireballs/
Bottom of NASA page shows cosmic vxyz but that’s for the big Apollo class orbit.
Just looked at the x axis shot again. It looks a perfect Che candidate orbit.
Andrew
Did you play around with the vz when you fired it off
Yes, I did a bit.
To get ac6c I moved around ac6b’s 8 feb 2013 start location, trying to get it somewhere near Chelyabinsk. After several attempts I found one impact which was in about right latitude but the wrong longitude (the east coast of Russia), but coming in at the right sort of angle. It was just that Chelyabinsk wasn’t underneath it.
This is when I thought that I could use my Chelyabinsk launch method by simply giving it the velocity state vectors of the impacting ac6c, while retaining the Chelyabinsk position vectors. However, one slight problem was that the ac6c impact vectors had a vz that was a small positive value. This would have meant, at Chelyabinsk, that the rock was coming out of the Earth. To fix this little problem, I simply flipped vz from positive to negative. But the vx and vy were left unaltered. This meant that the net speed was exactly the same, but the approach angle was a bit different.
All I was trying to do, really, was to find any old half-plausible state vectors to plug into my model which I could then use to reverse back out, because previous guesses of mine hadn’t been much good.
Anyway, it all worked much better than I had expected, and in this manner Che1 was born. So Che1 has a great deal of ac6 DNA in it, because it was created using the impact velocity vectors of a slight variant of it.
Anyway, there have been further interesting developments today, which I will recount in a separate comment.
Frank
That’s an interesting story. I was thinking, it reminded of that John Peel programme called, err, Rock Family Trees. All these iconic rock names acting as the foundation for new ideas. And I thought of the programme before I realised the actual double entendre. Would’ve been even more crass the other way around!
Continuing on from the last comment, you may remember that last night I mentioned that Che1’s orbital period was several days (in fact nearly 2 weeks) longer than the Earth/DA14 orbital period. And you may also remember that I then tried a slight variant which had an orbital period that was only slightly slower than that of Earth/DA14. This was Che2.
Today, I decided to see if I could get a rock that had more or less the same period as Earth DA14, simply by changing the launch velocity vectors a bit more in the same direction as from Che1 to Che2. Che3 had Earth vx x – 8081.815 m//s, and Earth vy + 8797.159 m/s, and Earth vz – 2733.171 m/s.
And as luck would have it, reversing this rock back a year to 16 feb 2012, I found that it was closer to the Earth than DA14 was, but on the other side of it. So Che3 would have been in the rock cloud fore and aft of DA14 on 16 Feb 2012 when the Earth punched a hole in it. But while DA14 was slightly slowed in this encounter, Che3 was speeded up.
I was rather excited by this, because Che3 is a rock that a) is in the rock cloud around DA14 on 16 feb, and will have been more influenced by the Earth than DA14 (which was quite a lot) during that encounter, and b) lands slap bang on top of Chelyabinsk at 3:20 am UT 15 Feb 2013, a year later.
However, while I’d be concentrating on getting a 16 feb 2012 close approach, I’d not been looking at the impact end of things. And when I eventually did look, I found that the approach angle had increased from 118 degrees E of N to 123 degrees E of N (both measured with a protractor).
But Che3 is a very interesting rock. It’s far and away the best shot yet. For while it’s in the rock cloud on 16 feb 2012, it’s really just passing through. Either that, or a slight variant of it could be a rock cloud rock which passed close enough to the Earth to get thrown into a Che3 orbit.
After finding that the approach angle had increased to 123 degrees, I went back to Che1 and started changing the velocity vectors in the opposite direction, reducing from 118 degrees downwards to about 115 degrees. But when I did this, the orbital periods got even longer than Che1. But what I want is something that comes in at about 105 degrees E of N at impact, but also passes near the Earth on 16 feb 2012. At the moment I don’t seem to be able to do that.
Andrew,
I was having a look earlier at the Z&F paper. They seem to have found the state vectors for their impactors, and used a numerical integrator (a simulation model like mine, I guess) to follow their rocks back 4 years before impact. But they don’t seem to have published the state vectors they used.
However, they have published the orbital elements, and it seems that it’s fairly straightforward to go from one to the other. I found an online converter which might be useful. Or maybe we could just email them to ask if the state vectors are available. I haven’t seen an ‘other details’ page. The page you linked to earlier has
Position vector at impact time (Ecliptic J2000.0):
x , y, z (AU) = -8.2296109700e-01 5.4628609900e-01 2.6028354800e-05
vx, vy, vz (AU/day) = -1.7823720100e-02 -9.2681019800e-03 -3.4069700000e-03
r (AU) = 9.8777197200e-01
Is that what you meant?
.
Frank
I summed your second Che 1 impact velocities (or was AC6c first attempt so Che 1 second attempt anyway? Confused by scrolling miles up and down old thread!) Anyway, its the one where you decreased vx and increased vy. You increased the one by about the same amount you decreased the other hence an absolute speed that was 12.244km/sec, only 10m/sec less so that’s why it was almost the same, a tiny bit slower period.
Have you got an idea of the inclination of the orbit once it’s extrapolated backwards out of the gravity well? If you changed the vz to positive I presume the inclination is quite a lot less than 7.83 degrees. That’s not a problem of course.
It didn’t actually look shallow in that sideways snapshot of the orbits along the y axis but there was some 56 deg of parallax making it difficult to gauge: apsides are at around 56 deg and 236 deg heliocentric ie almost bisecting the x and y axes. The nodes are about 90 deg to this, hence the ascending node (Earth, 15th Feb) also being 56 deg to the y axis (146 heliocentric).
Andrew
Frank
My comment crossed with two of yours so it looks a bit marooned in the light of your runaway successes. However it is relevant because slowing the absolute velocity is lengthening the orbital period. Therefore, in order to shorten the orbital period while keeping the radiant and inclination I reckon you need to increase the absolute velocity while keeping the same vx:vy:vz proportions.
So if you look at your best radiant (but too slow) candidate and multiply each velocity vector by the same amount you will get a faster orbit with the same radiant and inclination. I would guess adding 1% to all three? Can’t do any harm.
Are you able to measure the semimajor axis easily? You’d need to compute closest and farthest approach to sun, measure between and halve it. If it’s same as Earth’s it will be 365.25 days even if it’s more elliptical. If Che 3 is getting accelerated it might be faster than 365.25 so if for instance it is 364.6 days it will have a semimajor axis about 170,000 km less than Earth’s 145,598,261km.
I only have a vague memory of seeing the Z & F state vectors so I suppose what you’ve pasted must be them.
Andrew
Andrew,
I used the Z & F (if that’s what they are) velocity vectors with my Chelyabinsk launcher. The reversed rock went out in a very plausible direction, but a good deal faster than anything we’ve been looking at. The resulting orbit heads much further out, and is completely different from DA14 (no surprise there). What was slightly surprising was the direction, which is only a few degrees off one of my later attempts today (Che5).
Really I should put in their x,y,z vectors too. I’ve just supposed that they would be Chelyabinsk.
I really must also calculate correctly the approach azimuth (?) and elevation.
I’m too sleepy now, but tomorrow I’ll try slowing Z&F down to our sorts of speeds. I won’t be too surprised if it’s somewhere near the rocks we’ve been looking at, and in some sort of orbit near DA14. If that’s true, then we’ll maybe only be in disagreement with them over speed.
Frank
You and I are right in the same track. When you were struggling with the radiant vs speed, I knew that you needed to reduce the velocities while keeping the same proportions for vxyz.
It was a bit early to give you a polished up rock ready to fire for an east of sun radiant because I didn’t have the absolute velocity but I did laboriously calculate the geocenteric radiant angle: 13.166 degrees. This involved getting the orbit turn in the 171.333 hr run exactly right and the Earth position in the first place on 8th Feb 2013 00:00 hrs
When I added this to the Earth’s angle of 49.350 deg on Feb 8th 00:00 I got 62.517 degrees. I nearly chucked it out because it meant that the vx:vy was 1.9 to 1 (tan 62.517).
But then I looked more closely at Z and F’s vx vy and saw immediately that they had the same proportion. I divided one into the other and found the tan-1 angle. It was 62.526 deg! That’s less than 1/100th of a degree out.
So the Z and F proportions do chime with my geocentric radiant for a 105.1 degree azimuth. So I was going to use that proportion for my vx vy out at 8th Feb 00:00 but didn’t know the definitive absolute velocity yet (somewhere around 30.140 km/sec I think).
You have done exactly the same thing with the programme and can reduce the velocities in the same proportions (in z as well) to get to DA14 speeds. It’s going to be 12.5-12.9 km/sec absolute geocentric speed, divided into vx vy vz in the same proportions as Z and F. Then you can check on the 8th Feb to see what the velocity vectors are. But the proof is in where it goes back to in 2012. Exciting stuff.
The absolute v is the root of vx^2 +vy^2+vz^2 so you can keep checking your reductions to get them into that 12.5-9 window.
Andrew
Frank
I hope this saves you some time.
I took the Z and F velocities, squared them then proportioned them with their sum. Then I took the upper end of the sort of speed we want: 12.9km/sec, squared it and crunched the result according to the proportions I’d obtained. This left me with vx^2 vy^2 and vz^2 for our 12.9 km/sec rock. So I got the roots to obtain the actual vx vy vz. They are:
vx: -11.284037 km/sec
vy: -5.867552 km/sec
vz: -2.156922 km/sec
When I divide vx by vy and get the vxy radiant angle via tan-1, it is exactly the same as the comment above, 62.526 degrees so it should work.
The three minus signs for the velocities above are mimicking the original signs for the Z and F velocities.
I can do 12.5 km/sec via the same method tomorrow but you may get there first by reducing the velocities from 12.9 km/sec.
FYI the proportions of the vx^2 etc to the absolute v^2 are:
vx^2: 0.765155418
vy^2: 0.206887646
vz^2: 0.027956935
These proportions apply to any velocity.
So, for 12.7 km/sec, you square it, slice the result acc to the above proportions, then take the root of each. Those are your velocities.
I don’t know quite why I do the proportions for the squared vxyz’s. I think it works without squaring. But I do so much squaring that it feels safer seeing it in that format when converting. There are other instances where you can come a cropper without squaring. The 62 deg angle means it should work a treat.
Andrew
Frank
I was looking at your snapshots of Z AND F orbit and close approach. It mimics their diagram of the orbit very nicely.
As for the close approach, now that they have plumped for NASA’s 105 deg azimuth, instead of the 99 deg that was generally accepted earlier on, it is going over Hong Kong instead of Taiwan. If you produce that line down, southward, it goes right over East Timor and Victoria, Australia, where other sightings occurred on 17th and 16th Feb respectively. Also there was another in Australia on the 18th. The azimuth for these is either not given or off the line. However, the one given (Victoria) could easily be a line of sight issue, viewed from a distance.
http://thewatchers.adorraeli.com/2013/02/16/in-last-2-days-fireballs-were-reported-all-over-the-world-sweden-netherlands-russia-japan-cuba/
Andrew
Andrew,
It mimics their diagram of the orbit very nicely.
I’m glad to hear it. I haven’t seen their orbital map for a while, and I was slightly having doubts about my version, because it goes out beyond Mars. But then, beyond Mars is where the Asteroid Belt is supposed to be, from whence all asteroids are supposed to originate.
it is going over Hong Kong instead of Taiwan.
I think I may start dropping lines from incoming rocks down onto the Earth’s surface, so that it’s quite clear where they’re going over.
As for your link, whose title is “In the last two days fireballs were reported all over the world”, well, we now know that there was almost certainly a big cloud of rocks (our ‘fireball’) where the rock train has folded after the 2012 close approach, which the Earth passed through on 15 Feb 2012, so there darn well should have been fireballs all over the world.
I haven’t done any more today.as yet.
Frank
I have now done the velocity vectors for a 12.5km/sec launch from Chelyabinsk on the Zuluaga and Ferrin radiant:
vx: 10.93414533 km/sec
vy: 5.68561296 km/sec
vz: 2.09004093 km/sec
And I’ve also done the vxyz for a faster, 14.9km/sec launch if you feel like doing it. The reason I’ve chosen 14.9 is that it is implied as a minimum velocity in the Z and F details page (17.5 +\- 2.6 or thereabouts). Also, the Russians always said 15km/sec which is about 14.8 without Earth spin. It is also the higher end of plausible atmospheric travel for 32.5 secs of infrasound recording and there were other snippets of evidence for this speed which now elude me. Although it is high for a DA14 companion I think it could be approached with some eccentricity and perihelion changes.
For 14.9km/sec launch from Chelyabinsk on the Z and F radiant:
vx: 13.03350123 km/sec
vy: 6.77725064 km/sec
vz: 2.491328790 km/sec
Andrew
Frank
Link at bottom is for the z axis view of the latest Zuluaga and Ferrin orbit.
Zuluaga and Ferrin ask for their paper to be cited when using info from it. Strictly speaking, this is just a link and I have cited it before, but it’s time I cited it again in full and in the format they ask:
Zuluaga, J. I., I. Ferrin, and S. Geens (2013). The orbit of the Chelyabinsk event impactor as reconstructed from amateur and public footage. ArXiv e-prints. [ arXiv | download | stats ]
Andrew,
I tried out your 12.5 km/s suggestion as Che6-ac12_5. It’s no. 17. It was very slow going (in reverse) around the orbit, and nowhere near the Earth by the time it had got back to where it started.
I then thought I’d slow it down a bit. So multiplied your suggestions by 0.95, and this brought it round a bit faster. Multiplying by 1.01 slowed it. Then I tried 0.9, which brought it up not far behind the Earth a year later. And then I tried multiplying by 0.89.
What then happened was rather strange. Che9 (as I called it) stayed very close to the Earth more or less the whole way round. Wondering what was going on, I took a look at the near environs of the Earth throughout the previous year. And it emerged that Che9, after lifting off from Chelyabinsk, performed a slow spiral pirouette around the Earth, returning quite close to it in October, before moving further away. Other rocks I’d been trying yesterday, Che1 and Che5, move away in more or less straight lines, So Che9 was more or less in orbit around the earth. It had come up from somewhere, and spiraled slowly in before finally landing on Chelyabinsk at 3:20 on 15 Feb 2013.
Clearly this isn’t a DA14 companion rock. It’s also only doing 11.1 km/s over Chelyabinsk.
I didn’t try your 14.9 km/s rock, because clearly it was going to have an even slower orbital period than the 12.5 km/s rock
OK, Frank. Bit of a shame. I was wondering, would it be possible to look back at your idealised South Pole approaches. Those were geocentric but in orbit they would all be doing Earth speed and sitting just below. Then they could come up on any great circle (with 67 degrees S as the ‘south pole’). I find it interesting that DA14 is approaching 67-69 S but from an inside track and yet a rock that is circularised right under the Earth is heading for the same latitude but from directly below. If Chelyabinsk was much more circularised it would do this, and its great circle does run round to the ecliptic south.
We don’t have any DA14 companions that suddenly shadow the Earth, sitting underneath but a more circularised rock on a 1 degree inclination would mimic that effect. Not sure how to find it though.
Andrew
Frank
I just reread your comment from last night. I read it late and didn’t see the significance of the rock adjusted to 0.9 of my suggestion. It appears that works and so there’s something fairly positive here even if it’s a tad slow. So, I’m sorry I said ‘shame’ because that was unwarranted.
Andrew
Andrew,
I must say that I won’t be too bothered if we can’t find a rock that comes in along what now seems to be the ‘consensus’ 105 degrees. Or is it 105.1 degrees? The main reason is that I don’t see that such numbers can be calculated accurately using the evidence of a few bits of video footage.
If nothing else, it seems to me that the shift from 99 degrees to 105 degrees that you mentioned was a very large change. To me that says that there’s 6 degrees of play in these numbers, maybe more.
And I’m not a great believer in ‘consensus’ science. Just because everyone agrees about something doesn’t make it right, in my view. We’ve got too much ‘consensus science’ these days.
So my view is simply that the Chelyabinsk rock came in roughly from the east, slightly north of the rising sun (the Korkino video), and we have may 10 or even 20 degrees to play with. And if we can get a rock from DA14’s companion rock cloud to land on Chelyabinsk coming in from that sort of direction, we’re onto a winner.Or at least, a contender.
Also, what we’re doing is outside the consensus. The consensus view (from day one) has been that the Chelyabinsk rock couldn’t have been a DA14 companion. And all our efforts are being directed to show that it could have been. Although I currently think that, to do that, we’re going to have to invoke the close approach of 16 Feb 2012 to disturb some rock in the same sort of way that DA14 was disturbed.
And in this respect I’m still very pleased with Che3. It lands on Chelyabinsk at 3:20 on 15 Feb 2013 after having a close encounter with the Earth on 16 Feb 2012. The only problem is that it’s coming up from too far south. And Che3 was only the third rock that I fired out of Chelyabinsk. That says to me that there are plenty more where that came from.
My view is that there are hundreds of thousands of millions of rocks that could have landed on Chelyabinsk at that precise time. Z & F’s rock is just one of them. Che3 is another. And soon, I’m sure, we’ll have plenty more.
Frank
I’ve done vxyz for a (hopefully) 99 degree azimuth. I piggybacked on the 12.5 km/sec rock that you multiplied by 0.9. I haven’t multiplied it by 0.9 and in fact, by squidging the x and y, the absolute launch speed is now a bit more, 12.987 km/sec. However, because it is being fired off nearer to the line of the Earth’s orbit, I don’t think you will have to adjust it down to the speed you did for the other one (which was 12.5 x 0.9=11.25 km/sec).
Incidentally, multiplying by 0.89 to get Che9 gave you a speed that was almost exactly
Earth’s escape velocity. That’s why it hung around till October (backwards) and thats also why the 0.9 multiple stayed close.
They are both rather slow but they are doing something along the lines of my comment above where they hang around for long periods nearly underneath the ecliptic pole and are therefore in a position to drift up on any available great circle from that radiant.
This would explain all the different directions of the fireballs despite coming from the same radiant (ecliptic south). The one thing they would all have in common would be that their great circles intersected at roughly that point on the Earth, straight down the z from the geocentric, but in practice intersecting within a circle of maybe 1000-2000km across. That area of smudged intersection would reflect the various small disparities in the radiants from the DA14 cloud ‘fireball’ approaching from underneath. If DA14 had been slowed from 368 days to 366.2 days, there must have been cloud rocks that were slowed to 365 days and so approached slowly from the ecliptic south.
This all assumes slower speeds for
Chelyabinsk but as I’ve said before, they’ve worked on the assumption of no camera distortion whilst acknowledging it’s there. They say definitive info on specific cameras isn’t available so fair enough. But where they are measuring trail distances and shadows trams lines are ‘hanging’ upwards. Also all the other N hemisphere fireballs were slow skimmers, suggesting our cloud is lurking under the ecliptic south at almost the same orbital speed as the Earth.
Andrew
Frank
My last comment crossed with yours but the reason I did the 99 deg azimuth vxyz was for the very reason you stated- that the consensus is dubious. Some were dead sure it was 99 deg at first whilst NASA always said 105. Then Z and F jumped to 105 deg. Meanwhile, The videos are still suggesting nearer 99 deg.
My preference for 99 degrees is twofold. For one thing it drastically reduces the speed due to the trajectory-to-shadow intersection angle. Secondly it allows for a cloud/fireball to be sitting closer under the ecliptic south ie approaching more towards ‘down the z underneath 67 S’ rather than from the inside track towards 67 S. That sounds like the same thing but 67 S is a circle: DA14 was approaching a different point on it, from the sunward side. I’m suggesting cloud rocks approaching that same latitude line from nearer to directly down the z ie about 56 deg eastwards round on 67 south, that is, approaching mid South Pacific, not NZ.
Gotta go. Battery going!
Frank
The rock which hung around the Earth till October (in reverse) was launched right on the cusp of escape velocity. I presume it only just escaped the Earth’s influence, supposed to be 1.5 million km, (but as we know, there’s significant influence at 2.6 million km).
This is useful to remember for bounce events. Seeing as DA14 cloud rocks would be travelling just over the escape velocity before skimming it means that almost any and every bounce event rock would return sooner or later. It seems that the ‘later’ is probably measured in months, maybe 2 or 3, before they get far enough away to get affected more by the sun and plucked from that last lip of the gravity well. I still think most would return within days or weeks because of the non linear relationship between launch speed and orbital period but this is useful to know.
This weird ‘0.89’ rock shows the solidity of your programme because it is behaving exactly as it should as far as I can see.
On rereading my last comment (June 26th at 12:26) it might appear that I am stating categorically that all the meteor trajectories’ great circles intersect at the ecliptic South Pole. What I meant was that this may well be the case and would be a very strong signature of apparently disparate meteor locations and trajectories coming from the same cloud, loitering under the Earth.
As for the azimuth, 99 degrees, 105 degrees etc., I agree that it can’t be nailed right down and in fact I’m starting to wonder if some of the 97 degree statements made early on could be correct because, if we are right, the meteor was slower and 97 degrees shortens the path length. I wouldn’t say a 20 degree leeway but certainly 10. This is because it is clearly at least 5 degrees from the sun but not 25 degrees. Also, if we are to make a convincing argument that will make people sit up and think about Apophis I think our rock has to come noticeably from the correct side of the sun as do many of your Che family. My gut feeling is that, for Chelyabinsk, anything between 97 and 107 degrees, 10-20 degrees altitude and almost any speed above 11.4 km/sec is a DA14 cloud candidate. If it also comes close to the Earth in 2012, as with Che 3 it’s even better. If it can be shown to be interacting in 2012 in such a way as to perturb it from a more DA14 type orbit to a more Che3 type orbit then I think we are pretty well there- but it’s the 2012 interaction that is the real needle in a haystack. Do you remember that Apophis remained on the watch list until it was established that it wouldn’t be going through an 800 metre wide keyhole in 2029? If it had, it would’ve hit in 2036. 800 metres…gulp!
I just looked at the snapshot of Che 3 in 2012. It’s encouraging that it can be perturbed to hit Chelyabinsk from 1.2 million km out. However, I thought the rocks that went round the back were sped up and flung out into longer orbits and therefore slower to go round. Is there an explanation for that? I know the change is very subtle, only hours faster.
Andrew
Shockwave went twice around the world.
[…] Long Talkshop thread here […]
Andrew,
My current view is that the rock we’re looking for will have been disturbed enough by the Earth in February 2012 (or maybe February 2011) to have been shifted into the sort of orbit that would bring it into collision with the Earth in 2013, much in the same way as DA14 was disturbed.
Che3 has been the best I’ve managed so far. This rock lands on Chelyabinsk in 2013 after passing close to the Earth in 2012. But Che3 is very arguably not a member of the DA14 rock train. So I’ve been wondering if it’s possible for members of the rock train lying fore and aft of DA14 to be diverted into an orbit like Che3’s.
To assist the study, I’ve now set up a model in which one of my rock trains is built into the input file, rather than generated from a small cloud each time the simulation is run. I currently start the model on 1 Feb 2011, just before the 2011 fairly close approach of the Earth to DA14. And I’m now growing new rock clouds in and around the DA14 rock train. I now have rock clouds in rock clouds. It makes for a lot of rocks. About 700. And it slows the model significantly.
And I’ve been using this facility to hunt for something that will be diverted into an orbit that will bring it into collision with the Earth in 2013.
I haven’t had much success with this so far, but I have managed to get a rock cloud somewhere that I was beginning to think was impossible to do so – the gap in the rock train that opens up during 2012.
Here’s a snapshot of a little red rock cloud sitting.in the gap, just before the 2013 close approach of Earth (which can be seen up to the left a little) This little red rock cloud started life on 1 feb 2012 as a very small rock cloud right next to the rock train. (where I placed it -100,000 km in x and +200,000 km in y and 0 km in z relative to one of the rock train members (number 258), and with the same velocity vectors as it. This little rock cloud then gradually wanders up into the gap.
It’s of no particular significance, except to show that close relatives of the DA14 rock train can find their way into the gap, and that there are some interesting dynamics taking place.
Another interesting little thing I tried was to build a little red rock cloud around one of the lone rocks that occasionally appear in the ‘gap’ because the Earth has passed right over them, and not thrown them one way or the other. Trying this out, with the Earth passing through a small cloud of 343 rocks produced a rather spectacular implosion of the rock cloud, throwing rocks at high speeds in all directions.
Andrew,
In a further development, I’ve added a facility to add coloured ‘impact craters’ to my world map. Quite fun to run the Earth through a dense rock cloud, and watch the craters appear.
Frank
I read today’s comments with great interest and shall have much to say, I’m sure, including digging out an old, shelved comment re rocks coming from the gap because I thought it irrelevant.
I would have had all my remarks and questions ready but for the fact that I decided to do some actual calculation my end and I thought I’d better see it through.
On the way to trying to find a 365 day fragment orbit that would hang around under the Earth pole around the 15th Feb 2013 and cause havoc, I thought I would calculate the specific angular momentum of the Earth and DA14.
The specific angular momentum is the angular momentum per unit mass and allows comparison. Because both bodies orbit in ellipses and I had the vectors for 8th Feb I worked out, first of all, the angle of the instantaneous velocity tangential to their orbits to the perpendicular to the radial direction to sun (this angle is called phi). For DA14 it was 6.04 deg, away from the sun. This correlates well with the crossover angle of 5.86 deg on the 8th. For Earth it was 0.55 deg, towards sun because the Earth was just past perihelion.
I used these angles to calculate the perpendicular-to-radial component of v for each on 8th Feb which accounts for the component of v pertinent to angular momentum. I obtained the following two figures:
Earth: 4 455 903 625 km^2 sec -1
DA14: 4 436 806 300 km^2 sec -1
These values are to within 0.4% of each other and it leads me to believe that DA14 could be somehow more intimately related to the Earth than we thought. Its purported composition would suggest it wasn’t ejecta from some previous catastrophic meteor and the same goes for Chelyabinsk. But these two figures are very close. Seems too much of a coincidence.
I realised afterwards, I could check Earth’s specific angular momentum and found a web-based calc of 4 457 948 306 km^2 sec -1 so it’s the same by a nat’s whisker, the difference probably due to my Earth angle decimal places. That makes me confident about the DA14 calculation.
This same figure can be applied to cloud rocks so I think I can get the correct v for rocks (almost) sitting under the Earth. If r is the same in that scenario, as it would be, then the v for a cloud rock would be 0.4% slower than the Earth’s on the 15th Feb if outside the gravity well and with the same e as DA14. That translates to Earth catching up at 130 m/sec in pure ellipse comparison.
I say pure ellipse comparison because the 10.33 inclination causes an approach speed in z of 6 km/sec. But if the inclination is 4 degrees like for Che 3, it is 2.1 km/sec in z and then you add the miniscule 130 m/sec, predominantly in xy, which barely changes the overall approach speed.
So, I believe that any rock cloud coming up slowly from under the the Earth on a smaller inclination will be ahead of the Earth and be caught up at around 130 m/sec disregarding the gravity well. This would certainly cause fireballs in all directions in the northern hemisphere.
I reckon if Che 3 was shunted over so as to make a direct ecliptic polar pass from behind in 2012 and you gave it a close-knit cloud, you might see some nicely spread out fireballs around the Earth. Not sure but just a thought.
I like the craters BTW and I really liked that explosion snapshot. I need to review that and see what’s happening.
Andrew
Frank
I think you’ve already done it with that implosion cloud. Were there any hits? If the Earth threw them in all directions, any hits would be coming up any and every conceivable longitude line mimicking what we saw in February.
Andrew
The specific angular momentum is the angular momentum per unit mass
Do you remember me telling you that Leif Svalgaard asked me to use my model to calculate some stuff. Well, it was the angular momentum and spin momentum of the planets. And maybe just the momentum as well. At the end, after considerable efforts on my part (I’d not heard of angular momentum and spin momentum) he said, “Humph. Quite good.” Anyway, the code to work out these various momenta is embedded in my model, unused. But it’s probably fairly easy to reactivate.
it leads me to believe that DA14 could be somehow more intimately related to the Earth than we thought.
In what way?
I reckon if Che 3 was shunted over so as to make a direct ecliptic polar pass from behind in 2012 and you gave it a close-knit cloud, you might see some nicely spread out fireballs around the Earth.
Well, if you can tell me where to put it, I’d be more than happy to give it a try.
My current ‘one shot’ model, as you know, just starts and runs and stops. I’m currently working on trying to get the model to restart itself, So that it cycles round and round, repeatedly launching rocks on the same date. I want it to be able to also launch slight variations of rocks. And then I want to ‘score’ the rocks for how near they meet certain conditions. For example I might launch a cloud of rocks on 1 feb 2011, and give high scores to rocks that a) pass close to the Earth on 15 feb 2012, and b) pass close to the Earth on 15 feb 2013. I then let the model cycle round and round, trying new rocks, and building up a list of promising rocks.
This is a way of automating the search process. I let the computer do the searching, as it runs for hours and hours. I’ve sort of half got it working, although it’s requiring quite a lot of the old ‘one shot’ code to be rewritten. And I want it to work in reverse, so that it can fire rocks back out of Chelyabinsk with -dt. It’ll probably take me a while to get it working.
Were there any hits?
I didn’t look. Probably not. Rocks get sprayed in all directions, and it’s very unlikely that any of them got anywhere near the Earth. I was just trying to get a feel for what happened when a very small cloud of rocks within the DA14 rock train was disturbed on 15 feb 2012.
Frank
It would be great to get the momentum section working, especially if it’s worked so well in the past.
A: it leads me to believe that DA14 could be somehow more intimately related to the Earth than we thought.
F: In what way?
Well, I wasn’t quite sure at the time, but you know how, for example, the radiant of DA14 and all those fireballs suggested too much of a coincidence for there to be no link? I think the same about that 0.4% disparity in the specific angular momentum. I know I dismissed ejecta, that is, material of Earth origin, due to analysis showing their being of ancient asteroidal origin. But then I thought, hang on, if it was a biggish one, 200 metres, maybe 30k yrs ago there would probably have been bounced fragments as it ablated. This would be more likely with a larger asteroid that had gravitational integrity but little cohesive strength, ie a gravitationally bound rubble pile.
There could have been hundreds of bouncers, all with different vx vy and even KE. If it was from the asteroid belt, many of these fragments would have flown off into similar orbits and probably not been so restricted by returning to the node of the hit. I was thinking that it would be too neat for the asteroid fragments coming in at such speed to bounce away with momentum in perfect parity with Earth but if there were many bouncers, those remnants would be the only ones you would see hanging around the Earth- it’s like a filtering system.
Those fragments would return to the node periodically and become progressively circularised over tens of thousands of years, hence this spookily circular, 366 day orbit with identical specific angular momentum and identical semi major axis (1.001AU). Only those fragments with very similar SA momentum would be able to conduct this intimate waltz with the Earth over many years. If they had much greater or lesser SA momentum they would not be teased out into a ‘circle’ that mimicked the Earth’s such that they kept on interacting. Larger or smaller values may involve crossings but no possibility for sailing along together for days on end- that could never happen.
Presumably, whatever the fate of the original purported meteor and its multiple bouncers, the ones like DA14, as far as I can see must have been thrown at just above Earth escape velocity so that they ended up with the same SA momentum. I am thinking about your weird 11.25 km/sec rock that nearly got caught in orbit round the Earth but then escaped and shadowed the Earth on a nearly 365 day orbit.
“Well, if you can tell me where to put it, I’d be more than happy to give it a try.”
I’ll try to do that in due course. I’ll need to look up thread for the close approach and geocentric vxyz at that point. I think it’s there somewhere.
Your continuous cycle idea is neat. I’ve read a lot in the past about evolutionary approaches to computing especially for proteins and drugs as well as for AI. This would be a novel use and perfectly adapted to it. On a very loosely related note, did you notice how the rocks in the train in recent snapshots showed marked waves, mini crescents? I think that may be the remnants of each line of rocks in the original cube.
Andrew
Russians continue to be very concerned.
Russian scientist wants ‘Satan’ missile shield to shoot down meteorites in wake of Urals strike
Andrew,
After your suggestion that I took a look at my rock cloud ‘implosion’ to see whether any of the rocks got anywhere near earth on 15 feb 2013, I ran it again, and took a look. Rather to my surprise, one of the rocks actually did get quite near the Earth.
Here’s a snapshot of its (red) orbit around the sun after 15 feb 2012. Da14 is the blue orbit, and Earth’s the innermost black orbit. I haven’t looked at it closely, but it’s a rock that has come very near the Earth on 15 feb 2012 (all the rocks in the little rock cloud I set up were very near, and about 150 of them hit it). The orbit is also more oblique than DA14’s. I’ll get hold of its state vectors tomorrow.
Here’s another geocentric snapshot (with DA14 passing the Earth) showing it passing a little over 2 Moon radii away from the Earth, at almost exactly the right time.
Frank
I have a question about the imploding cloud in 2012 which was set up in the middle of the (presumably) 2011 gap.
I always thought gap rocks, though impressively reduced in inclination, always passed behind the Earth on the next pass. True, I think there are exceptions especially those near to the gap edge with a bit more vx than vy but the standard result tends to be that the gap passes behind the Earth and takes anything therein with it. You do say:
“all the rocks in the little rock cloud I set up were very near, and about 150 of them hit it”
So I presume it was in the 2011 gap but right near the edge? I’m still surprised that the cloud passed so close, with 150 hits. But maybe that’s what I’m missing. I’ve been schooled in 2012 gap behaviour. Perhaps the Earth passes through the 2011 gap. If so, I’ve forgotten though I do remember clicking on at least two snapshots of it a long time ago.
As you can see I’m slightly confused. I’m sure this little cloud will be even more interesting when I get it. It could be key to, first of all, reducing inclination in a 2011 pass, then orbit period/shape in 2012-3.
Andrew
Frank
Looking at your orbit snapshot of the imploding cloud escapee that returns close by on 15th February 2013, it crosses the opposite (descending) node at about 34 degrees past solstice. That translates to the July 26th Earth position. Seeing as, by definition, it has the same period as DA14 or within a few hours but probably increased eccentricity, it would be likely to cross the descending node at the same time as Earth- on July 26th.
This means that if the meteor shower of 2013 came from an imploding cloud similar to your 2012 one, perhaps we should book a two week Antarctic cruise, starting the week after next. 🙂
Andrew,
To get an imploding rock cloud, I took one of the rocks in the rock train (no. 191) which was going to come very close to the Earth on 15 feb 2012, and constructed a rock cloud around it on 1 feb 2012. This rock cloud had more or less exactly the same velocity vectors as rock 191, and so could be regarded as an unusually dense patch in the rock train fore and aft of DA14. You can see the cloud in this snapshot as a red dot in the rock train a few days before the close approach.
two week Antarctic cruise
Is that to look for meteors, …or to escape from them?;-)
Frank
Thanks for the answer to my question.
Antarctic cruise: escape from them (because it’s the descending node).
BTW, your reference to Antarctic cruise was in blue as if it was a link. It wouldn’t click and link to anything though. Was it a link?
Andrew
No it wasn’t supposed to be a link. I just put the wrong markup tags around the text. Nice red colour though.
Frank
Do you know if there are any similarities between the rock from the implosion that comes near in 2013 and Che 3?
You said Che 3 was possibly not identifiable as a DA14 companion. I wasn’t sure why that was because it looked quite a lot like other 2012 snapshots. Then again, I didn’t see the direction it came from before the 2012 pass.
I’ve been doing some stuff related to the cloud sitting under the Earth. As far as I can tell, it can get to as good as zero, but that means about 100-150m/sec.
This would be a more useful phenomenon if it were related to a cloud or train in a more circularised orbit so that it was 365 days, low inclination and sitting below or near-below for days on end. For that to be the case, I think DA14 would have to be one of the perturbed rocks from that cloud, by which I mean a rather more drastic perturbation in the distant past.
The only problem with this is that it would be arriving back every single February and not passing through for a few years before getting out of synch as with DA14. Mind you, is that such a problem, bearing in mind the February fireballs?
That link was blue for me. You said red. I wouldn’t point it out except that you use WordPress quite adeptly and might like the feedback.
Andrew
I wonder whether this will work?
That was good! I was wondering this afternoon whether it was possible to make YouTube videos showing my simulation. I had to find something to record a video to start with, and turned up a free utility called Debut that would record an AVI movie from my computer screen (or webcam). When I’d figured how to use it (which didn’t take long), I used YouTube’s upload facility. It was all very easy!
This will be a much better way of showing what’s going on.
I haven’t looked very much more at the new rocks today. I’ve been sorting out one or two things with the simulation, and then I got into Debut and YouTube.
I didn’t think that Che3 was a rock train member because while it passes through the rock train on 15 Feb 2012, it’s transitory.
BTW, have you signed up for that NASA thing? I haven’t yet. I think the expiry date is coming up soon.
Here’s another impact, of an even smaller cloud, hopefully.
Mmm! Very very cool.
Are you ready to be a YT sensation? It might go viral. Only half joking because if you one day get a run, as I said a few comments back, with all fireballs explained it will be a very interesting movie.
This is definitely the way to go.
I haven’t done anything about NASA. My only definitive suggestion would be to scan 10-20 arc seconds either side of known asteroids when they make a close pass but do so from 5 million or so km out. Ah, just thought of another one: seismometers on the moon. These would exploit the moon as a large collecting dish, detecting small hits in an otherwise dormant rock. I know the moon is active in a very small way but small hits would have a signature. If more occurred during a close pass, it would be an indication of clouds. I forget whether the Apollo ones are somehow still working but I know there are some or one up there working now. I don’t know how sensitive they are. Obviously the moon is smaller and draws in vastly fewer rocks and so depends on direct hits. However, that could be an advantage because the rock shadow would be more pronounced and betray the radiant. Behaviour of P waves and S waves might also betray an angle of attack signature. This should be a separate comment. I’ll shut up. Well done on the video and looking forward to more.
Andrew
Frank
I posted the same time as you. That’s another good one. I can see those northern outliers are at about 49 degrees at least. In fact in the first video there’s one that looks as if it’s off British Columbia which means it’s above the 49th parallel- a very useful guide.
I tried to find you on YouTube but to no avail. I searched for ‘small rock cloud’. I wanted to see how many views there were. Sometimes you can click through to YT but those icons don’t always click through for the iPhone.
Andrew
I’m cfrankdavis on YouTube, like on WordPress. The videos have only just been posted up, so I wouldn’t expect them to have been noticed. I doubt if anyone would be much interested in them anyway.
The Moon is something I might get interested in. I reckon it was a half moon on 15 Feb 2013, and that impacts on its surface should have been visible from Earth. They would have thrown up clouds of dust, but only in the visible southern hemisphere.
Huge Rock Impacts on the Moon (March 17 Mare Imbrium)
Frank
Right, well that’s definitive proof of clouds, then. It also shows that they are at least 500,000 km wide and arrive with a distribution of rock sizes (micrometeoroid to small boulder at least).
I was waiting to see whether they mentioned seismometers for the other 300 detections since 2005. I can’t believe they’ve been using only small telescopes and only since 2005. I’m not saying that’s wrong- I’m all for low key, cheap and innovative methods and it’s also a testament to your quick thinking re plumes visible in the Southern Hemisphere. I just would have thought that it was as a backup to a moon-based seismometer system, even one that may be low resolution as to size and velocity.
If they aren’t using anything moon-based and hi-res, there’s much scope for ideas.
Andrew
Andrew,
Gradually closing in.
I’ve spent a while today putting micro rock clouds in the path of the Earth on 15 Feb 2012, looking for ones that come near the Earth on 15 Feb 2013. It’s been quite successful. On 15 feb 2012 rocks that pass just about over the Earth’s north pole get sent down the right sort of path. I’ve got a whole bunch that show up very near the Earth (a few thousand kms) at 21:00 UT 14 Feb 2013 (about 6 hours before the Chelyabinsk rock). Here’s a snapshot of the view down the z axis onto their trails as they’re coming in. The 15 Feb 2012 ‘implosion’ rocks are red, and ordinary rock train members are blue. There’s a very distinctly different angle of approach between the red lines and the blue lines. Here’s a view along the x-axis as they approach Earth.
Some Moon meteor links:
http://www.heavy.com/news/2013/05/nasa-records-largest-meteor-impact-on-moon/
http://www.nasa.gov/centers/marshall/news/lunar/program_overview.html
http://science1.nasa.gov/science-news/science-at-nasa/2008/21may_100explosions/
These “off-shower” impacts come from a vast swarm of natural space junk littering the inner solar system. Bits of stray comet dust and chips off old asteroids pepper the Moon in small but ultimately significant numbers. Earth gets hit, too, which is why on any given night you can stand under a dark sky and see a few meteors per hour glide overhead—no meteor shower required. Over the course of a year, these random or “sporadic” impacts outnumber impacts from organized meteor showers by a ratio of approximately 2:1.
http://astrogeology.usgs.gov/HotTopics/index.php?/archives/231-Meteor-impacts-on-the-Moon.html
Meteor impacts on the Moon
“On Dec. 14, 2006, we observed at least five Geminid meteors hitting the Moon,” reports Bill Cooke of NASA’s Meteoroid Environment Office in Huntsville, AL. Each impact caused an explosion ranging in power from 50 to 125 lbs of TNT and a flash of light as bright as a 7th-to-9th magnitude star.
The explosions occurred while Earth and Moon were passing through a cloud of debris following near-Earth asteroid 3200 Phaethon. This happens every year in mid-December and gives rise to the annual Geminid meteor shower: Streaks of light fly across the sky as rocky chips of Phaethon hit Earth’s atmosphere. It’s a beautiful display.
Frank
That’s very promising, especially that very nice x axis shot which shows a perfect (almost too shallow) geocectric radiant. They must have a very low heliocentric inclination.
How far above the pole in 2012 did they pass, those close approachers?
Andrew
Andrew,
Yes, they have a low inclination.
I didn’t notice how far they passed over the pole, but the nearest they got to the Earth on 15 feb 2012 was about 2 Earth radii from the centre of the Earth.
Video of close approach to Earth of micro rock cloud on 15 Feb 2012 (Y axis and then Z axis)
Frank
That’s interesting. These videos are useful.
You did this close pass on Feb 15th 2012 but DA14’s close approach was on the 16th. What would’ve been the 16th this year became the 15th. So you have a 366 day orbit for these rocks.
So might I suggest trying to wiggle the cube over in 2012 so that it actually does go over the pole or a fraction behind. I think this will reduce the orbit period and swing the geo approach angle round by a degree or two. It might also cross in front of the Earth too early in 2013 as a result but I think that could be addressed by adjusting the cube pass date. I’m not sure I can give you the vectors to move it due to it being such a small shunt and also the pinching of the well. You said you had played around with many cubes and of course, I realise you might have done the above suggestion already but it sounds as if you are adept at adjusting.
The shunt would be in xy and I would guess it would be a 560 km shunt in -x and an 830 km shunt in +y at the firing point. Just a guess, from the Earth angle.
You are getting ever closer.
Andrew
Frank
I think I meant 830 km in plus y, all well and good, but 560 km in PLUS x (not minus x) ie towards x=0
Andrew
Andrew,
Here’s a view of the orbits of Earth, DA14 rock train, and a minicloud that passes close to Earth on 15 Feb 2012, and again on 14 Feb 2013.
Frank
You said that the mini cloud in that orbit snapshot passed about 15,000 km from the Earth in 2012. Is that the same cloud as the one with the video showing it pass almost over the pole and with the two stills of its 2013 pass? Or is it a subtely different version? That cloud had a close approach of “about two Earth radii”, you said, from centre which isn’t far off 15,000 km. However, I think we are now in ‘shower control’ territory.
Andrew
Frank
Notwithstanding all our ideas and calculations to get us this far, now that you are fine-tuning, isn’t it just a case of trying to get the angle of the cloud in 2012 at 60 or so degrees to the Y axis as it goes over and it will do the same in 2013? In other words, if you drop it down closer in z and it curves round to the left a bit (tending towards horizontal on the screen) so long as it is at 60 deg to Y axis as it passes near the pole it will do the same next year for Chelyabinsk. I realise this could change the period a bit but I think there’s something in it. 60 deg puts you 4 deg east of sun.
Andrew
Frank
Looking at your latest orbit diagram, it seems that the perihelion is shifted perhaps 10 degrees clockwise. The orbit may also be more eccentric such so that it has around a 365 day orbit but a closer perihelion.
I think this is why it’s coming in with more -x than +y and therefore 4 or 5 degrees further round. The perihelion would have to swing round a bit if more eccentric because otherwise the sun wouldn’t be at the focus of the new ellipse. This means that the ascending node, the Earth crossing, is further along the new ellipse
resulting in a bigger crossover angle.
Although this increase in angle appears to swing the wrong way, west, it is what all the experts are invoking because it means more sideways approach speed and when vector-summed produces a more easterly radiant. Theirs is a 13km/sec vector-summed approach speed, based on a 5.6 km/sec absolute velocity catch-up so they gain 6.4 km/sec just from the increased sideways angle.
If your new inclination is 2 or 3 degrees, it means you have in effect stolen the 6km/sec z approach and reinvested it in an xy sideways approach vector. Furthermore, if the eccentricity is greater, that extra sideways kick is coming even more from the side instead of from behind.
The only fly in the ointment is that an extra 10 degrees past perihelion means it’s slowed more. But I think some of this is on the right track.
Andrew
Andrew,
Nothing much to report right now. I’ve been slowly generating rocks that pass close to the Earth on 16 Feb 2012, and again on 15 feb 2013. Currently I have a bunch of rocks that comes within 15,000 km of the Earth at 3:05 UT 15 Feb 2013 after a close approach the previous year.
I’m hoping to get some impacts fairly soon.
To do this, I’ve found that I’ve needed the rocks to come in further away from the pole on 16 feb 2012, rather than closer.
Frank
Your current experiments reminded me of a particular rock you mentioned in early June. Looking back through the thread, it is rock number 321 of a particular cube. I’ve pasted the relevant comments and part-comments below and a summary of my thoughts below that. Rock 321 passes the Earth in 2012 in a similar manner to these recent ones you are doing and yet returned in September 2012! It seems you are approaching the ‘shower control’ fine-tuning. This is further evidenced by the fact that rock 322 (the next highest approaching rock, I think) was flung high and wide. 322 was a rock which I had suspected would behave like the ones you are finding now but it wasn’t kinked to the inside track enough. So we had two rocks, making similar close approaches in similar directions and at similar distances but one going to the extreme inside track and the other to the fairly extreme outside. Were they close but on opposite sides, one accelerated, one deccelerated? Or were they both behind and rock 321 bent inwards more? It’s difficult to keep their positions and speeds juggled in my head just using the close approach vectors.
I think these two rocks would be useful as examples of the upper and lower boundaries of close approach distances in 2012. Both were around the same sort of distance as your recent successes.
The culled comments are below and may be useful as a contextual backdrop to our thoughts at the time. However if you are able just to dig out 321 and 322 easily, it’s of less relevance because I’ve highlighted most of my thoughts above.
Frank Davis on June 6, 2013 at 10:55 pm
Andrew,
I found a rock that might interest you. Rock 321 appears in this snapshot as the innermost orbit. It’s inside the 7 I sent you yesterday, but after feb 2012 it has z positive (z = +4.75 million km on 15 March 2012) while they have z negative.
Here’s its closest approach in 2012:
FIGURES WERE INITIALLY WRONG. BELOW ARE CORRECT FROM A LATER COMMENT ON JUNE 7TH:
[321] x-31y17z-17e3km closest approach[7]: 10736.227 km on 16 Feb 2012 16:49:34 dt=1.0 s
x-31y17z-17e3km is -6565.8013,-6143.467,5866.41 km relative to 3
x-31y17z-17e3km is -0.4781135,7.5533676,7.3937335 km/s relative to 3
state vectors (km seconds)
[321] x-31y17z-17e3km, 10000.0, 0.025, 2456337.000053561, A.D. 13 Feb 2013 12:00:04 dt=64.0 s,
5.9214286848E7, -4.9552433152E7, 727665.792,
27.776240234375, 38.71473046875, -3.992997314453125,
[321] is quite near the sun, and on the other side of it from DA14, after doing about one and a half orbits since feb 2012, while DA14 has done almost 1 orbit..
Scute on June 7, 2013 at 1:26 pm
Frank
Rock 321 is situated just before the ‘August’ node on Feb 13th 2013. It’s at the equivalent of Earth’s July 31st or August 1st position. It has a speed of about 48km/sec and an inclination of 4.79 degrees. That is, a positive inclination according to the convention but angled down due to approaching the descending node.
So tight ellipses are 50% faster but we suspected it was something like that. It’s still useful info for working out what happens to lesser disturbed rocks.
[MY THOUGHTS ON SUNDAY 14th July (this comment): ROCK 321 appears to be approaching Earth in 2012 from the same direction as the close approacher(s) at around 15,000 km, found by you circa 10th July (do they have names?). However, its absolute close approach distance is 10,736 km not 15,000km. Frank, you mentioned ‘2 Earth radii’ for one of these close approachers which would make
It very similar to rock 321 and yet rock 321 arrives back in Sept/Oct 2012 and these others arrive back on Feb 15th 2013. Is the ‘shower control really that fine? Is it the distance of the close approach or is it the direction of the close approach and possibly therefore the radiant and subsequent acceleration and absolute speed?]
Andrew
Frank
Some general thoughts on the 2012 close approachers.
It does seem that rock 321 passes in front (sunward) in 2012 and rock 322 passes behind (night time). Their close approaches are 4300 km above the surface in front for rock 321 and 1300 km above the surface behind for rock 322.
It is tempting to think that 321 was pulled back and slowed because it was ahead, hence its dropping into a tighter, faster ellipse and 322 was pulled forward and accelerated because it was behind. I think that is the case for 321 but I’m not so sure about 322 being accelerated. These two rocks were so close that their close approaches were 5 minutes apart (16:49 for 321/ 16:54 for 322) and they were only 18,000 km apart, only 10,000 or so in xy. It’s natural to think that they were pinched together by the gravity well, one from ahead, the other from behind but I suspect that 322 was pulled back and slowed just like 321 but just happened to miss and be whipped round the back.
The reason for thinking this is that the DA14 rock train slows as it approaches the Earth (by some 70m/sec per day) to the point where rocks that had previously passed on the inside a few days earlier are now slowly being reeled in even without the gravity well influence. It’s like watching from your train as you pull into Victoria to see another train overtake you slowly but because it is decelerating at a faster rate you end up at the same speed momentarily and then slowly pass all the carriages that just passed you.
The reason for saying this is that I think that all rocks that got close to the Earth in 2012 were slowed because the entire rock train was already going backwards in relation to the Earth. All rocks that were behind the Earth on the 8th, say, were therefore even further behind on the 15th because they were like the carriages that never got to overtake us before the train slowed. That doesn’t mean they weren’t accelerated and flung out, we know that happened. But all that means is that they didn’t slow as much as their ellipse would have allowed if untrammelled by the Earths gravity well so they had to adopt a wider ellipse. They didn’t need to pass really near the Earth to accomplish that.
Therefore, if I’m right, all rocks that got close to the Earth in 2012 were slowed by a similar amount, coming up the same ahead-of-Earth radiant. That would mean the wildly different ellipses brought about by these close approaches are purely due to vector angle changes, not due to slowed versus accelerated rocks. This would be, in turn, due to whichever great circle they passed over as they were perturbed. So it would be all to do with the deflection vector because all approach speeds and radiants from a few hours out would be almost identical (but for their translational relation to each other as they zoom up your gravity well cylinder). The cylinder would have rocks flying up it, perfectly parallel to each other but because, by definition, it girdles the Earth it would account for all deflections, both from the sunward side and the dark side. The cylinder is wider than your 27,000km wide impact parameter cylinder because that cylinder only accounts for the hits. For near-miss, big deflections it would be more like 40-60,000 km diameter.
Although that is an encouragingly wide cylinder, it’s nowhere big enough to pull slowed rocks from ahead and accelerated rocks from behind and have them meet at the Earth in the middle. This is because the slowing of the DA14 rock train in xy is in the hundreds of metres per second (in 2013 at least: 448 m/s slower at 11:00 on 15/2/13). That would surely overwhelm any accelerations from behind. The accelerations would still occur but never catch up with the Earth.
I realise that some of that 448m/sec is due to the gravity well in the previous few days but we know DA14 was being ‘reeled in’ from ahead anyway because it came in on a west of sun radiant. If it was west of our midday line, it had to be ahead in its orbit. Its position vectors in previous days show this too and almost all your snapshots show this radiant. For accelerated rock train members from behind to come close to the Earth, we would have seen Chelyabinsk type radiants from 15 or 20 deg east.
That leads me to believe that all close approachers in 2012 were slowed, coming in from ahead and being deflected both on the night side as well as the day side. Some of the deflections on the night side would be flung out and appear to be accelerated rocks from behind in the train when really they got hiked right round from ahead and in front and up the back-just like DA14 in 2013.
Andrew
Andrew,
For the past week or so, I’ve been looking at rocks in the rock train fore and aft of DA14 that make a close approach to DA14 on 16 Feb 2012, and then make another close approach on 15 Feb 2013.
Several days ago, I had rocks which came within 15,000 km of the Earth in 2012, and about 15,000 km of the Earth in 2013 also.
I thought that it wouldn’t take too much effort to bring a rock down onto the Earth on 15 Feb 2013 to impact somewhere or other, largely by pulling it down the z-axis, so that it turned more and more sharply around the Earth below it on 16 Feb 2012.
But this has proved (so far) impossible to do. For when I got within 7,500 km of the Earth on 15 Feb 2013, my rocks were skimming over the Earth on 16 Feb 2012 at a radius of 6430 km – just 60 km above the surface of the Earth.
I could go no lower down the z-axis without impacting the Earth on 16 Feb.
The net result is that I haven’t managed to land a rock on the Earth on 15 Feb 2013 after making a close approach in 2012.
Here are a few snapshots. the first one shows a view down the x-axis of the close approach on 16 Feb 2012, showing a red microcloud performing a sharp turn around the Earth. The upper blue lines are other rocks which also come quite near the Earth in 2013
The second one is a view down the z-axis on the same occasion, with the microcloud passing over Belorussia and Greenland.
The third one shows the subsequent close approach of the microcloud (now greatly elongated) with the Earth on 15 Feb 2013 at about 5 am, at a radius of about 7500 km.
I’m not sure what to try next.
Andrew,
Here are the state vectors of b576, which passes over Chelyabinsk on 15 Feb 2013.
At initialisation in the DA14 rock train on 1 Feb 2012:
x12029y-14323z-60999km b[576], 1.0, 0.025, 2455958.76, date,
-9.5713107968E7, 1.0713980928E8, -6968281.088,
-24.924017578125, -17.749912109375, 5.1985341796875,
At closest approach on 16 Feb 2012:
b[576] x12029y-14323z-60999km closest approach[1]: 6432.4404 km on 16 Feb 2012 07:02:19 dt=1.0 s JD 2455973.793286402
x12029y-14323z-60999km is 2081.5718,-1470.5535,5906.0146 km relative to 3
x12029y-14323z-60999km is -8.720595,7.745497,5.026551 km/s relative to 3
At closest approach on 15 Feb 2013:
b[674] x12029y-14323z-60997km closest approach[3]: 8566.497 km on 15 Feb 2013 05:22:09 dt=1.0 s JD 2456338.7237153505
x12029y-14323z-60997km is -6712.123,1772.3851,5019.0664 km relative to 3
x12029y-14323z-60997km is -4.1937804,6.952971,-8.045227 km/s relative to 3
State vectors of b576 on 8 feb 2013 0:00
-1.09202284544E8, 9.3378347008E7, 1013355.968,
-24.501953125, -18.847927734375, -1.5915703125,
and vz = -1.59 km/s downwards. The orbit is inclined in the opposite sense to DA14 et al.
Also b576’s closest approach to the Earth on 15 Feb 2013 is over Newfoundland during the night. If it could have been seen, it’s one of those rocks which would have been moving N-S over the USA during the night of 14-15 Feb 2012 (there were several of these around that time).
I found it rather galling that I got so close to the Earth but couldn’t get any closer. It was like climbing up an apple tree to pick a particularly juicy apple, only to find that the nearest you can get to it is to brush it with your fingertips.
Andrew,
I don’t think the likes of b576 would be good candidates for Chelyabinsk. But they do seem to be good candidates for the fireballs that that were seen during the night over the USA around that time.- e.g. the California fireball after dark on 15 Feb 2013.-..that went over the pole and grazed the Earth’s dark side.
Also, since b576 comes within about 60 km of the surface of the Earth, there’s a case for saying that it might have been seen as a fireball. Or, if not, then it might arguably have been slowed slightly by atmospheric friction, which would have had an effect on its subsequent orbit. Perhaps I should re-introduce atmospheric drag forces.
And I’m beginning to think that there were several sorts of rock that passed close to the Earth on 15 Feb 2013. Most of them were members of the DA14 rock train (although within this train there would also have been rocks that had been disturbed in 2011 and 2012). But some of them would have been rocks like b576.
Most of the fireballs on 15 Feb 2013 were most likely from the rock train, and would have a had a signature direction. But others, including the Chelyabinsk rock, would have been rocks that had been disturbed in 2012 (much like DA14 itself was disturbed). These rocks would have a different signature direction, and could reach every latitude. But they were probably relatively few in number.
Frank
I thought you meant 7,500 km above Newfoundland. Now I realise you mean 7500 km from geocentre and above Newfoundland. That is incredibly close.
You gave me 3 sets of state vectors for rock b[576] and one set of state vectors for rock b[674] at close approach on 15th Feb 2013. Is it supposed to be different? It appears to be 2km offset in z from b [576] at launch date.
Anyway, assuming it’s not much different, the 1.59 km/sec in z translates to a 3 degree inclination which is negative on Feb 15th due to the high angle change over the pole in 2012.
Seeing as the z component of the 7500km close approach is 5019 km and the z distance of Newfoundland (assumed 50 deg N) above geocentre is roughly 4100 km, then taking into account latitude and tilt, the shortfall in z is only about 920 km. That means that any rock on a similar orbit eccentricity, inclination and period but arriving just 10 minutes later, would arrive at the Newfoundland z height some 17,000 km further along in its orbit. This is because its inclination (only measured in z, of course) allows it to slide down from 920 km too high to the right height for a skim. That would mean a skim just off the east coast of Newfoundland at 5:32.
So can you tweak your 7500 km close approacher (b[674] presumably?) so that it has a slightly longer period? I would suggest adding xy position offsets to your Feb 1st 2012 start vectors For b[674]. I would hazard 33km in plus y and 22km in plus x (towards x=0). I know that sounds ludicrously accurate but it’s just a wild guess of a 40km offset that corresponds to sliding the rock 40km back down its orbit and therefore arriving in the 2012 geocentric view as a parallel line 40km above its progenitor. The 33km and 22km values are simply the 40km offset vectored to the 56.2 degree Earth angle. I think I should be going to the DA14 angle but its 1 or 2 km fiddling on a wild guess. The initial z position vector on Feb 1st 2012 would remain the same as would all the velocity vectors.
By tracking back down in its orbit by 40km, I would hope to see this new rock trade a minuscule amount of inclination for a longer orbit period (10 minutes longer!). This is because it would be flung outward in xy a tiny but more (like rock 322) and it would not be quite so squarely over the ecliptic pole in z, reducing the potential for inclination reduction but only by a vanishingly small amount.
If we can get this right, I think we can shift to and fro across the pole in 2012, for any given close approach height increasing and reducing the inclination and orbit period on a whim.
If you have 2 km offsets at the moment, on a 343 rock cube, does that mean it is 14km across. If so, that means my suggestion is three cube side lengths away. Forgive me if your cube is bigger and encompasses my suggestion. If it does and that means the answer isn’t at 40km then perhaps we need to go further out along that line.
Andrew
You gave me 3 sets of state vectors for rock b[576] and one set of state vectors for rock b[674]
I must have copied the wrong text. I’ll do it again.
So can you tweak your 7500 km close approacher
I can try, but skimming in 60 km over the surface of the Earth on 16 Feb 2012, I have very little room for manoeuvre.
If you have 2 km offsets at the moment, on a 343 rock cube, does that mean it is 14km across.
There are 7 rocks along each side, so six intervals between them. My current spacing is 1 km. So the cube side length is 6 km.
Correction:
Close approach of b576 on 15 Feb 2013:
b[576] x12029y-14323z-60999km closest approach[1]: 7529.957 km on 15 Feb 2013 05:22:15 JD 2456338.7237950256
x12029y-14323z-60999km is -4651.378,3626.3362,4681.6836 km relative to 3
x12029y-14323z-60999km is -5.593754,5.024095,-9.324565 km/s relative to 3
I inadvertently copied the line above it to produce b674’s numbers. B674 is quite near b576.
Andrew,
I added +22 km to the initial microcloud x, and +33 km to the initial y, But this resulted in the microcloud of 343 rocks completely missing the Earth, after passing just 50 km over its surface on 16 Feb 2013 (although I don’t have oblateness built into this calculation, I don’t think). Very small changes have very large effects!
However, I didn’t spend any time playing with x and y. Clearly there’s scope for doing this. What I can’t do is decrease z.
Frank
Yes, small changes do indeed have very large effects.
I realise now, I wasn’t recognising fully the extent of those effects, made evident by your 2013 close approach of the cube rocks. That 6km on-a-side cube accounted for that entire xy spread across the Earth and beyond, so of course a 40 km offset would be out beyond the moon.
That 40 km offset cube appears to be 16 to 18 hours behind in its orbit because it is 1,700,000 to 2,000,000 km behind the Earth. However, seeing as the snapshot is on the 16th Feb 2013 at 16:02, this close approach of the adjusted cube rocks is crossing the Earth’s 15th Feb 20:00 to 22:00 position.
So we have learnt one small thing apart from the erratic nature of small tweaks 50 km above the Earth. That is that the offset in plus x and plus y did in fact bring the ecliptic crossing to a forward position along the cube’s/rock’s orbit. This is because the original cube arrived at the Earth at 5:22 on the 15th (and so its position was at that Earth position by definition) and the adjusted cube crossed the Earth’s orbit at a position which corresponded to the Earth’s position 15-17 hours later. The only problem was that it was slowed so it crossed that point ‘late’ ie 16:02 on the 16th.
I had wondered whether I’d conceived the process the wrong way round and was going to suggest offsets in minus x and y. But then I realised that would just produce a crossing that was too far ahead, in front of the Earth.
However, it is as well to remember that this is a niggling problem only for the particularly erratic case scenario of a 60 or 50 km altitude pass in 2012. I have a feeling that if you moved back up a few thousand km in z to the sort of height that gets you a 4-7 deg inclination (10,000 km from geocentre?) and only then start doing xy offsets I think it would be a much subtler scenario. It’s true you would be missing the Earth by, let’s say, 1.5 million km in 2013 without xy offsets but with the right xy offsets I think you could bring the rock in earlier, ie crossing at an Earth position of 1.5 million km up range in 2013. That corresponds to 14 hours earlier than the 2012 pass which is almost exactly Chelyabinsk time.
I think the answer is to offset the opposite way, in minus x, minus y and do so from that higher 2012 pass position. I’m thinking along the lines of rock 321. That rock, which arrived back so quickly in September 2012, had a 10,700 km close approach in Feb 2012 as well. But instead of its curvature around the Earth being over the ecliptic pole producing a pure 4 deg or so inclination and no orbit period change, it curved round the side causing the increased eccentricity and reduced orbital period. This was because a huge component of that 10,700 km was offsets in minus x and minus y (see close approach vectors below).
Therefore, I believe that if you slide back up z till you are 10,700 km from geocentre, retain a polar pass for all intents and purposes but steal some of rock 321’s DNA I think we might get closer. This is because by doing a small minus x minus y offset on 1st Feb 2012, you would keep the 4 degrees or so inclination but you would come in 14 hours earlier in 2013 (instead of 5 months earlier as 321 did!)
Rock 321 had these 2012 close pass vectors:
[321] x-31y17z-17e3km closest approach[7]: 10736.227 km on 16 Feb 2012 16:49:34 dt=1.0 s
x-31y17z-17e3km is -6565.8013,-6143.467,5866.41 km relative to 3
x-31y17z-17e3km is -0.4781135,7.5533676,7.3937335 km/s relative to 3
So what should the 1st Feb 2012 xy offsets be? I have no idea except that at 10,700 km they won’t produce such an erratic response and we know they have to produce a 2012 pass that is way less than rock 321’s minus x minus y close approach vectors. I would suggest starting with a y offset of -33 km and an x offset of -22 km on Feb 1st 2012. I suspect that won’t be enough at this new height of 10,700 from geocentre but I think you would be able to discern whether the rocks are crossing the Earth’s orbit earlier in 2013 and therefore up range. You could then do -66/-44; -99/-66; -132;-88 etc. for related rocks.
Andrew
Andrew,
When I started out (on Monday or Tuesday), my microcloud was passing about 15,000 km over the Earth on 16 Feb 2012, and about the same distance from the Earth a year later, on 15 Feb 2013. I found that I could get some rocks in the cloud closer to the Earth on 15 Feb 2013 if I moved the cloud in -z direction at its cubical initialisation on 1 Feb 2102. But to keep it near the Earth on 15 Feb 2013, I also needed to slightly adjust x and y.
Here are several sets of the x,y,z adjustments (km) I used:
1288, -1504, -6400
2570, -3008, -12800
6954, -8219, -35000
8908, -10567, -45000
9898, -11741, -50000
12031, -14323, -61000
z offset of -61000 was about as far as I could go, because it resulted in the microcloud passing just 50 or 60 km above the surface of the Earth on 16 Feb 2012, and 7500 km from the centre of the Earth on 15 Feb 2013.
I’ll see if I can interpolate between the numbers above (which maintain an almost constant ratio to each other, I think) to come 10700 km above the Earth on 16 Feb 2012. It’ll be a processs of trial and error. I’ll do it later today.
Frank
That’s very interesting, especially the constancy of the xy offset ratio. I think that may have something to do with the radiant.
This is becoming tricky to visualise. I think you have a point that interpolating between the previous figures will get you a pass at 10,700 km in 2012 but I have a feeling that you will simply be correspondingly high above the Earth on 15th Feb 2013. So what I think I’m suggesting is getting to that stage, yes, but then doing a further tweaking of x and y while keeping the 10,700 z height constant. This is so as to slow the rock (initially) like a very subtle version of rock 321 so that it becomes more eccentric, has a closer perihelion, shorter orbit period and crosses Earth on the 14th. I have a feeling that by being more eccentric and crossing earlier, it will not be flying above but coming in lower. It might be too low and so need adjusting to half a day forward again.
The further offsets in xy (while keeping z at 10,700) would be in -x, -y if rock 321 is anything to go by.
I realise that the 14th isn’t the date of Chelyabinsk but it would prove a principle if it hit or crossed Earth’s orbit nearby and at z=0 or z=5000km which is about Chelyabinsk z value.
If that proof of principle can be made, it can always then be argued that the 2012 pass was on the 17th, coming in early on the 15th Feb 2013. It’s already some 12 hours earlier (16:00 for the 2012 pass as opposed to 03:20 for the 2013 pass).
It may also look as though I’m chopping and changing, saying in one comment that you should try coming in ahead of the Earth down range and then saying come in early, up range but that’s because the last time the inclination was negative and sloping down past the Earth; at 10,700km I believe it is positive, 4 deg or so, and so sloping up. Therefore to catch it on the ecliptic, you have to come in something like a half-day to a day earlier than the 2012 pass time ie further down the slope. I hope that’s right anyway.
Andrew
Andrew,
interpolating between the previous figures will get you a pass at 10,700 km in 2012 but I have a feeling that you will simply be correspondingly high above the Earth on 15th Feb 2013.
Indeed I am. Here’s one interpolated result that comes in slightly higher than 10700 from the Earth on 16 Feb 2012. It’s also 10,800 km from the Earth at closest approach on 15 Feb 2013.
b[445] x5952y-7042z-30002km closest approach[0]: 10793.708 km on 16 Feb 2012 05:44:35 JD 2455973.7393049006
x5952y-7042z-30002km is 4550.7134,-3189.734,9255.19 km relative to 3
x5952y-7042z-30002km is -6.7256703,5.9355063,5.544462 km/s relative to 3
b[445] x5952y-7042z-30002km closest approach[2]: 10890.3125 km on 15 Feb 2013 04:20:07 JD 2456338.6806468423
x5952y-7042z-30002km is -4348.808,2405.2295,9690.681 km relative to 3
x5952y-7042z-30002km is -6.871977,6.401724,-4.752722 km/s relative to 3
b
Andrew,
I’ve found that these close approaches are much more sensitive to the time step that I use than I had imagined. The above results are not particularly meaningful.
Frank
Yes, I can see your point. Could you use 1 second time steps for the close approach and the usual steps outside the gravity well?
The interesting point to make though is that although the rock cloud is exiting the gravity well on the wrong ellipse, there probably would be a slightly different but viable close approach great circle that would’ve sent it on that path. What I’m saying is that although this is tricky, there is an underlying proof of principle at play which can probably be teased out eventually. Of course, once this slight problem is solved it may break through that 7500km floor in 2013 and plop into Lake Cherbarkul.
Frank
While thinking on the time step thing I wondered whether the rocks in the cloud were attracting each other like they did in your 1913 procession experiment. Clearly that experiment was crafted to produce that effect but I wonder if there are some slight gravitational attractions between the rocks at 1 km distances. If they are set to 1kg I think you would be totally safe. If they are 10 tons I think there would be a tiny vector direction effect which would be magnified on a very close approach. I can calculate the delta v involved but not right now.
Andrew
I wondered whether the rocks in the cloud were attracting each other
No, they’re not. At the moment I’m using a 1 kg mass threshold for calculating gravitational accelarations, and the rocks in this cloud are less than that.
Could you use 1 second time steps for the close approach and the usual steps outside the gravity well?
The time step I was using at the close approach could easily be set to that. But it slows down the model considerably. I’m now trying to work out a way of having most rocks have a large DT, and a few have a small DT, because most of the time there are only one or two rocks near the Earth. That way the model shouldn’t slow down too much.
Andrew,
Just a quick update.
Over the past week, I’ve slowly continued to look at rocks that pass close to the Earth on 16 Feb 2012, and again on 15 Feb 2013. The slowness is partly due to the fact that I’ve been using shorter time steps, and so the models are running slower. It’s also due to the heat. To counter this, I’ve reduced the number of rocks under consideration, cutting out the 343 rocks in the rock train lying fore and aft of DA14. I’m now starting the model on 14 Feb 2012, with a smaller rock cloud. And drinking more beer.
The real problem that I seem to be up against is that when rocks come close (9000 km radius) to the Earth in 2012, very slight differences in their positions get magnified into very large differences in positions a year later. For example, a difference of 1 km on 14 feb 2012 can easily translate into 12,000 km on 15 feb 2013. That’s the kind of sensitivity there seems to be.
I’m wondering if I’m up against the limits of my model. It’s pretty good at modelling the planets, and even rock clouds, producing pretty consistent results. But it’s not so good when it’s used to consider close approaches. I have a suspicion that NASA may have a similar problem, and they had to wait until they got more accurate data on DA14 after its 2013 close approach before they could update its previous and subsequent orbit.
Anyway, there’s not too much to report right now.
I found the position of a single rock – b111 – after it had passed about 9000 km from the Earth’s centre on 16 feb 2012, and was again approaching the Earth on 14 Feb 2013.
I then changed its start position (x0,y0,z0) by just 1 km in x, y, and z, and found the x,y,z offset of the subsequent final position.
x0 + 1 km produces b[111] offset 14070.869324684143, 13337.764898911119, 226.3084490639958 km
x0 – 1 km produces b[111] offset -14050.582378938794 -13351.409877762198 -224.94391704686132 km
y0 + 1 km produces b[111] offset 18489.883924022317, 17527.659530565143, 321.21614704101376 km
y0 – 1 km produces b[111] offset -18452.669614851475, -17547.409361585975, -317.9337975710441 km
z0 + 1 km produces b[111] offset -1649.4677260816097, -1567.8108752816916, -35.252821334273904 km
z0 – 1 km produces b[111] offset 1653.1230672001839, 1568.6391854733229, 35.301969345619 km
Pretty sensitive, eh?
Frank
Yes, very sensitive indeed, especially at 9000km as opposed to 6430 km.
I presume b111 sailed just in front of the Earth in 2012 and not behind and was therefore slowed a bit? If I’m guessing correctly it would mean it dropped into a tighter orbit with a shorter period of orbit. It’s approaching on the 14th which seems quicker unless it’s still a long way out.
Anyway, could you get b111’s vxyz? Even on the 14th I think it would show much the same geocentric approach angle as on the 8th. Most of these rocks seem to have very steady radiants owing to the similar orbital period and consequently small relative drift as they sail along with us.
I’ve been giving it some thought in the hiatus tsomelast week. I really think there’s something in these 364 to 365 day rocks, coming in at a 4-7 deg inclination to land on Chelyabinsk , earlier, so lower and not overflying and hopefully a more easterly radiant due to vxvy changes.
I suspect that within this highly sensitive 2012 region, there is a keyhole through which all rocks have to fly so as to hit in 2013. I think it might be centred on 9-10,000 km and offset in -x,-y,-z from most of the recent snapshots i.e. kept at 10,000km but swung round and down towards the bottom left of the screen by a few km.
Andrew
Andrew,
Here’s a shot of b111’s orbit., which lies almost exactly on the ecliptic.
Here are its initial vectors on 14 feb 2012 ( Julian Date 2455971.6):
x-2y-2z2km b[111], 1.0, 0.025, 2455971.6, -1.20578695168E8, 8.4762279936E7, -1027497.216, -19.765974609375, -22.367162109375, 5.43393994140625,
Here’s its close approach on 16 Feb 2012
b[111] x-2y-2z2km closest approach[0]: 9090.058 km on 16 Feb 2012 05:45:35 dt=8.0 s JD 2455973.7399999853
x-2y-2z2km is 3593.802,-2533.8823,7955.96 km relative to 3
x-2y-2z2km is -7.3219333,6.468765,5.440793 km/s relative to 3
Here’s its close approach on 15 Feb 2013
b[111] x-2y-2z2km closest approach[1]: 36042.8 km on 15 Feb 2013 04:29:51 dt=8.0 s JD 2456338.687406651
x-2y-2z2km is -27843.65,-19256.682,12369.299 km relative to 3
x-2y-2z2km is -4.6804523,6.043803,-1.0871022 km/s relative to 3
As you can see from the 2013 close approach, it’s quite a long way (36000 km) from the Earth. Other rocks in the microcloud are a lot nearer. So it’s not really a very interesting rock. But at the moment I’m really only looking at the sensitivity of the model, and b111 – which is at one extreme end of the microcloud – is a convenient rock to look at.
I should have mentioned that b111’s orbit is shown in red. DA14 is between it and the Earth as they approach the Earth in Feb 2013.
You wanted the vectors on 14 feb 2013
x-2y-2z2km b[111],jd 2456338.1352301436, (14 Feb 2013 15:14 UT)
x,y,z: -1.22482597888E8, 8.192344064E7, 23208.712, km
vx,vy,vz: -21.9445625, -20.7749453125, -0.37554226684570313, km/s
This is an arbitrary date I’m using to stop the run and print out all the state vectors. The actual date of the close approach on 2013 is a day or two later (see post above).
Frank
Thanks for the vectors. The velocity vectors show that b111 is coming in at 46.56 degrees to the y axis on the 14th Feb 2013 which means that with the Earth at about 55 degrees at that time, it is around 8.5 degrees west of the sun so it’s much the same as DA14’s 11 deg or so but swung round a bit. Although you mention it’s not very interesting due its 36,000 km close approach, it is probably very similar to the other cube rocks with regard to its vxvy vectors and consequent radiant. A 3.5 degree swing is interesting and in the right direction.
I did some measuring of your orbit snapshot. Of course, it’s not really very accurate but more so than one might think due to my flat screen. Before starting I checked the parallax due to DA14’s inclination: 1.5 % and that of b111:less than 0.05%.
I measured DA14’s major-to-minor axis ratio, correcting for the parallax and did the same for b111. It seems that b111 is very slightly more eccentric than DA14 by which I mean that it has much the same major axis but a slightly squashed minor axis. It’s squashed by about 1.5%. So although the semi major axis is almost unchanged, as it should be for a 12-24 hour faster orbit, the semi minor axis is about 2.5 million km shorter. Playing around with an ellipse calculator this corresponds to about a 5 million km shorter orbit than DA14. There seems to be a 5 or 10 degree clockwise swing in perihelion (accounting for parallax beforehand) and the perihelion is now closer. The squashing and perihelion swing account for the 10-12 degree clockwise swing of the August crossover.
I’m not sure I can glean much from this behaviour but it does seem to point to the idea of faster, shorter orbits. If the 2012 pass was on 17th or 18th Feb and slightly faster/shorter than b111 it might come in on 15th Feb 2013 with the right characteristics. We already know how to tweak the inclination in 2012 by shifting up and down in z around the 10,000 km mark. We may get proficient at tweaking the eccentricity too.
Andrew
it is probably very similar to the other cube rocks with regard to its vxvy vectors and consequent radiant
That’s true. The closest rock to the Earth on 15 Feb 2013 was about 9300 km (so only about 3000 km above its surface), and would have had very similar vx, vy, and vz.
it does seem to point to the idea of faster, shorter orbits.
Well, we are constrained by events. The Chelyabinsk fireball was at 3:20 UT on 15 Feb 2013, and that’s one constraint. The constraint at the other end is that the Earth punches through DA14’s rock train on 16 Feb 2012.
But if the 2013 Chelyabinsk fireball provides a very tight constraint, the 2012 constraint is quite a lot looser, in that there is a period of time (rather than a moment in time) around 16 Feb 2012 during which the Earth is passing through the rock train and throwing a succession of (theoretical) rocks onto 2013-earth-interception orbits.
it’s worth noting that the DA14 rock train that I’ve been using to provide close approaches in 2012 is actually one which is slightly ahead of DA14 itself, and arrives on 15 Feb 2013 some 15 or so hours sooner than DA14. If DA14 is taken as lying in the centre of the rock train, then we can think about its rocks being disturbed at any time between about 0:00 UT 16 Feb 2012, and 12:00 UT 17 Feb 2013. If we want to use 17 Feb 2012 close approach rocks, we’ll need even shorter orbital periods than b111’s.
there is a keyhole through which all rocks have to fly so as to hit in 2013.
I think that’s right. And I have no idea where (or when) it is!
Frank
“I think that’s right. And I have no idea where (or when) it is!”
I think you just described where it is: between the 16th and 17th Feb Earth position, probably a long sausage! Or maybe that’s in heliocentric. Maybe a giant half-doughnut girdling the sunward side of the Earth, sort of hanging down by 50 degrees. That would extrude into a 2.6 million km long, slightly flattened sausage over a day’s Earth travel. Hilarious, but I’m 95% serious.
Andrew
PS I’m in the middle of a longer comment but I’m glad you are thinking the same way on the 2012 rock train constraints (or freedom depending on its width).
Frank
“If DA14 is taken as lying in the centre of the rock train, then we can think about its rocks being disturbed at any time between about 0:00 UT 16 Feb 2012, and 12:00 UT 17 Feb 2013.”
I would agree with that on the whole because at a geocentric velocity of just under 6km/sec it implies a rock train that is about 750,000km wide. That probably is an upper limit but we have talked about clouds being wider than that. I really would like to stay within your sensible constraints but I’m just wondering if you can see something that makes it out of the question to extend another few hours/100,000 km?
Andrew
Frank
[This is my longer comment.]
b111 is doing 30.220 km/sec absolute speed which is about 20 metres per second faster than the Earth on the 14th February. That’s about how fast I drive to the shops. This means that the geocentric radiant is entirely due to the crossover angle of 8.5 degrees: if you slid back a day or so down b111’s orbit, produced a gigantic ruler and rested it against the Earth’s position on the 15th, it would kiss the Earths surface on the sunward side at 8.5 degrees past the midday line. Even though the Earth is alongside b111 at this earlier time, the target for crossover is the 15th. If you add in an increase in absolute velocity at that same crossover angle the geocentric radiant will swing round east as we managed with AC6 and AC7. If you reduce it, it swings round west.
I’m not describing anything very new here but with the fact that the speeds of the Earth and the rock cloud, including b111, are so perfectly matched, I’m wondering if you can displace some rocks on the 14th of Feb 2013 so that they fly just behind the Earth, get caught in the well and get pulled in. They wouldn’t land near Chelyabinsk because of the flat inclination but if they did get pulled in, they would experience a shunt in -x,-y which would magnify the -x and reduce the +y of the radiant vector. That is what is needed for Chelyabinsk so if this worked you could do the same pass behind the Earth when you manage to increase the inclination a few degrees. You could run a rock positioned a tiny way over in +x,+y and, due to the lower z approach, get it to come in over NZ rather than E. Timor. It would then be on course for Chelyabinsk.
I realise that if you extrapolate the orbit back it might be mutated from the cloud DNA enough to come in to the Feb 2012 environment a couple of days either side of the 16th but that wouldn’t matter and in fact might show up a new region in which to generate 2012 clouds.
Andrew
probably a long sausage!
Or equally it could be a short ribbon, and a very small keyhole indeed.
I’m just wondering if you can see something that makes it out of the question to extend another few hours/100,000 km?
No, it’s not out of the question. In fact, since we know that in 2011 the rock train had a hole punched in it, and the sides of this hole folded over, we could use ‘folded’ rocks to argue for a double thickness rock train in 2012, just like there’s a double-thickness fold through which the Earth passes in 2013.
But the further away we get from DA14, the less we can say that the Chelyabinsk meteor was a DA14 companion. At the moment, b111 is a regular (if rather advanced) member of the DA14 rock train, and that’s not a problem.
I’m wondering if you can displace some rocks on the 14th of Feb 2013 so that they fly just behind the Earth, get caught in the well and get pulled in.
This was a nice little exercise.
The microcloud was making a closest approach of about 9300 km in 2013. So at the 14 feb 2013 stop date, I wrote a bit of code to move the entire cloud 4000 km down the z axis. Here’s the approach of the displaced microcloud. Of these rocks, b96, b72, and b117 will impact around Greenland.
I then ran the model again, and this time ran the displaced microcloud back in time to see where it started. And this proved very interesting. Because many of the rocks impacted on the Earth, going in the opposite direction, in a nice pattern on Russia.
I noted the numbers of the impact bodies.
12 impact at lat 64.62529900825679 lon 45.57703147346439
37 impact at lat 64.28858278622582 lon 43.87714680272906
62 impact at lat 63.94041404816855 lon 42.20043920516593
86 impact at lat 62.96269574601968 lon 41.903139363297534
111 impact at lat 62.6107853492743 lon 40.29285482945823
116 impact at lat 65.297355889315 lon 54.55506186439729
67 impact at lat 65.02730951935963 lon 59.043372137698135
91 impact at lat 64.79190460242638 lon 57.33529730878947
112 impact at lat 65.22733374273642 lon 60.13982856383863
17 impact at lat 64.6626968218981 lon 63.30977029618805
42 impact at lat 64.45604733887612 lon 61.64624113769198
87 impact at lat 64.64490487326432 lon 62.71609511165087
38 impact at lat 64.21196459670838 lon 66.76239800248801
63 impact at lat 64.02604387869717 lon 65.15007808882548
13 impact at lat 63.549883825188296 lon 69.00037526516566
96 impact at lat 61.4102760401749 lon 74.64667061887768
72 impact at lat 60.161523555608085 lon 78.73079575864132
117 impact at lat 59.41255614472648 lon 79.84028084192944
47 impact at lat 58.95486456423826 lon 80.37777277913791
92 impact at lat 58.20269837715921 lon 81.38967732404052
22 impact at lat 57.74370694619248 lon 81.87685776804852
68 impact at lat 56.54085380430355 lon 82.87673353395373
113 impact at lat 55.78957907458882 lon 83.7353634865299
43 impact at lat 54.87862473944577 lon 84.19980633462635
88 impact at lat 53.67783994011773 lon 85.01828850813175
18 impact at lat 53.222102172976555 lon 85.37972189101494
64 impact at lat 51.47979713757998 lon 86.97285545710596
39 impact at lat 48.47304293575807 lon 87.92687209528493
14 impact at lat 43.42790704621221 lon 88.72900953834062
101 impact at lat 41.345535205554015 lon 88.80112248875712
And compared these with the labelled rocks in the 2013 approach (see above). They matched up almost perfectly with most of the labelled rocks.
In short, these rocks would have had to have been launched from Russia in 2012!
And this is the converse (or mirror image) of what I found a few weeks ago, you may remember. The closer I pulled rocks to the Earth in 2012, in order to get them closer in 2013, I could never get any rocks to impact in 2013, however close to the Earth I took the rock cloud..
I think that this means that it’s simply not going to be possible to get this particular bunch of rocks to hit the Earth. They are outside the keyhole. We’ll have to look elsewhere.
Frank
That was pretty cool even if it did show a negative result. And that list along with the crescent-shaped ‘hits’ was interesting. I forgot they approach in a line in 2013 so that +x,+y position offset is sort of spoken for. I’m not too vexed by it though because you are only a razor width off and that’s while dutifully coming up the exact DA14 radiant for 2012 and I don’t think you have to. If rocks are overflying in 2012, they probably did to a certain extent in 2011. Even if they were ahead of the Earth they would have been slowed more and their inclination changed to a greater extent than DA14’s velocity and inclination was at that time. That means that if you add an inclination drop to your 2012 approach, you may have more scope to clear the Earth in 2012 and hit in 2013. I’m not so familiar with the 8th Feb 2012 vectors although I think you posted them up above somewhere. The process would be just like AC4 onwards where we reduced the vz from 5.339 km/sec to 4.170 km/sec and shunted it up 828,000 (?) in plus z to account for the finer approach angle. It ended up approaching at around 7 deg instead of 10.33 deg. You could reduce the 2012 approach by maybe 2 degrees of inclination meaning that you can get quite near the Earth on your 2012 pass without hitting it and also have some sideways scope.
Andrew
The process would be just like AC4 onwards where we reduced the vz from 5.339 km/sec to 4.170 km/sec and shunted it up 828,000 (?) in plus z to account for the finer approach angle.
Not sure about that. It seems to me that one of the constraints we now have is that the rocks approaching the Earth in 2012 must be members of the DA14 rock train. And this means that they’ll all have much the same velocity vectors. Our choice is limited to the x,y,z positions of these rocks in the rock train. We didn’t have such constraints with AC4.
What I think we might consider is other paths of getting from close approach in 2012 to impact in 2013. Differing orbital inclination or period are two possibilities.
But at the moment I’m still trying to concentrate on the accuracy and sensitivity of the model.
[…] thanks to contributor ‘Scute’ whose fascinating thread on the companionship of the meteor which burst over Chelyabinsk in Russia and the large asteroid […]
Frank
I got behind with my comments as ideas were superseded and became obsolete due to your progress in ruling out b111 et al. I’ve got some thoughts about the behaviour of the orbits in the Earth’s neighbourhood though. It might shed light onto where to go next. I should finish that tomorrow.
Andrew
Andrew,
No further progress to report.
But it occurred to me that when some asteroid breaks up due to tidal forces, most of its constituent rocks will gradually get spread out along its orbit. And if some assumptions are made about the mass of the asteroid, it should be possible to work out what the mean rock density along the circumference of the orbit is likely to be.
Guys, have you seen this ?
“The meteor that exploded over Russia on 15 February, scattering debris across the Chelyabinsk region and injuring hundreds, came as a complete surprise. Since then researchers have traced it to the Apollo asteroid family, but no one had matched it to a particular member of the group.
Now Carlos de la Fuente Marcos and his brother Raul, both of the Complutense University of Madrid, Spain, are pointing the finger at asteroid 2011 EO40. Roughly 200 metres wide, it is a rock – or cluster of rocks – previously listed as potentially hazardous by the International Astronomical Union’s Minor Planet Center in Cambridge, Massachusetts.”
From,
http://www.newscientist.com/article/dn23976-asteroid-pinpointed-as-likely-source-of-russian-meteor.html?cmpid=RSS|NSNS|2012-GLOBAL|online-news#.UfrFWawTUpU
J Martin
No, I hadn’t seen that. Thanks.
Andrew,
I was beginning to wonder if I’d ever manage it. Today, I moved the cloud we were considering a few days ago +1000 km in x, -1000 km in y, and – 2000 km in z, and started generating new microclouds that had a close approach to the Earth on 16 Feb 2012, and another close approach on 15 Feb 2013.
And this time I got some impacts. This snapshot is taken just after impact. The arc around Iceland is a set of impact craters. And the other rocks to the upper left have skimmed very low over the Earth and are heading past.
Here are some vectors from initialisation in 2012 as a member of the DA14 rock train.:
x1052y-977z-1950km b[78], 1.0, 0.025, 2455971.6,
-1.205776384E8, 8.4761305088E7, -1029449.216,
-19.765974609375, -22.367162109375, 5.43393994140625,
And here are some vectors for 14 Feb 2013
x1052y-977z-1950km b[78], 1.0, 0.025, 2456338.1352301436,
-1.2244946944E8, 8.1950121984E7, 18124.048,
-21.9638046875, -20.779994140625, -0.2544303131103516,
And here’s the impact on 15 Feb 2013
b[78] impact at lat 69.38250772676874 lon -15.058278521570571 on 15 Feb 2013 04:17:27 dt=0.5 s
The closest approach to the Earth in 2012 is about 9360 km (radius)
Here are the rock trails as they approach from distance. The black circle is the Moon’s orbit. The approaching black curve is DA14. And the red curve is the microcloud containing b78 above.
I ran the model with a 10x smaller time step and got the same result.
I haven’t thought what to try next. Maybe bring it in an hour or two earlier, and see what the impact approach altitude and azimuth look like.
From Andrew Cooper
Thanks J. Martin. That was an interesting read. Always interested in parallel studies!
Andrew
Frank, you’ve been a busy bee. That was a good shot. Very exciting to see and even more so because it’s only about 50 minutes late, when you subtract the travel time from its Chelyabinsk flyover. And I could see that it had swung round and really was closer to a Chelyabink pass, just from the distribution of the hits. Sure enough, the red geocentric approach is 6 deg round from DA14, a big jump. Also, assuming Iceland hits are grazers, it means you are on course for around the right altitude when you draw it down 2 or 3 thousand km for a hit at 55 deg N.
I agree to try and bring it in an hour earlier. The reason I think that’s a good idea is that it will be lower down its 2 degree (at a guess?) inclination. That translates to a 3,765 km drop in z, or 3,140 km if 50 minutes. That would be exquisitely close if it doesn’t fly off to the side for some reason.
I want to say more but I’m a bit busy hence my other comment not done. So, to bed, and likely to sleep happier.
Andrew
Frank
It’s 0.5 degrees or a shade under for inclination which means about 900 to 1000km in -z for every hour you go back.
If it has to come in 3 hrs too early to get a hit, you could then try to increase the inclination to two degrees so that it comes in on time: for the same time on 14th Feb 2013 as your supplied vectors, the vz for that would be adjusted to 1.055 km/ sec so your vxvy would then need to be adjusted to -20.767 km/sec for y and -21.948 km/sec for x (using your vectors on 14th to nearest metre/sec and calculating azimuth angle to 3 decimal places).
Andrew