Overview
More than a year after “Part II” of a guest post from Talkshop contributor ‘Galloping Camel’ on the Moon’s equatorial temperature here is “Part III”. Peter actually sent this to Tim Channon last year, but Tim became to ill to deal with it and forgot to throw it my way. In current discussion of Ned and Karl’s new paper, the issue of planetary surface temperature variation due to speed of rotation arose. Ned thinks it makes no difference. Peter’s model says it does, so now is a good time for discussion, as this impacts theoretical estimates for the temperature of ‘Earth with no atmosphere’.
Modeling the Moon
It has been claimed that the GHE (Greenhouse Effect) is 33 Kelvin because the Earth’s average temperature is 288 K compared to a temperature of 255 K assumed for an “Airless Earth”. The Diviner LRO showed that the Moon’s average temperature is 197.3 K which makes one wonder how an estimate based on impeccable mathematics could be so wrong? Vasavada et al. published a paper in 2012 that mentioned a one-dimensional model of the Moon’s regolith. As I was unable to obtain details of this model I attempted to replicate it using Quickfield, a powerful FEA (Finite Element Analysis) program. Results obtained using my model were published here.
It is now clear that the 255 K estimate was wrong because it made unrealistic assumptions about the thermal properties of planetary surfaces. In contrast Vasavada used properties of lunar regolith measured by Apollo astronauts. There are now several other models that can reproduce the Diviner LRO measurements.
Rate of rotation
Scott Denning, Monfort Professor of Atmospheric Science at Colorado State University has such a model and he sent me temperature data for a wide range of rotation rates. I tried to replicate his results using my FEA model.
The above graph shows good agreement between our models and supports the idea rotation rate has a significant effect on the average temperature of airless bodies. Here is a graph showing how temperatures should vary with rotation periods ranging from 14 minutes to 7,300 days:
MOON0.01 shows how temperatures should vary if the rotation period is 0.01 of an Earth day or roughly 14 minutes. At the other extreme MOON7300 corresponds to one rotation every 20 years. This graph shows that daytime temperatures remain constant until rotation rates can be measured in hours, whereas night time temperatures are more sensitive to rotation rates:
Tallbloke's Talkshop 
A few points to note as a preface to discussion:
1) Peter’s model is for the Lunar equator, not the global temperature. Diurnal swings are much smaller near the Lunar poles, so we would expect less of a ‘rotation effect’ on a global simulation than the ~15K equatorial difference between 29 day and 1 day lunar rotation rates estimated by Peter’s equatorial model.
2) Peter mentions a surface temperature theoretical estimate of 255K for an airless Earth (and Moon?) and talks about a difference in surface conditions between lunar regolith and Earth’s surface as a factor.
3) A much bigger factor causing the difference between the 255K estimate and the modeled 197.5K derived from DIVINER measurements is the misapplication of the S-B equation to a ‘flat disc’ Earth or moon. Ned and Karl corrected this with a properly done spherical integration in their 2014 paper.
Hum?…always an intriguing question with regard to a planet with no atmosphere. From the article…
“the thermal properties of planetary surfaces”
matters!
If one were to attempt to isolate our water world from the atmosphere, via ONLY considering the thermal properties of water, and pretending water vapor to the atmosphere was not a reality, then only the greatly varied residence time of disparate solar W/L entering the oceans would be considered. In considering that it is not deniable that the amount of surface energy absorption would vary tremendously over disparate solar cycles, as various decadal changes in insolation W/L would greatly affect how much solar energy is contained in the oceans. As energy can not be destroyed, then long term surface residence time changes due to long term flux in TSI W/L flux, even if total TSI remains the same, would be additive or subtractive to total surface energy. This addition or subtraction to total surface energy would accumulate or reduce in exact proportion daily for hundreds or thousands of days, in exact proportion to the length of residence time of disparate solar insolati on flux!
It appears obvious that a water surface, prior to its frozen state, would move all the different temperature curves in the one day T chart closer to the .01 length of day curve, and IMV raise the mean daily T.
BTW, that is precisely what I do not see in the above charts and brings in additional flaws in determing the mean daily T. Do you take just the high and the low, or is the average better found via taking many more measurements each one day period, the more measurements the more accurate the mean T?
If I am confusing mean and average feel free to correct me, yet I think both my supposition and intention is clear.
David, this comment from Peter on a recent thread amy help:
Roger, I do not understand the Information privided with the second graph at all, how can you have 24 local hours?
Surely that should be rotation divided in to 24 increments, as you can’t have 24 hours in a 14 minute rotation.
ACO: Yes, 24 ‘increments’ or ‘local hours’ just means the division into 24 segments of whatever the rotation period is. 0.01 days is 14 minutes.
Tallbloke, follows the link to GC comments, but did not see Peter comments.
Regardless it did not address my assertions in my long first paragraph, all of which would, AFAICT, affect the Earth’s mean T sans consideration of any GHG.
As to mean T dividing each day, very short to very long, into 24 periods, I follow that now I think.
In earlier coments you quoted….
“The Law of Energy Conservation dictates that a change in rotational speed may only affect the magnitude of the diurnal temperature amplitude at the surface but not the diurnal mean, i.e. rotation solely acts to redistribute the total available energy between daytime and nighttime hemispheres through the planet’s thermal inertia.”
The only counterargument to this I can think of is that a slowly rotating body will get hotter on the day side and that will make it emit more radiation back to space at a T^4 rate. But there again, the slower rotation will lead to greater heating and retention of energy at depth which might counteract that higher emission by keeping the surface warmer in the twilight zone?
The last chart in this post, if correct reflects that with the night time T being more senstive to length of day, and changing the mean T.
My post was making an analogy to that; asserting that a liquid water surface, such as we primarily have, mimics this affect and must be potentially considered as part of the T difference now attributed almost so let to GHG.
This is the comment from Galloping Camel I linked:
gallopingcamel says:
June 3, 2017 at 8:23 am (Edit)
wildeco2014 says, June 2, 2017 at 2:31 pm
“Could the difference between 33k and 90k be accounted for by the kinetic energy originally conducted from the surface which is thereafter stored by convection in the vertical column but in potential energy form?”
“Consensus” science says that the GHE is 33 K. Everyone agrees that the Earth’s average temperature is 288 K so the interesting question is what would Earth’s temperature be “Sans Atmosphere”?
“Consensus Scientists” calculate 255 K as the temperature of an airless Earth using impeccable mathematics. It follows that the GHE is 288 – 255 = 33 K.
“Consensus Scientists” made the unrealistic assumption that an airless Earth would be at a uniform temperature.
So what is a realistic assumption? Thanks to the Diviner Lunar Radiation Experiment we know that the moon’s average temperature is 197.3 K. That would be the temperature of an airless Earth if it rotated at the same rate as the moon and if its surface was made of regolith (moon dust).
At least five people including this camel and the late Tim Channon have independently modeled the moon’s surface temperature with excellent accuracy. My model came out at 196 K or roughly 1 K lower than the Diviner observations. I used my model to answer questions such as:
1. What would the moon’s temperature be if it rotated in 24 hours?
2. How would the temperature change if the moon’s surface was ice instead of regolith?
My model estimates the average temperature of an airless regolith covered Earth as 209 K. For an icy surface the temperature should be 234 K. So my GHE estimate is 79 K (regolith) or 54 K (ice). I have no idea how that 90 K GHE was calculated.
I have long argued that the speed of rotation is material, particularly of a body that is devoid of an atmosphere..
Of course, heat capacity and thermal lags play a part, but as I see it, as the speed of rotation tends towards infinity, the entire surface tends towards being constantly illuminated/irradiated by solar irradiance, such that the temperature of the moon tends towards that which we see on the warm side of the moon.
Gradually the whole body warms up, and there is insufficient time to produce a cold side of the moon.
4.5 billion years is plenty of time to see this effect.
Whether the speed of rotation has any effect on a body that has a dense atmosphere such as Venus, is moot. I can see that the speed of rotation of Venus has limited (if any) impact on its temperature.
I feel that as with many things from ‘climate science™’ the time element has been lost with the bathwater.
How long a area of a globe stays in the sun and warms is important, just as how long part of the surface stays in shadow and cools. Surely it is the rate of warming and cooling that matters as much as the absolute surface temperatures, and thus impacts the average temperature attained?
What heat capacity is being used for the surface? It isn’t the rotation rate directly that will matter, it depends how quickly the object warms and cools. The model for a point on the object’s surface should just be something like:
C*dT/dt = (1-a)*I – e*s*T^4
T = Surface Temperature (K)
dT = Temperature change (K)
dt = time step (s)
C = Heat Capacity (W*s*K^-1*m^-2)
I = Insolation (W/m^2)
s = Stefan-Boltzmann Constant (W*K^-4*m^-2)
a = Albedo
e = Emissivity
The insolation, albedo, etc should vary by latitude, longitude, and time of day. Eg:
I(lat, long, tod) = S_0*cos(lon + tod)*cos(lat)
Here S_0 is the solar constant (insolation at the subsolar point) and tod is the distance (in degrees longitude) of the subsolar point from a given point at that time of day. Then just start T=0 and update it with dT each time step until it stabilizes.
This is very interesting work. My understanding is this (comments/corrections welcome):
For an airless celestial body the amount of energy from the sun is fixed by only distance from the sun and albedo. Rotational speed cannot change this energy input. Any celestial body will be in thermal equilibrium i.e. energy in = energy out. As clearly rotational speed does affect equatorial day/night temperatures then it must also affect day/night temperatures on all parts of the globe.
The average temperature of the rotating globe must however be the same irrespective of rotational speed as energy input to the whole globe is constant. However in view of the thermal properties of the surface (regolith) the true average temperature calculated from energy input will NOT be the average temperature of the surface. Some energy will be within regolith below the surface and the amount that is there will vary with rotational speed. Energy within regolith below the surface is not warming the surface in the day, it’s cooling it. At night energy within the regolith is warming the surface.
Blob says, June 6, 2017 at 9:51 pm
“What heat capacity is being used for the surface?”
My model slices the Moon’s surface into 50 layers each 10 mm thick. Each layer has different thermal properties as measured by Apollo missions…..see Figure 8 and supporting text here:
https://www.researchgate.net/publication/230766556_Lunar_equatorial_surface_temperatures_and_regolith_properties_from_the_Diviner_Lunar_Radiometer_Experiment-JGR_in_press
With regard to radiative properties, the Moon is non-Lambertian for incoming (short wavelength) radiation so Vasavada used an 8th order polynomial to model it. Tim Channon and I used slightly different functions as shown here:
https://tallbloke.files.wordpress.com/2014/08/image-416-small.png?w=550&h=367
Outgoing (long wavelength) radiation is easy to model. I used a uniform emissivity of 0.95.
Thus the model uses two radiative channels whereas the Lunar Radiometer Experiment uses a minimum of seven. Even so the agreement between model and measurements was remarkable with RMS errors as low as 0.67 K. Tim Channon’s model based on Berkeley PSPICE was even closer!
The model is explained in some detail here:
The 2 graphical representations of lunar equatorial temperature vs rotational rate deserve careful consideration. In both it appears that the minimum night time temperature is trending towards a limit of around 75K. This seems counter intuitive as one would expect it to trend towards 3K as given long enough all daytime heat should be radiated away to space. If we assume the model is a reasonably correct representation of reality then this 75K limit deserves further investigation. I suggest the model be extended to very much longer times, the minimum night temperature be plotted graphically and various parameters of the model adjusted to see which specifically affect this 75K apparent limit. This may reveal something about the actual physical process involved (or something about the model!).
GC: I realise it would be a time consuming exercise, but is it in theory possible to rerun your model adjusted for, say, 10 degree latitudinal increments so the results could be integrated to get a global estimate for the rotation effect? Is there any way to do a ‘rough and ready’ engineer’s estimate that would be within a reasonable margin of error?
At very low temperatures, not a lot of energy is required to make quite a big difference to absolute temperature, so if it turns out the difference between you and Ned for Earth’s Tna is only a few degrees, it means that in energy terms, there’s little separating your positions.
Tallbloke quotes, “Could the difference between 33k and 90k be accounted for by the kinetic energy originally conducted from the surface which is thereafter stored by convection in the vertical column but in potential energy form?”
Is this not apples to the oranges of much greater thermal capacity of deep oceans and their capacity to hold vastly different quantities of energy relative to the varied WL of insolation, even if the TSI remains constant? Could not the oceans themselves account for some of the increased GHG effect?
This portion from my post perhaps most cogent is this…. “…it is not deniable that the amount of surface energy absorption would vary tremendously over disparate solar cycles, as various decadal changes in insolation W/L would greatly affect how much solar energy is contained in the oceans. ”
If This logic is followed then any change of any bodies T, ( with or without an atmosphere) assuming no change in total insolation, is due to something within that body affecting the residence time of energy entering the system. Any material increasing residence time, say deep waters, vs white sand, warms said body, and likewise any material decreasing residence time, say white sand, cools said body.
Now the energy beneath the ocean surface does not count as surface energy until it reaches the surface. In a rotating body that energy moving to the surface warms the surface even though the surface is rotated away from its primary energy source. This warmth, accumulated over time, then released to the surface, while insolati on continue uncanged, thus increases the total energy of the surface, as well as moderating the day night flux.
gallopingcamel wrote: “My model slices the Moon’s surface into 50 layers each 10 mm thick…”
Thanks, I will take a look. I am interested to see why such a complicated model is necessary.
The 255 K estimate is wrong because the Earth and the Moon are not heated uniformly. The (bogus) 255K figure includes 0.3 albedo, airless bodies do not have clouds.
https://www.linkedin.com/pulse/earths-surface-temperatures-using-hemispherical-rather-ulric-lyons
@The Badger,
“……..it appears that the minimum night time temperature is trending towards a limit of around 75K. This seems counter intuitive as one would expect it to trend towards 3K as given long enough all daytime heat should be radiated away to space.”
In the absence of any radiation for an extended period one would expect a body to reach equilibrium with the universe’s background radiation which implies a temperature of 2.73 K:
https://en.wikipedia.org/wiki/Cosmic_microwave_background
However, in the case of the Moon, there is another source of heat which was quantified by the Apollo missions. See figure 7 and supporting text here:
https://www.researchgate.net/publication/230766556_Lunar_equatorial_surface_temperatures_and_regolith_properties_from_the_Diviner_Lunar_Radiometer_Experiment-JGR_in_press
The Apollo missions only measured the basalt bed rock temperature in the lunar equatorial regions as mentioned by Ned Nikolov on another thread:
“3) The subsurface (below 0.3 m depth) temperature at the lunar equator is nearly constant at about 250K. However, this does not mean that subsurface temperatures at higher latitudes have the same value! As the surface temperature declines from equator towards the poles, so does the subsurface temperature. Unfortunately, we do not have direct subsurface measurements outside the lunar tropics (the Apollo missions were within 22 degrees of the equator).”
Owing to heat conduction from the bedrock the lowest temperature at the Moon’s equator cannot fall below 75 K.
As Ned Nikolov points out, we don’t have direct measurements of the lunar bedrock temperature at the poles but we do know that the temperature in polar craters permanently shielded from solar radiation can be as low as 35 K. This is strong evidence for the idea that the bedrock temperature there is very low. My “back of the envelope calculation” says the southern polar bedrock is at 43 K.
For folks who like useless information. AFAIK the temperature in those southern lunar craters is the lowest measured in our solar system to date. I look forward to someone finding somewhere cooler.
@Tallbloke,
“GC: I realise it would be a time consuming exercise, but is it in theory possible to rerun your model adjusted for, say, 10 degree latitudinal increments so the results could be integrated to get a global estimate for the rotation effect? Is there any way to do a ‘rough and ready’ engineer’s estimate that would be within a reasonable margin of error?”
There is a huge source of error once you stray outside the Moon’s equatorial regions. To predict the Moon’s night time surface temperature accurately you need to know the bedrock temperature. See my earlier comment to “The Badger”.
There may be ways to estimate the bedrock temperature at all latitudes. Maybe someone already did it by carefully studying the night time LRE data. I will look into it.
‘The Lunar Reconnaissance Orbiter measured temperatures of minus 396 F (minus 238 C) in craters at the southern pole and minus 413 F (minus 247 C) in a crater at the northern pole. That is the coldest temperature ever recorded in the solar system, colder even than Pluto. Scientists think water ice may exist in those dark craters that are in permanent shadow.’
http://www.space.com/18175-moon-temperature.html
Slow rotation speed + lack of an atmosphere + low tilt (no seasons) = coldest poles?
– – –
Update:
Water Ice Discovered on Mercury (2012)
Evidence of big pockets of ice is visible from a latitude of 85 degrees north up to the pole, with smaller deposits scattered as far away as 65 degrees north.
. . .
Researchers also believe the south pole has ice
http://www.space.com/18687-water-ice-messenger-discovery.html
Slow rotation speed + lack of an atmosphere + low tilt
gallopingcamel said:
“..is it in theory possible to rerun your model adjusted for, say, 10 degree latitudinal increments so the results could be integrated to get a global estimate for the rotation effect?”
The surface temperature profile of the sunlit side could be roughly calculated with a parabolic function. With the pole to pole profile a bit steeper:
“To predict the Moon’s night time surface temperature accurately you need to know the bedrock temperature.”
Assuming that the sunlit side of the Moon is in thermal equilibrium with solar irradiance, its average temperature would be: 394K*0.5^0.25 = 331.3K
331.3K, plus the average temperature of the dark side (the mid point as it’s virtually a straight line), divided by two, gives a figure very close to the Diviner average Lunar surface temperature.
[Moderator’s note] So far as I know, DIVINER has not measured an “average Lunar surface temperature”. Vavasada has made an estimate of lunar equatorial average temperature using DIVINER data. – TB
TB, good point thanks, so I have the average surface T of the dark side too high. The sunlit side figure is still valid, as the sunlit side is roughly in thermal equilibrium with the solar flux.
http://www.sciencedirect.com/science/article/pii/S0019103516304869
“So far as I know, DIVINER has not measured an “average Lunar surface temperature”. Vavasada has made an estimate of lunar equatorial average temperature using DIVINER data. – TB”
They don’t report an average (interestingly they seem uninterested in this value), but clearly you can do so if you wanted from the data used for this paper (Williams et al. The global surface temperatures of the Moon as measured by the Diviner Lunar Radiometer Experiment. 2017): http://www.sciencedirect.com/science/article/pii/S0019103516304869
I haven’t checked but presumably you can get the data from here:
http://pds-geosciences.wustl.edu/missions/lro/diviner.htm
DAILY NEWS 8 June 2017
The mystery xenon in Earth’s atmosphere came from icy comets
On Earth, we presumed the presence of this isotope was primarily a result of decaying iodine. We know the rate at which iodine decays, so we use the quantity of xenon-129 to gauge the time at which planetary events took place. But if 22 per cent of the xenon in Earth’s atmosphere was transported by comets, models based on iodine decay aren’t accurate. They overestimate the age of Earth’s atmosphere and the moon. [bold added]
http://www.newscientist.com/article/2134120-the-mystery-xenon-in-earths-atmosphere-came-from-icy-comets/
In order to calculate global average temperatures my method involves slicing the Moon by latitude, 0+/-5, 10+/-5, 20+/-5 and so on to 80+/-5 and 87.5+/-2.5.
Thus it takes ten FEA runs to calculate a global average compared to one for an equatorial analysis. Even with highly efficient Russian software a single FEA run can take several hours on my HP 2000 laptop.
Here is a table showing averages for the Moon “As is” from an earlier post:
DIVINER MODEL
Global average (K) 197 196
Equatorial av. (K) 212 212
My model shows decent agreement with reality as measured by the Diviner LRE. It seems likely that the difference between global and equatorial averages would be around 10 K for all rates of rotation.
Even though the global average results take ten times as long as the equatorial calculations I may decide to do it if I can make some improvements. For example I used a constant temperature of 250 K for the bedrock whereas that is clearly much too high for the high latitudes. The bedrock temperature may be as low a 43 K in deep craters at the Moon’s southern pole.
@Oldbrew,
Thanks for that interesting information about Xenon. I used to be an avid reader of the “New Scientist” but that was 30 years ago!
@Ulric,
That plot you displayed includes a bar at -18 C (255 K) for the “S-B avg Earth No Atmosphere Temp”.
That temperature estimate is not even close to reality because it is based on false assumptions about the thermal properties of Earth’s surface.
gallopingcamel says: June 9, 2017 at 9:37 am
Assuming a still hot interior for the moon, the sub surface temperature will establish around the average surface temperature for that location. The geothermal gradient starts from that temperature towards the hot core.
@Ulric,
I would have to agree with Galloping Camel, -18 C (255 K) for the “S-B avg Earth No Atmosphere Temp” is not correct. That figure is derived by simply imputing 240 w/m2 into the S-B equation with emissivity and absorptivity set to unity.
The reality is that instantaneous radiative balance equations (like S-B) cannot be used to find “avg Earth No Atmosphere Temp”. The reason is that 71% of earth’s surface is ocean, an extreme SW selective surface. It is translucent to sunlight (SW + SWIR) yet opaque to LWIR. For materials like this the Stefan-Boltzmann equation fails as demonstrated by this simple experiment:
The experiment involves 2 acrylic blocks with equal ability to adsorb solar SW and emit LWIR. The only difference is that block A absorbs SW at its base and block B at its upper surface. Illuminate with 1000 w/m2 for 2 hours and block A will average around 20C hotter.
Here’s a simple CFD run of what is happening as the two blocks heat at the same time stamp:
Solar energy absorbed by block B is more readily lost as LWIR. Solar energy absorbed by block A must slowly conduct back to the upper surface before it can be lost as LWIR. This allows greater energy accumulation for the same solar illumination.
The surface properties of this planet mean that the only way of calculating “avg Earth No Atmosphere Temp” is via empirical experiment or CFD. Instantaneous radiative balance equations simply won’t work. Given 71% of this planets surface is ocean, “avg Earth No Atmosphere Temp” should be 312K or above. This means our radiatively cooled atmosphere is cooling surface temperatures by around 24C ie: there is no net atmospheric radiative GHE.
@Ben Wouters,
We have direct measurements of the bedrock temperature to a latitude of 22 degrees. Above that latitude it may be possible to estimate the bedrock temperature by looking at how the Diviner night time temperature profile changes with latitude. While I would like to try that my laptop and software has some limitations.
While I have modeled the entire Moon, the Diviner data I have covers only the equatorial region. My laptop is incapable of handling the full raw data set as reported in an earlier post:
“………my spreadsheet could not handle even the “Level 3” data. The Diviner team did much better and showed that the Moon’s average temperature is 197.3 Kelvin.”
gallopingcamel
“That temperature estimate is not even close to reality because it is based on false assumptions about the thermal properties of Earth’s surface.”
It does not address the thermal properties of the surface. It is just an equivalent black-body temperature for a planet at this distance from the Sun, but including 0.3 albedo for no logical reason:
394K * 0.25^0.25 = 278.6K.
Minus the albedo:
278.6K * 0.7^0.25 = 254.833K
The real hemispheric black-body heating figure without albedo is:
394K * 0.5^0.25 = 331.313K
allopingcamel says: June 10, 2017 at 3:56 am
I would take the max and min temperatures for the latitude you”re looking at, average them and use that number as the bedrock temperature for that latitude.
Use the Diviner plot:
eg for the equator this would give ~240K,
I wrote up the model I described above[1] in R. If you play with the heat capacity (parameter C) and rotation rate (parameter W) you can clearly see the mean temperature at equilibrium depends on both and can vary between ~150 and ~270 for an object with the moons albedo:
https://pastebin.com/GFbM22cd
I recommend setting C to between 1e4 and 1e8 W*s*K^-1*m^-2. Also it will go much faster if you plot/save results less often (parameter nSave).
Ulric Alexander Lyons, June 10, 2017 at 2:42 pm,
There is nothing wrong with your mathematics……you (and Consensus Climate Scientists) are getting the wrong answer because you are assuming a body at uniform temperature.
Also you assume that albedo is a simple number as it would be for a Lambertian body. However the Moon is non-Lambertian which means the albedo is a function of the angle of incidence. Failure to take account of this leads to major modeling errors in lunar daytime temperatures. These issues were explained in the first and second posts in this series of three:
In Figure 2 in the link that follows there is a plot of the effect of angle of incidence on absorbance (1-albedo). Note that the three different models used slightly different functions. FWIW Tim Channon got the closest “Fit” to the Diviner equatorial data. All three models have RMS errors of less than 1 K and it is hard to do better owing to “Noise” on the Diviner data.
Konrad June 10, 2017 at 3:44 am,
Good to hear from you again.
Did you build that thing shown on your drawing?
gallopingcamel:
“There is nothing wrong with your mathematics……you (and Consensus Climate Scientists) are getting the wrong answer because you are assuming a body at uniform temperature.”
Leaving aside albedo, we get very different black-body answers, as I do not assume that the Moon is uniformly heated.
Konrad
Very nice to see more details on your experiments.
This has evolved quite a bit since we both took issue with Willis (from WUWT) regarding his article Radiating the Oceans.
I seem to recall that Dr Spencer did some sort of experiment regarding the so called GHE, and DWLWIR, but as I recall this was not to simulate the position over oceans, and oceans cover some 71% of our planet and account for almost 95% of all the energy stored in the system.
I would recommend that you take a look at his experiments (they are set out somewhere on his site) although since they do not deal with the situation over the oceans, in my opinion, they have only limited relevance. It is the oceans (and the resultant water cycle) that control the climate on this water world on which we live.
gallopingcamel says:
June 11, 2017 at 2:10 am
“Did you build that thing shown on your drawing?”
Yes:
The results for constant illumination were:
Block A – surface 80C base 120C
Block B – surface 80C base 80C
The first time I tried it, the base temperature for Block A rose to 80C then started to drop. I disassembled the experiment to find if the thermocouple had failed and blistered my fingers. (The heat sink tape had failed and the thermocouple had detached while the block had heated above 100C)
If you intermittently illuminate the two blocks with sunlight, temperatures for both blocks run lower, however the surface temperature of Block A then exceeds that for Block B.
You can also try a similar experiment with water:
The deeper you build it, the better the results. (If you try this one outdoors, pick a day with low wind speed to reduce confusion from evaporative cooling).
The message from simple experiments like these is that the S-B equation is inapplicable to solar heating of the surface materials of our planet. CFD or empirical experiment is required to determine “avg Earth No Atmosphere Temp”. Joseph Fourier made a huge mistake in the 1820s, and the whole AGW conjecture is built with that mistake as its very foundation.
richard verney says:
June 14, 2017 at 7:03 am
Richard Hi,
Good to see you are still injecting some reason into WUWT (AKA lukewarmer central).
After being banned I have not sought to re-enter the fray there (despite moderators privately emailing). I wanted all my work to appear under one web handle.
I don’t hold any animosity to Anthony, and I do care about the personal toll running WUWT has cost him. But “lukewarmer” means part of the problem, not part of the solution. It is sad that someone who entered the fray on the basis of empirical experiment, like myself, ended up rejecting empirical experiment in favor of the blackboard scribbling of Willis and Monckton.
I got to the answer regarding solar heating of the oceans in a very roundabout way. In hindsight empirically checking the foundation claim of the climastrologists: “The sun alone could only heat the surface materials of this planet to an average of -18C in the absence of radiative gases” should have been the first step. But I went the wrong way
The savage opposition to the Makarieva 2010 paper led me to experiments proving diabatic energy loss could indeed cause horizontal winds. From that to experiments proving surface air pressure determined surface heating of the atmosphere. From that to empirical experiments that showed that surface incident LWIR could neither heat nor slow the cooling rate of water free to evaporatively cool. Finally, finally I thought to check if some property of liquid water made it respond to solar illumination in a way dramatically different to a theoretical “near blackbody”.
Right there is the stunning 80K error for 71% of this planet’s surface in all of the climastrologist’s foundation calculations. AGW is not “less than we thought”, it is actually a physical impossibility.
Konrad – Re LWIR, would ocean surface water behave like this man’s glasses?
http://en.wikipedia.org/wiki/Infrared#CIE_division_scheme
@Konrad.
I am impressed by your experiments but am having great difficulty with your conclusions. In situations like this my experience is that nothing can be achieved by “on line” discussions. I am hoping that you will want to exchange information “off line”.
In this three part examination of the temperature of airless bodies I have relied on Stefan-Boltzmann radiation theory. I remain a “Believer” in SB theory because my model (based on SB) replicates reality with amazing precision.
My public email is info(at)gallopingcamel.info
@oldbrew,
What an impressive pair of photos. It reminds me of observations of stellar events such as GRBs (Gamma Ray Bursters) that look different at different wavelengths.
oldbrew says:
June 15, 2017 at 12:10 pm
Water is more complex than glass, but yes, both are opaque to LWIR.
The difference is this, glass is a solid while water is fluid.
LWIR absorption and emission for solid glass is from within the first micron thickness.
But for water LWIR absorption and emission is from the first 100 microns depth. This leads to the dramatic difference in absorption and emission LWIR response of water depending on angle from zenith.
Climastrologists use 0.98 a/e for water, but when in reality the hemispherical LWIR emissivity for water is only 0.67.
You may have seen the handle “Will Janoschka” on climate blogs. He’s an insufferable bastard, but when it comes to field theory of radiative physics that insufferable bastard is right, and the climatologists utterly wrong. His work for the US military was the foundation for the modtran software so abused by climastologists. Under intense provocation, I empirically checked his claims:
That old bastard turns out to be right. After 55 degrees from vertical, LWIR emissivity for water falls off a cliff and LWIR reflectivity goes through the roof.
PS. just as you may think I’m being evil to you, but Will was evil to me. He made me replicate his 1970’s empirical experiments. He is evil, but he is not wrong.
Tell me I’ll forget. Show me I’ll understand. Let me do it I will KNOW.
Do you want to know?
Do you know the man who sold me the Moon for $500? Tell him I want my money back.