A new Lunar thermal model based on Finite Element Analysis of regolith physical properties

Posted: April 18, 2014 by gallopingcamel in Analysis, Geology, methodology, radiative theory

Article by Peter Morcombe (gallopingcamel) with some assistance from Tim Channon.

Image

While investigating Nikolov & Zeller’s “Unified Theory of Climate” it seemed odd that professional scientists could not agree what the temperature of an airless Earth should be. Given that one needs to know this in order to compute the Greenhouse Effect (GHE), I tried to settle the question by analyzing the Diviner LRE data that accurately mapped the Moon’s surface temperature. This effort failed as 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.

While the temperature of the Moon is now known with impressive precision, would an airless Earth have the same temperature or would the different rates of rotation have an effect?

If one had a mathematical model that corresponded closely with the Diviner observations it should be possible to answer that question. At least two such models have been constructed namely Ashwin Vasavada’s one dimensional heat transfer model and Tim Channon’s PSPICE model.

Tim found that rotation had very little effect on average temperature. While that made sense I wanted to confirm it by building a simple model using freeware or a spreadsheet. I was making slow progress using my spreadsheet when Tim suggested Quickfield. This is a powerful Finite Element Analysis program that can solve many kinds of field problems. The “Student Version” is available as a free download, so I decided to try it. The student version is limited to 255 nodes which might be enough for a problem with simple geometry.

Tim adds:- The conceptual overview of how we are treating an illuminated sphere will help. Imaging a small square patch of the Moon directly under the sun. Day will last 27 earth days. Unlike most physics maths models we are doing dynamic modelling (it moves in time) so here we rotate the moon for day and night except we can’t so we reverse reality and gradually turn the sun off to night then back on again, same effect. In addition the moon is a sphere so if modelling a whole body is wanted this can be approximated by further sun modulation to take into account the double curvature, and sum. This was done with the SPICE model. The DiVINER satellite orbit facilitates slowly scanning the entire surface as the moon rotates. Work has also been done on eclipse results.

Peter continues:- Most of the Moon’s surface is covered in debris created from the impact of meteorides. This material is called lunar regolith and it can be several meters deep. The first model represented the regolith as a single layer. This crude model (29 nodes) was able to reproduce the temperature at the Moon’s equator with good accuracy using credible values for the thermal properties of regolith. The same model worked well for Mercury too.

Eventually the model evolved into 50 layers (102 nodes) each 10 millimetres thick. Vasavada has estimated the thermal properties of lunar regolith and when those values were plugged into the model the night time temperatures were an almost perfect “fit” with an RMS error of 0.06 Kelvin. Close enough for government work!

The daytime temperatures proved much more difficult. While it was trivial to get a perfect “fit” for the maximum temperature, the Moon’s Albedo varies according to the angle of incidence. Vasavada modelled this using an 8th order polynomial. I could not figure out how to do this in QuickField so I used trigonometric functions that did not fit nearly as well. Consequently my daytime RMS error was a horrid 0.66 Kelvin. Here’s how the model shaped up against Diviner LRE observations at the lunar equator:

Diviner Model Difference
Maximum (K) 388.0 387.4 -0.6
Average (K) 212.1 213.2 +1.1
Minimum (K) 93.6 95.8 +2.2

 

Tim writes:- Back when I was creating the SPICE model, from first principles as a dual between electrical and thermal I was unaware of Vasavada’s work. I hit the same problem of almost good. A difference showed some kind of smooth error function during daylight. Earlier I wrote about lunar rotation, modulating the sun, this is about altering the profile.

Empirically, I used my synthsiser software to match both data but as continuous waves. The answer was trivial, raise insolation to power 1.3, in SPICE via a maths function block. Naturally having an answer without a reason is disturbing, why the difference and that particular number?

Image

I searched and asked blog readers if anyone had any ideas, nothing. As an idea I looked at lunar photographs, Rog Tallbloke providing a usable example with known processing, none and I am familiar with camera technology. This showed a light falloff to the same shape. With nothing else to justify what I was doing I blog published anyway, theorising on possible detail reasons for the effect.

The solution I now know is about non-Lambertian surfaces, roughness, beyond this text.

Image

Credit as image and borrowed from PDF here

The moon is non-Lambertian.

Peter continues: If you are wondering how the night time minimum error can be so much larger than the RMS error (0.06 Kelvin) the answer is “noise”. The Diviner data has “spikes” while the model is relatively smooth.
Respectable climate scientists say that the temperature of an airless Earth would be 255 Kelvin which is roughly 58 Kelvin too high. Their mathematics was fine but they used unrealistic assumptions about the properties of regolith. They assumed that the surface of an airless Earth would maintain a uniform temperature. The regolith consists mostly of basalt but it has quite different thermal properties from the bedrock:

Surface @0.5m Basalt
Specific heat 840 840 840 J/kg/K
Density 1,300 1,800 2,900 kg/m3
Conductivity 0.001 0.007 1.69 W/K/m

An airless Earth might have a higher Albedo than the Moon (~0.11) which would make it colder than the Moon’s average of 197 Kelvin but over time the regolith would build up until it was comparable with the Moon. How much regolith would be needed? The model shows that 0.2 meters (~8 inches) is sufficient owing to its exceptionally low conductivity. While there are hot spots on the Moon corresponding to rocks with little or no regolith cover, Vasavada found this had no significant effect on the average temperature of the Moon.
Overall, the model did a good job of reproducing the Diviner LRE equatorial data without introducing any questionable assumptions or fudge factors. However it did not match the precision of Tim’s PSPICE model. Vasavada points out that emissivity in the thermal Infra Red is a function of wavelength and therefore of temperature. In the model a fixed emissivity of 0.95 was used.

Tim adds:- I used empirical matching so of course it matches. Odd perhaps is that I did not use any concept of albedo, the development was more fundamental making no assumptions about values except for solar radiative flux and measured temperature as deduced from the output of remote sensing. At its most fundamental the surface is converting thermal flux into actual heat in a thermal entity, the loss of resisting a flux. This heat applies at the end of a bidirectional thermal delay line into a thermal capacitance. Add surface roughness, it matches.

Part two is in preparation.

 


Ref1

History of Quickfield

Developed by Tor Cooperative, a Russian firm, the program is marketed outside Russia by Tera Analysis. It first appeared in an MS-DOS version (under the name Elcut) early in the 1990s [Spectrum, December 1993, p.64].

http://www.quickfield.com/publications/ieee_pub.htm

Tim confirms this, was in contact with Tor in that era and yes it was called Elcut. Tera Analysis are US based. I have no involvement today.

Quickfield is a Microsoft Windows program, versions from DOS through 8.1. BSD, Linux, Mac are not supported but there are reports it did at one time operate under simulation.

You need to register to download the Student version. I’ve heard of no reports of bad behaviour by the US based company. I doubt they are interested in selling to students or individuals, CAD software is expensive.

Link to Quickfield web site, link there to download Student edition. (don’t recall the file size, large, maybe 150MB)

Ref2

Lunar model, zip archive of Quickfield design files, link is here (344kB)

There is no need to run the solver, is supplied with results ready. Redo as you wish.

Comments
  1. Konrad says:

    Peter,
    this is a great result, the difference between Diviner and your model is tiny compared to previous lunar temperature modelling.

    With regard to “roughness” there is an effect that is known to engineers using thermal imaging for non-contact inspection of machinery. When a material of emissivity <1 (every material know to man) has holes or pockets in its surface, to thermal imaging they appear as hot spots, despite the fact that the hole or pocket is no hotter than the base material. Images of this effect can be seen here –
    http://www.reliableplant.com/Read/14134/emissivity-underst-difference-between-apparent,-actual-ir-temps

    This effect should theoretically occur at all scales above the wavelength of emission.

    In other news, the LADEE lunar mission has been successfully de-orbited into the far side of the moon. LADEE was observing the tenuous gas and dust atmosphere of the moon. When data becomes available it may shed light on the process of deposition of fine dust over regolith.

  2. Peter, Konrad mentions emissivity. I was thinking that you should not be using the word albedo because that is not properly defined. Some authors (non-engineers) uses the word in reference to light but do not define the wavelength. Emissivity of a surface is the ratio of emission at a temperature for all wavelengths compared to the emission of a blackbody at the same temperature. “According to Kirchhoff’s law, the emissivity and absorptivity of a surface IN SURROUNDINGS AT ITS OWN TEMPERATURE are the same for both monochromatic and total radiation. When the temperature of the surface and its surroundings differ, the total emissivity and absorptivity of a surface often are found to be different.”
    A real problem arises with emissivity of the earth surface because about 70% is water mostly being salt water. In the temperature range 0 -100 C water has an emissivity of 0.95 with salt water having lower emissivity (partly when frothed -as white cap waves). With no atmosphere there would be no water and no vegetation on land.
    Secondly, the atmosphere has an effect on in coming radiation. There is an ozone layer which removes UV wavelengths, Then there are clouds. No one has determined the radiation absorption and reflection of clouds which consist of droplets of water and particles of ice. It will be probably impossible to determine because of varying thickness and composition.
    You probably know that the Stefan-Boltzmann equation adjusted for emissivity and view factors only applies to surfaces in a vacuum. However, it is useful for calculating heat transfer in furnaces where high temperatures are involved.
    I have my doubts that the Diviner results are as accurate as indicated because the surfaces of the moon have many variations. Is it not very rugged on the side not seen from the earth? That will change the view factor. Your results could be just as accurate.

  3. Peter, I should add that the radiation from the sun you use may also not be correct. The sun has an atmosphere, It has an inner surface or core of unknown size (area) and temperature. Calculations of radiation energy flux are a mathematical construct assuming a) the Stefan-Boltzmann applies b) that the sun is a black body at the assumed temperature (not correct) c) that is has a surface equal to the visible projection (not correct) d) the temperature at the assumed surface is constant (not correct) e) the is no energy absorbing material between the sun and earth (there are some indications of an aether or something)

  4. oldbrew says:

    cementafriend says: ‘Is it not very rugged on the side not seen from the earth? ‘

    Someone who has been there (Collins) says:
    ‘This gives you some idea of the rough surface available on the Moon. (Photo 35.) Of course, the maria on the front side are smoother than this, but in general the back side of the Moon is quite rough.’

    http://history.nasa.gov/ap11ann/FirstLunarLanding/ch-6.html

  5. Steve C says:

    Interesting. If that equilibrium temperature of 197.3K is about right, and disregarding for the sake of simplification any differences between Earth and Moon in albedo/emissivity, then presumably the “thermal equilibrium height” of the Earth/atmosphere system is somewhat higher than the 5km or so usually derived from assuming an equilibrium temp of 255K. Rough calculation from the known lapse rate suggests that it would have to be around 15-16km, three times as far above the surface as generally reckoned, if the lapse rate calculations still apply unchanged over that extra 10km. (Do they? If not, where do the differences occur?)

    Could someone with a more mathematical brain than mine take a few minutes to work out what, if any, implications that would have for all the usual arguments based on the 33°C figure ‘twixt Down Here and Up There? (For a start, if there were a “Greenhouse Effect” responsible for our cosy surface temps, it would appear to need to be about three times fiercer than it was claimed to be – not that I want to give Them any ideas here, you understand, just saying.)

  6. Steve C,

    There is a prolonged discussion of the concept of “Effective Emission Height” on another thread with over 850 comments so far:
    https://tallbloke.wordpress.com/2014/03/11/effective-emission-height/

    Personally I see the concept of EEH as an attempt to simplify a complex problem so I am more interested in the approach of Robinson & Catling who have developed a model for planetary atmospheres based on physical parameters that can be directly measured. For more information here is a link to a letter published in Nature:
    http://faculty.washington.edu/dcatling/Robinson2014_0.1bar_Tropopause.pdf

  7. cementafriend,

    As with most models this one has some simplifications built in. For example, it uses a single parameter (0.95) to describe emissivity even although we know the lunar surface is non-Lambertian.

    While the dependence of Albedo on the angle of incidence is included it is only a rough approximation.

    The next step is to use the model to predict the surface temperature of Mercury and the temperature of the Moon if it rotated once every 24 hours.

  8. tchannon says:

    A lot of research is taking place into the lunar surface and Mineralogy flowing from the new generation of exploration satellites, after basic been-there comes a better look.

    I could point at a variety of papers but it’s better I step aside, suggest some search keywords.

    mineralogy
    lunar
    lambertian
    Clementine
    Diviner
    Moon Mineralogy Mapper (M 3 )

  9. Steve C says:

    Gallopingcamel – Thanks for the link to the EEH thread, I thought I remembered seeing one but didn’t have time to look this morning as I was racing around prior to getting the weekend shop done. Also the pdf (for later!).

    Yes, I appreciate that the EEH is a simplification, but I like to have a reasonably mathematically easy example in my head so that, when some idiot starts banging on at me about the 33°C as “proof” of AGW, I can nonchalantly mention that, of course, if you want 255K, then about 5km up at a lapse rate of six and a half K/km … etc. (Usually the phrase “lapse rate” all by itself does the job … err, what? … 🙂 ) My “quick comment” was the result of realising I needed to recalculate my mental model, though now after looking at that thread it sounds like a complete redesign is needed.

    Or of course now, as soon as they say “33°”, we can start, look seriously at them and say “No, no, it’s ninety, ninety-one degrees, if you mean what I think you mean …”

  10. Konrad says:

    Steve C says:
    April 19, 2014 at 10:42 am
    ———————————
    Peter has pointed to the EEH thread where this was indeed discussed. The die-hard AGW defender there tried to dance from the 255K figure to 197K when it became impossible to argue against the Diviner evidence. At that point I did point out that EEH would then be raised above the tropopause.

    While EEH is a mathematical fiction, it is supposed to be the median level of IR opacity in the atmosphere. An EEH above the tropopause is clearly ludicrous as almost all radiative gases exist below this level.

    Peter asks the question –
    “.. would an airless Earth have the same temperature or would the different rates of rotation have an effect”

    My answer is most definitely not. Only the driest deserts would come close to the 197K number. 71% of the our planet surface is ocean and a great deal of the remaining 29% is vegetated. Empirical experiment tells us that if the oceans could be retained in the absence of an atmosphere, they would become a giant evaporation constrained solar storage pond with temperatures topping 80C.

    Standard blackbody calculations cannot be applied to transparent oceans or a moving gas atmosphere. It is not just the AGW hypothesis that is in error, in is the very concept of a net radiative greenhouse effect that is in error.

  11. tchannon says:

    konrad, I think I asked for evidence, measurements. Haven’t seen anything, might have missed it.

    You claim water is transparent.

    Firstly please remember that substance can reflect which does not involve SB. I will later on be saying more about this in a very different context.

    So lets assume water really is totally transparent. In that case you have to deal with the bottom.

    It means that if a large block kept at the same temperature as the surface water was suspended over the water the block would lose heat to a 4C bottom.

    Is that what happens?

  12. Konrad,

    If the Earth was magically made airless its average temperature would be well below the freezing point of water so the oceans would be solid ice.

    Over time meteoride impacts would build up a layer of regolith similar to what is found on the Moon so this layer would lie on top of the oceans of ice. It would only take 8 inches of regolith to insulate an airless Earth to the point that the surface would have thermal properties similar to the Moon.

    That raises an interesting question. Could there be large bodies of ice under the lunar regolith? Let’s hope so as that would have major implications for our impending colonization of the Moon.

  13. Q. Daniels says:

    gallopingcamel,

    Thanks for the link to Robinson 2014. That looks like a very interesting paper.

  14. Konrad says:

    tchannon says:
    April 20, 2014 at 1:40 am
    ———————————–
    Tim,
    I am doing a write up for Roger on the acrylic block experiment, demonstrating the science of “selective coatings”. I am currently working on some build diagrams as all I have are photos of the assembled experiment.

    In my experiments with transparent materials I am using a black SW absorption surface at the base of the material. Our oceans would take about 200m to absorb all UV/SW and I do not have the money to build experiments that big. Compromises must be made. The important thing is to study the difference between SW absorption at the surface (blackbody instantaneous calculation) and SW absorption at depth (selective coating iterative calculation).

  15. Robert JM says:

    Galloping camel,
    Who cares about average temp?
    The peak solar heating is more than enough to melt water.

    My question is will energy storage/buffering such as occurs in oceans and latent heat cycle lead to a higher temp.
    It should based on the exponential relationship between temp and IR emissions.

  16. Konrad says:

    gallopingcamel says:
    April 20, 2014 at 1:55 am
    ——————————–
    “If the Earth was magically made airless its average temperature would be well below the freezing point of water so the oceans would be solid ice”

    Peter,
    This is the most critical error in all of climate science. If the oceans could be retained in the absence of an atmosphere they would become a giant evaporation constrained solar pond. Basic blackbody calculations cannot be used on transparent/translucent materials. These act as what are known as “selective coatings” for the lithosphere of the planet.

    Remember the lab verification experiment in the second Thomas Diviner paper? It is possible to run a similar experiment for water. The Diviner experiment was run in vacuum which would cause water to boil off, but it is possible to remove all conductive and evaporative cooling and allow only SW heating and IR cooling of a water sample –

    I have not built this one yet due to cost, but I have run three verification experiments, two with liquid, one with solid acrylic blocks. We also have data from the engineering of solar thermal storage ponds.

    The basic answer is this – without atmospheric cooling our oceans would heat far beyond what blackbody calcs indicate. If our oceans were only a 1mm layer over a black surface, they would conform closely to black body calcs for an emissivity ~1. But the deeper they get, the more their equilibrium temperature will diverge.

    When I get the acrylic block experiment up, I hope you will be able to model it using your multi layer FEA analysis. A brief description of the experiment is here –

    https://tallbloke.wordpress.com/2014/03/11/effective-emission-height/comment-page-1/#comment-71772

    For block A layer 1 can heat by conduction and cool by IR radiation. Layers 2 to 49 conduction only. Layer 50 heating by SW and cooling by conduction.

    For block B layer 1 can heat by SW and cool by conduction and IR radiation. Layers 2 to 50 conductive transport only.

    But in the meantime, here is a pretty picture of an evaporation constrained solar pond –

    Now why don’t engineers make layer 2 black? They could then have fillers and UV stabilisers in the plastic. Maintenance costs would be greatly reduced. Why does it work so much better with layer 2 transparent?

  17. oldbrew says:

    Research satellite LADEE has been deliberately crashed into the moon as its mission was over.

    ‘LADEE gathered detailed information about the structure and composition of the thin lunar atmosphere. In addition, scientists hope to use the data to address a long-standing question: Was lunar dust, electrically charged by sunlight, responsible for the pre-sunrise glow seen above the lunar horizon during several Apollo missions?’

    http://www.sciencedaily.com/releases/2014/04/140419193729.htm

  18. Ben Wouters says:

    Robert JM says: April 20, 2014 at 4:42 am

    ” The peak solar heating is more than enough to melt water”

    Could we have some calculations to support this position?

    The peak equatorial temp is ~390K, reached after 7 earth days of increasing solar radiation.
    This peak temperature is reached only in the upper 10 cm of the lunar surface, below 30 cm no solar signal anymore.
    Heat capacity of this stuff is given as 840 J/kg/K.
    Heat capacity of water is roughly 5 times higher: ~4200 J/kg/K.
    Sun penetrates up to 100 metres into water iso just 30 cm.

    Start with lets say1000m ice at 0K and show how the sun can melt this ice and heat the resulting water to the boiling point, assuming no evaporation. Remember above I referred to the equator, towards the poles solar decreases to zero.

  19. tchannon says:

    konrad, I’m not expecting you to do grand experiments.

    My guess is that the work has been done but is not made obviously available to strangers.

    A snippet I would like to see is high resolution net radiation balance over the ocean at sea level. At night what happens?

    My guess is that a dense water vapour layer is always present and that this is dominant.

    Apparently temperature inversions are very common over the ocean.

    You might find the following interesting if you can access the link, from the following select Evaporative Duct Height (goes to current forecast)
    https://www.fnmoc.navy.mil/wxmap_cgi/cgi-bin/wxmap_DOD_area.cgi?area=15km_europe&set=All

    This might explain a little
    http://www.dtic.mil/dtic/tr/fulltext/u2/a213841.pdf

    Naval (and aeronautic) interest comes from EM wave ducting , optical mirage writ large.

    Tim Ball could probably add some more because this is about layers you can’t see, both underwater and in the air.

    Figuring what goes on for real is not easy.

  20. Konrad says:
    “For block A layer 1 can heat by conduction and cool by IR radiation. Layers 2 to 49 conduction only. Layer 50 heating by SW and cooling by conduction.”

    The free “Student” version of QuickField will make short work of conduction and convection problems with simple geometry.

    However, when I tried to use it to set up a model of Earth’s atmosphere I ran into trouble dealing with radiative heat transfer between the layers. It must be possible to model this but it may need the full version of QF which costs several thousand dollars. That is why I took an interest in the Robinson and Catling model.

    If we can’t solve your problem using the free version of QF, it might be worth trying the R&C model which is also free.

  21. Robert JM, April 20, 2014 at 4:42 am

    “Who cares about average temp? The peak solar heating is more than enough to melt water.”

    That is an interesting question. We could settle it by using QuickField to model surface temperature with the lunar regolith replaced by a sheet of ice.

    I am preparing to leave on a business trip so it will be at least a week before I can get around to it. As Tim has published the model maybe someone here will answer the question before I can.

  22. br says:

    “Respectable climate scientists say that the temperature of an airless Earth would be 255 Kelvin which is roughly 58 Kelvin too high.”

    I recently became interested in the question of equilibrium temperature, so am interested in this thread. Thanks!

    I found the following statement on the Diviner website http://www.diviner.ucla.edu/science.shtml , which says that “Equator Average Temperature (K) ~206K” and :”The mean temperature measured 35cm below the surface of the Apollo sites was 40-45K warmer than the surface” and didn’t suffer the same variations as the surface.

    As the interior of the moon is hugely more massive (literally) than the surface regolith, that would make the average temperature of the ball of rock (as opposed to the surface) to be about 250 K (assuming no interior heating, unlike the Earth which contains a huge fission reactor).

    This disagrees with Tim’s model at https://tallbloke.wordpress.com/2012/04/02/a-model-of-lunar-temperature/ .

    As a first step to find out why, can I ask the simulators to speed up the lunar day/night, at least by a factor of 30 (to make it one Earth day long)? Maybe even x100, to smooth out the day/night ripples? And obviously run the sim long for enough days/nights for the temperature to long-term stabilise.

  23. tchannon says:

    Nothing is clear br.

    I did speed it up. That is not an answer.
    Simulation starts at zero time with prebias to speed up settling. That is not an answer.

    Peter’s result agrees, in that case using the material properties vs. depth.

    This is an open end if anyone wants to dig. There have been many publications and interpretation published but the original reports need to be found with information on what exactly they did.
    When Peter is available again maybe we can look at this.

  24. Ben Wouters says:

    br says: April 22, 2014 at 3:53 pm

    I think a small geothermal flux is still present on the moon. The moon used to be volcanically active:
    http://en.wikipedia.org/wiki/Lunar_mare
    A small flux like 50 -110 mW/m^2 can explain the relatively high temperature in the Hermite crater
    (25K iso the expected 2,7K) and the levelling of the temperature at lat 89 towards 50K, after well over 100 days of no radiation.
    If this flux exists, a temperature gradient will also exist, with increasing temperature towards the hot core. Starting temp will be roughly the average surface temp of the spot where you start drilling, increasing towards the core.

    For radiative balance temperatures two situations should be considered imo.
    First no rotation (actually one rotation / orbit) which results in one sphere receiving all solar radiation, and the other side nothing.
    Simple calculation results in a balance temp for the moon of ~161K. More accurate calcs like N&Z did give ~155K. The surface doesn’t really matter, as long as emissivity is ~ 1.0.

    At the other end rotation once every second. This results in the 300.000 km long tube of solar that hits the moon every second to be spread around evenly over all longitudes. BUT the difference between the equator and the poles remains, so the temperature will be again independent of the surface structure (given emissivity 1.0). For the moon this temperature will be below its Effective temperature of 270K, perhaps 250K or so.

    For all rotational speeds in between the surface becomes relevant, including it’s heat storage capabilities.

    Earth and our moon are both slow rotators in this respect.

  25. br says:

    OK, I’ve written my own sim, from scratch in Matlab. This divides a band of moon into segments of a certain latitude/longitude angular width, with a depth, and works out the heat flows between segments laterally and radially going internal into the moon. Solar irradiation with angle, and SB emission/emissivity are taken into account for the surface layer.

    I get very similar results to you guys, average T approx 200 K, give or take, and the temperature with depth is the same as the average surface temperature (not 40 K hotter, as the actual moon seems to be).

    However, when I speed up the rotation of the moon from a 28 (Earth) day period to a 0.1 day period, the extremes in temperature converge and the average temperature at the equator increases to 291 K! Temperature towards the poles is lower, but also fairly flat, somewhere below 200 K depending on latitude. These faster rotation results don’t seem to depend so much on regolith properties. I’ll work out a global average shortly.

    Now I’ve done that, I’m not sure what it means, but I’m trying to get a handle on what an expected 255 K average temperature means.

    Where can I find the actual Diviner data the you guys have been plotting? It wasn’t obvious from a quick search.

  26. br says:

    Hi Ben,

    I posted before reading your comment. I’ll chew on that for a bit.

    One phrase jumps out: “Earth and our moon are both slow rotators in this respect”. But the Earth has much lower day/night temperature variation than the moon, so presumably should be much closer to its ‘expected’ temperature (=infinite spin rate?).

    Hence the fact that the moon seems to have a much lower ‘real’ temperature than its ‘expected’ temperature actually doesn’t make any difference as far as Earth is concerned. The moon can have a ‘bare’ ‘real’ temperature of 200 K at the equator, while the Earth could have a ‘bare’ ‘real’ temperature of 250 K?

  27. Ben Wouters says:

    br says: April 23, 2014 at 4:36 pm

    Have a look here: http://bartonpaullevenson.com/Albedos.html

    Effective temperature for Earth is ~255K, for the moon ~270K.
    Actual ~290K vs ~197K. Something seriously wrong obviously.

  28. tchannon says:

    The Apollo report mentions an excess of radioactive material. They also mention dodgy probes.

    So far I can find no information on the water walking implicit in recent papers.

    br, looks like you are following the same path through the same series of issues. I think it is better I say little, see where you get without being led.

    Data? You can’t find it because it isn’t there. Welcome to science.
    Admin magic should appear shortly.

  29. br says:

    Hi all,

    I put my model through its paces and came up with the linked presentation: http://www.slideshare.net/brslides/lunar-vs-earthtemperature
    [moderator adds inline image]
    Slide 10

    Slide 10 is probably the most important. I find that the global temperature of a body depends on its temperature difference between day and night. For the moon this is 280 K, and gives it an average temperature much less than its effective temperature. For the Earth, the day/night variation is only about 20 K, which makes almost no difference to its average/effective temperature.

    Basically, the moon is a ‘slow rotator’ while the Earth is a ‘fast rotator’. The rate of rotation does indeed make a big difference to the average temperature.

    Comments?

  30. br says:

    Hi Tim,

    You wrote ‘I did speed it up. That is not an answer.’

    I think there must be some shortcoming of the simulation there, unless I misunderstand you – a fast spin speed should bring one up to the effective temperature (my model showed this, as it should).

    Did you post the data for that around here? The closest ones I found were for temperature vs depth, but that is not the same thing.

    p.s. Thanks for sending on the Diviner data, really useful! Strange thing is, the ‘lat00’ plot at http://diviner.ucla.edu/science.html is subtly different! No idea why…, that website does seem quite messy.

  31. tchannon says:

    When I did the original work I did play around with amongst other things the rotation rate. As far as I can remember I thought it _did_ have a profound effect and said as much since.

    Year or so later I was challenged, something like that so went back and looked again and reached a different conclusion.

    Gallopingcamel seems to have reached the little effect conclusion.

    Okay, grin, these things come to pass.

    One of what I thought were more important effects came with increasing surface conductance spin rate, leads to a mean maximum around earth mean but writing about this is asking for a deluge of blog commenters doing mental jumps to human head height, not what it is about so I have kept away.

  32. Ben Wouters says:

    br says: April 25, 2014 at 3:51 pm

    “p.s. Thanks for sending on the Diviner data, really useful! Strange thing is, the ‘lat00′ plot at http://diviner.ucla.edu/science.html is subtly different! No idea why…, that website does seem quite messy.”

    Seems we have 2 versions of that page:
    http://www.diviner.ucla.edu/science.shtml
    and
    http://diviner.ucla.edu/science.html

    The second doesn’t have a “Home” page (yet?), but the data on that page hides the very low lunar temperatures.
    Perhaps better to make a copy of the first version before it “disappears”.

  33. Ben Wouters says:

    br says: April 25, 2014 at 3:38 pm

    Hi br

    thanks for your work on this.

    I’m not sure what you mean with ” I find that the global temperature of a body depends on its temperature difference between day and night.”
    Sounds like you use the day/night temp difference as an input?

    It would be useful imo to have a baseline “model” using pure black (actually grey) body behaviour with a given TSI and albedo. Plots could be average temperature vs rotational speed.

    My simple greybody calculation gives 161K average for albedo .11 (no rotation)
    Expect the result to be 5-10K lower with better calculations, like N&Z did.

    Rotation once every second should come out 10-20K below the effective temperature, since the difference between equator and poles will still be there.

  34. Home at last! Now to try to understand br’s plots and explore the question of how much effect the rate of rotation has.

    Initially, I found the Moon’s equatorial temperature to be insensitive to the rate of rotation. One of my dumb mistakes was to work in radians rather than in degrees so my model was rotating 57.3 times too slowly! Even so the plots came out very close to observations and Tim’s model.

    I will re-examine my calculations on the lunar equatorial temperatures (Tmax, Tave & Tmin) with different rates of rotation. Thus far my model has agreed with Tim’s conclusion of a “Small Effect”. I hope we can nail this down to everyone’s satisfaction!

  35. Ben Wouters,
    I followed your first link and found something that puzzles me:
    http://www.diviner.ucla.edu/science.shtml

    How can the night temperatures at Lat89 be higher than at Lat75?

    Could this be a “Deliberate Mistake” to see if anyone is paying attention!

  36. Ben Wouters says:

    gallopingcamel says: April 26, 2014 at 4:04 pm

    “How can the night temperatures at Lat89 be higher than at Lat75?”

    That’s midsummer for the hemisphere in this plot. Moons axis has ~1,5 degree tilt, so lat 89 will have a midsummer sun of sorts 😉
    In wintertime lat89 temps approach 50K for many days in a row.

  37. Ben Wouters says:

    Question about the 3 models;
    how is the average temperature calculated?
    You can calculate the solar radiation for the various latitudes, but we have more square metres between equator and lat10 than between lat80 and the pole.
    Is this taken into account?

  38. Ben,

    Thus far I have only calculated for Lat00 so the average is simply the sum of all temperatures plotted divided by the number of points.

    If I was to calculate an average temperature for the entire Moon the integrals for each latitude would be weighted according to the areas of each slice.

  39. br says:

    Hi all,

    I copied Tim’s Spice circuit, rewriting it in (free!) LTSpice. I emailed this to Tim, so anyone who wants it can simply ask and he’ll send it on. I didn’t spend long on tweaking the RC time constants, but it looks quite good from a distance. I also have a ‘latitude’ parameter if anyone wants to see what that does.

    Most importantly, perhaps, I think there is a mistake in Tim’s placement of the surface SB emission current sink. This should come *after* the first RC time constant, which represents the surface temperature. Otherwise the surface seems to have an infinite speed response, and the result won’t depend on spin rate. This is not right, as all thermal masses have a thermal time constant, and this can only been seen at the output of an RC, not the *input*.

    Increasing the spin rate (frequency of the sine wave voltage source) now converges the temperature to the effective temperature. The *equatorial* average temperature is about 285 K, but the global average temperature is near 270 K, as the colder poles bring that number down.

    If Tim approves of the repositioning of the SB emission sink, then his and my model now agree.

  40. br says:

    Ben:
    ‘how is the average temperature calculated?
    You can calculate the solar radiation for the various latitudes, but we have more square metres between equator and lat10 than between lat80 and the pole.’

    Yes, I take the area into account. The distance between latitudes stays the same, but the distance between longitudes goes down as cos(theta), so the area goes down by cos(theta). One can then take the weighted average.

  41. br says:

    Ben:
    ‘Sounds like you use the day/night temp difference as an input?’
    Not really. I changed the spin speed, then looked at what the day/night difference was and plotted mean global temperature against that. I have an intuition that day/night difference is much more fundamental than spin speed because of the way the SB emission works, and the way effective temperature is calculated.

    ‘It would be useful imo to have a baseline “model” using pure black (actually grey) body behaviour with a given TSI and albedo. Plots could be average temperature vs rotational speed.’
    The problem is that one needs surface thermal time constants. The moon has very small thermal mass with bad conduction on the surface, whereas Earth’s surface seems to have a much slower thermal time constant, as presumably a greater mass is effected by radiation (the ocean’s have relatively deep solar penetration, for example).

    ‘My simple greybody calculation gives 161K average for albedo .11 (no rotation)
    Expect the result to be 5-10K lower with better calculations, like N&Z did.’
    I’m not sure what the purpose of a no rotation calculation is – why do N&Z look at this? For sure it will come out lower than the effective temperature, but what will it tell you?

    ‘Rotation once every second should come out 10-20K below the effective temperature, since the difference between equator and poles will still be there.’
    if you get below the effective temperature, I would suggest speeding up the rotation! My calculations, at the ‘slideshare’ link above, get the equatorial temperature above the effective temperature, and the poles below the effective temperature. The average comes out bang on the effective temperature (assuming Lambertian absorption and small day/night variation).

  42. Ben Wouters says:

    br says: April 26, 2014 at 9:47 pm

    “I’m not sure what the purpose of a no rotation calculation is”

    The no-rotation case is imo the case that should be used as the baseline radiation only temperature iso the effective temperature (Te).
    see http://bartonpaullevenson.com/Albedos.html at the end:
    “For planets with A) no atmosphere, and B) no internal heat source, and C) surface emissivity equal to 1, effective temperature is the whole story.”
    All these effects all will increase the actual temperature ABOVE the Te, yet the moon is actually ~73K BELOW its Te.
    A realistic Te (Ter) should have taken into account the fact that a sun can only warm one hemisphere at a time. For the moon this results in a Ter of 161K (simple method) or even slightly lower.
    On top of the Ter we can now see the warming effect of rotation together with heat storage, an atmosphere etc. etc.
    The Te is based on spreading incoming solar equally over the entire sphere. This is physically not possible, but results in the highest possible temperature using radiation only, due to the fourth power in SB. Physically possible is a rotation of once every second. This spreads the incoming solar equally over the circumference of the sphere, but the difference between equator and poles remains, so even with this extreme rotation the Te will not be reached.
    On the scale between no rotation and one rotation every second the rotation speeds of earth and our moon are close to the no rotation end, so I see them as slow rotators, and expect the effect of the rotation on the radiation only temperature (greybody) to be minimal.

    To me the Ter for earth is ~151K, so we have to explain why our average surface temperature (~290K) is almost 140K higher. And yes, I do have an explanation 😉

  43. br says:

    Ben:
    ‘A realistic Te (Ter) should have taken into account the fact that a sun can only warm one hemisphere at a time.’
    I agree, which is why I did the numbers and posted them at http://www.slideshare.net/brslides/lunar-vs-earthtemperature

    ‘The Te is based on spreading incoming solar equally over the entire sphere. This is physically not possible’
    I agree

    ‘This spreads the incoming solar equally over the circumference of the sphere, but the difference between equator and poles remains, so even with this extreme rotation the Te will not be reached.’
    Incorrect. See my presentation. The global average reaches Te so long as the difference between day/night temperatures is below about 50 K.

    ‘On the scale between no rotation and one rotation every second the rotation speeds of earth and our moon are close to the no rotation end, so I see them as slow rotators’
    Incorrect. The moon is a fast slow rotator, the Earth is a slow fast rotator. There is a balance between surface thermal time constant and rotation rate. The Earth not only spins faster than the moon but also has a longer surface thermal time constant, as evidenced by the fact that the temperature difference between day and night is globally less than 10 K (I think NASA puts it as low as 5 K). Certainly much less than 50 K. In this case, Te applies.
    [mod edited]

  44. br says:

    gallopingcamel:

    Reading around this topic a little more, I came across your post at http://diggingintheclay.wordpress.com/2014/03/23/challenging-arrhenius-again/

    You said:
    ‘The great thing about models is that you can do crazy things so I increased the rate of rotation by a factor of 100,000. This raised the average temperature to 290 Kelvin.’

    290 K is approx correct! The equatorial average temperature is higher than the Te of 270 K. The colder poles are lower, and the global average will come out at 270 K.

    Keep going!

  45. br says:

    I said:
    ‘The moon is a fast rotator, the Earth is a slow rotator.’

    grrr! obviously I meant that the other way around…
    [mod: corrected]

  46. Ben Wouters says:

    br says: April 27, 2014 at 10:44 am

    “‘This spreads the incoming solar equally over the circumference of the sphere, but the difference between equator and poles remains, so even with this extreme rotation the Te will not be reached.’
    Incorrect. See my presentation. The global average reaches Te so long as the difference between day/night temperatures is below about 50 K”

    I did see your presentation, and unless I miss something it is about our physically existing moon. The Te and my Ter are about idealized blackbody calculations, using ao emissivity 1.0 and no heat storage/distribution. I don’t see how in that case a temperature above the Te can be reached with an uneven distribution of incoming radiation. The Te is the absolute maximum for radiative balance temperatures, using physically impossible equal distribution of incoming radiation around the whole sphere.
    For real, physical bodies the whole game changes of course.

    I’m convinced the Ter should be used as the “base” temperature, on which real world effects will INCREASE the temperature, not DECREASE it as with our moon from 270K to 197K.

    The no rotation case (actually one rotation / orbit) will be the situation for our earth sometimes in the not so near future. We already slowed down from once every 6 hrs (from memory) to once every 24 hrs.
    Our moon is a no rotation case relative to the earth.

  47. tchannon says:

    Ben
    “Question about the 3 models;
    how is the average temperature calculated?”

    I assume we are all doing…

    Weighted average.
    http://mathworld.wolfram.com/WeightedMean.html

    In the simplistic case weight is cosine(latitude).
    In this case the mean for a latitude is computed first, then weighting is done. Same maths for gridded to global on earth.

  48. br says:

    Ben:
    ‘The Te and my Ter are about idealized blackbody calculations, using ao emissivity 1.0 and no heat storage/distribution.’
    ah, ok. So I just did this idealised calculation for myself. For a stationary, non-thermally-conducting sphere at 1 AU I get:

    Albedo 0 (blackbody): 157.8 K
    Albedo 0.11 (moon): 153.3 K
    Albedo 0.306 (Earth): 144.0K

    These sound the same as N&Z, and close to the values you got.

    ‘I’m convinced the Ter should be used as the “base” temperature, on which real world effects will INCREASE the temperature’
    In one sense you are right, these are the lowest possible global temperatures for such bodies.

    The real world effects, *without atmosphere* that increase these temperatures are surface heat storage and spin rate. Including these effects, but leaving out the atmosphere, I get:

    Albedo 0.11 (moon): 199 K
    Albedo 0.306 (Earth): 254.5 K

    The value for the moon is within 2 K of its measured value, close enough for a simple model. The reason the Earth’s value is higher than the moon, despite the worse albedo, is due to the increased heat storage of the surface and its faster spin rate.

    The effect of the atmosphere will be on top of that.

    Happy?

  49. br says:

    tchannon:
    ‘In the simplistic case weight is cosine(latitude).’
    yes, that works for a sphere.

  50. Ben Wouters says:

    tchannon says: April 27, 2014 at 4:06 pm

    Thanks Tim. I expected it had been taken care of, just wanted to make sure 😉

  51. J Martin says:

    http://www.tak2000.com/data/Satellite_TC.pdf

    Don’t know if parts of this pdf may be of interest. I was pointed towards slide 32 in particular.

    “See slide 32 compare sphere shape black body and sandblasted Al both equilibrate at 5degC.”

    http://judithcurry.com/2014/04/11/curry-versus-trenberth/#comment-520895

  52. Ben Wouters says:

    br says: April 27, 2014 at 6:30 pm

    “Albedo 0 (blackbody): 157.8 K
    Albedo 0.11 (moon): 153.3 K
    Albedo 0.306 (Earth): 144.0K

    These sound the same as N&Z, and close to the values you got.”

    Absolutely. I used a very simple calculation, knowing the result would be slightly too high.
    Good enough for a rule of thumb calculation.
    http://www.principia-scientific.org/moons-hidden-message.html

    “The real world effects, *without atmosphere* that increase these temperatures are surface heat storage and spin rate. ”
    Don’t forget geothermal heat, the major factor on earth imo.

    “Happy? ”
    If we can get the idea across that the 255K Te for earth is total nonsense and a Ter of ~151K (or slightly lower) should be used as base temperature to arrive at the actual temperature by including the above mentioned effects, yes, I would be happy.
    Focusses the discussion on an explanation for the ~140K higher temperature on earth than the Ter indicates. No chance the atmosphere alone can achieve this.

  53. Given that there are several models at our disposal that agree with Diviner it will be interesting to find out whether they still agree with different rates of rotation!

    I am hoping Tim will sort it all out in “Part 2”.

  54. br says:

    Ben Wouters says: April 27, 2014 at 10:22 pm

    Glad we agree on the ‘low’ temperatures, and I am happy to take these as ‘lowest baseline’ temperatures.

    ‘Focusses the discussion on an explanation for the ~140K higher temperature on earth than the Ter indicates. No chance the atmosphere alone can achieve this.’

    I also agree, there is no chance the atmosphere alone achieves this.

    The thing is, I just calculated that the thermal storage and spin rate of the Earth bring the temperature up from 144.0 K to 254.5 K. As the measured average temperature is 288 K, then the atmosphere+geothermal flux need to only bring up the remaining 33.5 K. That this is the same as current climate models may be a lucky coincidence, but I find that they are not wrong.

    Geothermal flux can be seen here: http://geophysics.ou.edu/geomechanics/notes/heatflow/global_heat_flow.htm
    and Wikipedia quotes the average flux as: http://en.wikipedia.org/wiki/Geothermal_gradient#Heat_flow
    ‘This is 0.087 watt/square meter on average (0.03 percent of solar power absorbed by the Earth[14] )’
    which places it in the ‘negligible’ category as far as global averages go. That it affects deep ocean temperature is granted, but not that it contributes significantly to global average. It sounds like you don’t accept these values, but they would have to be wildly different to make a significant contribution.

    I was pleasantly surprised that the moon’s average temperature is *not* equal to its Te, but I guess we both agree that it is above it’s Ter of 153 K. So how do you explain that? If you agree that it is due to surface storage and spin rate, then I think you will eventually have to accept that the same effects on Earth bring its temperature up to 254.5 K. If not, why not?

  55. br says:

    For those interested in Spice equivalent models, here is what I came up with. This copies Tim’s one at https://tallbloke.wordpress.com/2012/04/02/a-model-of-lunar-temperature/ , so all dues to him for putting in the hard work of sorting out what goes where. The code below should be copy-pasted into a ‘name.asc’ file, which can then be loaded and run in LTSpice, which is free to download here: http://www.linear.com/designtools/software/#LTspice

    As I said above, I didn’t spend too long tweaking the RC time constants, but the voltage out at the node labelled ‘Tmoon’ (voltage=temperature of the moon surface in Kelvin) looked close enough. If anyone wants to do more tweaking, feel free.

    There is a ‘.param latitude=0’ text box at lower left. You can plug a value in here, from 0 to 90 in degrees, to get the temperature profile at a different latitude.

    To get a global average takes a bit of manipulating of the output trace. I guess the thing to do is choose a latitude, then save the V(Tmoon) data, then average the last few cycles to get the average T at that latitude. Then change the latitude, and get the average at a different latitude. If you choose 5,15,25, …, 85 degree latitudes, then multiply the ave T at each latitude by cos(theta), with theta=5, 15, 25, …, add up all those values then divide the total by (cos(5)+cos(15)+cos(25)+…). That’s a bit of work, but I think should be right. The answer to that average is then the *global* average.

    Spin speed is set by right clicking the leftmost voltage source labelled ‘LunarRotationRate’. At the moment the frequency of its sine wave is 0.1 Hz (as Tim points out, all values simply scale, so not too important that it is not ‘absolute’ frequency). To speed up the rotation, set a value of 10 Hz or 1000 Hz, etc, as this represents cycles per second.

    If anyone gets this working, it would be nice to hear from them!

    Code follows: {moderator: better code posted in a later comment –Tim}

    Version 4
    SHEET 1 1760 680
    WIRE 688 -96 0 -96
    WIRE 352 -16 160 -16
    WIRE 400 -16 352 -16
    WIRE 352 0 352 -16
    WIRE 352 48 336 48
    WIRE 688 96 688 -96
    WIRE 640 112 544 112
    WIRE 0 128 0 -96
    WIRE 160 128 160 -16
    WIRE 400 192 400 64
    WIRE 688 224 688 176
    WIRE 832 224 688 224
    WIRE 976 224 912 224
    WIRE 1104 224 1104 160
    WIRE 1104 224 976 224
    WIRE 1248 224 1104 224
    WIRE 1360 224 1360 160
    WIRE 1360 224 1328 224
    WIRE 1408 224 1360 224
    WIRE 1520 224 1520 160
    WIRE 1520 224 1488 224
    WIRE 1104 256 1104 224
    WIRE 976 272 976 224
    WIRE 1360 272 1360 224
    WIRE 1520 272 1520 224
    WIRE 400 304 400 192
    WIRE 0 400 0 208
    WIRE 80 400 0 400
    WIRE 160 400 160 208
    WIRE 160 400 80 400
    WIRE 336 400 336 48
    WIRE 336 400 160 400
    WIRE 400 400 400 384
    WIRE 400 400 336 400
    WIRE 544 400 544 192
    WIRE 544 400 400 400
    WIRE 640 400 640 160
    WIRE 640 400 544 400
    WIRE 976 400 976 336
    WIRE 976 400 640 400
    WIRE 1104 400 1104 336
    WIRE 1104 400 976 400
    WIRE 1360 400 1360 336
    WIRE 1360 400 1104 400
    WIRE 1520 400 1520 336
    WIRE 1520 400 1360 400
    WIRE 80 416 80 400
    FLAG 80 416 0
    FLAG 1104 160 Tsurface
    FLAG 400 192 HalfSine
    FLAG 1360 160 Tshallow
    FLAG 1520 160 Tdeeper
    SYMBOL voltage 160 112 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -48 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName LunarRotationRate
    SYMATTR Value SINE(0 1 0.1)
    SYMBOL res 384 288 R0
    SYMATTR InstName R1
    SYMATTR Value 1e6
    SYMBOL sw 400 80 M180
    WINDOW 3 24 -12 Invisible 2
    WINDOW 0 39 56 Left 2
    SYMATTR Value MYSW
    SYMATTR InstName DayNightSwitch
    SYMBOL g2 688 192 M180
    WINDOW 0 44 58 Left 2
    SYMATTR InstName AbsorbedFlux
    SYMATTR Value 1213
    SYMBOL res 928 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R2
    SYMATTR Value 10
    SYMBOL bi 1104 256 R0
    WINDOW 0 37 37 Left 2
    WINDOW 3 24 80 Invisible 2
    SYMATTR InstName SBemission
    SYMATTR Value I=0.98*(5.67e-8)*pwr(V(Tsurface),4)
    SYMBOL cap 960 272 R0
    SYMATTR InstName C1
    SYMATTR Value 0.133
    SYMBOL cap 1344 272 R0
    SYMATTR InstName C2
    SYMATTR Value 0.233
    SYMBOL res 1344 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R3
    SYMATTR Value 10
    SYMBOL res -16 112 R0
    SYMATTR InstName R7
    SYMATTR Value 1
    SYMBOL cap 1504 272 R0
    SYMATTR InstName C3
    SYMATTR Value 0.33
    SYMBOL res 1504 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R4
    SYMATTR Value 100
    SYMBOL bv 544 96 R0
    WINDOW 0 35 40 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName Sun
    SYMATTR Value V=pwr(V(HalfSine),1.3)*cos(latitude*pi/180)
    TEXT -80 512 Left 2 !.tran 500
    TEXT -80 568 Left 2 !.model MYSW SW(Ron=0.001 Roff=1000Meg Vt=0 Vh=0)
    TEXT -80 616 Left 2 !.ic V(Tmoon)=210 V(Tshallow)=210 V(Tdeeper)=210
    TEXT -80 464 Left 2 !.param latitude=0
    
  56. br says:

    small comment: I think I relabelled ‘Tmoon’ as ‘Tsurface’. Whatever, it should still work.

  57. tchannon says:

    The source loads and runs.

    In private we have a disagreement over a detail difference between how I connected SB and how br connects SB. Lets leave that as an open end.

    Trying to handle latitude is not quite trivial which is a warning. Lambertian applies to a sphere so latitude ought to hand this. However, emission faces space not the sun so this does not apply in the same way. (if you can figure what I mean)

    When I did latitude using different software a massive array was read from output file into a spreadsheet where post processing could be done.

  58. I thought about compensating for the fact that the sun is not a point source (0.53 degrees) but found it made little difference given that my FEA was set to steps almost twice as large.

    I downloaded the Windows version of LTSPICE. It installed flawlessly in Linx Mint 14 (MATE version). I ran br’s file and got this error message:

    Circuit: * Z:\home\p\PHM\CORRGEN\Climate\Tallbloke\br.asc

    Unknown node in .ic card: “tmoon” — ignoring
    WARNING: Node TDEEPER is floating.

    Direct Newton iteration for .op point succeeded.

    Date: Mon Apr 28 12:41:19 2014
    Total elapsed time: 36.079 seconds.

    tnom = 27
    temp = 27
    method = modified trap
    totiter = 1759324
    traniter = 1759317
    tranpoints = 413112
    accept = 309858
    rejected = 103254
    matrix size = 10
    fillins = 0
    solver = Normal
    Matrix Compiler1: 346 bytes object code size 1.3/1.1/[1.0]
    Matrix Compiler2: 580 bytes object code size 1.1/2.1/[1.0]

  59. The SPICE simulations have advantages compared to my FEA in terms of tweaking the variables. Perhaps we should regard the FEA approach as a way of validating the SPICE which appears to be more accurate and easy to use.

  60. tchannon says:

    I’m less lucky in Windows so maybe it is ported from ‘nix, Windows front end is stone age on install.

    1. Installed fine and then ran fine.
    2. Exit.
    3. Can’t find where it has hidden itself.
    4. Do a system wide search.

    Good grief, it has put an .ini file in an operating system directory. (ie. doesn’t know how to find anything so play dumb as Windows 3)
    Ignore, doesn’t matter as such provided uninstall removes it.

    Turns out it installed under “Program Files/LTC/LTspiceIV/” an unnecessary directory layer.

    And the executable is named… as expected… not
    scad3.exe

    And no links from user places. (not that I can see)

    An option if all users need access is create a shortcut (in the directory with scad3), rename as you wish. Then find the “All Users” directory and in there “Start Menu”, drag the shortcut there (in a subdir there if you want). Snag is you might need admin login and the OS will try to hide those directories from you.
    Or to the menu for you only.

    Or your desktop or pin or whatever.

  61. br says:

    gallopingcamel says:April 28, 2014 at 5:36 pm

    I see the code at line 113 reads
    ‘TEXT -80 616 Left 2 !.ic V(Tmoon)=210 V(Tshallow)=210 V(Tdeeper)=210’
    whereas it should be
    ‘TEXT -80 616 Left 2 !.ic V(Tsurface)=210 V(Tshallow)=210 V(Tdeeper)=210’

    So you can change that (sorry, my version of LTSpice didn’t pick this up, but this is ignorable as ‘.ic’ just means ‘initial condition’, which is optional. Better to change it though,either in the .asc file or on the LTSpice layout window. You are also free to set these values to 0 or delete the line altogether).

    The other warning
    ‘WARNING: Node TDEEPER is floating.’
    sounds wrong, the wires don’t connect for some reason (again, it worked for me!). I presume you can fix this by drawing wires on your layout window just to make sure everything connects around there.

    So despite those warnings, it sounded like it produced an output. Did it look ok?

    I agree Spice may be easier to use, but would be wary of saying it is more accurate. After all, it only has a few nodes (the RC elements, you can add more if you want), whereas Quickfield presumably has a mesh and ‘real’ material parameters. Still, I expect the two methods will give very similar outputs if tweaked properly. It is also good to keep both methods going, as we still have to agree on what happens with lunar spin rate. Your fast spin result in Quickfield of 290 K sounds correct for the equator!!! Running the Spice model I posted at 1000 Hz gives 285.3 K for the equator, using the cos(theta)^1.3 Lambertian factor, or 289.9 K using the straight Lambertian cos(theta), which corresponds almost exactly to your Quickfield result. As I noted above, the poles will be colder, and in my Matlab model brings the global average down to 270 K. I expect the Spice model to do the same, though I haven’t actually checked that. So it seems like the sims are converging!

  62. Ben Wouters says:

    br says: April 28, 2014 at 10:03 am

    “The thing is, I just calculated that the thermal storage and spin rate of the Earth bring the temperature up from 144.0 K to 254.5 K”
    I find it hard to believe this is correct. The Te for earth (255K) is the HIGHEST achievable radiative balance temperature given albedo 0,3. This is only achievable when the surface temperature is equal over the whole sphere. I’d like to see your input parameters for the earth situation that can create a max. radiative temperature with large temperature differences between equator and poles and also marked day/night differences.

    My ideas about the reason for our average temperature being ~140K above the Ter are posted here: https://tallbloke.wordpress.com/2014/03/03/ben-wouters-influence-of-geothermal-heat-on-past-and-present-climate/

    “I was pleasantly surprised that the moon’s average temperature is *not* equal to its Te, but I guess we both agree that it is above it’s Ter of 153 K. So how do you explain that?”
    Why “pleasantly” surprised ? 😉
    My explanation is part geothermal, part surface storage /spin rate.

    Assuming the same geothermal flux as for ocean floor: 100 mW/m^2.
    Moon in outer space will have an equal surface temp. of ~40K (SB)
    Moon now in earth like orbit around the sun, one rotation/orbit.
    Sunny side goes to ~322K average temperature. The 40K for the dark side will now add 20K to the average. So a small geothermal flux will have a significant effect in this case.
    So imo ~25K of moons actual temperature is due to geothermal flux.
    Also in reality we see the 89lat winter temperature approach 50K when over 100 days without sun.

    Other effect is that due to the geoflux we also have a temp.gradient towards the core.
    On the dark side the surface temp (40K) is the starting point, and going towards the core the temperature will increase every km. On earth the value is ~25K/km.
    Assuming the same for our moon 1km deep the temp will be 65K.

    On the equator the highest surface temp. will be ~380K continuously. Due to the flux eventually the temperature gradient will have adjusted to the actual surface temp., and the temp. at 1km deep will be ~405K. (assuming no thermal transfer laterally)
    Starting rotation now the geo gradient start temp. for each latitude will settle around the average surface temperature there. So with an average surface temperature for the equator of 200K, the 1km deep temp will be 225K.

  63. tchannon says:

    The warning is normal and correct. It may or may not be benign such as erratic failure to stabilise starting conditions.

    There is no resistive path whatsoever between part of the circuit and ground.

    Wot yer do is add in a resistor set to an extremely high value so it has negligible effect. 1Tohm or whatever. Try left hand end R2 to ground.

    If really necessary there are trick ways to reduce the effect further.

  64. Tim & br,

    I will take the evasive action recommended just to kill the error message but the output temperature plot is exactly as expected.

    Fifteen years ago I was using MicroSim to optimize 100 MW Pulse Forming Networks. Now I am a little rusty so your help is greatly appreciated.

    The models seem to be consistent when it comes to the Moon’s equatorial temperature. How about independently calculating the following:
    1. The Tmax, Tave and Tmin averaged over the entire surface of the Moon.
    2. Repeat the above for a Moon rotating once every 24 hours.

    How about slicing the Moon up into strips each 10 degrees of latitude wide so that our numbers can be directly compared to Diviner?

  65. tchannon says:

    Perhaps us two are rusty, I’ve not done heavy work for much the same time, br is fairly current why he made short work of implementation in the ltspice for which I am very grateful.

    A number of things can be done given time and patience.

    Diviner data poses a problem. Best guess is having to deduce where level 2 data files exist and then write code for decoding. All obfuscated. Could be the data is too large for the meagre resources I have here.

    Previously I computed a mean lunar and this more or less matched the Diviner figure. Seemed about right so I left it there.

  66. gallopingcamel says:

    Tim,

    My spreadsheet choked on the Diviner “Level 3” data!

    The only way I have to compare my model to observations is the Diviner data that you sent me.

  67. gallopingcamel says:

    Using the FEA approach I can adjust the emissivity to correctly model the non-Lambertian emissivity of the Moon at ten (if that is the number of slices we agree) latitudes.

    Likewise with the Albedo.

    Thus when it comes to an average temperature for the entire Moon I expect the FEA approach to be closer to observations than was achieved for Latitude “00”.

  68. br says:

    Hi all,

    As Tim said, offline we were having a disagreement about where to place the SB emission in the equivalent circuit. So I checked it out, and found… we were both wrong!

    See Slide 57 of this presentation: http://www.cs.virginia.edu/~skadron/thermal_tutorial_coolchips08.pdf

    This shows how to do the RC equivalent model in the presence of a power source. The resistor goes in parallel with the capacitor, and they give the equations to prove it.

    This resolves our dispute, as now there is only one node where the R, C, and input power meet, so the SB emission has to go there too.

    The rest of the ‘deeper’ RC elements are ok, as they are driven by temperature (=voltage), so the usual RC connection applies.

    So I updated my Spice code, see below, and while I was at it should have removed the two warnings that GC got.

    Version 4
    SHEET 1 1760 680
    WIRE 688 -96 0 -96
    WIRE 352 -16 160 -16
    WIRE 400 -16 352 -16
    WIRE 352 0 352 -16
    WIRE 352 48 336 48
    WIRE 688 96 688 -96
    WIRE 640 112 544 112
    WIRE 160 128 160 -16
    WIRE 400 192 400 64
    WIRE 688 224 688 176
    WIRE 848 224 688 224
    WIRE 1104 224 1104 160
    WIRE 1104 224 848 224
    WIRE 1200 224 1104 224
    WIRE 1248 224 1200 224
    WIRE 1360 224 1360 160
    WIRE 1360 224 1328 224
    WIRE 1408 224 1360 224
    WIRE 1520 224 1520 160
    WIRE 1520 224 1488 224
    WIRE 1584 224 1520 224
    WIRE 1696 224 1696 160
    WIRE 1696 224 1664 224
    WIRE 1104 256 1104 224
    WIRE 848 272 848 224
    WIRE 1200 272 1200 224
    WIRE 1360 272 1360 224
    WIRE 1520 272 1520 224
    WIRE 1696 272 1696 224
    WIRE 400 304 400 192
    WIRE 0 400 0 -96
    WIRE 80 400 0 400
    WIRE 160 400 160 208
    WIRE 160 400 80 400
    WIRE 336 400 336 48
    WIRE 336 400 160 400
    WIRE 400 400 400 384
    WIRE 400 400 336 400
    WIRE 544 400 544 192
    WIRE 544 400 400 400
    WIRE 640 400 640 160
    WIRE 640 400 544 400
    WIRE 848 400 848 352
    WIRE 848 400 640 400
    WIRE 1104 400 1104 336
    WIRE 1104 400 848 400
    WIRE 1200 400 1200 336
    WIRE 1200 400 1104 400
    WIRE 1360 400 1360 336
    WIRE 1360 400 1200 400
    WIRE 1520 400 1520 336
    WIRE 1520 400 1360 400
    WIRE 1696 400 1696 336
    WIRE 1696 400 1520 400
    WIRE 80 416 80 400
    FLAG 80 416 0
    FLAG 1104 160 Tsurface
    FLAG 400 192 HalfSine
    FLAG 1360 160 Tshallow
    FLAG 1520 160 Tdeeper
    FLAG 1696 160 Tdeep
    SYMBOL voltage 160 112 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -48 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName LunarRotationRate
    SYMATTR Value SINE(0 1 0.1)
    SYMBOL res 384 288 R0
    SYMATTR InstName R1
    SYMATTR Value 1e6
    SYMBOL sw 400 80 M180
    WINDOW 3 24 -12 Invisible 2
    WINDOW 0 39 56 Left 2
    SYMATTR Value MYSW
    SYMATTR InstName DayNightSwitch
    SYMBOL g2 688 192 M180
    WINDOW 0 44 58 Left 2
    SYMATTR InstName AbsorbedFlux
    SYMATTR Value 1283
    SYMBOL res 1120 352 R180
    WINDOW 0 36 76 Left 2
    WINDOW 3 36 40 Left 2
    SYMATTR InstName R2
    SYMATTR Value 50
    SYMBOL bi 848 272 R0
    WINDOW 0 37 37 Left 2
    WINDOW 3 24 80 Invisible 2
    SYMATTR InstName SBemission
    SYMATTR Value I=0.98*(5.67e-8)*pwr(V(Tsurface),4)
    SYMBOL cap 1184 272 R0
    SYMATTR InstName C1
    SYMATTR Value 0.1
    SYMBOL cap 1344 272 R0
    SYMATTR InstName C2
    SYMATTR Value 0.25
    SYMBOL res 1344 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R3
    SYMATTR Value 7
    SYMBOL cap 1504 272 R0
    SYMATTR InstName C3
    SYMATTR Value 0.18
    SYMBOL res 1504 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R4
    SYMATTR Value 10
    SYMBOL bv 544 96 R0
    WINDOW 0 35 40 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName Sun
    SYMATTR Value V=pwr(V(HalfSine),1.3)*cos(latitude*pi/180)
    SYMBOL cap 1680 272 R0
    SYMATTR InstName C4
    SYMATTR Value 0.3
    SYMBOL res 1680 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R5
    SYMATTR Value 50
    TEXT -80 512 Left 2 !.tran 50
    TEXT -80 568 Left 2 !.model MYSW SW(Ron=0.001 Roff=1000Meg Vt=0 Vh=0)
    TEXT -80 616 Left 2 !.ic V(Tsurface)=210 V(Tshallow)=210 V(Tdeeper)=210 V(Tdeep)=210
    TEXT -80 464 Left 2 !.param latitude=0
    
  69. tchannon says:

    Egg on chips, oh yes, excellent first page.

    How the .. did you find that PDF!
    Good subject too as if it isn’t obvious I’ve had to handle thermal stuff. The doc is dated much later than when various manufacturers had messed up, failures and fires. Design stupidity but that came from management stupidity, sales wag reality, reality speaks up.
    I note it doesn’t get critical issues which still bite.
    Mentions Arrhenius Equation, good, we use that in design.

    Slide 57. Ah yes a transformation. This is always a bit mind bending. Equivalence. I think this is the same thing where you and I are looking from different perspectives.

    Hopefully we are all happy.

    Odd thing, the last serious design I did, quite some time ago involved constant current, thermal and optics in a small space. Funny how things appear again.

  70. br says:

    Ben Wouters says: April 28, 2014 at 8:48 pm

    ‘I’d like to see your input parameters for the earth situation that can create a max. radiative temperature with large temperature differences between equator and poles and also marked day/night differences.’

    The plots are on slide 8 of the slideshare I linked above.

    Things to note:
    The equatorial temperature profile (left hand figure) has an average of 267 K, which is above Te. The temperature variation, peak-minimum, is 22 K.

    It took a while to find any sort of reliable reference on what ithe day-night difference should be, but http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
    says:
    ‘Diurnal temperature range: 283 K to 293 K (10 to 20 C)’
    in other words, a global average of 288 K, with a peak-minimum difference of only 10 K, which is less than my model.

    Next to note is the plot on the right of my slide 8. This shows the average temperature near the poles to be 150 K, which compares to 267 K at the equator!

    So…

    The model contains a day-night temperature swing of 22 K, and an equatorial-polar temperature difference of 117 K.

    If you look at the title of the right hand figure, slide 8, you see the global average temperature for this case is 253.7 K.

    As the real day night swing is less than the one I used, I reduced it a bit in slide 9 and the global average increased to 254.5 K. The equatorial-polar difference is still over 100 K!

  71. tchannon says:

    I was looking through ltspice vs. what might be available.

    Possibly there are suitable delay lines in there which are equivalent to a thermal delay line. If we are lucky this can bring closer alignment, ie. easier to put in thermal properties, a homogeneous rock layer can be lumped.

    I don’t think this needs to be more than ball park unless inner detail matters a lot.

  72. br says:

    Hi all,

    Thought I’d let you know why I put the number ‘1283’ for absorbed flux. Start with solar flux of 1366 W/m2, we then have the albedo of the moon at 0.11, so one would expect 1366*(1-0.11) = 1216 W/m2 to be the absorbed flux.

    The problem is that the cos(theta)^1.3 which seems to be needed to model the daily profile properly, acts to reduce the absorbed flux, especially at higher angles. As the difference gets reflected, this effectively increases the albedo, which would then apparently be greater than 0.11, so disagreeing with the value we started with. To restore the total albedo to what it ‘should’ be, I performed an integral over cos()^1.3 and found that we need to add 5.58% back again. So the final number for flux is 1366*(1-0.11)*1.0558 = 1283, which now gives a total albedo of 0.11, but includes the cos()^1.3 absorption profile.

    I think that’s right, but reserve the right to change my mind if necessary 🙂

  73. tchannon says:

    Real world is messy but none of us want fudge factors at least without a good basis.

    Lets wind the clock back.
    One of the issues where I am long wary is albedo, a vague and abused concept. I won’t but could also point at equations where this is abused by politicals, such as only including half of the effect then declaring they are right about the answer.

    When I toyed with the moon I deliberately did not use or want albedo, would the expected result appear without this? Yes, which means the effect if any is encoded in the empirically determined values.

    From my point of view for any particular wavelength (or frequency) the radiative interaction cancels to unity, acceptance and emittance are the same. This also varies in little understood ways with wavelength whilst still unity.

    There is also the dreaded quantum world where gallopingcamel and br are both familiar. This in my experience lead to difficulty with discussions generally, lot of arguments about much the same thing. I tend to fall silent.

    Although an aside I’m interested in the long and far IR, hence the work on pyrgeometers and the con-trick played with these. Don’t actually measure what is claimed but add it on afterwards because we know.
    Total darkness, photons if they exist below ionising capability. Into cryogenics which rarely seem to get used and evidence, very little is concrete. What does Diviner actually sense, left as a rhetorical question.

    This lot loops back to what is going on with atmospheric water observed from below. I contend there is IR reflection, in effect a partial mirror as well as the claimed molecular effects. This used to be acknowledged but has fallen out of usage. I suppose too there is the deep water problem which quite a few on the Talkshop wonder about.

  74. tchannon says:

    Add the directive
    .step param latitude list 0 15 30 45 60 75 89

    The model needs additions.

    There is no measurement facility in the program so far as I can see so external post processing it needed. This difficult because forcing regular output times is not generally part of spice with transient analysis.

    Doing this portably for others is not feasible unless there is a common programming language or whatever. Worry about that later.

  75. br,

    That fix worked…..no more error messages!

    Thanks for that fascinating review of chip design issues. I was gobsmacked by the predicted 39.7 GHz clock considering that only forty years ago as project manager of the ICL ME29 mini-computer project I was struggling with a 39 MHz clock! There was 1.5 MB of DMA RAM and up to 16 hard drives each with 200 MB capacity.

  76. Tim,
    That latitude plot looks good! Do you a have a Tave calculated for it?

    I guess it is time to see how well the FEA model agrees.

  77. tchannon says:

    Just spent an hour swearing at the software author. I’d saved data to disk. Go to do this again, no way could I find any facility to save data. Classic idiocy, menu items change and in this case although the data is showing on screen the export item only appears if the plot window is foreground as far as the operating system is concerned. That is a warning for others.
    (file / export, only present if a plot is focus window)
    What X will do I have no idea, focus behaves very differently, usually focus is mouse pointer location but configurable.

    I’ve just done a hand linear interpolation of the output data, guessed time offset by eye and taken the common average (mean) for the regularly sampled interpolation output, zenith data 214.3K
    As a warning for the unwary, the same average on irregular data over the same time interval is 199K

    Awkwardness has appeared, output is truncating floats which messes up interpolation code.
    Have to think about this. No force plot step either, just minimum step size, different thing.

  78. tchannon says:

    Wrestling continues. won’t go into the dead ends of trying to use directives, it squirms out.

    You can get a mean but only for a single trace, not parametric.
    A starting point to figure how, hold down control and click the top centre located name of the trace, a popup should appear.

  79. br says:

    Hi all,

    I figured out the directives, after reading http://www.electro-tech-online.com/threads/how-to-measure-using-meas-with-ltspice.120640/ and playing about with stuff.

    The short story is to change .tran line to something like:

    .tran 0 100 80

    This runs the simulation up to t=100 seconds, but only records after t=80. It’s important to keep the difference between these two values equal to a whole number of periods, which for the sim above is 10 s. So this will record the last two periods.

    Then you can set ‘.meas’ directives. To measure average temperature over the two recorded periods, insert

    .meas Tave avg V(Tsurface)

    and just to make sue this has converged, add another one

    .meas Tdeeperave avg V(Tdeeper)

    If you like, you can add two more to measure max and min:

    .meas Tmax max V(Tsurface)
    .meas Tmin min V(Tsurface)

    so you now have four outputs. Finally, you can replace the ‘.param’ directive I had with the parametric sweep that Tim found:

    .step param latitude list 5 15 25 35 45 55 65 75 85

    and then run the simulation. I chose these values to divide the hemisphere into 10 degree steps, using the middle value as the typical value (so latitude 0-10 degrees is taken as latitude 5 degrees). With a voltage probe on Tsurface, it will look like nothing is happening for a few seconds, but when simulation time reaches 80 s it will start plotting the output, and then it will step the latitude and do it again. To read the measured outputs, you have to go to the ‘View’ menu and click on the ‘SPICE error log’. A bit strange that LTSpice puts the output here, but you should see all the values you want. If Tave and Tdeeperave are within about 0.5 K, then the results should be pretty good.

    You can then copy-paste the tables into a spreadsheet/analysis prog of your choice and perform the cosine averaging to get the global average. As mentioned above, this goes as:

    Tglobal=(Tave1*cos(5)+Tave2*cos(15)+…+Tave9*cos(85))/(cos(5)+cos(15)+…+cos(85))

    You may need to change degrees to radians so the cos works, and might even be possible in Spice, but hey presto! Global average temperature!

  80. br says:

    here is the closest I could get:

    Add two more .meas directives:

    .meas Tlatareavals avg Tave*cos(latitude)

    and

    .meas cosvals avg cos(latitude)

    Strangely, LTSpice takes values in degrees for the cos function here, while it takes them in radians elsewhere. Whatever. Now to get a global average, all you have to do is go to the SPICE error log, sum up all the Tlatareavals and divide by the sum of the cosvals.

    I don’t think LTSpice has a way to perform these sums over successive runs to do that automatically, but I would be delighted to see how! I tried a couple of ways without success, and seems not to be possible. They mention script files, but haven’ got that far.

  81. br,

    I think you just saved me some serious hand cranking of data files.

    On second thoughts I will do the hand cranking using the same intervals you used to see how close I can get to your model. Don’t spoil it by telling me your answer for Tave!

    IMHO the FEA should do better for the entire Moon (as opposed to the equator) because I will be able to use different values of emissivity for each latitude in accordance with its measured non-Lambertian properties. Likewise Albedo.

    Given my propensity for making dumb mistakes it could take several days as I have visitors this week!

  82. tchannon says:

    This will probably run a little hot. The power 1.3 modifier applied the latitude both directions if you see what I mean.

  83. tchannon says:

    Dumb mistakes drive the world.

  84. tchannon says:

    For clarity

    sum of products of weights and values, divided by sum of weights

    where weight is cosine(radians(latitude in degrees))

    Good enough
    http://en.wikipedia.org/wiki/Weighted_arithmetic_mean

  85. br says:

    Hi Tim,

    You were wondering where the factor of 1.3 in cos(theta)^1.3 might come from.

    It is well known in optics that reflection is a function of angle, and depends on refractive index and polarisation. See http://iopscience.iop.org/2040-8986/15/1/014014/article , eqns(14-15) for example.

    I wrote these equations up, and plotted what that does to absorption as a function of tilt for a 1 metre squared area. This is what I got:

    http://s1257.photobucket.com/user/brspics/media/Absorption_vs_angle_zpsc4af7fb8.png.html

    In this plot, a material with refractive index=1.8 is chosen, and unit power is incident on a 1 metre squared surface which is tilted at some angle given on the X axis. The blue line is the ‘viewing factor’ = cos(theta), the green line is the absorbed TM polarised light, the black line is absorbed unpolarised light, and the red line is cos(theta)^1.3.

    You will note that the TM line in green is quite close to the cos(theta)^1.3 in red. This doesn’t entirely explain where the 1.3 comes from, as last I heard the Sun gives unpolarised light, which should be absorbed according to the black line (TE+TM), which never really approaches the cos(theta)^1.3 line no matter what refractive index is chosen.

    The refractive index of 1.8 is not unreasonable, as according to this site http://www.reade.com/Products/Minerals_and_Ores/basalt.html has an index of 1.62, which gives a curve not so far away from the one I plotted (only slightly closer to the cos(theta) line), and who know what else is up there and what its spectral dependence is. The above plot is for a smooth surface, while there is surface roughness to consider, as you have noted. Presumably the roughness should bring one closer to the cos(theta) curve, so I don’t have a full explanation, but it is not a totally unjustifiable fudge.

  86. br says:

    tchannon says: April 30, 2014 at 3:25 pm
    ‘This will probably run a little hot. The power 1.3 modifier applied the latitude both directions if you see what I mean.’

    Albedo is something I know little about. A quick trip to our good friend Wikipedia 😉 http://en.wikipedia.org/wiki/Albedo found some interesting quotes:

    ‘The overall albedo of the Moon is around 0.12, but it is [b]strongly directional and non-Lambertian[/b], displaying also a strong opposition effect. Although such reflectance properties are different from those of any terrestrial terrains, they are [b]typical of the regolith surfaces[/b] of airless Solar System bodies.’
    So noticeably non-Lambertian, and interesting that they quote the albedo as 0.12, rather than 0.11, showing there may still be some wiggle room as to what it actually is.

    ‘Bond albedo (measuring total proportion of electromagnetic energy reflected)’
    which is the one NASA quotes, and is the relevant one for our calculations. This takes into account the spherical nature of the object, so any strange reflectance at high angles (limbs and poles) is included in this number. This shows that my ‘fudge-factor’ of 1.0558 is justified, taking the cos^1.3 absorption, which would otherwise increase albedo. To be honest, I didn’t include the polar region as non-Lambertian (except longitudinally) – that can all be included if needed, depending on what detail we want.

    Opposition effect: http://en.wikipedia.org/wiki/Opposition_effect
    ‘the reflected light from the Moon and Mars appear significantly brighter than predicted when at astronomical opposition’
    Not sure what to do about that, or what it really implies!

    Anyway, while all of that may change the average value by a [i]relatively small[/i] amount, it won’t strongly affect the dependence on [i]spin rate[/i], which is one question we all still need to agree on.

  87. tchannon says:

    It’s close enough. Your new variant of the spice model gives very close to the figure I had.
    As Peter M. wants, not giving numbers.

    All could do with detail checking when the basics are done.

    Two effects: spin rate; surface conductance.

    My original conclusion where I didn’t save the work well enough was elevation to roughly earth mean was possible. Subsequently on challenge I looked again and on spin found little effect.

  88. br says:

    ok guys, it’s dirty hack time.

    I did manage to get LTSpice to do the global average in one click of the run button, see code at end of post. This involved removing the ‘.step param latitude’ directive, and instead using a voltage source with an internal step, labelled ‘latstep’, not actually connected to the main circuit. The voltage stays at one value (=latitude) for a while, the value being read by the ‘Sun’ voltage source, and waits for the temperature on the moon to stabilise. Then it has a fast ramp which increases to the next value (=latitude), and waits again for the ‘temperatures’ to stabilise to the new ‘latitude’. At the end of the sim, all the temperatures are in the V(Tsurface) trace, equally spaced in time. The averages of each section can then be taken as we know when the latituudes were changed, and are combined into a global average. This result can then be found directly in the log file under the name ‘Tglobalave’! The average temperature at the equator can be found under the name Tave1.

    This is a dreadful hack, because if you want to change the latitude spacing (for example, step in 5 degrees instead of 10) you don’t just change one parameter, you have to write a whole stack of new .meas commands. In that case, it is better to go back to the previous version and just sum the outputs yourself.

    In the version in this post, there are now two main parameters to tweak, both given in directive boxes at the lower left of the layout window. The first is ‘.param lunarday’. This is set to 29.5, which is the number of Earth days between sunrises if you are standing on the moon, representing a standard lunar day. If you set it to 1 then the moon will turn at the same rate as Earth, or if you set it to 0.01 then it will spin a hundred times as fast as the Earth, etc. The second is ‘.param convergetime’. This is to keep the results meaningful. The value represents how long to wait for the temperatures to stabilise before taking an average (the average is only taken over the last two complete cycles at any particular latitude, the early parts of the trace are ignored). A value of 40 is good and is not too slow when lunarday=29.5, but can be painfully slow if you shorten the lunarday. So you can change this to compromise between accuracy and speed, or go get a cup of coffee. If it is too small, then if you put voltage probes on the ‘deeper’ or ‘deep’ nodes, you might see they haven’t have settled before the ‘latitude’ is changed. Sometimes it makes little difference.

    I also tweaked a couple of RC values, and added the non-Lambertian absorption with latitude as well as longitude.

    Have fun!

    Version 4
    SHEET 1 4200 1136
    WIRE 688 -96 0 -96
    WIRE 1008 -80 1008 -112
    WIRE 1008 -80 944 -80
    WIRE 1040 -80 1008 -80
    WIRE 352 -16 160 -16
    WIRE 400 -16 352 -16
    WIRE 352 0 352 -16
    WIRE 944 0 944 -16
    WIRE 1040 0 944 0
    WIRE 352 48 336 48
    WIRE 688 96 688 -96
    WIRE 640 112 544 112
    WIRE 160 128 160 -16
    WIRE 400 192 400 64
    WIRE 688 224 688 176
    WIRE 848 224 688 224
    WIRE 1104 224 1104 160
    WIRE 1104 224 848 224
    WIRE 1200 224 1104 224
    WIRE 1248 224 1200 224
    WIRE 1360 224 1360 160
    WIRE 1360 224 1328 224
    WIRE 1408 224 1360 224
    WIRE 1520 224 1520 160
    WIRE 1520 224 1488 224
    WIRE 1584 224 1520 224
    WIRE 1696 224 1696 160
    WIRE 1696 224 1664 224
    WIRE 1104 256 1104 224
    WIRE 848 272 848 224
    WIRE 1200 272 1200 224
    WIRE 1360 272 1360 224
    WIRE 1520 272 1520 224
    WIRE 1696 272 1696 224
    WIRE 400 304 400 192
    WIRE 0 400 0 -96
    WIRE 80 400 0 400
    WIRE 160 400 160 208
    WIRE 160 400 80 400
    WIRE 336 400 336 48
    WIRE 336 400 160 400
    WIRE 400 400 400 384
    WIRE 400 400 336 400
    WIRE 544 400 544 192
    WIRE 544 400 400 400
    WIRE 640 400 640 160
    WIRE 640 400 544 400
    WIRE 848 400 848 352
    WIRE 848 400 640 400
    WIRE 1104 400 1104 336
    WIRE 1104 400 848 400
    WIRE 1200 400 1200 336
    WIRE 1200 400 1104 400
    WIRE 1360 400 1360 336
    WIRE 1360 400 1200 400
    WIRE 1520 400 1520 336
    WIRE 1520 400 1360 400
    WIRE 1696 400 1696 336
    WIRE 1696 400 1520 400
    WIRE 80 416 80 400
    FLAG 80 416 0
    FLAG 1104 160 Tsurface
    FLAG 400 192 HalfSine
    FLAG 1360 160 Tshallow
    FLAG 1520 160 Tdeeper
    FLAG 1696 160 Tdeep
    FLAG 944 0 0
    FLAG 1008 -112 latitude
    SYMBOL voltage 160 112 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -48 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName LunarRotationRate
    SYMATTR Value SINE(0 1 {2.95/lunarday})
    SYMBOL res 384 288 R0
    SYMATTR InstName R1
    SYMATTR Value 1e6
    SYMBOL sw 400 80 M180
    WINDOW 3 24 -12 Invisible 2
    WINDOW 0 39 56 Left 2
    SYMATTR Value MYSW
    SYMATTR InstName DayNightSwitch
    SYMBOL g2 688 192 M180
    WINDOW 0 44 58 Left 2
    SYMATTR InstName AbsorbedFlux
    SYMATTR Value 1283
    SYMBOL res 1120 352 R180
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    SYMATTR Value 40
    SYMBOL bi 848 272 R0
    WINDOW 0 37 37 Left 2
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    SYMATTR InstName SBemission
    SYMATTR Value I=0.98*(5.67e-8)*pwr(V(Tsurface),4)
    SYMBOL cap 1184 272 R0
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    SYMATTR Value 0.1
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    SYMATTR Value 0.25
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    SYMATTR Value 10
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    SYMATTR InstName Sun
    SYMATTR Value V=pow(V(HalfSine),1.3)*pow(cos(V(latitude)*pi/180),1.3)
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    SYMATTR Value 0.3
    SYMBOL res 1680 208 R90
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    SYMATTR Value 50
    SYMBOL voltage 944 -96 R0
    WINDOW 123 0 0 Left 2
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    SYMATTR InstName LatStep
    SYMATTR Value PWL(0 5 {twait} 5 {twait+1} 15 {2*twait} 15 {2*twait+1} 25 {3*twait} 25 {3*twait+1} 35 {4*twait} 35 {4*twait+1} 45 {5*twait} 45 {5*twait+1} 55 {6*twait} 55 {6*twait+1} 65 {7*twait} 65 {7*twait+1} 75 {8*twait} 75 {8*twait+1} 85 {9*twait} 85)
    SYMBOL res 1024 -96 R0
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    TEXT 696 496 Left 2 !.tran {9*twait}
    TEXT 696 536 Left 2 !.model MYSW SW(Ron=0.001 Roff=1000Meg Vt=0 Vh=0)
    TEXT 696 584 Left 2 !.ic V(Tsurface)=240 V(Tshallow)=240 V(Tdeeper)=210 V(Tdeep)=240
    TEXT 696 672 Left 2 !.meas Tave1 avg V(Tsurface) TRIG at {twait-2*lunarday/2.95} TARG at {twait}
    TEXT 104 496 Left 2 !.param lunarday 29.5
    TEXT 696 456 Left 2 !.param twait=(ceil(convergetime*2.95/lunarday)+2)*lunarday/2.95
    TEXT 696 712 Left 2 !.meas Tave2 avg V(Tsurface) TRIG at {2*twait-2*lunarday/2.95} TARG at {2*twait}
    TEXT 696 752 Left 2 !.meas Tave3 avg V(Tsurface) TRIG at {3*twait-2*lunarday/2.95} TARG at {3*twait}
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    TEXT 704 1072 Left 2 !.meas cossum avg (cos(5)+cos(15)+cos(25)+cos(35)+cos(45)+cos(55)+cos(65)+cos(75)+cos(85))
    TEXT 696 1112 Left 2 !.meas Tglobalave avg Tlatareasum/cossum
    TEXT 104 552 Left 2 !.param convergetime=40
    
  89. tchannon says:

    Lots is possible.

    Goodness 🙂
    Watching people at work can be fun as unexpected routes are followed.

    You are arriving at a slightly lower figure than the previous version. Can’t figure out why. Pertinent parameters look the same.

    Tried bumping up time for settling hugely, trivial effect so it’s not that.

    Puzzled.

  90. br says:

    tchannon says: May 2, 2014 at 1:56 am
    ‘You are arriving at a slightly lower figure than the previous version.’
    there are a few relatively minor differences.

    First I tweaked the surface and deep RC values. I needed to tweak the surface because I was getting the equatorial average a teensy bit too high, and now it seems right. Then I tweaked the deep level because the RC time constant was taking too long to stabilise, and now it runs a bit faster without any noticeable effect at the surface. Thirdly was that I included a cos(theta)^1.3 in the ‘Sun’ source for latitude as well as the one we had previously for longitude. This lowers the global average a little.

    So, for our purposes, these numbers now seem spot on! Of course, everyone is free to use their own parameters, but this is my best shot for the moment.

  91. My figure for the average lunar temperature came out six Kelvin high. I sent my calculations to Tim in the hope he can spot my my dumb mistake(s).

  92. tchannon says:

    Roughly we agree. Diviner is unlikely to be accurate anyway moreover the only Diviner data we have is without parentage. We are all within 5%

    The lunar temperature is not constant, varies with orbit. There is also axial tilt.

    Roughly 200K, or 200K +-5K

    I’ll take a look at what you have sent.

  93. tchannon says:

    Ahem, not sure I wanted to know. The differences are subtle but these are very different animals.

    With FEA you are taking into account the rock structure whereas the spice models at the moment are approximating “something”.

    The Spice can be extended to more closely take into account the rock. I think this means a more representative delay line using the dual of the rock conductance and thermal capacity as parameters.

    The only thing I spotted in the spreadsheet which might be wrong but has negligible effect is maybe one time slot too many for 24 hours, something seems not quite right.

    A few things can be simplified, I expect they are like that for historic reasons.

    I’ve not looked at the ^1.3 vs. latitude which br has included.

  94. br says:

    and here is my Spice model for Earth.

    A few things needed to be adjusted. First is obviously the length of day, which is now equal to 1. We need to change albedo (which is now a .param to make it easy to find). All the RC values needed to be retweaked – I chose values that give a surface 15 K day/night variation at the equator, to bring it close to NASA’s value of 10 K for a global average (the poles have smaller variation in this model, so lower the global day/night swing average). The ‘shallow’ and ‘deeper’ RC values were chosen so that they had a small but noticeable day/night variation, while the ‘deep’ values give practically no variation, so we have some day/night trend going to a stable underground temperature. As Earth is not covered in rubble like the moon, I removed the ^1.3 value on the cos(theta) functions, which brings it more in line with the black ‘unpolarised’ curve in my figure a few posts up.

    Finally, I changed emissivity, which is now also a .param value as I couldn’t find an ‘average emissivity of planet Earth’ value, so may need adjusting. From http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html we see that water has an emissivity in the range 0.95-0.963, ice is 0.966-0.985, sand is 0.76. As the surface is mostly water, I chose a value of 0.96, but appreciate that others might prefer a different value. Playing around with this parameter is very interesting!!!

    Code below:

    Version 4
    SHEET 1 4200 1136
    WIRE 688 -96 0 -96
    WIRE 1008 -80 1008 -112
    WIRE 1008 -80 944 -80
    WIRE 1040 -80 1008 -80
    WIRE 352 -16 160 -16
    WIRE 400 -16 352 -16
    WIRE 352 0 352 -16
    WIRE 944 0 944 -16
    WIRE 1040 0 944 0
    WIRE 352 48 336 48
    WIRE 688 96 688 -96
    WIRE 640 112 544 112
    WIRE 160 128 160 -16
    WIRE 400 192 400 64
    WIRE 688 224 688 176
    WIRE 848 224 688 224
    WIRE 1104 224 1104 160
    WIRE 1104 224 848 224
    WIRE 1200 224 1104 224
    WIRE 1248 224 1200 224
    WIRE 1360 224 1360 160
    WIRE 1360 224 1328 224
    WIRE 1408 224 1360 224
    WIRE 1520 224 1520 160
    WIRE 1520 224 1488 224
    WIRE 1584 224 1520 224
    WIRE 1696 224 1696 160
    WIRE 1696 224 1664 224
    WIRE 1104 256 1104 224
    WIRE 848 272 848 224
    WIRE 1200 272 1200 224
    WIRE 1360 272 1360 224
    WIRE 1520 272 1520 224
    WIRE 1696 272 1696 224
    WIRE 400 304 400 192
    WIRE 0 400 0 -96
    WIRE 80 400 0 400
    WIRE 160 400 160 208
    WIRE 160 400 80 400
    WIRE 336 400 336 48
    WIRE 336 400 160 400
    WIRE 400 400 400 384
    WIRE 400 400 336 400
    WIRE 544 400 544 192
    WIRE 544 400 400 400
    WIRE 640 400 640 160
    WIRE 640 400 544 400
    WIRE 848 400 848 352
    WIRE 848 400 640 400
    WIRE 1104 400 1104 336
    WIRE 1104 400 848 400
    WIRE 1200 400 1200 336
    WIRE 1200 400 1104 400
    WIRE 1360 400 1360 336
    WIRE 1360 400 1200 400
    WIRE 1520 400 1520 336
    WIRE 1520 400 1360 400
    WIRE 1696 400 1696 336
    WIRE 1696 400 1520 400
    WIRE 80 416 80 400
    FLAG 80 416 0
    FLAG 1104 160 Tsurface
    FLAG 400 192 HalfSine
    FLAG 1360 160 Tshallow
    FLAG 1520 160 Tdeeper
    FLAG 1696 160 Tdeep
    FLAG 944 0 0
    FLAG 1008 -112 latitude
    SYMBOL voltage 160 112 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -48 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName EarthRotationRate
    SYMATTR Value SINE(0 1 {2.95/earthday})
    SYMBOL res 384 288 R0
    SYMATTR InstName R1
    SYMATTR Value 1e6
    SYMBOL sw 400 80 M180
    WINDOW 3 24 -12 Invisible 2
    WINDOW 0 39 56 Left 2
    SYMATTR Value MYSW
    SYMATTR InstName DayNightSwitch
    SYMBOL g2 688 192 M180
    WINDOW 0 44 58 Left 2
    SYMATTR InstName AbsorbedFlux
    SYMATTR Value {1366*(1-albedo)}
    SYMBOL res 1120 352 R180
    WINDOW 0 36 76 Left 2
    WINDOW 3 36 40 Left 2
    SYMATTR InstName R2
    SYMATTR Value 200
    SYMBOL bi 848 272 R0
    WINDOW 0 37 37 Left 2
    WINDOW 3 24 80 Invisible 2
    SYMATTR InstName SBemission
    SYMATTR Value I=emissivity*(5.67e-8)*pwr(V(Tsurface),4)
    SYMBOL cap 1184 272 R0
    SYMATTR InstName C1
    SYMATTR Value 4
    SYMBOL cap 1344 272 R0
    SYMATTR InstName C2
    SYMATTR Value 0.1
    SYMBOL res 1344 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R3
    SYMATTR Value 1
    SYMBOL cap 1504 272 R0
    SYMATTR InstName C3
    SYMATTR Value 0.2
    SYMBOL res 1504 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R4
    SYMATTR Value 1
    SYMBOL bv 544 96 R0
    WINDOW 0 35 40 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName Sun
    SYMATTR Value V=pow(V(HalfSine),1.0)*pow(cos(V(latitude)*pi/180),1.0)
    SYMBOL cap 1680 272 R0
    SYMATTR InstName C4
    SYMATTR Value 5
    SYMBOL res 1680 208 R90
    WINDOW 0 0 56 VBottom 2
    WINDOW 3 32 56 VTop 2
    SYMATTR InstName R5
    SYMATTR Value 1
    SYMBOL voltage 944 -96 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -99 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName LatStep
    SYMATTR Value PWL(0 5 {twait} 5 {twait+1} 15 {2*twait} 15 {2*twait+1} 25 {3*twait} 25 {3*twait+1} 35 {4*twait} 35 {4*twait+1} 45 {5*twait} 45 {5*twait+1} 55 {6*twait} 55 {6*twait+1} 65 {7*twait} 65 {7*twait+1} 75 {8*twait} 75 {8*twait+1} 85 {9*twait} 85)
    SYMBOL res 1024 -96 R0
    SYMATTR InstName R6
    SYMATTR Value 1
    TEXT 696 496 Left 2 !.tran {9*twait}
    TEXT 696 536 Left 2 !.model MYSW SW(Ron=0.001 Roff=1000Meg Vt=0 Vh=0)
    TEXT 696 584 Left 2 !.ic V(Tsurface)=240 V(Tshallow)=240 V(Tdeeper)=210 V(Tdeep)=240
    TEXT 696 672 Left 2 !.meas Tave1 avg V(Tsurface) TRIG at {twait-2*earthday/2.95} TARG at {twait}
    TEXT 104 496 Left 2 !.param earthday 1
    TEXT 696 456 Left 2 !.param twait=(ceil(convergetime*2.95/earthday)+2)*earthday/2.95
    TEXT 696 712 Left 2 !.meas Tave2 avg V(Tsurface) TRIG at {2*twait-2*earthday/2.95} TARG at {2*twait}
    TEXT 696 752 Left 2 !.meas Tave3 avg V(Tsurface) TRIG at {3*twait-2*earthday/2.95} TARG at {3*twait}
    TEXT 696 792 Left 2 !.meas Tave4 avg V(Tsurface) TRIG at {4*twait-2*earthday/2.95} TARG at {4*twait}
    TEXT 696 832 Left 2 !.meas Tave5 avg V(Tsurface) TRIG at {5*twait-2*earthday/2.95} TARG at {5*twait}
    TEXT 696 864 Left 2 !.meas Tave6 avg V(Tsurface) TRIG at {6*twait-2*earthday/2.95} TARG at {6*twait}
    TEXT 688 904 Left 2 !.meas Tave7 avg V(Tsurface) TRIG at {7*twait-2*earthday/2.95} TARG at {7*twait}
    TEXT 696 952 Left 2 !.meas Tave8 avg V(Tsurface) TRIG at {8*twait-2*earthday/2.95} TARG at {8*twait}
    TEXT 696 1000 Left 2 !.meas Tave9 avg V(Tsurface) TRIG at {9*twait-2*earthday/2.95} TARG at {9*twait}
    TEXT 704 1040 Left 2 !.meas Tlatareasum avg (Tave1*cos(5)+Tave2*cos(15)+Tave3*cos(25)+Tave4*cos(35)+Tave5*cos(45)+Tave6*cos(55)+Tave7*cos(65)+Tave8*cos(75)+Tave9*cos(85))
    TEXT 704 1072 Left 2 !.meas cossum avg (cos(5)+cos(15)+cos(25)+cos(35)+cos(45)+cos(55)+cos(65)+cos(75)+cos(85))
    TEXT 696 1112 Left 2 !.meas Tglobalave avg Tlatareasum/cossum
    TEXT 104 552 Left 2 !.param convergetime=50
    TEXT 96 616 Left 2 !.param albedo=0.306
    TEXT 96 672 Left 2 !.param emissivity=0.96
    
  95. br says:

    I was thinking about the relationship between albedo and emissivity. As Tim pointed out, one expects absorption and emission to be equal, Kirchhoff’s law and all that, so maybe a high albedo should correspond to a low emissivity. The problem is the wavelength dependence – the albedo is an average dominated by the wavelengths at the temperature of the sun, whereas emissivity is an average dominated by wavelengths at the temperature of the planet. As they are weighted by different spectral regions, there is no simple correlation. It is possible to have a high albedo and high emissivity, or high albedo and low emissivity, etc.

    So I did some more reading around this – seems like a value of 0.96 is often used (I actually didn’t know that, honest, guv!). And ‘effective emissivity’ is used as a greenhouse parameter. Well this is fun!

  96. tchannon says:

    You are due for a bravery award… I have deliberately not touched earth but I have to face the consequences of herding cats.

    A difference between what I was doing and what you are doing is including factors do with the assumed surface, which I wanted to be whatever works without inserted figures, then later try and figure out what it means.

    The earth has no surface as such, let’s say the matter is up-in-the-air with so much argument room, so little actual data that Mark Twain put it well on science.

    For where the ground is involved the real properties, in Ireland, have a clue in this data

    I’ve not done more than start to look at the whole problem of any linkage between air temperature and subsurface temperature. Daily Armagh data won’t answer the question of diurnal response.
    An estimate might be feasible since there is unlikely to be a complicated structure, is probably little more than single time constant.

    The ground effect gets mixed up with UHI which most is I think one and the same: an increase in thermal coupling and hence the phase delay. Very little high resolution data is available so doing work on this is not feasible.

    I have no trustworthy high resolution data. Chilbolton Observatory does not seem to include soil data, it does however now have a net radiative instrument. The only result is quick look
    https://tallbloke.wordpress.com/2014/04/02/surface-thermal-balance-central-southern-england/
    This poses a question: if the calculated albedo is correct and the instruments do respond suitably, the curve may imply a grass surface is non-Lambertian. I think this is the case for most of earth, water too.
    Clouds and direct atmospheric interaction (Rayleigh etc.) might be Lambertian.
    Worse still is the myriad optical effects, reflection and at IR too, polarisation, goodness knows what.

  97. br, said:
    “Finally, I changed emissivity, which is now also a .param value as I couldn’t find an ‘average emissivity of planet Earth’ value, so may need adjusting. From http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html we see that water has an emissivity in the range 0.95-0.963, ice is 0.966-0.985, sand is 0.76. As the surface is mostly water, I chose a value of 0.96″

    I assumed that an airless earth would be covered in regolith just like the Moon. However it is not something worth arguing about as the emissivity of water is similar to that of regolith. Likewise for ice.

    Thus I plan to keep the emissivity I used for the Moon (0.95) and the Albedo too as there will be no clouds to reflect SW radiation. Once I am convinced there are no more dumb mistakes in my lunar model I will simply change the rate of rotation to 24 hours.

  98. With regard to the lack of correlation between Albedo and emissivity I found the link provided upthread by J Martin useful:
    http://www.tak2000.com/data/Satellite_TC.pdf

  99. Ben Wouters says:

    br says: April 29, 2014 at 3:47 pm

    Sorry for the slow response, thanks for your reply.

    Seeing your parameters it is imo impossible to reach the Te, since this is based on spreading incoming radiation equally over the whole sphere and using emissivity 1,0.
    Only in this case the 270K or 255K Te for the moon and earth can be reached.
    All other cases with uneven radiation, if opnly between equator and poles must result in a lower actual temperature.

    I still suspect that in the modelling some geothermal influence has been included.
    see http://www.diviner.ucla.edu/science.shtml
    The lat89 plot around day 100 cools with sunset from ~100K to ~60K, and then for the next ~130 earth days very slowly approaches 50K. Is this reasonable with the little heat that was stored the last day, or is a small geothermal flux more likely?

    The 100 mW/m^2 flux through oceanic crust would already result in a SB temp. of ~36K.

    This would explain your results approaching or even exceeding the Te.

  100. br says:

    Hi Tim,

    Sure the Earth is complicated, and a Spice model with four nodes will hardly do it justice. However, it works well enough for the moon, so should give a similar ballpark for the Earth, at the same level of coarseness. After that one would need something a bit more heavy duty!

  101. br says:

    gallopingcamel says:May 3, 2014 at 2:28 pm
    ‘ However it is not something worth arguing about as the emissivity of water is similar to that of regolith.’
    as far as emissivity goes, it is not worth arguing about, but as far as heat conductivity goes, there is a huge difference. The result of this difference is that the surface of the moon can’t store heat well, and the day/night temperature differences are huge (>380 K). On Earth, water stores heat very well, and the ocean surface has very little temperature difference between day and night (<10 K). This matters a lot as to what the global average temperature is, and changes the situation entirely due to the T^4 radiation factor.

  102. br says:

    Ben Wouters says:May 3, 2014 at 3:48 pm
    ‘Seeing your parameters it is imo impossible to reach the Te, since this is based on spreading incoming radiation equally over the whole sphere and using emissivity 1,0.
    Only in this case the 270K or 255K Te for the moon and earth can be reached.
    All other cases with uneven radiation, if opnly between equator and poles must result in a lower actual temperature’
    I agree, which is the very reason I started these simulations – to see for myself what the values would be. For the Earth I find the result is around 253-254 K, so yes this is lower than Te.

    ‘I still suspect that in the modelling some geothermal influence has been included.’
    Not in my Matlab model, or in the Spice simulation. The spice one is posted, so you can check it out for yourself, I can post the Matlab one, but that is much harder to read so not so sure if it is useful. Also please note that an extra 0.1 W/m2 will not ‘add 36 K’ to a temperature of 200 K! SB applies and radiates at T^4 – this makes the extra 0.1 W/m2 increase the 200 K to 200.05 K. At higher temperature, the effect is even less.

  103. br says:

    tchannon says:May 3, 2014 at 1:10 pm
    ‘A difference between what I was doing and what you are doing is including factors do with the assumed surface, which I wanted to be whatever works without inserted figures, then later try and figure out what it means’
    well, in the same way we were tweaking the assumed lunar surface to match experimental values (you have to insert figures for this!), we should also tweak the Earth surface to match experimental values. As far as I am concerned, the relevant factor is the day/night difference, because the equivalent factor in the sim is thermal time constant.

  104. tchannon says:

    Peter, I wish people plotted those curves properly, makes the law obvious

    Image from Talkshop last year. Not sure which post or comment it is attached to,

  105. tchannon says:

    Four in a row br, grin.

    I must not get too involved, the space is for you two and any others who actually do, don’t think it can be made much easier. (pulling myself back from trampling)

    I’ll add a link to an existing image file since this may help readers trying to understand what is being done.

  106. br says:

    gallopingcamel says:May 3, 2014 at 2:34 pm
    ‘I found the link provided upthread by J Martin useful:http://www.tak2000.com/data/Satellite_TC.pdf
    yes, slide 12 clearly shows why albedo is relevant from 0.15 to 5 um, whereas emissivity is relevant from 3 to 100 um. Not necessarily the same things at all!

  107. br says:

    wow, read the pdf a bit more. Slide 17 spells out the albedo=\=emissivity issue, fine, but slide 25 was an eye opener for me. It lists the albedo as 0.33 +-0.13, with contributions as:
    clouds = 0.4 to 0.8
    forests = 0.05 to 0.1
    oceans = 0.05

    so nearly all the albedo comes from clouds! I wasn’t aware of that, learning a lot in this thread, thanks guys. So if GCM models fudge cloud cover badly, then that could change things a lot, as I’m sure someone may have mentioned :). This is probably not the thread to get into that though! Must stay on topic…

  108. Ben Wouters says:

    br says: May 3, 2014 at 4:29 pm

    “Also please note that an extra 0.1 W/m2 will not ‘add 36 K’ to a temperature of 200 K! SB applies and radiates at T^4 – this makes the extra 0.1 W/m2 increase the 200 K to 200.05 K. At higher temperature, the effect is even less”

    see http://www.diviner.ucla.edu/science.shtml
    I referred to “lat89”, probably should have used “89 degrees latitude”
    It is the dark blue plot in the graph. Latitude 89 is close to the pole, and due to the ~1,5 degrees axis tilt of the moon this latitude will see no sunshine for long periods each year (called winter on earth).
    The last sunshine for lat89 before winter sets in is around day 90, and only brings the temperature to ~100K. Sun is barely above the horizon at lat89, so not much heat storage possible at all.
    Yet roughly from day 100 to day 230 the temperature is very slowly approaching ~50K.
    Solar radiation in that period 0 W/m^2 => SB 0k.
    Where does this 50K temperature come from in your opinion if not from geothermal?

    Same for the no rotation case. If a geothermal flux does exist, it won’t make a difference on the sunlit side, BUT on the dark side it most certainly will.

  109. Ben Wouters says:

    br says: May 3, 2014 at 6:23 pm

    “clouds = 0.4 to 0.8
    forests = 0.05 to 0.1
    oceans = 0.05

    so nearly all the albedo comes from clouds!”

    see http://en.wikipedia.org/wiki/File:Solar_Spectrum.png
    Shows nicely that the reflection is mostly in UV and visible, but also in IR up to ~1300nm.
    The large bites out of the clean TOA spectrum when it reaches the surface are also due to clouds.
    Supposedly the IR from the ground is heating up our greenhouse gasses, but in reality the sun is doing this at least during daytime.

  110. br says:

    Ben:
    ‘Where does this 50K temperature come from in your opinion if not from geothermal?’
    At that low temperature, geothermal could have an influence, but there are three things to note. One is that subsurface *storage* can still have an influence, the time periods are longer for the lunar winter, so the subsurface temperature has longer to appear at the surface (this does not need to be geothermal *generation* though!). Two, the models we have developed actually don’t have a winter, as the polar axis is taken to be vertical rather than 1.5 degrees off. This is for sure an approximation, but as far as the global average go shouldn’t affect things much, as the polar area is relatively small. Three, geothermal generation in the moon is much lower in the moon than Earth. I was interested to read that the moon does indeed have a liquid core, but I didn’t get a clear answer as to geothermal flux for the moon – do you have a reference?

    ‘Same for the no rotation case. If a geothermal flux does exist, it won’t make a difference on the sunlit side, BUT on the dark side it most certainly will.’
    I agree, for the no rotation case that geothermal flux will have a more significant effect. But once the body rotates, the day/night temperatures start to converge, and heat storage dominates the temperature.

    ‘Shows nicely that the reflection is mostly in UV and visible, but also in IR up to ~1300nm’
    Some of those differences are due to absorption, and some due to reflection. I expect the visible part to be mostly reflection (clouds), and the IR part mostly due to absorption (CO2 and water). I suspect the UV is mostly absorption (ozone). Albedo must be the part that doesn’t contribute to ‘thermal budget’, so I guess the IR chunks that are absorbed would not be included as albedo (even if they subsequently reradiate back out into space!).

  111. br says:

    I was thinking about where these models are going, in order to figure out what they mean at each stage of building up the layers of refinement. It would be nice to separate out what temperature rise is due to what underlying factor, for the moon and the Earth. To this end I came up with the following list:

    1. Non-rotating, non-conducting, no heat storage, no oceans, no atmosphere. What is the global average temperature here? I calculated that for the moon (albedo 0.11) that this average should be 153 K, and for Earth (albedo 0.3) 144 K. This is the lowest possible global average at this distance from the Sun. Calculating these averages is interesting, as it shows that while the average temperature is lowest, the *peak* temperature is highest, which for the moon is around 400 K! I bring this up because while one can calculate the ‘standard’ value of Te=270 K for the moon, which is the max global average possible, the value of 270 K does not represent the max temperature possible.

    2, Some rotation, some heat storage, no oceans, no atmosphere. For the moon albedo of 0.11, this could have any global average in the range 153 – 270 K, depending on how fast the rotation is and how much heat storage there is (these factors go together due to thermal time constant). Observationally, the average seems to be 197 K or thereabouts and the simulations can reproduce this. For an albedo of 0.3, the average could have any value from 144 – 255 K, again depending on thermal time constant and spin rate. One can take the bare rock Earth surface as similar to the moon if you like, we don’t have much data to go on here. The answer will be similar enough to the moon’s value (slightly lower if considering albedo, though maybe one should note that as Earth’s albedo mostly comes from clouds it may not be appropriate at this stage of the calculation! Leaving out albedo and just changing spin rate will gave an answer very similar to the moon).

    3, Some rotation, some heat storage, some oceans, no atmosphere. The moon has no oceans or atmosphere, so the moon analysis stops at the previous step. For the Earth, the oceans constitute the majority of the surface, and contribute a huge thermal storage and thermal time constant. Without considering ocean circulation, I calculate the average here to be 253-254 K.

    4, Some rotation, some heat storage, some oceans, some atmosphere. I didn’t get this far yet, but one could start including the atmosphere (without water vapour/CO2!) as a set of thermal time constants with altitude.

    5, Ocean and atmospheric circulation.

    6, Gas absorption lines and radiative transfer.

    7, Topographic structure (mountains, continents, etc).

    8, Hydrological cycle and feedback effects.

    I spell this out for clarity, so we can compare the different stages the models are at. I got up to step 3, and don’t think I will get any further with a Spice model (presumably possible, but a PITA!). If one wants to see what the effect a basic *atmosphere* adds, one needs to go at least to step 3, preferably to step 4. To see the basic effect of CO2/H20 vapour, one would need step 6. Obviously one would ultimately like to get to step 8 and deconstruct all the previous steps to see the contributions, but I don’t think we’ll get to that in this thread!

  112. Ben Wouters says:

    br says: May 4, 2014 at 10:17 am

    “At that low temperature, geothermal could have an influence, but there are three things to note. One is that subsurface *storage* can still have an influence, the time periods are longer for the lunar winter, so the subsurface temperature has longer to appear at the surface (this does not need to be geothermal *generation* though!). ”

    Without geothermal we should consider the subsurface temperature to be 0K.
    (subsurface to me is below solar influence)
    At the equator the sun barely reaches 30cm deep, near the poles where solar radiation is practically parallel to the surface think millimetres, perhaps even microns.
    So where is the energy stored near the poles to sustain a 50K temperature for ~130 earth days?

    In deep craters near the poles, where the sun never shines, the surface temperature never drops below ~25K. Switching of the sun would let the moon cool down pretty fast, surface temperatures stabilising around that 25K mark, and maintain that temperature for as long as geothermal heat is available.

  113. Ben Wouters says:

    Some references at last 😉

    http://www.exp-geochem.ru/JPdf/01…/16_01_Eng.pdf
    has a temperature profile of the moon, resulting in a subsurface temp around 350-400 C.

    http://adsabs.harvard.edu/abs/2012AGUFM.P52A..08S
    has a surface heat flux around 16-20 mW/m^2, consistent with the 25K polar crater temperatures.

  114. br says:

    Ben Wouters says:May 4, 2014 at 2:03 pm
    ‘So where is the energy stored near the poles to sustain a 50K temperature for ~130 earth days?’
    I fully accept that the moon has a liquid core, transferring heat to the surface since the moon was formed, so I don’t really have a different view from you on that one. Still, the sun has been shining on the moon for billions of years, plenty of time for bad conductivity to accumulate subsurface solar energy to some significant depth. That can be called geothermal heat if you like. I guess we might have a different view on what these do to the lunar global average – I agree it would affect the non-rotating average, but found it will hardly affect the rotating global average, which will be weighted towards the non-polar area at higher temperatures where it is not so noticeable.

    ‘Switching of the sun would let the moon cool down pretty fast, surface temperatures stabilising around that 25K mark, and maintain that temperature for as long as geothermal heat is available.’
    I agree with this also, but one can’t then conclude that 25 K of the moon average is due to geothermal heat, because the T^4 dependence washes out the geothermal contribution. In that case, the small surface heat capacity driven by the sun dominates and the average geothermal increase in temperature is practically negligible, IMHO. The few remaining local cold spots don’t contribute much to the global average.

  115. tchannon says:

    Ben addressing br
    “Where does this 50K temperature come from in your opinion if not from geothermal?”

    Since you are interested in deep heat two factors come to mind.

    1. Moon is radioactive. Depth profile is unknown.

    2. Conduction from midhrif. _

  116. tchannon says:

    A sanity check is probably within the capability of the student FEA since this is a separate problem.

    A hemisphere on a 200K constant temperature plate.

    The flux at 40K is very small but also the Diviner instrument capability ought to be considered.

  117. The FEA model would show very low temperatures at the lunar poles if one removed the thermal conduction from the Moon’s bedrock which I have assumed to be 0.5 meters below the surface and at a temperature of 240 K. See Vasavada’s Figure 7.

  118. br says:

    Ben Wouters says:May 4, 2014 at 2:29 pm
    ‘http://adsabs.harvard.edu/abs/2012AGUFM.P52A..08S
    has a surface heat flux around 16-20 mW/m^2, consistent with the 25K polar crater temperatures.’
    Excellent!

    OK, so let’s take 20 mW/m2.

    On its own, that corresponds to a black-body temperature (with emissivity of 0.98) of 24.5 K. Very good.

    Now let’s see the effect on a rotating moon. We have from measurement that the average temperature on the day side of the moon is about 300 K, and on the night side about 95 K, close enough. The global average is then (300+95)/2=197.5 K, spot on.

    The average power emitted on the day side is 0.98*(5.67-8)*300^4 = 450.085 W/m2.
    On the night side is 0.98*(5.67-8)*95^4 = 4.526 W/m2.

    Let’s add a uniform 0.02 W/m2 to both these to get radiated power of 450.105 and 4.546 W/m2 respectively. Inverting the SB formula to find T=(P/(0.98*5.67e-8))^0.25 gives the new day temperature as 300.0033 K and the new night temperature of 95.1048 K.

    The new global average is then (300.0033+95.1048)/2 = 197.554 K.

    So while geothermal can increase 0 K to 24.5 K, it can only increase 197.50 K to 197.55 K.

    Consider again the night time radiated power of 4.526 W/m2. Even if that already included geothermal, one still needs 4.506 W/m2 to come from solar powered heat storage. That is 225 times the geothermal flux, even when the surface receives no direct sunlight.

    Also consider that the relative area drops with latitude. At 89 deg latitude, the contribution to global average area is only cos(89)=0.0175 times that of a similar strip at the equator.

    And an occasional individual crater will have even less contribution to area, so no matter how you look at it, the geothermal contribution is not really significant to the global average.

  119. tchannon says:

    Nice comment.

  120. Ben Wouters says:

    br says: May 4, 2014 at 4:02 pm

    I see only radiative calculations for the surface temperatures. Incoming solar should imo first be converted to J/s per m^2, then used to calculate the temperature increase this energy can achieve using the depth into the material the sun is reaching and the specific heat of the material it is heating.

    I did this calculation for 10m water and the temperature rise during one day of sunshine in the tropics is < 0,5K.

    Looking at Vasada's equator graph I estimate the temperature increase during one solar morning to be ~290K. So the start temperature decides how high the noon temperature will be.
    And yes, 100K plus ~290K does explain the ~390K noon temperature.
    During the night all daytime heat is lost to space again.
    I do not assume heat creep towards the core from solar to be a factor, seeing the low surface temperatures relative to the high(er) crust and mantle temps.

    SB for TSI 1364 and albedo 0,11, emissivity 1,0 is only ~383K. Lower emissivity to eg 0,95 and you won't even reach 380K.

  121. Robert JM, April 20, 2014 at 4:42 am

    “Galloping camel,
    Who cares about average temp? The peak solar heating is more than enough to melt water.”

    Robert,
    It took a while but I can address your point. According to my FEA model, an airless Earth would have an average surface temperature of ~261 K at the equator. Temperatures are likely to exceed 273 K for about seven hours per day assuming the surface is composed of ice. This means there could be some water but only at low latitudes.

    Over time the frozen oceans on an airless Earth would be covered by regolith that would prevent even local melting at the equator. This suggests to me that there is plenty of water on the Moon.

  122. br says:

    Ben Wouters says:May 4, 2014 at 6:38 pm

    I agree with your first two paragraphs, all taken into consideration.

    ‘So the start temperature decides how high the noon temperature will be.’
    Not in the case of the moon. The surface pretty much reaches radiative equilibrium temperature, as it has such a low thermal capacity and poor conductivity. Vasavada says as much.

    ‘And yes, 100K plus ~290K does explain the ~390K noon temperature.’
    OK, but where did the 100 K come from???

    ‘During the night all daytime heat is lost to space again.’
    Sure, but you can see from the shape of the night time temperature that it is still dropping. This is due to it still losing heat that it had collected during the previous day, all the way till sunrise dawns on a bright new day.

    ‘SB for TSI 1364 and albedo 0,11, emissivity 1,0 is only ~383K. Lower emissivity to eg 0,95 and you won’t even reach 380K.’
    with a lower emissivity, the temperature reached is *higher*. For TSI 1364, albedo 0.11 and emissivity 0.95, the equilibrium temperature is 387.5 K. No need to invoke anything else.

  123. tchannon says:

    Sublimation is rapid except at very low temperatures.

    This paper may help for lunar water ice which is only expected at the poles, shaded craters.

    New estimates for the sublimation rate for ice on the Moon
    Edgar L Andreas
    U.S. Army Cold Regions Research and Engineering Laboratory
    Received 6 April 2006; revised 18 August 2006
    http://www.nwra.com/resumes/andreas/publications/Icarus_Moon.pdf

  124. br, May 4, 2014 at 11:04 am

    “1. Non-rotating, non-conducting, no heat storage, no oceans, no atmosphere. What is the global average temperature here? I calculated that for the moon (albedo 0.11) that this average should be 153 K.”

    Nikolov & Zeller came up with 154.3 K for the average temperature for the Moon in sharp contrast with the consensus average for Earth (255 K).

  125. br says:

    gallopingcamel says:May 5, 2014 at 5:56 pm

    ‘Nikolov & Zeller came up with 154.3 K for the average temperature for the Moon in sharp contrast with the consensus average for Earth (255 K).’

    Sure, but what is implied by that? There is also an unacceptable contrast between N&Z’s calculated 154 K for the moon vs the measured value of 197.5 K.

    This is why I wrote that list from 1 to 8, to show what goes in to these calculations and what the different temperature contributions are. All the different numbers floating around can be happily reconciled when one sees how many factors have been included at each step.

    For example, for the moon N&Z calculate 154 K when it is ‘non-rotating, non-conducting, no heat storage, no oceans, no atmosphere’, and I agree with them. Add in some thermal capacity and spin, and the number goes up to 197.5 K. I guess we would all agree to that? Obviously there is no atmosphere taken into account on the moon, yet the average temperature increases purely due to thermal capacity and spin.

    For the Earth, I calculate 143 K for the baseline lowest temperature. Add some thermal capacity and spin and that number goes up to 254 K. No atmosphere included (though I admit the cloud albedo is there. Removing the cloud albedo means more heat is absorbed and gives an even *higher* temperature!). Why does the moon temperature only increase by 44 K, while the Earth temperature increases by 111 K??? The answer is in the different thermal capacities and spin rates!

  126. br,
    I have tweaked my FEA model (hopefully for the last time) and here is what I got for Lat00, all temperatures in the Kelvin scale:

    Moon (surface properties as in Vasavada et al, 2012)….Tmax = 386.6, Tave = 212.1, Tmin = 94.6

    Earth (assuming surface is composed of ice)…………..Tmax = 297.4, Tave = 261.5, Tmin = 236.9

    I did not include the effect of ice melting that would presumably lower the Tmax to 273 K and thereby reduce Tave to ~257 K.

    Are these numbers consistent with your model?

  127. br says:

    gallopingcamel says: May 5, 2014 at 7:45 pm

    Your numbers look good from here, but I feel the Earth numbers are very sensitive to albedo. What did you use in this case?

    ‘I did not include the effect of ice melting that would presumably lower the Tmax to 273 K and thereby reduce Tave to ~257 K.’
    that will depend on how you include albedo. Water has a far lower albedo than ice, so will absorb more solar radiation which in turn increases the average temperature significantly. Definitely a positive feedback there! Furthermore, light will penetrate deeper into water than ice, and this will increase the thermal time constant. That’s another positive feedback, as we have seen how the average temperature depends on time constant (otherwise it wouldn’t have needed tweaking!). I guess you mean that water will be pinned at 273 K as it is in contact with ice. My feeling is that if one starts with an ice planet, it will remain an ice planet, with quite a low average temperature well below freezing. However, once conditions heat up (atmospheric greenhouse effect, anyone? Or indeed geothermal at planet formation), then the contribution of the ice becoming to water will be a very significant positive feedback for the above reasons. If you just simulate a liquid water planet, what values do you get?

    That of course is only the effect of the oceans and doesn’t include clouds, which will become a negative contribution by increasing albedo once they start forming.

    Also note that the quoted value of Te=255 K is the max global average, not the max equatorial average. The equatorial average can be significantly higher than Te.

  128. Ben Wouters says:

    gallopingcamel says: May 4, 2014 at 3:34 pm

    “The FEA model would show very low temperatures at the lunar poles if one removed the thermal conduction from the Moon’s bedrock which I have assumed to be 0.5 meters below the surface and at a temperature of 240 K. See Vasavada’s Figure 7.”

    If I’m correct about the geothermal flux being the cause for the subsurface temperature, then the 240K is roughly halfway the min/max temps for the lunar equator (90 + 390)/2 = 240K as expected.
    The subsurface temp at the poles would be around 90 – 100K.

  129. Ben Wouters says:

    br says: May 4, 2014 at 11:02 pm

    “with a lower emissivity, the temperature reached is *higher*. For TSI 1364, albedo 0.11 and emissivity 0.95, the equilibrium temperature is 387.5 K. No need to invoke anything else.”

    As I understand emissivity, it shows how much less radiation a material will emit relative to the blackbody number. Less radiation means a lower emitting temperature.
    TSI 1364 albedo .11 em. 1.0 SB > 382,5K (BB case)
    TSI 1364 albedo .11 em 0,95 SB > 377,6K which is lower than the BB result.

  130. Ben Wouters says:

    gallopingcamel says: May 5, 2014 at 5:56 pm

    “1. Non-rotating, non-conducting, no heat storage, no oceans, no atmosphere. What is the global average temperature here? I calculated that for the moon (albedo 0.11) that this average should be 153 K.”

    “Nikolov & Zeller came up with 154.3 K for the average temperature for the Moon”
    N&Z used TSI 1362, albedo 0,11 and em. 0,955.
    They also included the Cosmic Background Radiation resulting in 2,7K.
    In a non-rotation situation this could explain the 1,3K difference.

  131. Ben Wouters says:

    br says: May 6, 2014 at 10:06 am

    “My feeling is that if one starts with an ice planet, it will remain an ice planet, with quite a low average temperature well below freezing. ”

    Agree. My estimate would be some 10K below the Te of 255K, due to the uneven temperature distribution between equator and poles and the day/night difference.

  132. br, May 6, 2014 at 10:06 am

    “….that will depend on how you include albedo. Water has a far lower albedo than ice, so will absorb more solar radiation which in turn increases the average temperature significantly. Definitely a positive feedback there! Furthermore, light will penetrate deeper into water than ice, and this will increase the thermal time constant.”

    The numbers I quoted above were with the same Albedo as the Moon (~0.12) which might be correct if the ice was dirty enough! Then I tried and Albedo of ~0.5 for clean ice and got this:
    Tmax = 262.8 K, Tave = 241.6 K, Tmin = 227.7 K

    I will submit my calculations to Tim in the form of a spreadsheets and notes later today. He will be able to pass them on to you for comment.

  133. br says:

    Ben Wouters says:

    ‘If I’m correct about the geothermal flux’
    What is your position here? That 0.02 W/m^2 adds significantly to average surface temperature? I think you have trouble with the SB formula. Radiation is the only way a body in space can lose heat, unless it is a star blowing off matter, so for planets and moons SB is the only game in town.

    ‘Less radiation means a lower emitting temperature.’
    Now I know you have trouble using the SB formula.

    The formula for power emitted per unit area is

    P=e*SB*T^4

    In your example, 1364*(1-0.11)=1214 W/m^2 is absorbed, so that much power needs to be emitted. The two equations are then
    1214 = 1*(5.67e-8)*T^4 -> T=382.5 K
    and
    1214 = 0.95*(5.67e-8)*T^4 -> T=387.5 K

    ‘N&Z used TSI 1362, albedo 0,11 and em. 0,955.
    They also included the Cosmic Background Radiation resulting in 2,7K.
    In a non-rotation situation this could explain the 1,3K difference.’
    I also included 2.7K on the dark side, but used an emissivity of 0.98, which is what the Diviner crowd seemed to say. As I’m sure we all agree, a higher emissivity will result in a lower temperature 😉

  134. Ben Wouters, May 6, 2014 at 3:44 pm

    “My estimate would be some 10K below the Tave of 255K, due to the uneven temperature distribution between equator and poles and the day/night difference.”

    You are definitely in the right ball park although my guess is a little lower still. It all depends on what assumptions one makes for the surface properties of an airless Earth. My calculations based on a surface made of ice (Albedo = 0.5) suggests an average of ~242 K at the equator and ~210 K averaged over all latitudes.

    It should be noted that Europa appears to have an icy surface and its Albedo is higher still
    (~0.67).

  135. Ben Wouters says:

    br says: May 6, 2014 at 7:23 pm

    “‘If I’m correct about the geothermal flux’
    What is your position here? That 0.02 W/m^2 adds significantly to average surface temperature?”
    and
    “I also included 2.7K on the dark side”

    I do hope you realise the flux for this 2,7K is something like 0,0001325 W/m^2.
    The geothermal flux is ~150x larger than the CBR, yet you include the CBR and claim the geothermal flux is insignificant ??
    If we could tow the moon to outer space it would have a temperature of at least ~25K all over its surface. To me this is pretty significant in explaining the difference between the basic radiative temp of ~154K and the actual temp of ~197K.

    “In your example, 1364*(1-0.11)=1214 W/m^2 is absorbed, so that much power needs to be emitted.”
    NEEDS to be emitted? The moon is a pretty large heatsink, During one day the sun warms the upper ~30 cm, which cools again during the night. We have probably only two short moments during a lunar day when the incoming and outgoing fluxes are equal.

  136. Ben Wouters says:

    gallopingcamel says: May 6, 2014 at 9:44 pm

    I assume you changed my Te into Tave.
    I consciously used Te (Effective temperature) which I consider the absolute maximum temperature possible, given a certain TSI and albedo.

    Could you use “magic” water, that never freezes and always has the same specific heat capacity to see what the global temperature would be using TSI 1364 and albedo 0,30?
    Solar influence would reach down 10 or perhaps even 100 meters.

    Just noticed Triton, albedo 0,89. Seems like a mirror 😉

  137. gallopingcamel says:

    Ben Wouters,
    My apologies for editing your comment and getting it wrong!

    At present my model does not have the ability to cope with liquids. I will need to look at that to see whether it can be included without exceeding the limited capabilities of the free version of Quickfield,

  138. br says:

    Ben Wouters says: May 7, 2014 at 1:01 pm
    ‘I do hope you realise the flux for this 2,7K is something like 0,0001325 W/m^2.’
    No I don’t. I realise it is something like 0.0000030133 W/m^2.

    ‘The geothermal flux is ~150x larger than the CBR, yet you include the CBR and claim the geothermal flux is insignificant ??’
    No, the geothermal flux is ~6637x the CBR, yet it is still insignificant. They are both insignificant.

    ‘If we could tow the moon to outer space it would have a temperature of at least ~25K all over its surface. To me this is pretty significant in explaining the difference between the basic radiative temp of ~154K and the actual temp of ~197K.’
    No it isn’t, I posted the maths above already. You might find this useful: https://www.youtube.com/watch?v=nfZ12UGiisM#t=23

    ‘“In your example, 1364*(1-0.11)=1214 W/m^2 is absorbed, so that much power needs to be emitted.”
    NEEDS to be emitted?’
    Yes, needs to be emitted. I was discussing your statement ‘TSI 1364 albedo .11 em. 1.0 SB > 382,5K (BB case)’. You didn’t take a rotating sphere into account here, so neither did I.

  139. br says:

    gallopingcamel says: May 7, 2014 at 1:38 pm
    ‘At present my model does not have the ability to cope with liquids.’
    Well, it won’t do circulation (point 5 of my 8 point list above), and probably won’t do penetration depth, and for a start we can ignore evaporation, but so long as you give it a thermal capacity, conductivity and albedo appropriate to water, then you will get some first-order ball-park. I still find that informative, as it is one step in the right direction (of having a complete model, which I appreciate we are all a long way from). Doing any of this ‘properly’ is a tough job, but so long as you keep a clear note of what approximations go in to each step, then there is some chance it may still mean something.

  140. Ben Wouters says:

    br says: May 7, 2014 at 2:50 pm

    ” I realise it is something like 0.0000030133 W/m^2.”
    You’re fully correct. My bad. I used the “fudge factor” N&Z used in their formula to get to 2,7K for situations without any solar radiation.

  141. Ben Wouters says:

    br says: May 7, 2014 at 2:50 pm

    “‘If we could tow the moon to outer space it would have a temperature of at least ~25K all over its surface. To me this is pretty significant in explaining the difference between the basic radiative temp of ~154K and the actual temp of ~197K.’
    No it isn’t, I posted the maths above already”

    Planet with TSI 1364, albedo 0,11, non rotating.
    Average “sunny” side temperature 308K (or 306K, whatever)

    Dark side no solar, so 0K. Average temperature (308 + 0 ) / 2 = 154 K.
    Dark side 2,7K due CBR. Average temperature (308 + 2,7) / 2 = 155,35 K.
    Dark side 25K due geoth. Average temperature (308 + 25 ) / 2 = 166,5 K.

    What is wrong in the geothermal calculation?
    If nothing, why would a very slow rotation all of sudden wipe out the 12,5K effect of the geo flux?

  142. Ben Wouters says:

    br says: May 7, 2014 at 2:50 pm

    “NEEDS to be emitted?’
    Yes, needs to be emitted. I was discussing your statement ‘TSI 1364 albedo .11 em. 1.0 SB > 382,5K (BB case)’. You didn’t take a rotating sphere into account here, so neither did I.”

    I tried to point out that the equatorial max temp. is above the max. achievable SB temperature.

    Imo emissivity doesn’t increase the surface temperature, unless the not emitted radiation warms the heat sink above the radiative balance temperature for the incoming radiation.
    In the above case it means that the surface temp. will be 382,5K, but with em 0,95 will emit like a BB at 363.4K. The not emitted energy will warm the subsurface heat sink, which is pretty large on the moon 😉

    I assume the Diviner measured IR temp of 363,4K is corrected to the actual 382,5K (as a thermometer on/in the surface would measure.)

  143. br says:

    Ben Wouters says:

    ‘What is wrong in the geothermal calculation?’
    Nothing

    ‘If nothing, why would a very slow rotation all of sudden wipe out the 12,5K effect of the geo flux?’

    A very, very, very slow rotation wouldn’t. The problem is that the moon doesn’t rotate very, very, very slowly, it only rotates very slowly. The difference is that the dark side of the moon has a temperature around 95 K, as it is still getting rid of the daytime temperature. In that case, geothermal increases the temperature to 95.1048 K. As a global average, including the daytime side where geothermal makes practically no difference, results in a 0.05 K contribution. As posted above already.

    ‘I tried to point out that the equatorial max temp. is above the max. achievable SB temperature.’
    But that is incorrect, as you have the SB formula wrong.

    The correct version per unit area (please check it out!) is:
    P=e*SB*T^4

    whereas you have
    P.e=SB*T^4

    Please use the correct form and try again, as your calculations are presently incorrect and causing confusion.

    ‘I assume the Diviner measured IR temp of 363,4K is corrected to the actual 382,5K (as a thermometer on/in the surface would measure.)’
    The daytime maximum Diviner measured temperature is 386.6 K, from their data. The radiative equilibrium temperature (TSI 1364, albedo 0.11, emissivity 0.95) is 387.5 K. No need to invoke anything else.

    A body with lower emissivity must go to higher temperature to get rid of a given amount of absorbed power. Emissivity and T are on the same side of the equation so they compensate each other. This is basic physics, please check on it to your own satisfaction.

  144. Ben Wouters says:

    br says: May 9, 2014 at 9:50 am

    “‘If nothing, why would a very slow rotation all of sudden wipe out the 12,5K effect of the geo flux?’
    A very, very, very slow rotation wouldn’t”

    Moons day lasts ~28 earth days. Is a slow rotation in my book.

    `The difference is that the dark side of the moon has a temperature around 95 K`
    And due to this temperature and the much higher day temps the geothermal gradient will be such that the subsurface temp is roughly equal to the average temperature, and increases towards the core. Just 30 cm below the surface on the equator the temp is 240K, facilitating the rise to high temps at noon, and preventing very low temps during the night.
    see http://en.wikipedia.org/wiki/Geothermal_gradient especially the part under `Variations`

    [Moderation note] Rescued from spam bin. Sorry for the delay Ben, I’m busy getting 9000 campaign leaflets delivered for the Euro and local elections. TB

  145. Ben Wouters says:

    “[Moderation note] Rescued from spam bin. Sorry for the delay Ben, I’m busy getting 9000 campaign leaflets delivered for the Euro and local elections. TB”

    No problem. Is it the embedded link or something else that triggers the spam filter?

  146. Ben Wouters says:

    br says: May 9, 2014 at 9:50 am

    Maybe we’re talking past each other.

    Simple thought experiment.

    2 surfaces in outer space, that can only lose energy by radiation at their surface, 1 m^2 each and electrically heated.
    Several embedded temp. probes make sure the surface has an equal temperature.
    Tile A has a (non existent) perfect blackbody surface, Tile B concrete or similar with em 0,60
    We heat both tiles to 290K and maintain that temp by regulating the electrical current.

    Tile A will radiate 400 W/m^2 to space and requires 400 J/s to maintain 290K.
    Tile B will radiate 240 W/m^2 to space and requires 240 J/s to maintain 290K.

    Measuring the temperature from above with an IR temperature sensor for Tile A we’ll read 290K,
    for Tile B we’ll read 255K.

    What of the above do you not agree with and why not?

  147. br says:

    Ben Wouters says:May 12, 2014 at 6:28 pm
    ‘Tile A will radiate 400 W/m^2 to space and requires 400 J/s to maintain 290K.
    Tile B will radiate 240 W/m^2 to space and requires 240 J/s to maintain 290K.’
    Agreed

    ‘Measuring the temperature from above with an IR temperature sensor for Tile A we’ll read 290K,
    for Tile B we’ll read 255K.’
    So long as the sensor is not calibrated for surface emissivity, yes. If the operator of the sensor knows that the emissivity is 0.6, then that can be factored in and the sensor will read 290 K.

    All good so far. What next?

  148. Ben Wouters says:

    br says: May 12, 2014 at 8:10 pm

    “So long as the sensor is not calibrated for surface emissivity, yes. If the operator of the sensor knows that the emissivity is 0.6, then that can be factored in and the sensor will read 290 K.”
    I guess the Diviner sensors are calibrated for em. 1.0, so for every location on the moon the emissivity has to be established to arrive at the actual temperature.

    “All good so far. What next?”
    A second moon, without geothermal heat, so the body temperature is 0K throughout.
    Temp. profile for the equator shows 0K from surface to core for the first rotation. At sunrise the surface temp. will start to rise, always lagging the radiative balance temp., since the underlying heat sink has to heat up first.
    At noon with max. radiation, the temp. will not be at max. radiative temp. and the heat sink most certainly hasn’t heated up yet. Only the upper 1 or 2 cm reach peak temp. and start to cool down when the radiation reduces to zero again during the afternoon.

    Question is what temperature will be left after a night long cooling. If above 0K the average temperature will continue to rise until an equilibrium is reached between incoming and outgoing energy over the whole day.
    Imo this avg. temp. will be lower than on the actual moon with geothermal heat.

  149. br says:

    Ben Wouters says:May 14, 2014 at 11:31 am
    ‘If above 0K the average temperature will continue to rise until an equilibrium is reached between incoming and outgoing energy over the whole day.’
    That’s for sure. At the end of rotation number 1 on day 1, there will of course be a little bit of stored solar heat left over. After billions of years, that can add up to whatever it takes for daily equilibrium. I guess we are calling that different from *geothermal* heating, which seems to be everything that isn’t solar storage (gravitational collapse at formation + nuclear reactions + tidal(?)).

    ‘Imo this avg. temp. will be lower than on the actual moon with geothermal heat.’
    I agree, but you would need to do some hard sums to figure out by how much.

    The sums I did came out at about 0.05 K difference. Anything wrong with them?

  150. tchannon says:

    Ben, if you want a fair answer you can get it by modifying one of the lunar simulations.

    The sources of ground heat / geothermal heat whatever you want to call it comprises

    Heat from Lunar formation a long time ago
    Radioactive decay
    Mechanical loss from crustal tidal flexure

    Add any others you think are there and estimate a total flux value per unit area.

    With the model remove solar input and put the above flux at the bottom of the stack.

    Run the model until stabilised. This doesn’t need the body spin except best not change too much, later can add dawn.

  151. br says:

    tchannon says:May 14, 2014 at 2:29 pm
    ‘Run the model until stabilised.’
    I know you were directing this at Ben, but I couldn’t resist. Please find updated LTSpice code below.

    I now have a geothermal heat source, with a ‘.param geothermal=0.02’ option for easy changing. This heat source is applied to the surface node, as that was how the number was specified in the reference. I also tried it at the ‘deep’ node, and it made almost no difference to the surface but it does make the underlying layers noticeably hotter! So that’s nice. The model includes solar incidence, lunar rotation rate, several material layers going deeper under the surface, heat capacitance, heat conductivity, albedo and emissivity, all consistent with everything we know from Diviner data.

    Result for global average with ‘.param geothermal=0’ is:
    tglobalave: AVG(tlatareasum/cossum)=198.366 K

    Result with ‘.param geothermal=0.02’ is:
    tglobalave: AVG(tlatareasum/cossum)=198.423 K

    so an increase of 0.057 K. Compared to my rough ballpark calculation of 0.055 K above, is pretty good!

    Ben can of course install LTSpice and change the parameters as he wishes.

    Version 4
    SHEET 1 4200 1136
    WIRE 688 -96 0 -96
    WIRE 1008 -80 1008 -112
    WIRE 1008 -80 944 -80
    WIRE 1040 -80 1008 -80
    WIRE 352 -16 160 -16
    WIRE 400 -16 352 -16
    WIRE 352 0 352 -16
    WIRE 944 0 944 -16
    WIRE 1040 0 944 0
    WIRE 352 48 336 48
    WIRE 688 96 688 -96
    WIRE 640 112 544 112
    WIRE 160 128 160 -16
    WIRE 400 192 400 64
    WIRE 688 224 688 176
    WIRE 848 224 688 224
    WIRE 1088 224 848 224
    WIRE 1360 224 1360 160
    WIRE 1360 224 1088 224
    WIRE 1456 224 1360 224
    WIRE 1504 224 1456 224
    WIRE 1616 224 1616 160
    WIRE 1616 224 1584 224
    WIRE 1664 224 1616 224
    WIRE 1776 224 1776 160
    WIRE 1776 224 1744 224
    WIRE 1840 224 1776 224
    WIRE 1952 224 1952 160
    WIRE 1952 224 1920 224
    WIRE 1360 256 1360 224
    WIRE 848 272 848 224
    WIRE 1088 272 1088 224
    WIRE 1456 272 1456 224
    WIRE 1616 272 1616 224
    WIRE 1776 272 1776 224
    WIRE 1952 272 1952 224
    WIRE 400 304 400 192
    WIRE 0 400 0 -96
    WIRE 80 400 0 400
    WIRE 160 400 160 208
    WIRE 160 400 80 400
    WIRE 336 400 336 48
    WIRE 336 400 160 400
    WIRE 400 400 400 384
    WIRE 400 400 336 400
    WIRE 544 400 544 192
    WIRE 544 400 400 400
    WIRE 640 400 640 160
    WIRE 640 400 544 400
    WIRE 848 400 848 352
    WIRE 848 400 640 400
    WIRE 1088 400 1088 352
    WIRE 1088 400 848 400
    WIRE 1360 400 1360 336
    WIRE 1360 400 1088 400
    WIRE 1456 400 1456 336
    WIRE 1456 400 1360 400
    WIRE 1616 400 1616 336
    WIRE 1616 400 1456 400
    WIRE 1776 400 1776 336
    WIRE 1776 400 1616 400
    WIRE 1952 400 1952 336
    WIRE 1952 400 1776 400
    WIRE 80 416 80 400
    FLAG 80 416 0
    FLAG 1360 160 Tsurface
    FLAG 400 192 HalfSine
    FLAG 1616 160 Tshallow
    FLAG 1776 160 Tdeeper
    FLAG 1952 160 Tdeep
    FLAG 944 0 0
    FLAG 1008 -112 latitude
    SYMBOL voltage 160 112 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -48 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName LunarRotationRate
    SYMATTR Value SINE(0 1 {2.95/lunarday})
    SYMBOL res 384 288 R0
    SYMATTR InstName R1
    SYMATTR Value 1e6
    SYMBOL sw 400 80 M180
    WINDOW 3 24 -12 Invisible 2
    WINDOW 0 39 56 Left 2
    SYMATTR Value MYSW
    SYMATTR InstName DayNightSwitch
    SYMBOL g2 688 192 M180
    WINDOW 0 44 58 Left 2
    SYMATTR InstName AbsorbedFlux
    SYMATTR Value 1283
    SYMBOL res 1376 352 R180
    WINDOW 0 36 76 Left 2
    WINDOW 3 36 40 Left 2
    SYMATTR InstName R2
    SYMATTR Value 40
    SYMBOL bi 848 272 R0
    WINDOW 0 37 37 Left 2
    WINDOW 3 24 80 Invisible 2
    SYMATTR InstName SBemission
    SYMATTR Value I=0.98*(5.67e-8)*pwr(V(Tsurface),4)
    SYMBOL cap 1440 272 R0
    SYMATTR InstName C1
    SYMATTR Value 0.1
    SYMBOL cap 1600 272 R0
    SYMATTR InstName C2
    SYMATTR Value 0.25
    SYMBOL res 1600 208 R90
    WINDOW 0 0 56 VBottom 2
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    SYMATTR InstName R3
    SYMATTR Value 7
    SYMBOL cap 1760 272 R0
    SYMATTR InstName C3
    SYMATTR Value 0.18
    SYMBOL res 1760 208 R90
    WINDOW 0 0 56 VBottom 2
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    SYMATTR Value 10
    SYMBOL bv 544 96 R0
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    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName Sun
    SYMATTR Value V=pow(V(HalfSine),1.3)*pow(cos(V(latitude)*pi/180),1.3)
    SYMBOL cap 1936 272 R0
    SYMATTR InstName C4
    SYMATTR Value 0.3
    SYMBOL res 1936 208 R90
    WINDOW 0 0 56 VBottom 2
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    SYMATTR Value 50
    SYMBOL voltage 944 -96 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 0 0 Left 2
    WINDOW 0 -99 6 Left 2
    WINDOW 3 24 96 Invisible 2
    SYMATTR InstName LatStep
    SYMATTR Value PWL(0 5 {twait} 5 {twait+1} 15 {2*twait} 15 {2*twait+1} 25 {3*twait} 25 {3*twait+1} 35 {4*twait} 35 {4*twait+1} 45 {5*twait} 45 {5*twait+1} 55 {6*twait} 55 {6*twait+1} 65 {7*twait} 65 {7*twait+1} 75 {8*twait} 75 {8*twait+1} 85 {9*twait} 85)
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    SYMATTR Value I={geothermal}
    TEXT 696 496 Left 2 !.tran {9*twait}
    TEXT 696 536 Left 2 !.model MYSW SW(Ron=0.001 Roff=1000Meg Vt=0 Vh=0)
    TEXT 696 584 Left 2 !.ic V(Tsurface)=240 V(Tshallow)=240 V(Tdeeper)=210 V(Tdeep)=240
    TEXT 696 672 Left 2 !.meas Tave1 avg V(Tsurface) TRIG at {twait-2*lunarday/2.95} TARG at {twait}
    TEXT 104 496 Left 2 !.param lunarday 29.5
    TEXT 696 456 Left 2 !.param twait=(ceil(convergetime*2.95/lunarday)+2)*lunarday/2.95
    TEXT 696 712 Left 2 !.meas Tave2 avg V(Tsurface) TRIG at {2*twait-2*lunarday/2.95} TARG at {2*twait}
    TEXT 696 752 Left 2 !.meas Tave3 avg V(Tsurface) TRIG at {3*twait-2*lunarday/2.95} TARG at {3*twait}
    TEXT 696 792 Left 2 !.meas Tave4 avg V(Tsurface) TRIG at {4*twait-2*lunarday/2.95} TARG at {4*twait}
    TEXT 696 832 Left 2 !.meas Tave5 avg V(Tsurface) TRIG at {5*twait-2*lunarday/2.95} TARG at {5*twait}
    TEXT 696 864 Left 2 !.meas Tave6 avg V(Tsurface) TRIG at {6*twait-2*lunarday/2.95} TARG at {6*twait}
    TEXT 688 904 Left 2 !.meas Tave7 avg V(Tsurface) TRIG at {7*twait-2*lunarday/2.95} TARG at {7*twait}
    TEXT 696 952 Left 2 !.meas Tave8 avg V(Tsurface) TRIG at {8*twait-2*lunarday/2.95} TARG at {8*twait}
    TEXT 696 1000 Left 2 !.meas Tave9 avg V(Tsurface) TRIG at {9*twait-2*lunarday/2.95} TARG at {9*twait}
    TEXT 704 1040 Left 2 !.meas Tlatareasum avg (Tave1*cos(5)+Tave2*cos(15)+Tave3*cos(25)+Tave4*cos(35)+Tave5*cos(45)+Tave6*cos(55)+Tave7*cos(65)+Tave8*cos(75)+Tave9*cos(85))
    TEXT 704 1072 Left 2 !.meas cossum avg (cos(5)+cos(15)+cos(25)+cos(35)+cos(45)+cos(55)+cos(65)+cos(75)+cos(85))
    TEXT 696 1112 Left 2 !.meas Tglobalave avg Tlatareasum/cossum
    TEXT 104 552 Left 2 !.param convergetime=40
    TEXT 1008 192 Left 2 !.param geothermal=0.02
    
  152. br,
    The FEA model has the Moon’s internal heat source built in by means of the 240 K temperature assumed for the bedrock. In my model the regolith is 500 mm thick. Increasing the depth of the regolith has very little effect on the surface temperature.

  153. tchannon says:

    Ah, already there. What-if is why we do these things.

    A question is whether Ben is convinced, perhaps for the Lunar case, the earth being a very different situation without good cross check data.

  154. br says:

    gallopingcamel says:May 14, 2014 at 10:33 pm

    ‘The FEA model has the Moon’s internal heat source built in by means of the 240 K temperature assumed for the bedrock.’
    great, seeing as the average equatorial surface temperature is about 214 K, that temperature difference will provide a heat flux to the surface.

    ‘In my model the regolith is 500 mm thick. Increasing the depth of the regolith has very little effect on the surface temperature.’
    I guess the question is what a change in underground temperature will do. If you run the model with the bedrock at 214 K (solar storage temperature), what difference does it make?

  155. Ben Wouters says:

    The geoflux at the equator is only relevant when the surface temp is below 240K. Otherwise the geoflux is zero.
    Surface temp. below 240K will “expose” the lunar subsurface rock, which is at 240K just 30cm below the surface and warmer deeper down. During the night this can create a surface temp. higher than the leftover heat from the previous day can account for.
    We have already seen that in the no-rotation case the geoflux adds ~12,5K to the average temperature for the whole moon. A rotation rate of once every 28 earth days is pretty close to no rotation.

    “I guess the question is what a change in underground temperature will do. If you run the model with the bedrock at 214 K (solar storage temperature), what difference does it make?”
    On a moon without geothermal heat and energy in/out at the surface in equilibrium, how deep will the solar influence reach? 50cm, 1m, 10m? There is no energy surplus to heat the moon deeper down.
    So the question should be what the surface temp. will be with the bedrock at 0K.

    What the geothermal gradient does is decrease the volume the sun has to heat during the day. Less volume to heat, same energy in will result in a higher temperature. This is imo the reason, the noon temperature is higher than the pure radiative maximum can explain.

  156. br says:

    Ben Wouters says: May 15, 2014 at 10:15 am
    ‘During the night this can create a surface temp. higher than the leftover heat from the previous day can account for.’
    by about 0.1 K.

    ‘A rotation rate of once every 28 earth days is pretty close to no rotation.’
    no it isn’t.

    ‘On a moon without geothermal heat and energy in/out at the surface in equilibrium, how deep will the solar influence reach?’
    after billions of years, all the way to the core.

    ‘the noon temperature is higher than the pure radiative maximum can explain.’
    no it isn’t, you had emissivity the wrong way around.

    In your tile example you had:
    ‘Tile A will radiate 400 W/m^2 to space and requires 400 J/s to maintain 290K.
    Tile B will radiate 240 W/m^2 to space and requires 240 J/s to maintain 290K.’
    But that doesn’t correspond to the moon. The moon absorbs TSI*(1-albedo). Note that emissivity does not appear in that, it is albedo. This absorbed power acts as the heater power in your tile example. If Tile A and Tile B both have an albedo of 0.11, then they both have the same ‘heater power’, and need to radiate that power away again in order for their respective temperatures to be in equilibrium. When Tile A and Tile B both have to radiate 400 W/m2, what would their respective temperatures be???

  157. Ben Wouters says:

    br says: May 15, 2014 at 11:10 am

    ” need to radiate that power away again in order for their respective temperatures to be in equilibrium.”
    The equatorial surface temperature is not in equilibrium with the incoming radiation, at least not in the morning. TSI 1364, albedo 0,11 em 1,0 > SB gives 382,5K.
    Max temps go up to 400K in some places.
    Max radiation is at noon, and lasts only an instant. The max temp is only reached in the upper centimetre or so. To reach equilibrium the whole heat sink below the surface has to heat up first.
    Is not happening.
    Only effect of emissivity 0,95 is that 5% more solar is available to heat up the heat sink in the morning, since it is not radiated away according the actual surface temp.

  158. br asked:
    “I guess the question is what a change in underground temperature will do. If you run the model with the bedrock at 214 K (solar storage temperature), what difference does it make?”

    With 0.5 meters of regolith, changing the bedrock temperature from 240 to 214 K lowered Tmin by 0.4 K and Tave by 0.2 K. No significant effect on Tmax.

    I tried increasing the regolith thickness from 0.5 to 0.75 meters and found -0.2 and -0.1 K respectively.

  159. br says:

    Ben Wouters says: May 15, 2014 at 6:39 pm
    ‘The equatorial surface temperature is not in equilibrium with the incoming radiation, at least not in the morning. TSI 1364, albedo 0,11 em 1,0 > SB gives 382,5K.’
    The moon doesn’t have an emissivity of 1.0.

    ‘Max temps go up to 400K in some places.’
    local spots with lower emissivity might indeed have higher temperatures.

    ‘Max radiation is at noon, and lasts only an instant. The max temp is only reached in the upper centimetre or so. To reach equilibrium the whole heat sink below the surface has to heat up first.
    Is not happening.’
    Agreed. The surface temperature at noon is slightly lower than for radiative equilibrium.

    ‘Only effect of emissivity 0,95 is that 5% more solar is available to heat up the heat sink in the morning, since it is not radiated away according the actual surface temp.’
    Incorrect, that is not the only effect of emissivity. I see you didn’t answer my ‘Tile A vs Tile B’ question. Tut, tut.

  160. br says:

    gallopingcamel says: May 15, 2014 at 9:59 pm
    ‘I tried increasing the regolith thickness from 0.5 to 0.75 meters and found -0.2 and -0.1 K respectively.’
    Sounds good to me, though it sounds like the model hadn’t quite converged. After all, at some (relatively small) depth there should be no fluctuations between day and night temperature. Increasing the depth beyond that thickness should make no difference. So to check for convergence, make the depth deeper until there is no further difference. That might also require running the simulation for longer, as it has to calculate more transport, I’ll leave that up to you.

  161. Ben Wouters says:

    br says: May 16, 2014 at 11:26 am

    “‘Only effect of emissivity 0,95 is that 5% more solar is available to heat up the heat sink in the morning, since it is not radiated away according the actual surface temp.’
    Incorrect, that is not the only effect of emissivity. I see you didn’t answer my ‘Tile A vs Tile B’ question. Tut, tut”

    Didn’t answer because the question is totally irrelevant. The surface temperature of the moon is never in equilibrium with incoming radiation. As long as the subsurface heat sink hasn’t heated up fully the surface temp. will not reach it’s maximum potential temperature.
    The reason for the high noon temperature is the high starting temp at sunrise, ~100K.
    (and 10 cm below the surface the temp is still ~210K at sunrise)

    Solar is barely able to increase the temperature of the upper 1 cm. from ~100K to ~380K at noon.
    Just 10 cm below the surface the temperature then is only ~250K, barely 10K above the 240K at ~30cm, which is caused by the geothermal flux.

  162. br says:

    Ben Wouters says: May 16, 2014 at 1:39 pm
    ‘Didn’t answer because the question is totally irrelevant’
    The fact that lower emissivity materials have higher temperature when radiating a given amount of power is totally relevant.
    The answer to the question: ‘When Tile A and Tile B both have to radiate 400 W/m2, what would their respective temperatures be?’ is that tile A would have a temperature of 290 K, while tile B would have a temperature of 329 K. Would you agree with those numbers, to nearest integer? The lower emissivity material in this case has a 39 K higher temperature. I’m only trying to find common ground.

    ‘As long as the subsurface heat sink hasn’t heated up fully the surface temp. will not reach it’s maximum potential temperature.’
    Agreed. The maximum potential temperature for the moon is 387.5 K (albedo 0.11, emissivity 0.95), and it only reaches 386.6 K, no argument there.

  163. tchannon says:

    Posted by moderator on behalf of gallopingcamel who had a spot of bother getting a PDF export. Kicked up at me too! Reformatted a little. And of course I can upload.

    I hope that is right.

    br: –

    “The chart below shows how temperature varies with depth in my FEA model
    with 0.5 meters of regolith and a bedrock temperature of 240 K. It
    looks quite similar to “Figure 7″ in the Vavasada et al., 2012.

    Adding more regolith will lower the Tmin and Tave by a few tenths of a
    degree Centigrade but I am running out of nodes in the student version
    of Quickfield.”

    https://tallbloke.files.wordpress.com/2014/05/depth.pdf

    (loaded into openoffice calc, changed wall colour and trace colours, then dragged a copy to draw, export selected to pdf… sheesh, how to avoid trace jaggies and other sillies)

  164. […] article is part II  of “A new Lunar thermal model based on Finite Element Analysis of regolith physical properties“,  written primarily by gallopingcamel (Peter Morcombe), edited and prepared for WordPress […]