Would the Apollo Moon astronauts have been frozen then cooked like chicken-in-a-bag - One Star Warrior unveiled - page 28

great title

chicken in a bag? Is that in fashion these days?

It looks like it is over here..

[link to www.dailymail.co.uk]

seems like we can't brush our teeth or wash our hands

K

Astronauts are not spherical, well not unless they have really let themselves go but as an approximation to find the average amount of insolation over their spacesuits, over time, how good is it?

We will use a value of 1362 W/m^2 for the solar constant.

The proposal for a spacesuit design from B.F. Goodrich [link to commons.erau.edu] modelled the astronaut as a cuboid ( box ) but we can do better than that. Finding a regular shape that reflects what a space-suited astronaut looks like is difficult but an elliptical prism [link to www.efunda.com] is better than most.

Taking the height of the prism to be 2m
the average width to be 50 cm ( he will be broader at the shoulders and narrower at the ankles )
the average depth to be 30 cm

we can find the surface area of the curved surface to be 200cm * 128cm ( circumference ) = 2.56m^2
and the ends to be 2 * 1180cm^2 = 0.24m^2
gives a total of 2.8m^2 surface area, in the right ball park for a suited astronaut and in line with our figure of 2.5 to 3 m^2 we had been using.

A sphere of 2.8m^3 would always present the equivalent of 2.8/4=0.7 m^2 of flat surface area to the Sun and so its insolation will always be 1362 W/m^2 * 0.7 m^2 = 953 W

The astronaut will be turning to present different aspects to the sun as he goes about his work so we can find the approximate average diameter like so D = sqrt ( ( a^2 + b^2 ) /2 ) giving us a figure of 0.41 m and therefore an average area perpendicular to the surface of 0.41m * 2m = 0.82 m^2 Remember the top surface parallel to the lunar surface is 0.12 m^2

for a given sun elevation "e" then the total insolation will be

1362 W/m^2 *( (cos(e) * 0.82 ) + ( sin(e) * 0.12 )

[link to www.wolframalpha.com]

So for lunar dawn ( never during Apollo ) the figure will be 1117 W, 17% higher than a sphere.

The lowest Sun elevation during landing was Apollo 12 at 5 deg = 1127 W 18% higher than a sphere.

The highest elevation at landing was Apollo 16 at 15 deg = 1121 W 18 % higher than sphere.

The lowest elevation at EVA termination was 15 deg for Apollo 11 = 1121 W 18% higher than a sphere.

The highest elevation at EVA termination for which we have a regolith temperature is Apollo 15 at 41 deg = 950 W the same value as a sphere.

The highest elevation at EVA termination was Apollo 16 at 48 deg = 869 W 9% lower than a sphere.

So overall, compared to a sphere, the eliptical prism will tend to smooth out the temperature variations as the suit will have more insolation in the morning when the regolith is cold and less at the highest Sun angles when the regolith is warm. Crucially the value for insolation at the time of Apollo 15 EVA termination ( which we have the highest recorded regolith temperature ) is almost exactly the same as the value derived from using a sphere as a model.

K

Man. Is there _anything_ IDW ever gets right?

IUPAC accepts both spellings -- but worldwide, few countries outside of the US and Canada use "aluminum."

Speaking solely from personal observation, i.e. those posts by him I've read, the answer seems to be: No.

Next up: rant by KBW on why the insignificant part of the world that isn't the US and Canada (including England) got it all wrong.

Here is a very useful site for calculating radiation view factors. [link to www.thermalradiation.net] I will be using this one instead of the radiation wizard in the next run through.

K

The idea that static on a tell lie vision or a homemade radio telescope proves the "big bang" is perhaps one of the stupidest proofs ever used in "science". It's fucking stupid to the point of ludicrous. To prove such a link is impossible and therefor not a proof.

The static could actually be coming from any of MILLIONS if not BILLIONS of sources.

Bullshit. There was no apollo EVA. Everything else is fantasy based on that bullshit.

Pity none of those "billions and billions" of sources are isotropic.

Edit : just found the context of the quote I was replying to.

K

Called it!

I had to change the thread title to something more unwieldy as some people don't get past the title before giving karma. On the plus side being told "Insane conspiracy theorist is insane. -1" from someone on GLP counts as proof of non compos mentis should I ever be on trial, it doesn't get crazier than that ;)

K

Anyone who may have trouble sleeping can look into this list of standard radiation view factors.

[link to webserver.dmt.upm.es]

I needed something better than the radiation wizard [link to www.thermal-wizard.com] I used first time around ( quick and easy but insufficient for us ) but if I wanted to find just the view factor between two finite rectangles, things start to get complicated, here is the relevant model:
[link to www.thermalradiation.net]
and its formula
[link to www.thermalradiation.net]

So just to find the view factor on a small patch of spacesuit would involve finding the double integral of that function for the entire portion of visible Moon surface and that doesn't take into account that the patch of spacesuit could be angled at any angle to the lunar surface.

K

Maybe KeyBoardWarrior could check those formulae for us



K

He'd have to figure out how exponential notation works first.

Zing!

10^5 = one million

Yes I saw that in 74444's list. One thing I find hard to believe is that he has been doing this for years ! I thought he was about 16 years old, going by his first post on the thread and the way he carried on like a keyboard warrior kid after that.

[link to misterknuckles.com]

K

IDW is quite insane, of course, but still provides tremendous entertainment.

He should never be banned from GLP.

I sometimes think the "IDW" persona is actually a collective - a group of trolls who keep the insane personality going over the years, even though none of them believe a word they post about.

Reading through the posts you can almost recognize several different styles coming through. Some of them are even almost polite :-)

I've never considered that possibility, quite honestly.

It does make sense, though.

Regardless of who or what "IDW" is, I hope the persona keeps posting. It's worth it for the lulz.

Maybe, I am new to this whole IDW thing but he did say:


in this thread. That implies it is one person, whose real name is know in Apollo hoax circles.

K

Yes, he does make that claim, although whether it's true or not is disputed as well.

I'm only peripherally interested in debunking the hoaxies, but from what I've read, if IDW is a single person, he suffers from some serious delusions about both his competence and his influence on NASA and the more serious debunkers.

Most responses I've read to his claims say that they actually have no idea who he is.

Again, whether he's actually contacted NASA is disputed, and may be another of his fantasies.

Oh well, he will be along later to tell us all about himself.

NASA weren't the last people to address the problem of traversing the lunar surface whilst minimising energy expenditure and heating and cooling requirements for astronauts. A software tool called SEXTANT was developed at MIT for just such a purpose. SEXTANT makes use of the lunar landscape to plot a course, taking best advantage of sunshine and shade whilst minimising gradients. [link to dspace.mit.edu]

You would think that SEXTANT would have a carefully modelled view factor that accurately reflects the complex shape of a spacesuit, and it has such a representation. In the paper, that author says:

"Within SEXTANT the only two view factors are between the spacesuit and the lunar surface and between the spacesuit and deep space...each of these is assumed to be..."

some double integral of a function of the surface maybe ?

K

"Within SEXTANT the only two view factors are between the spacesuit and the lunar surface and between the spacesuit and deep space...each of these is assumed to be 0.5"



That's it, a half ??

Then I found this [link to www.thermalradiation.net]

Using an infinite plane is one of those nice examples where a lot of complicating factors cancel each other out leaving the equation for finding the view factor from a rectangle (1) to an infinite plane (2) of

F sub(1,2)= ( 1 - cos(eta))/2

Integrating that function between 0 and pi to find the sum of all values for an external surface at an angle to the infinite plane ( Moon's surface in our case ) and finding the mean of the integral by multiplying it by 1 over the difference of its limits we get ( I can't do it in ascii so you will have to look at Wolfram Alpha )

[link to www.wolframalpha.com]

Yes it's a half ( or as WA puts it - about 0.50000 )

What is more this holds true for any surface you can construct between the base and an eqivalent upper face, so it really doesn't matter how bulging or wrinkled the spacesuit is. The view factor of the portion of the space suit between the base and upper surface will be ( nearly ) 0.5

K

That's strange he must have been too busy shouting at the pigeons in the street again.

The actual figure for the spacesuit view factor is smaller than 0.5 for a couple of reasons. Firstly the Moon's surface is not a flat infinite plane and the lunar surface itself is convex. At 1m above the surface the horizon is 1.9 km away. The second reason is that the summation of view factors tells us that the upper surface of the spacesuit ,equal in area to the boot prints, has a view factor of zero. The soles of the boots themselves won't be involved in this model because their main thermal process is conduction. So our overall view factor for the radiative part of the spacesuit is 0.5 * ( 2.56 / 2.68 ) * 0.98 = 0.46

K

Eh, but wouldn't that only hold for a basically smoothly spherical Moon? I suspect local relief is going to play a part. My first guess is that it averages out on the whole, but I'd be interested in seeing what the extremes were.

Still does seem at first-order approximation the external contribution is swamped by internal sources. Nice work on figuring out the contribution of internal electronics. Came close to my gut estimate, which when I worked on this (at a much much coarser grain!) I used 200 watts as the metabolic base (aka, folded in contribution of suit electronics and used a figure for active human instead of resting BMR).

You have to understand that actually *doing* math is IDW/A.A's kryptonite. It gives him nowhere to go, and no room to wriggle.

You are certainly right. The SEXTANT software does route planning to smooth out thermal variations. Standing in front of a sunlit ridge would increase your view factor and thermal input, standing on the ridge would reduce it. I guess craters are a bit more complicated from a thermal perspective because some of the crater floor could be in shadow, even though your view factor of the lunar surface is high, your thermal input might be lower. I could try a "crater of death" scenario later to see how that works out.


Yes I had a very detailed technical document on the spacesuit and PLSS. The radio transceiver was bolted directly onto the sublimator. The oxygen circulator fan was cooled by the gas flow. The water pump didn't originally have dedicated cooling but they found that it overheated so they used the cooling water flow to cool it. I think the pump at high flow would use an extra 4W but that's about it for extra electrical load.

The metabolic output ( for later missions A15-A17 ) was taken to be half driving ( 144 W ) and half brisk walking ( 360 W similar to "Forestry -cutting across the grain with a one-man power saw" ) but I am happy to take any reasonable value, in fact I have been asking people for their thoughts on this value, using this table to help [link to www.engineeringtoolbox.com]


K

Ooh, nice resource! And I guess I shouldn't be surprised that the active outputs are about twice what I was guestimating.

Heh. Zeroth-order approximation, human-powered aircraft studies give something like 400 watts for peak power at the crank, and assuming a biological efficiency below 50%, that translates out to peak exercise values nearing a kilowatt.

I'm happy enough to stay within OOMs, though, especially for the IDWs around. The claim that an astronaut would die in minutes from the heat isn't one that requires nailing down the third or fourth digit of precision. A power of 2 is close enough! I mean -- the hoaxies think film should MELT. That's oven temperatures. That's not overheating during strenuous exercise, that's a pot roast.