[Split]Moon hoaxes - split from: Mars Voyage too Soon?

[PetersCreek]

You might want to retry that calculation with the correct sun angles.

Edit to add: DOH! Tunnel vision on the angles led me to miss the very good catches by Shrinker and CurtC, below. Taking a tangent on the angular difference is a whopper of an error.

 

[Shrinker]

Well this might not be the most elegant of calculations, but here goes:

Assumptions:



Sun angle above horizon = 30 degrees (this itself, I think, presents an anomaly that might be worth looking into separately!). This has been calculated approximately by comparing the relative height of the flagpole and length of shadow cast by it as measured on the computer monitor (Tan-1 Sun angle = 49mm / 28mm). It can only be approximate because the video camera is not positioned perpendicular to the direction of the shadow. The shadow lengths can be clearly seen to be longer than the objects casting them, so the sun angle must be well under 45 degrees (the exact sun angle is not critical to the calculation, within reason, although it should be noted that the lower the Sun angle the more pronounced the shadow movements will be. I believe that the 30 degrees calculated is lower than actual!).

Flagpole is approximately 2,200mm tall. This has been assessed by comparing it to the height of the astronauts during placement at the start of the video.

Flagpole is planted vertically (this is not critical to the calculation, within reason).

Axis of Moon's rotation is perpendicular to the direction of the shadow, i.e. the shadows will simply elongate and shorten.

Calculations:

First, using trogonometry, let's calculate the length of the flagpole shadow; call it 'X':

X = 2200mm / Tan 30 degrees = 3,811mm

Now, using triangulation, let's calculate the angle formed by the flagpole and an imaginary line running from the tip of the flagpole to the tip of the shadow on the lunar surface; call it 'Y':

Y = 180 degrees - 90 degrees - 30 degrees = 60 degrees

Now, let's calculate the angle of rotation of the Moon over the duration of the video, namely 40 minutes:

40 minutes / (29.5 Earth days x 24 hours x 60 minutes) x 360 degrees = 0.34 degrees

So, after 40 minutes the Moon will have rotated about its axis by 0.34 degrees. In other words, the flagpole will have rotated from the vertical 90-degree position to either 89.66 degrees or 90.34 degrees, depending on which direction the Moon is rotating in relation to the Sun (think of it like a hand on a clock moving away from the 12 o'clock position by 0.34 degrees in either direction).

This rotation or 'tilting' from the 'vertical' results in an apparent reduction in the height of the flagpole from the lunar surface, which will manifest itself as a shortening or lengthening of the shadow. If the flagpole 'tilts' towards the Sun the shadow length will tend towards zero as the 'tilt angle' approaches 30 degrees (i.e. the one o'clock position, with the flagpole pointing directly at the Sun). If the flagpole 'tilts' away from the Sun the Shadow length will initially lengthen as the flagpole tend towards a 60-degree tilt (i.e. the 10 o'clock position, at which point the flagpole will be perpendicular to the 30-degree Sun angle). Thereafter, the shadow length will shorten as the flagpole angle tends towards 0 degrees, i.e. is laying flat on the ground (the 9 o'clock position), at which point there is no shadow!

So, the question is, what does a rotation of the Moon, or 'tilting' of the flagpole, by 0.34 degrees equate to in a shortening or lengthening of the shadow length? Let's calculate it (we need only calculate one and not both as we're only interested in the change in length from the original, so let's assume a shortening, i.e. the flagpole rotates or 'tilts' towards the Sun):

As the light from the Sun can be considered to be travelling in parallel 'rays' the shortening of the shadow length will equate to the horizontal distance that the tip of the flagpole moves from its original position; call it 'Z'. This can be calculated thus:

Z = Tan 0.34 degrees x 2200mm = 13mm.

Now, we could assume that instead of rotating perpendicular to the direction of the shadow the Moon is rotating parallel to it (in reality it is somewhere between the two). I do not believe that this will derive a movement of the shadow across the ground significantly more than the 13mm calculated above, but I'd be happy to have a go, just to test it.

So, 13mm appears to be the maximum 'movement' of the shadow across the lunar surface, which is supported by the Post-it note test.

CurtC - I think you need to check your calculation to identify where you've gone wrong, but may I suggest that you try the Post-it note test first, which will demonstrate that the shadow movement is nothing like 18cm! Everybody else, I think you need to get your eyes tested!

Now, which of you are going to try to convince me that you can detect a 13mm movement (that's about the width of an average finger nail!) of the tip of the shadow across the lunar surface from watching the linked video?! Who's going first?!

Priceless! Such confidence. First of all, who was talking about the flagpole? Not me, or Curt. Read my posts again. I don't know about anyone else, but I was looking at the shadow of the LM. It's moving much faster than the flag shadow since it's taller, but if you want to go for the shorter shadow because it helps your case then go ahead. I also suggested a shadow length to object height ratio of 2, so Curt was using 26.57 degrees as the angle of the sun. Guess you couldn't work that out so you went with your own numbers huh?

Second of all, 18mm!? Are you serious? What the heck was all that junk you typed?


With your assumptions...
Shadow length with sun at 30 degrees above the horizon= 2.2m * tan(90-30) = 3,811mm
Shadow length with sun at 30.35 degrees above the horizon = 2.2m * tan(90-30.35)=3,757mm
Difference = 54mm

Unlucky.

If we assume 20 degrees at the start of the film then the difference over 40 minutes becomes a quite easily detectable 112mm.

Edit: and if we use 15 degrees, the difference is around 200mm

 

[CurtC]

As the light from the Sun can be considered to be travelling in parallel 'rays' the shortening of the shadow length will equate to the horizontal distance that the tip of the flagpole moves from its original position; call it 'Z'. This can be calculated thus:

Z = Tan 0.34 degrees x 2200mm = 13mm.

You can't calculate just the tangent of 0.34 degrees to get the change in the lengths of the sides! The tangent function is very non-linear.

I did the calculations according to Shrinker's conditions: that the shadows are twice as long as they are tall, and that the LM is 6 m tall. So I was figuring how far the top of the LM's shadow will move. I'll review that here.

If the shadow is twice as long as it is tall, that's an angle of 26.57 degrees (atan(0.5)). The LM is 6 m tall, the shadow is 12 m long. I agree with your 0.34 degrees in 40 minutes figure, so at the end, the Sun's angle will be 26.90 degrees. For the 6 m LM, its shadow will now be 6/(tan(26.90)), or 11.82 m. Before, it was 12 m, so it moved by 18 cm.

CurtC - I think you need to check your calculation to identify where you've gone wrong

Nice tone. But if you do the same calculation with a 2.2 m flag pole, you get that it moved by 6.4 cm, not 1.3.

Maybe you can check your calculation to identify where you've gone wrong. Oh wait, I already did that for you - you assumed that the tangent function was linear, that it would change the same amount in going from 0 to 0.34, as it would from 26.57 to 26.90. (Note that the last figures look like they changed by 0.33 instead of 0.34, but I carried the calculations out with more significant digits and rounded off here).

 

[PetersCreek]

Which 'multiple directional light sources' would those be, then?!

Perhaps the two you specified in this post.

 

[uruk]

Well this might not be the most elegant of calculations, but here goes:

Assumptions:

Sun angle above horizon = 30 degrees (this itself, I think, presents an anomaly that might be worth looking into separately!). This has been calculated approximately by comparing the relative height of the flagpole and length of shadow cast by it as measured on the computer monitor (Tan-1 Sun angle = 49mm / 28mm). It can only be approximate because the video camera is not positioned perpendicular to the direction of the shadow. The shadow lengths can be clearly seen to be longer than the objects casting them, so the sun angle must be well under 45 degrees (the exact sun angle is not critical to the calculation, within reason, although it should be noted that the lower the Sun angle the more pronounced the shadow movements will be. I believe that the 30 degrees calculated is lower than actual!).

Flagpole is approximately 2,200mm tall. This has been assessed by comparing it to the height of the astronauts during placement at the start of the video.

Flagpole is planted vertically (this is not critical to the calculation, within reason).

Axis of Moon's rotation is perpendicular to the direction of the shadow, i.e. the shadows will simply elongate and shorten.

{snip}

Math is not my strong suit so I am just going to address your initial assumptions.
The sun angles for the Moon missions are listed here:
http://www.hq.nasa.gov/pao/History/alsj/alsj-sunangles.html

If the video you are refering to is from the Apollo 11 mission then the starting and ending sun angles are :
Apollo 11
Landing: 102.75 GET
EVA
Start: 109.00 GET, 14.0 deg.
Finis: 111.75 GET, 15.4 deg.

edited: Damn! Peters beat me too it.

Your 30 degree estimate seems a bit too generous by a factor of 2. And as you can see the angles shift by 1.4 degrees during the whole mission.

http://www.cs.indiana.edu/sudoc/image_30000061709352/30000061709352/pdf/11post.pdf
According to the above mission report briefing the flag pole was 8 ft tall or 2438.4 milimeters. Not bad. you even got the eight inches to account for the part embeded in the soil.

Try running the numbers with the corrected sun angles.

Now if you could be off by a factor of 2 for the sun angles by making an estimate based on an image. How about Groves? Think he might be off on one or two things?

 

[PhantomWolf]

Anybody care to have a go at debunking this famous shot then (obvious too-low Earth image insertion aside!)? Now who do you figure are the two guys who took that photo (one on camera the other on fill-lighting?!)?!?

You know, after calling other posters here clowns and dumb, I really hope you aren't seriously suggesting this is a real Apollo image.

 

[uruk]

Which 'multiple directional light sources' would those be, then?!

Well didn't you say in this post: http://www.internationalskeptics.com/forums/showpost.php?p=3218561&postcount=647 that the "lux" was the same as with stadium ligting?

You do know that stadium lighting consists of several "banks" of lamps. Right?
One single light source is insuficient, you need several.
an object being illuminated by that type of lighting is going to cast multiple shadows from all those little lamps. Ever notice all the shadows being cast on the stadium ground? Never? Convenient for you huh?

 

[CurtC]

You might want to retry that calculation with the correct sun angles.

Which mission and day was this? I think it was Apollo 15, and if they just put the flag out, that would be the first day, right? In that case, the Sun's angle was around 20 degrees, and the shadow of the top of the flag pole would move by 11 cm in 40 minutes.

Also, you could consider the tangent of just 0.34 degrees, but you'd need to apply that to the distance from the flag pole's top, to the shadow of the flag pole top, not just the height of the flag pole. I'm not sure which is the way Southwind was trying to do it, but either way he made an error that throws his answer way off.

 

[PhantomWolf]

Which mission and day was this? I think it was Apollo 15, and if they just put the flag out, that would be the first day, right? In that case, the Sun's angle was around 20 degrees, and the shadow of the top of the flag pole would move by 11 cm in 40 minutes.

Also, you could consider the tangent of just 0.34 degrees, but you'd need to apply that to the distance from the flag pole's top, to the shadow of the flag pole top, not just the height of the flag pole. I'm not sure which is the way Southwind was trying to do it, but either way he made an error that throws his answer way off.

The DAC footage I posted was from Apollo 11.

 

[Vincent Vega]

I just hope his screen name doesn't imply that there are 16 more like him.

 

[Shrinker]

I just hope his screen name doesn't imply that there are 16 more like him.

More likely just 16 lost passwords.

 

[uruk]

Edit to add: DOH! Tunnel vision on the angles led me to miss the very good catches by Shrinker and CurtC, below. Taking a tangent on the angular difference is a whopper of an error.

Hey Southwind, Do you think Dr. Groves might have made a similar error in his calculations?

 

[Hindmost]

This thread is soooooooooo over. Southwind doesn't appear to be interested in a reasonable exchange; otherwise the real questions would be answered....

glenn

two suggestions for southwind: take an optics course and learn what perspective is

 

[Southwind17]

The DAC footage I posted was from Apollo 11.

You guys don't even know what Apollo Mission video footage we're looking at?!

 

[uruk]

You guys don't even know what Apollo Mission video footage we're looking at?!

Ouch! Bad recovery. Try addressing your faulty math first.

 

[PhantomWolf]

You guys don't even know what Apollo Mission video footage we're looking at?!

I know exactly which mission the footage is from. I think CurtC may have confused the footage I posted with the footage of the flag being brushed/attracted that someone else posted, which was from Apollo 15. As far as I know the shadows discussion is based on the Apollo 11 footage, so I was letting him know so he had the right one. It's quite likely he was looking at the Apollo 15 footage thinking that was what was being discussed.

 

[Southwind17]

You might want to retry that calculation with the correct sun angles.

Thanks. See below.

Edit to add: DOH! Tunnel vision on the angles led me to miss the very good catches by Shrinker and CurtC, below. Taking a tangent on the angular difference is a whopper of an error.

riceless! Such confidence. First of all, who was talking about the flagpole? Not me, or Curt. Read my posts again. I don't know about anyone else, but I was looking at the shadow of the LM. It's moving much faster than the flag shadow since it's taller, but if you want to go for the shorter shadow because it helps your case then go ahead. I also suggested a shadow length to object height ratio of 2, so Curt was using 26.57 degrees as the angle of the sun. Guess you couldn't work that out so you went with your own numbers huh?

The shadow length to object height ratio that I've assumed is 1.75, which, as I wrote, equates to a Sun angle around 30 degrees. There was no guesswork involved; only the acknowldgement that this figure is approximate because the video camera wasn't placed perpendicular to the shadow direction. Seems Curt and I were pretty close to each other, albeit both wrong, but see later!

Second of all, 18mm!? Are you serious? What the heck was all that junk you typed?

If you read carefully I think you'll find I wrote 13mm, so you're only 46% out with this sloppy observation!

With your assumptions...
Shadow length with sun at 30 degrees above the horizon= 2.2m * tan(90-30) = 3,811mm
Shadow length with sun at 30.35 degrees above the horizon = 2.2m * tan(90-30.35)=3,757mm
Difference = 54mm

Unlucky.

Er ... I don't think luck has anything to do with this, and here's the rub: EVERYBODY QUOTED IN THIS POST (with the possible exception of PetersCreek, although I can't figure out why he hasn't pointed it out himself, if so!) IS MAKING THE GRAVE MISTAKE OF BELIEVING THAT THE ANGULAR ROTATION OF THE MOON EQUATES TO AN EQUIVALENT CHANGE IN SUN ANGLE. WRONG!!! THAT'S ONLY THE CASE WHEN OBSERVING THE SUN DIRECTLY OVERHEAD, I.E. FROM A POINT ON THE LUNAR SURFACE THAT'S PERPENDICULARLY TANGENTIAL TO THE DIRECTION OF THE SUN. Think about it: the lower the Sun angle, i.e. the closer the Sun is to the horizon, the more the rotation of the Moon is manifest in movement of an object on the lunar surface DIRECTLY TOWARDS OR AWAY FROM THE SUN. Movement directly toward or away from the Sun has ABSOLUTELY NO EFFECT on Sun angle! This would be most pronounced at a position on the lunar surface that is tangentially parallel to the direction of the Sun, i.e. exactly on the 'top' or 'bottom' of the Moon with the Sun out at the 3 O'clock position (if you could still see the Sun then, which you can't as it's dropped below the horizon).

So, given that the Sun angle was only around 15 degrees, a 0.34 degree change in rotation of the Moon will result in a MUCH SMALLER change in Sun angle!

If we assume 20 degrees at the start of the film then the difference over 40 minutes becomes a quite easily detectable 112mm.

Edit: and if we use 15 degrees, the difference is around 200mm

But it's a shame these figures are hopelessly wrong! Intuition alone should inform you that for an equivalent period of only 83 seconds on Earth you shouldn't really be able to detect the rotation of the Moon by observing changes in a 22m-long shadow from a distance!

You can't calculate just the tangent of 0.34 degrees to get the change in the lengths of the sides! The tangent function is very non-linear.

I'm not sure you're sure what I did. I did the equivalent calculation that you've done hereunder, but without the error that you've made:

I did the calculations according to Shrinker's conditions: that the shadows are twice as long as they are tall, and that the LM is 6 m tall. So I was figuring how far the top of the LM's shadow will move. I'll review that here.

If the shadow is twice as long as it is tall, that's an angle of 26.57 degrees (atan(0.5)). The LM is 6 m tall, the shadow is 12 m long. I agree with your 0.34 degrees in 40 minutes figure, so at the end, the Sun's angle will be 26.90 degrees. For the 6 m LM, its shadow will now be 6/(tan(26.90)), or 11.82 m. Before, it was 12 m, so it moved by 18 cm.

No - see above. You've woefully over-calculated the change in Sun angle.

Nice tone. But if you do the same calculation with a 2.2 m flag pole, you get that it moved by 6.4 cm, not 1.3.

See above!

Maybe you can check your calculation to identify where you've gone wrong. Oh wait, I already did that for you - you assumed that the tangent function was linear, that it would change the same amount in going from 0 to 0.34, as it would from 26.57 to 26.90. (Note that the last figures look like they changed by 0.33 instead of 0.34, but I carried the calculations out with more significant digits and rounded off here).

Not sure what you mean here. I calculated the length of the flagpole shadow as a function of the Sun angle (Tan 30 degrees) and the height of the flagpole (2,200mm), assuming the flagpole to be planted vertically. Do you believe there's something wrong with that?

Math is not my strong suit so I am just going to address your initial assumptions.
The sun angles for the Moon missions are listed here:
http://www.hq.nasa.gov/pao/History/alsj/alsj-sunangles.html

If the video you are refering to is from the Apollo 11 mission then the starting and ending sun angles are :
Apollo 11
Landing: 102.75 GET
EVA
Start: 109.00 GET, 14.0 deg.
Finis: 111.75 GET, 15.4 deg.

edited: Damn! Peters beat me too it.

Your 30 degree estimate seems a bit too generous by a factor of 2. And as you can see the angles shift by 1.4 degrees during the whole mission.

http://www.cs.indiana.edu/sudoc/image_30000061709352/30000061709352/pdf/11post.pdf
According to the above mission report briefing the flag pole was 8 ft tall or 2438.4 milimeters. Not bad. you even got the eight inches to account for the part embeded in the soil.

Try running the numbers with the corrected sun angles.

See below.

Now if you could be off by a factor of 2 for the sun angles by making an estimate based on an image. How about Groves? Think he might be off on one or two things?

Let's review this statement when we're all agreed on the change in shadow lengths, shall we!

Which mission and day was this? I think it was Apollo 15, and if they just put the flag out, that would be the first day, right? In that case, the Sun's angle was around 20 degrees, and the shadow of the top of the flag pole would move by 11 cm in 40 minutes.

Also, you could consider the tangent of just 0.34 degrees, but you'd need to apply that to the distance from the flag pole's top, to the shadow of the flag pole top, not just the height of the flag pole. I'm not sure which is the way Southwind was trying to do it, but either way he made an error that throws his answer way off.

I don't think I did, as described above.

Hey Southwind, Do you think Dr. Groves might have made a similar error in his calculations?

A similar error to you lot, is that?!

OK, let's re-run the calculation with the correct sun angle and for the LM shadow instead of the flagpole:

Assumptions:

Sun angle above horizon = 15 degrees.
LM is approximately 6,000mm tall.
LM stands vertically.
Axis of Moon's rotation is perpendicular to the direction of the shadow, i.e. the shadows will simply elongate and shorten.

Calculations:

First, using trogonometry, let's calculate the length of the LM shadow; call it 'X':

X = 6000mm / Tan 15 degrees = 22,392mm

Now, using triangulation, let's calculate the angle formed by the vertical axis of the LM and an imaginary line running from the top of the LM to the tip of the shadow on the lunar surface; call it 'Y':

Y = 180 degrees - 90 degrees - 15 degrees = 75 degrees

Now, let's calculate the angle of rotation of the Moon over the duration of the video, namely 40 minutes:

40 minutes / (29.5 Earth days x 24 hours x 60 minutes) x 360 degrees = 0.34 degrees

So, after 40 minutes the Moon will have rotated about its axis by 0.34 degrees. In other words, the LM will have rotated from the vertical 90-degree position to either 89.66 degrees or 90.34 degrees, depending on which direction the Moon is rotating in relation to the Sun (think of it like a hand on a clock moving away from the 12 o'clock position by 0.34 degrees in either direction).

This rotation or 'tilting' from the 'vertical' results in an apparent reduction in the height of the LM from the lunar surface, which will manifest itself as a shortening or lengthening of the shadow. If the LM 'tilts' towards the Sun the shadow length will tend towards zero as the 'tilt angle' approaches 15 degrees (i.e. between the two and three o'clock position, with the LM top pointing directly at the Sun). If the flagpole 'tilts' away from the Sun the Shadow length will initially lengthen as the LM tend towards a 75-degree tilt (i.e. between the 11 and 12 o'clock position, at which point the LM will be perpendicular to the 15-degree Sun angle). Thereafter, the shadow length will shorten as the LM angle tends towards 0 degrees (the 9 o'clock position), at which point there is no shadow!

So, the question is, what does a rotation of the Moon, or 'tilting' of the LM, by 0.34 degrees equate to in a shortening or lengthening of the shadow length? Let's calculate it:

As the light from the Sun can be considered to be travelling in parallel 'rays' the shortening of the shadow length will equate to the horizontal distance that the top of the LM moves from its original position; call it 'Z'. This can be calculated thus:

Z = Tan 0.34 degrees x 6000mm = 36mm.

So, 36mm appears to be the maximum 'movement' of the shadow across the lunar surface.

OK, a slightly greater movement than the flagpole shadow, but I'll ask again, which of you are going to try to convince me that you can detect a 36mm movement (that's less than the diameter of a golf ball!) of the tip of the LM shadow across the lunar surface from watching the linked video?!

 

[Southwind17]

I know exactly which mission the footage is from. I think CurtC may have confused the footage I posted with the footage of the flag being brushed/attracted that someone else posted, which was from Apollo 15. As far as I know the shadows discussion is based on the Apollo 11 footage, so I was letting him know so he had the right one. It's quite likely he was looking at the Apollo 15 footage thinking that was what was being discussed.

Sorry, I didn't mean it to seem that I was including you, obviously, as you're the one that flagged it.

 

[PetersCreek]

Er ... I don't think luck has anything to do with this, and here's the rub: EVERYBODY QUOTED IN THIS POST (with the possible exception of PetersCreek, although I can't figure out why he hasn't pointed it out himself, if so!) IS MAKING THE GRAVE MISTAKE OF BELIEVING THAT THE ANGULAR ROTATION OF THE MOON EQUATES TO AN EQUIVALENT CHANGE IN SUN ANGLE. WRONG!!! THAT'S ONLY THE CASE WHEN OBSERVING THE SUN DIRECTLY OVERHEAD, I.E. FROM A POINT ON THE LUNAR SURFACE THAT'S PERPENDICULARLY TANGENTIAL TO THE DIRECTION OF THE SUN. Think about it: the lower the Sun angle, i.e. the closer the Sun is to the horizon, the more the rotation of the Moon is manifest in movement of an object on the lunar surface DIRECTLY TOWARDS OR AWAY FROM THE SUN. Movement directly toward or away from the Sun has ABSOLUTELY NO EFFECT on Sun angle! This would be most pronounced at a position on the lunar surface that is tangentially parallel to the direction of the Sun, i.e. exactly on the 'top' or 'bottom' of the Moon with the Sun out at the 3 O'clock position (if you could still see the Sun then, which you can't as it's dropped below the horizon).

So, given that the Sun angle was only around 15 degrees, a 0.34 degree change in rotation of the Moon will result in a MUCH SMALLER change in Sun angle!

Wrong. Profoundly wrong. Diametrically wrong. The word "wrong" simply is not big enough for this one. And it's not even consistently wrong. For a moment there, it looked like you had rightness within your grasp but you then contradicted yourself in the next breath.

Rotation toward or away from the Sun serves to point the object at or away from (respectively) the Sun, thus changing it's angle. Rotation is the very reason the angle changes. The degree of change, of course, depends upon where one is on the sphere.

Changes in shadow length are more pronounced as one approaches the equator and are much less so at the poles. Shadow movement is also much more apparent at low sun angles. As a photographer and as an Alaskan and former Southerner, I'm quite familiar with these realities.

Try this on for size: assume for argument that the Moon's axis of rotation is exactly perpendicular to the ecliptic. Now assume that there's really a pole at the pole...either pole. This pole will effectively* cast no shadow on the ground...because...where it would fall is in Lunar night. So much for a greater degree of change.

*There would be some shadow because the terminator is not a perfect razor-sharp demarcation between light and dark...even on the Moon.

 

[Thabiguy]

ETA: sorry, I didn't notice that PetersCreek already addressed this, so some of the contents of this post may seem repetitive.

EVERYBODY QUOTED IN THIS POST (with the possible exception of PetersCreek, although I can't figure out why he hasn't pointed it out himself, if so!) IS MAKING THE GRAVE MISTAKE OF BELIEVING THAT THE ANGULAR ROTATION OF THE MOON EQUATES TO AN EQUIVALENT CHANGE IN SUN ANGLE. WRONG!!! THAT'S ONLY THE CASE WHEN OBSERVING THE SUN DIRECTLY OVERHEAD, I.E. FROM A POINT ON THE LUNAR SURFACE THAT'S PERPENDICULARLY TANGENTIAL TO THE DIRECTION OF THE SUN. Think about it: the lower the Sun angle, i.e. the closer the Sun is to the horizon, the more the rotation of the Moon is manifest in movement of an object on the lunar surface DIRECTLY TOWARDS OR AWAY FROM THE SUN.

I already said I'm not interested in the general debate, but this is so silly that it hurts.

Listen well: angular movement of the Sun on the sky does not result from the change of the observer's location as he physically moves around the planet's center. That does cause tiny wobbly parallax motion, the magnitude of which in the case of the Moon is slightly over 0.001 degrees, or in other words, entirely negligible.

Angular movement of the Sun results from the observer's rotation with respect to the Sun.

Movement directly toward or away from the Sun has ABSOLUTELY NO EFFECT on Sun angle!

That's correct. Rotation of the observer has absolutely all effect.

This would be most pronounced at a position on the lunar surface that is tangentially parallel to the direction of the Sun, i.e. exactly on the 'top' or 'bottom' of the Moon with the Sun out at the 3 O'clock position (if you could still see the Sun then, which you can't as it's dropped below the horizon).

In that position, the Sun is above the (hypothetically perfect) horizon with more than half of its angular diameter. And the higher you are above the ground, the more you see of it.

The silliest thing is that you could find out all of this from a simple experiment with a ball and a flashlight. But instead of doing that, you just make absurd claims that completely contradict reality.

So, given that the Sun angle was only around 15 degrees, a 0.34 degree change in rotation of the Moon will result in a MUCH SMALLER change in Sun angle!

This is utter nonsense. Rotation of the Moon by X degrees will cause angular movement of the Sun for a Moon-based observer by X degrees. Your claims of "much smaller change" are falsified by the simplest tests.

The only factor that was not considered here is that this angular movement of the Sun isn't generally perfectly perpendicular to the horizon, and so generally doesn't result in 100% corresponding elevation change. This depends on the observer's latitude and the angle between the planet's equatorial and orbital plane. On the other hand, even angular movements that don't change Sun's elevation at all will result in movement of an object's tip's shadow.