Yet, if you watch the video, you can see the shadows move. How can this be?
Perhaps you could show us your calculations? Mine show that the shadow of the LM will move visibly over 40 minutes on the moon. To help you out, the LM is a little over 6m tall, and the shadows appear to be about twice as long and the height of the casting object. How far will the LM shadow move?
Shadows move, here on Earth, in the space of 90 seconds. You don't normally notice this, because it's quite slow. But you can indeed see it if you have a long, sharp-edged shadow (the exact sort of shadow you'd get in a vacuum!) and a clear reference point. Filming it and then speeding up the film makes it that much easier to detect.
And look, that's exactly what we do see in this sped up film. How about that!
Perhaps doing the calculations would make things clearer for you? Give it a go...
Now you're asking him to come up with something concrete...that's just so unfair!
Ooh! ooh! ohh! Can I answer?
I get about 18 cm.
Gosh interesting. I wonder what Southwind will come up with?
Well, I understand how long a Lunar day is. I also understand that shadows cast upon a rotating sphere, by objects fixed upon that sphere, do not move at a linear rate when the light source is fixed in space. When the Sun is low on the horizon, shadows are longer and move faster as the Sun rises, than they do at when the Sun is more directly overhead.
Simply prorating the rotation period will not give you a fair approximation at low Sun angles. It's a circular function...not linear.
Pretty basic senior school physics trigonometry, really.
In fact I did make a mistake here; I thought the original film was longer than that, but -
Do the calculation, Southwind. The point stands. The shadow will move, visibly, by a distance consistent with the video.
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?!
Detentions all round!
