Even though I didn't want to, I have to write yet again, as I must correct a mistake that I had made.
My calculation of the distance of Aldrin's visor was, indeed, flawed. Specifically, at this point:
As the visor spans the horizontal center of the image, we'll use an approximation similar to Groves' to determine the distance.
I don't know what I was thinking; even though we measure the visor horizontally and it is near the horizontal center of the image, the distance from the image center cannot be ignored, as it will affect the result. I apologize to the readers.
Let's do a better calculation. We'll consider points A and B on the photograph:
A = (-2.04 mm, 21.35 mm)
B = (3.64 mm, 22.24 mm)
These points are arbitrarily selected to correspond to edges of Aldrin's visor.
(Note about accuracy: Although the points are listed with two digits after the decimal point, to allow readers to trace them to individual pixels on the high resolution scans, it would be unrealistic to assume greater accuracy in measurement than about 0.1 mm. Such inaccuracy will translate to less than 0.1 m in the resulting distance.)
Point P (the model pinhole) is at 3D coordinates (0, 0, (photo width/2)/tan(FOV/2)).
Assuming photo width of 52 mm and FOV of 49.2 degrees, the sides of the triangle PAB are then (values rounded for listing, unrounded for subsequent calculations):
|AB| = 5.75 mm
|PA| = 60.7 mm
|PB| = 61.1 mm
From the cosine law, the angle APB (the calculated angular width of Aldrin's visor) is about 5.4 degrees, which, assuming actual width of 280 mm, would give distance of 2.97 m, or about 3 m between the visor and the camera. This differs by 30 cm, or 10% from my previous estimate.
This change affects my subsequent comments on Groves' analysis at one other point: rederivation of Groves' result of 1073 mm yields 1.15 m, not 1.05 m, increasing the difference from about 1 inch to about 3 inches. The rest of my comments are not affected in any way.
Some forum members apparently have trouble comprehending written text, so I will point out once again:
- This is the distance between the edges of Aldrin's visor and the camera; the distance that Groves tried to calculate. Aldrin is a large object and different parts of his body will have different distances from the camera; his feet will be considerably farther away, for example. Distances from other points, such as the front of Armstrong's left foot, or back of Armstrong's right foot, will be different still.
In other words, the distance between Aldrin's visor and the camera cannot be used to estimate the length of Aldrin's shadow. Distance between astronauts' feet will be greater than that.
Finally, I would like to address the silly claim that Groves' figure of 37.987 degrees is a better estimate of the FOV than NASA's figure of 49.2 degrees. But how could we verify the FOV? If only we had a picture of an object whose distance and size are both known...
Well, guess what?
We do. This picture of the Earth taken from lunar orbit is very conveniently at the center of the image. Its edge goes almost directly through the optical axis, minimizing distortions and making the distance-angle conversion simple and straightforward.
The size of the Earth on the photograph is about 1.96 mm, or 1.9-2.0 mm if you wish to account for inaccuracy.
NASA's FOV of 49.2 degrees, image width 52 mm:
estimated angular size = 1.98 degrees (1.92-2.02 degrees)
Groves' FOV of 37.987 degrees, image width 52 mm:
estimated angular size = 1.49 degrees (1.44-1.52 degrees)
Angular diameter of the Earth, as seen from the Moon during the Apollo 11 mission, was about 1.9 degrees. I think it's unnecessary to say more.