Some of you asked how large the ufo was, here's some size analysis I found by Bruce Maccabee (
brumac.8k.com...) :
How large is the "object"? If we knew the distance of the camera from the
beam convergence and the focal length of the camera we could calculate the approximate size. This requires knowing what portion of the city the
object was over, where the cameraman was, and the altitude of the "object."
An alternative method is to estimate the diameter of a spotlight beam at
some distance from the spotlight and use that width as a reference size.
I found a research article by Dr. Louis Eltermann that reports research
in the latter 1940's in which he used an army searchlight to probe the
upper atmosphere in order to determine the vertical distribution of dust
in the atmosphere. (Note: Eltermann was the author of the infamous Project
Twinkle Report in November, 1951, which ignored or "covered up" or, at the
very least, misrepresented, the White Sands movie film that proved
unidentified objects were flying around. See THE UFO-FBI CONNECTION by Bruce
Maccabee [Llewellyn, St. Paul, MN, 2000.
Also,
brumac.8k.com...)
Eltermann described the
searchlight as being 5 ft in diameter and with a divergence of about 1.25
degrees or about 20 milliradians. This means that the diameter
at a distance d from the mirror would be about D = 5'+0.02d. Thus at 1000
ft the diameter would be about 25 ft. Of course, the beam is not uniformly
bright across its diameter, so the effective diameter might be closer to 20
feet.
Consider the beam at the right side of the photo.
It protrudes upward at some angle, probably not the angle
in the photo. Suppose the elevation angle were 30 degrees. The "object" width
is oriented horizontally (parallel to the ground) whereas the beam is
assumed to be tilted at about 30 degrees. Hence the horizontal width
of the beam, W,(not perpendicular to the beam axis) would be
W = D/sin(angle of elevation) = D/sin(30) = 2D for the assumed 30 degree
elevation angle. Hence if the object were
1000 ft from the projection lens it was about 2 x 25 = 50 ft wide. If at 2000 ft
the calculation yields D = 45 ft and W = 90 ft.
One estimate of the height of the object was 8,000 ft. For a 30 degree slant
angle of the beam from ground level up to 8,000 ft the distance along the
beam would be about 8,000/sin 30 = 16,000 ft. If this were so, then the beam diameter
at that height would have been about 165 ft and the horizontal width of the object
would have been about 330 ft.
If the slant angle of the beam was less than 30 degrees then the calculated sizes
would have been larger. Conversely, if the slant angle was greater the
calculated sizes would have been smaller.
Based on the above calculations, and realizing that a much better estimate
could be made if we had more accurate information on the spotlights,
camera, etc., I would hazard a guess that the width of the illuminated
"object" is on the order of 100 ft or more in size.
Without more solid information to go on this has to be no more than
a WAG (wild...rear-end... guess) (but I bet its close to right!)
[edit on 25-1-2008 by Leto]
