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OK, this is WAY geeky, but I just thought I'd share... This was triggered by a
cloud visible at sunset a couple of evenings ago. It was long and triangular,
and took up about 5 "hand-widths" at arms length. I started thinking, how close
(or far away) would a "super star destroyer" have to be to look like that? (Of
course, my 8-year-old son might have helped the discussion along). And could I
Photoshop such a scene with a model (LEGO) SSD?
So, I started calculating, thinking back to Trig. Tangent, that's what I need,
I know how long the SSD is (17.6 km, according to some sources), and the
altitude (depends, but known). The angle subtended by the object is twice the
arc-tangent of half the length of the object divided by its altitude.
So, here's what I got (keep in mind that the Moon and Sun subtend about 1 degree
of arc in the sky):
Geostationary orbit (22,223 mi.): 0.028 deg.
Hubble Telescope orbit (320 mi.): 1.958 deg. - twice the width of the full
Moon.
"Edge of Space" (62 miles): 10.08 deg. - now we're getting somewhere!
50,000 ft: 60.01 deg.
I'm not sure what to do with this, but I was surprised at how small something so
(fictionally) large would appear in the sky.
Like I said, it was WAY geeky.
James Wilson
Dallas, TX
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