maggiekb at January 30th, 2014 12:44 — #1
lorq at January 30th, 2014 12:57 — #2
ldobe at January 30th, 2014 21:05 — #3
I decided to do a little recreational math with this.
Assuming the human binocular FOV is ~120 degrees and the photo was taken at a distance of 5.5 miles, and the orbiters were moving in perpendicular directions at 3600mi/h, I wanted to figure out how quickly the orbiter would cross my FOV.
To calculate how wide my FOV is in miles at that distance Use Tangent=opposite/adjacent. Adjacent=5.5mi tan(60 degrees)~1.73205
9.526275 is half my FOV, so double it to make a total width of 19.05255 miles. 3600 mi/h = 1 mi/sec, so it takes roughly 19 seconds for the orbiter to cross my FOV if I were stationary relative to the moon.
Since These things are moving in perpendicular angles you have a fraction of that time to actually see each other.
maggiekb at February 4th, 2014 12:44 — #4
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