If I know the actual diameter of an object (the Milky Way for example), does anyone know the math to calculate how far away it would have to be to subtend 45 degrees of the sky?
I know there is a formula that will calculate apparent size from distance and actual size, but I just can't find it.
2007-09-08 07:36:50 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Astronomy & Space
There are formulae for determining how much sky is covered by an object, but there are two factors which make it somewhat hard to give a simple answer. The first is that objects are three-dimensional - the Milky Way, for example (although it's not a good example, since we're *in* the Milky Way) is vaguely cylindrical - so size varies for both angle and distance. The other is that how large something looks isn't measured by one angle - it's a square angle.
You might be able to do it with calculus, I suppose, but any way you look at it, it's tricky.
2007-09-08 08:06:19 · answer #1 · answered by peri_renna 3 · 3⤊ 0⤋
OK, the formula is Theta=Arctan (width/distance). So just use the inverse tangent function on your scientific calculator. Another way of stating the formula is tan(theta)=width/distance, where width and distance are measured in the same units, such as light years. I hope this helps.
2007-09-08 07:40:52 · answer #2 · answered by Sciencenut 7 · 5⤊ 0⤋