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Trying to relate to the orbit of the Moon, how far away would a 29" diameter ball be in order to appear to be .5 angular degrees across?
Given the Wiki page formula's here: http://en.wikipedia.org/wiki/Angular_diameter
How do I determine the distance I would need to place a 29" ball out to the point where it would be .5 degrees or 30 Arcminutes in angular diameter? I'd like to be able to talk about the difference in size between a "Supermoon" an "Minimoon" to people and would like to be able say if you put a 29" diameter ball "X" far away and changed it into a 34" ball, could you tell the difference.
I stink at math. Thanks for the help.
2 réponses
- PopeLv 7il y a 7 ansRéponse favorite
See the sketch. Point A is the observer, B is the center of the ball, and T is the point of tangency of a tangent sight line.
AB = 14.5/sin(0° 15') inches
- ChazInMTLv 4il y a 7 ans
I get 154 yards or so, Thanks for simplifying!!! I can muddle through the trig here and I'll try to remember this in the future.
