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An object has diameter of 1m; it appears to have a diameter of 1mm. How far away is it?
(1) It is far more convenient (and accurate) to discuss the apparent size in terms of an angle. When you describe a cloud as having the same apparent size as a tennis ball, the listener does not know if you mean a tennis ball held close to the face, at arm's length, 10m away, etc...
(2) I will assume that when you say the apparent size is 1mm, you measured this with a ruler held at arm's length. I assume the ruler is 60cm from your eye. Then the angle in the sky can be found:
** Draw an isosceles triangle; label the vertex angle as `alpha` , the altitude as d (for distance), and the base as w (for actual width). Then from trigonometry we have `tan(alpha/2)=(w/2)/d` . Then `d=1/(2tan(alpha/2))` . Also `alpha=2tan^(-1)(w/(2d))` **
The angle subtended by a 1mm object at .6m is `alpha=2tan^(-1)(.001/1.2)~~.095` , so your object covers about `.1^circ` of the sky. (Under the assumptions I made)
(3) Then the distance of the 1m object is `d=1/(2tan(alpha/2))=1/(2tan.05)~~572.96`
Thus the object was approximately 573m in the air.
Again, a better approach is to measure the angle, as this is more consistent.
thanks heaps. i havent done any maths since passing my hsc by 1/2 mark in 1989!
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