# Absolute sizes of a two-dimensional square in perspective

Given is the following square in perspective view:

1. The square has an edge length of 500 feet.
2. The square is rotated by 45 degree.
3. The viewer is 1600 feet away from the square's center.

Needed are formulas for the following sizes in a two-dimensional view (trapezium):

• width w
• height h1
• height h2

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You still need to specify some things. Where is the $1600$ foot distance measured to? The near edge, the center, the far edge? You also have nothing to set the vertical scale: think of looking at it through a zoom lens. Assuming the $1600$ feet is to the near edge, the angle subtended by the near edge is $2 \arctan \frac{250}{1600}= 17.76^\circ$. The far edge is then $1600+250\sqrt{2}=1953.55$ feet away and the angle is then $14.59^\circ$. The horizontal angle subtended can be found from the law of cosines: $500^2=1600^2+1953.55^2-2*1600*1953.55 \cos \theta$, giving $11.46^\circ$
Can you update what I gave for that case? The near edge will then be $1600-250/\sqrt{2}=1422.23$ feet away. – Ross Millikan Dec 22 '11 at 17:27
That is where the $\sqrt{2}$ comes from. $\tan45^\circ=\sqrt{2}$, so the 500 foot width means the near edge is $500/\sqrt{2}$ feet nearer us than the rear edge. – Ross Millikan Dec 22 '11 at 18:00