Absolute sizes of a two-dimensional square in perspective Given is the following square in perspective view:



*

*The square has an edge length of 500 feet.

*The square is rotated by 45 degree.

*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




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 $11.46^\circ$
