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Given is the following square in perspective view:

Plain square

  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

enter image description here

1 Answers 1

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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 $11.46^\circ$

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    @Elias: no, I don't know what you mean by absolute size. The angles as seen from your eye can be translated into a number of pixels at whatever scale you would like. You just decide that 1 degree is 15 pixels, or 100, or whatever makes the image the size you want.2011-12-23