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See the approximate cube in two point perspective below. It is approximate because I eyeballed the horizontal placement of the two vertical lines off-center representing the two far edges of the near faces. I wish to draw a "perfect" cube with all dimensions scaling appropriately given the vanishing points.

Question: Given two arbitrarily located (but restricted by the line segment between them will be perpendicular to the near edge vertical line segment) vanishing points, calculate all line segment dimensions (specifically only need horizontal placement of two off-center vertical line segements) for the cube.

I have tried using similar triangles, law of cosines, etc. without success.

cube in two point perspective

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  • $\begingroup$ Here are some methods how to achieve it. $\endgroup$ – t.b. May 23 '12 at 1:07

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