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Math Help - Rotated projected plans

  1. #1
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    Rotated projected plans

    Linear algebra was a couple years ago, so I'm a bit rusty on the steps I need to take for this problem. We have a top-down view of houses, and want to determine the area of the roof. We outline each plane of the roof, so that each separate polygon represents a contiguous part of the roof which has the same pitch. To make this example simple, just think of a shed with a single roof plane at 45 degrees. We know the orientation of each plane, and the pitch. Now, from a top-down view, the roof area may look square, but because it is pitched, in reality the roof is rectangle. For instance, if from a top-down view, the roof looks to be 1 meter by 1 meter, and the pitch is 45 degrees, in reality the roof is 1 meter by 1*srt(2) meters. It's easy in a trivial case, but I need a general solution for an arbitrary polygon with an arbitrary pitch on an arbitrary orientation (ie pitch is 45 degrees facing north or 0 degrees, etc). Since we're transforming the polygon in 3D space into a 2D viewing plane, I take it I'll need some sort of reverse transformation, but I'm not sure how to work in the orientation and angle into all of that. Any thoughts?
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  2. #2
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    If a polygon (or any shape) lies on a plane forming an angle \theta with the horizontal plane (which also means that a vector orthogonal to the polygon forms an angle \theta with a vertical vector (and \theta is computed using a dot product)), then its area is equal to the area of its projection on the horizontal plane (the "top view") divided by \cos\theta. If \theta is 45, this gives the \sqrt{2} factor.

    I don't know if this fully answers your question.
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  3. #3
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    Thanks! In the end all I want is the area, and that's MUCH easier than how I was trying to do it.
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