# Thread: Rotated projected plans

1. ## 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?

2. If a polygon (or any shape) lies on a plane forming an angle $\displaystyle \theta$ with the horizontal plane (which also means that a vector orthogonal to the polygon forms an angle $\displaystyle \theta$ with a vertical vector (and $\displaystyle \theta$ 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 $\displaystyle \cos\theta$. If $\displaystyle \theta$ is 45°, this gives the $\displaystyle \sqrt{2}$ factor.

I don't know if this fully answers your question.

3. Thanks! In the end all I want is the area, and that's MUCH easier than how I was trying to do it.