# How do you find the moment of inertia of a polygon?

#### epaik91

I'm working on an engine right now, and I'm having trouble calculating the moment of inertia for a polygon. Is there any way to easily do this without decomposing the polygon into triangles?

edit: I've looked at the wikipedia page with examples on the subject (http://en.wikipedia.org/wiki/List_of_moments_of_inertia) and I'm having trouble understanding the last example, which seems to be what I need.

Last edited:

#### arildno

Homework Helper
Gold Member
Dearly Missed
That last example is, indeed, the moment of inertia formula for the polygon, and is the final result of having decomposed the polygon into triangles.

I derived it once for myself, I'm sorry that I'm not inm the mood to do it once again.

#### kenewbie

Any particular reason your polygons are not triangles in the first place? Polygons of 4 or more sides complicates everything.

4 or more sides means your polygon does not need to have all vertexes on the same plane, which complicates shading.

Building a BSP-tree gets more complicated the more sides your polygons have (you will end up splitting a whole lot).

and so on.

k

#### epaik91

Any particular reason your polygons are not triangles in the first place? Polygons of 4 or more sides complicates everything.

4 or more sides means your polygon does not need to have all vertexes on the same plane, which complicates shading.

Building a BSP-tree gets more complicated the more sides your polygons have (you will end up splitting a whole lot).

and so on.

k
I'm sorry if I wasn't clear, but I'm doing everything on a 2D plane. I'm not sure what you mean by stating that the vertexes won't be on the same plane, but I think you may be confusing 3D graphics with my problem (shading?). In addition, finding a way to solve this problem for convex polygons will achieve what I'm going for, which is why I'm avoiding BSP-trees or any other kind of decomposition of the polygon.

#### kenewbie

Yep, I was thinking 3D, sorry.

k

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