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.

 

[arildno]

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