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

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Discussion Overview

The discussion revolves around calculating the moment of inertia for polygons, particularly in the context of a 2D plane. Participants explore methods for determining this without decomposing the polygon into triangles, as well as the implications of polygon complexity on calculations.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses difficulty in calculating the moment of inertia for a polygon without decomposing it into triangles, referencing a Wikipedia example that is unclear to them.
  • Another participant confirms that the moment of inertia formula for polygons is derived from decomposing them into triangles but does not provide a detailed derivation.
  • A participant questions the necessity of using polygons with four or more sides, suggesting that such shapes complicate calculations and shading, particularly in 3D contexts.
  • The same participant later clarifies that their work is focused on a 2D plane and expresses a desire to find a solution specifically for convex polygons without resorting to decomposition methods like BSP-trees.
  • A later reply acknowledges the misunderstanding regarding the dimensional context of the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for calculating the moment of inertia for polygons, and there are differing views on the implications of polygon complexity.

Contextual Notes

The discussion highlights the challenges associated with calculating moment of inertia for polygons with varying numbers of sides, particularly in relation to dimensionality and the need for decomposition methods.

epaik91
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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.
 
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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.
 
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
 
kenewbie said:
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.
 
Yep, I was thinking 3D, sorry.

k
 

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