Moment of Inertia of a Polygon (2D)

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3 comments, last by LonelyStar 18 years, 7 months ago
Hello together, OK, I have a convex polygon in a 2D plane. I want to calculate the moment of inertia of this polygon when it is rotating around the origin. Of course a 2D polygon can not really have a moment of inertia, so lets say it is a body with the ground-area being the polygon and the height of h=1. It shouldn't really matter because with a given mass, the moment of inertia should be independent from h. I have the mass and can get the densitiy without problems. Since the polygon is convex (and it is ensured, that the origin is within the polygon), one could split the polygon into a triangle fan around the origin ... but How could I calculate the moment of inertia of one of those triangles? Thanks for any help!
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oliii's tutorial has moment of inertia calculation plus a link to the formula.

yes, it can have inertia, lookup google moment of inertia thin disc or something else.
double integral over the 2D region of (p*(x^2+y^2)) where x and y are relative to the point about which it is rotating and p is the density function
from what I gathered,

// taken from // http://www.physicsforums.com/showthread.php?s=e251fddad79b926d003e2d4154799c14&t=25293&page=2&pp=15float PolyColl::CalculateInertia(const Vector* A, int Anum, float mass){	if (Anum == 1)	return 0.0f;	float denom = 0.0f;	float numer = 0.0f;	for(int j = Anum-1, i = 0; i < Anum; j = i, i ++)	{		Vector P0 = A[j];		Vector P1 = A;		float a = (float) fabs(P0 ^ P1);		float b = (P1*P1 + P1*P0 + P0*P0);		denom += (a * b);		numer += a;	}	float inertia = (mass / 6.0f) * (denom / numer);	return inertia;}


edit :
A ^ B is A.CrossProduct(B) = A.x*B.y - A.y * B.x;
A * B is A.DotProduct(B) = A.x*B.x + A.y * B.y;

Everything is better with Metal.

Thanks all of you!
I knew it had to be the integral over sqrt(x^2+y^2) but I do not know how to solve it.
Well, if the code from oliii works, I am quite happy!!!

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