Re: Moment of inertia calculations

Posted by Robert Dougherty on Feb 23, 2006; 6:50pm
URL: http://imagej.273.s1.nabble.com/Moment-of-inertia-calculations-tp3703568p3703572.html

If the 3rd direction is perpendicular to the image, then the macro computes
the (3,3) element of the inertia tensor.  If the object is really a 2-D
slice, then I33 will turn out to be one of the principle moments, since I13,
I23, I31, and I32 are 0 (since the CM is clearly in the plane).  In this
case, assuming the angular velocity vector is perpendicular to the image,
the rotational kinetic energy is (1/2)I33*(omega3)^2 and the angular
momentum in the third direction is I33*omega3.  For rotations with the
angular velocity parallel to the image, or with a real 3D object, it would
be necessary to compute more of the tensor.  Before putting more effort into
this, it would be good to understand the 3D problem better.

Bob  

Robert P. Dougherty, Ph.D.
President, OptiNav, Inc.
Phone [phone]
Fax [phone]
www.optinav.com
 

> -----Original Message-----
> From: ImageJ Interest Group [mailto:[hidden email]] On Behalf Of
> Pablo Domínguez García
> Sent: Thursday, February 23, 2006 4:02 AM
> To: [hidden email]
> Subject: Re: Moment of inertia calculations
>
> Only a comment,
>
> The algorithm below is for the component 1,1 of the tensor of inertia.
> The other components must be also calculated (20, and 02). These terms
> are analogous to the 1,1 component but using a sum with (x-x0)*(x-x0)
> and (y-y0)*(y-y0).
>
> Then, the matrix can be diagonalized for obtaining the main moments of
> inertia (in the main directions).
>
> Am I right?
>
>