How is the Cartesian orientation of body inertia "ximat" calculated #2495
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Hi, Though I do not have an exact answer to your problem, the following might help: If you were looking at the visualize section, and found During simulation, the orientation of bodies is computed (as the forward kinematics propagates) in Then, the dimensions of the equimomental inertia box/ellipsoid are computed as you have seen already... |
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Intro
Hi!
I am a graduate student using MuJoCo for my research on simulating quadruped robots.
My setup
I'm using MuJoCo version 3.1.5.
My question
When I tried to visualize the inertia box of a robot part which is non-symmetric in MuJoCo GUI I noticed a tilted inertia box.

I figured out that the tilted box could be corrected if I deleted the non-diagonal elements of the inertia matrix. But I'm still very confused because the non-diagonal elements of the inertia matrix is less than 1% of the diagonal ones.
I tried to look into the code to check how the inertia box was visualized in MuJoCo, which seemed to be in 3 steps:
It seemed that the size of the inertia box size is depended on the diagonal elements and the orientation is dependent on the "cartesian orientation of body inertia matrix". But I could not find how this matrix ("cartesian orientation of body inertia matrix") was computed in the code. Could anyone please give a pointer?
Thanks!
Minimal model and/or code that explain my question
No response
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