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Scaling Intertia Tensor

by alexanderBuzz
Tags: intertia, scaling, tensor
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alexanderBuzz
#1
May2-14, 03:22 AM
P: 2
Hi everyone,

I have the following problem in my hands, which I don't know how exactly to address.

Let's assume that from any CAD(Solidworks, Catia), I obtain the inertia tensor of my model (impossible to calculate by hand btw).

I_full=[Ixx Ixy Ixz
Ixy Iyy Iyz
Ixz Iyz Izz]

I know if I change the mass of my model, the inertia tensor will scale linearly with it.

But what If I scale my model to half-size, all dimensions? The mass probably goes by 1/8, since it's proportional to volume. Maybe the other factor would be 1/4 (α r^2).

So would the correct Inertia scaling factor be?:

I_half=1/4*1/8*I_full

Cheers!!
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UltrafastPED
#2
May2-14, 04:23 AM
Sci Advisor
Thanks
PF Gold
UltrafastPED's Avatar
P: 1,908
It must vary jointly with the mass and the distribution; for example consider how the simple inertia of a thick rod changes under your conditions: http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html
alexanderBuzz
#3
May2-14, 05:01 AM
P: 2
Quote Quote by UltrafastPED View Post
It must vary jointly with the mass and the distribution; for example consider how the simple inertia of a thick rod changes under your conditions: http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html
Thanks you!

It proves my derivation, if assumed constant density between the scaled object and full-size object.

by definition:

[tex]I=\int_V \rho r^2 dV[/tex]

if r1-> ar ( scaled by a factor a) dV1->a^3dV

replacing on the above equation:
[tex] I1=\int_V \rho a^2 r^2 a^3 dV = a^5 \int_V \rho r^2 dV -> I1=a^5 I[/tex]

again, assuming that the mass distribution remains constant.

Correct?


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