Skip to main content
2025-01-18

Parallel Axis

2 min read
section

Consider the inertia tensor Ix\bm{I}_x defined by

Ix≜∫x⋅x 1−x⊗x dα\bm{I}_x \triangleq \int \bm{x} \cdot \bm{x} \, \mathbb{1} - \bm{x} \otimes \bm{x} \, \mathrm{d} \alpha

The transformed tensor Ix+ρ\bm{I}_{x+\rho} corresponding to a translation by ρ\bm{\rho} is obtained by replacing x\bm{x} with x+ρ\bm{x} + \bm{\rho}, furnishing

Ix+ρ=Ix+2A ρ⋅ρˉ 1−A(ρˉ⊗ρ+ρ⊗ρˉ)+A(ρ⋅ρ 1−ρ⊗ρ)\bm{I}_{x+\rho} = \bm{I}_x + 2 A \,\bm{\rho}\cdot \bar{\bm{\rho}} \, \mathbb{1} - A (\bar{\bm{\rho}}\otimes\bm{\rho} + \bm{\rho}\otimes\bar{\bm{\rho}}) + A (\bm{\rho}\cdot\bm{\rho}\,\mathbb{1} - \bm{\rho}\otimes\bm{\rho})

where the centroid ρˉ\bar{\bm{\rho}} and area AA are defined as

ρˉ≜1A∫x dα and A≜∫ dα\bar{\bm{\rho}} \triangleq \frac{1}{A}\int \bm{x}\, \mathrm{d}\alpha \qquad \text{ and }\qquad A \triangleq \int \, \mathrm{d} \alpha

In the plane one often defines

J≜∫r^⊗r^ dα\bm{J} \triangleq \int \hat{\bm{r}} \otimes \hat{\bm{r}} \, \mathrm{d}\alpha

In terms of I\bm{I} this is

J=−i×Ii×\bm{J} = - \mathbf{i}^{\times}\bm{I}\mathbf{i}^{\times}

and

I=I0i⊗i−∫r^×(i⊗i)r^×\bm{I} = I_0 \mathbf{i}\otimes \mathbf{i} - \int \hat{\bm{r}}^{\times} (\mathbf{i}\otimes\mathbf{i})\hat{\bm{r}}^{\times}

or

I=I0i⊗i−i×∫r^⊗r^ dα i×\bm{I} = I_0 \mathbf{i}\otimes \mathbf{i} - \mathbf{i}^{\times}\int \hat{\bm{r}}\otimes\hat{\bm{r}}\,\mathrm{d}\alpha \, \mathbf{i}^{\times}

so that

I=I0i⊗i−i×J i×\bm{I} = I_0 \mathbf{i}\otimes \mathbf{i} - \mathbf{i}^{\times} \bm{J} \, \mathbf{i}^{\times}

It is convenient to note that with coordinates r=[x,y,z]\bm{r} = [x, y, z]

[r×][r×]t=−[0−zyz0−x−yx0]2=[y2+z2−xy−xz−yxx2+z2−yz−zx−zyx2+y2][\bm{r}^\times][\bm{r}^\times]^{\mathrm{t}} =-\left[\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{array}\right]^2=\left[\begin{array}{ccc} y^2+z^2 & -x y & -x z \\ -y x & x^2+z^2 & -y z \\ -z x & -z y & x^2+y^2 \end{array}\right]