# 平行軸定理

${\displaystyle I_{C}\,\!}$ 代表剛體對於質心軸的轉動慣量、${\displaystyle M\,\!}$ 代表剛體的質量、${\displaystyle d\,\!}$ 代表另外一支直軸 z'-軸與質心軸的垂直距離。那麼，對於 z'-軸的轉動慣量是

${\displaystyle I_{z'}=I_{C}+Md^{2}\,\!}$

${\displaystyle I_{z}=I_{x}+Ad^{2}\,\!}$

## 進階理論

${\displaystyle \mathbf {I} ={\begin{bmatrix}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{bmatrix}}\,\!}$

${\displaystyle I_{xx}\ {\stackrel {\mathrm {def} }{=}}\ \int \ y^{2}+z^{2}\ dm\,\!}$
${\displaystyle I_{yy}\ {\stackrel {\mathrm {def} }{=}}\ \int \ x^{2}+z^{2}\ dm\,\!}$
${\displaystyle I_{zz}\ {\stackrel {\mathrm {def} }{=}}\ \int \ x^{2}+y^{2}\ dm\,\!}$

${\displaystyle I_{xy}=I_{yx}\ {\stackrel {\mathrm {def} }{=}}\ -\int \ xy\ dm\,\!}$
${\displaystyle I_{xz}=I_{zx}\ {\stackrel {\mathrm {def} }{=}}\ -\int \ xz\ dm\,\!}$
${\displaystyle I_{yz}=I_{zy}\ {\stackrel {\mathrm {def} }{=}}\ -\int \ yz\ dm\,\!}$

${\displaystyle I_{xx}=I_{G,xx}+m({\bar {y}}^{2}+{\bar {z}}^{2})\,\!}$
${\displaystyle I_{yy}=I_{G,yy}+m({\bar {x}}^{2}+{\bar {z}}^{2})\,\!}$
${\displaystyle I_{zz}=I_{G,zz}+m({\bar {x}}^{2}+{\bar {y}}^{2})\,\!}$
${\displaystyle I_{xy}=I_{yx}=I_{G,xy}-m{\bar {x}}{\bar {y}}\,\!}$
${\displaystyle I_{xz}=I_{zx}=I_{G,xz}-m{\bar {x}}{\bar {z}}\,\!}$
${\displaystyle I_{yz}=I_{zy}=I_{G,yz}-m{\bar {y}}{\bar {z}}\,\!}$

a) 參考右圖 ，讓 ${\displaystyle (x\,',\ y\,',\ z\,')\,\!}$ ${\displaystyle (x,\ y,\ z)\,\!}$  分別為微小質量 ${\displaystyle dm\,\!}$  對質心 G 與原點 O 的相對位置：

${\displaystyle y=y\,'+{\bar {y}}\,\!}$ ${\displaystyle z=z\,'+{\bar {z}}\,\!}$

${\displaystyle I_{G,xx}=\int \ y\,'\,^{2}+z\,'\,^{2}\ dm\,\!}$
${\displaystyle I_{xx}=\int \ y^{2}+z^{2}\ dm\,\!}$

{\displaystyle {\begin{aligned}I_{xx}&=\int \ (y\,'+{\bar {y}})^{2}+(z\,'+{\bar {z}})^{2}\ dm\\&=I_{G,xx}+m({\bar {y}}^{2}+{\bar {z}}^{2})\ .\\\end{aligned}}\,\!}

b) 依照慣性張量的慣性積定義方程式 ，

${\displaystyle I_{G,xy}=-\int \ x\,'y\,'\ dm\,\!}$
${\displaystyle I_{xy}=-\int \ xy\ dm\,\!}$

{\displaystyle {\begin{aligned}I_{xy}&=-\int \ (x\,'+{\bar {x}})(y\,'+{\bar {y}})\ dm\\&=I_{G,xy}-m{\bar {x}}{\bar {y}}\ .\\\end{aligned}}\,\!}

## 實例

${\displaystyle I_{G}={\begin{bmatrix}{\frac {1}{12}}m(w^{2}+h^{2})&0&0\\0&{\frac {1}{12}}m(h^{2}+d^{2})&0\\0&0&{\frac {1}{12}}m(w^{2}+d^{2})\end{bmatrix}}\,\!}$

${\displaystyle I_{xx}={\frac {1}{12}}m(w^{2}+h^{2})+m\left(\left({\frac {w}{2}}\right)^{2}+\left({\frac {h}{2}}\right)^{2}\right)\,\!}$
${\displaystyle I_{yy}={\frac {1}{12}}m(h^{2}+d^{2})+m\left(\left({\frac {h}{2}}\right)^{2}+\left({\frac {d}{2}}\right)^{2}\right)\,\!}$
${\displaystyle I_{zz}={\frac {1}{12}}m(w^{2}+d^{2})+m\left(\left({\frac {w}{2}}\right)^{2}+\left({\frac {d}{2}}\right)^{2}\right)\,\!}$
${\displaystyle I_{xy}=-m\left({\frac {w}{2}}\right)\left({\frac {d}{2}}\right)=-{\frac {mwd}{4}}\,\!}$
${\displaystyle I_{xz}=-m\left({\frac {h}{2}}\right)\left({\frac {d}{2}}\right)=-{\frac {mhd}{4}}\,\!}$
${\displaystyle I_{yz}=-m\left({\frac {w}{2}}\right)\left({\frac {h}{2}}\right)=-{\frac {mwh}{4}}\,\!}$

${\displaystyle I_{G}={\begin{bmatrix}{\frac {1}{3}}m(w^{2}+h^{2})&-{\frac {1}{4}}mwd&-{\frac {1}{4}}mhd\\-{\frac {1}{4}}mwd&{\frac {1}{3}}m(h^{2}+d^{2})&-{\frac {1}{4}}mwh\\-{\frac {1}{4}}mhd&-{\frac {1}{4}}mwh&{\frac {1}{3}}m(w^{2}+d^{2})\end{bmatrix}}\,\!}$

## 參考文獻

• Beer, Ferdinand; E. Russell Johnston, Jr., William E. Clausen (2004). Vector Mechanics for Engineers. 7th edition. USA: McGraw-Hill, ISBN 0-07-230492-8