平行軸定理能夠很簡易的，從對於一個以質心為原點的座標系統的慣性張量 ，轉換至另外一個平行的座標系統。
對於三維空間中任意一參考點 Q 與以此參考點為原點的直角座標系 Qxyz ，一個剛體的慣性張量
I
{\displaystyle \mathbf {I} \,\!}
I
=
[
I
x
x
I
x
y
I
x
z
I
y
x
I
y
y
I
y
z
I
z
x
I
z
y
I
z
z
]
{\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}}\,\!}
這裏，對角元素
I
x
x
{\displaystyle I_{xx}\,\!}
I
y
y
{\displaystyle I_{yy}\,\!}
I
z
z
{\displaystyle I_{zz}\,\!}
慣性矩 。設定
(
x
,
y
,
z
)
{\displaystyle (x,\ y,\ z)\,\!}
d
m
{\displaystyle dm\,\!}
I
x
x
=
d
e
f
∫
y
2
+
z
2
d
m
{\displaystyle I_{xx}\ {\stackrel {\mathrm {def} }{=}}\ \int \ y^{2}+z^{2}\ dm\,\!}
I
y
y
=
d
e
f
∫
x
2
+
z
2
d
m
{\displaystyle I_{yy}\ {\stackrel {\mathrm {def} }{=}}\ \int \ x^{2}+z^{2}\ dm\,\!}
I
z
z
=
d
e
f
∫
x
2
+
y
2
d
m
{\displaystyle I_{zz}\ {\stackrel {\mathrm {def} }{=}}\ \int \ x^{2}+y^{2}\ dm\,\!}
而非對角元素，稱為慣性積 , 可以定義為
I
x
y
=
I
y
x
=
d
e
f
−
∫
x
y
d
m
{\displaystyle I_{xy}=I_{yx}\ {\stackrel {\mathrm {def} }{=}}\ -\int \ xy\ dm\,\!}
I
x
z
=
I
z
x
=
d
e
f
−
∫
x
z
d
m
{\displaystyle I_{xz}=I_{zx}\ {\stackrel {\mathrm {def} }{=}}\ -\int \ xz\ dm\,\!}
I
y
z
=
I
z
y
=
d
e
f
−
∫
y
z
d
m
{\displaystyle I_{yz}=I_{zy}\ {\stackrel {\mathrm {def} }{=}}\ -\int \ yz\ dm\,\!}
假若已知剛體對於質心 G 的慣性張量
I
G
{\displaystyle \mathbf {I} _{G}\,\!}
(
x
¯
,
y
¯
,
z
¯
)
{\displaystyle ({\bar {x}},\ {\bar {y}},\ {\bar {z}})\,\!}
I
{\displaystyle \mathbf {I} \,\!}
I
x
x
=
I
G
,
x
x
+
m
(
y
¯
2
+
z
¯
2
)
{\displaystyle I_{xx}=I_{G,xx}+m({\bar {y}}^{2}+{\bar {z}}^{2})\,\!}
I
y
y
=
I
G
,
y
y
+
m
(
x
¯
2
+
z
¯
2
)
{\displaystyle I_{yy}=I_{G,yy}+m({\bar {x}}^{2}+{\bar {z}}^{2})\,\!}
I
z
z
=
I
G
,
z
z
+
m
(
x
¯
2
+
y
¯
2
)
{\displaystyle I_{zz}=I_{G,zz}+m({\bar {x}}^{2}+{\bar {y}}^{2})\,\!}
I
x
y
=
I
y
x
=
I
G
,
x
y
−
m
x
¯
y
¯
{\displaystyle I_{xy}=I_{yx}=I_{G,xy}-m{\bar {x}}{\bar {y}}\,\!}
I
x
z
=
I
z
x
=
I
G
,
x
z
−
m
x
¯
z
¯
{\displaystyle I_{xz}=I_{zx}=I_{G,xz}-m{\bar {x}}{\bar {z}}\,\!}
I
y
z
=
I
z
y
=
I
G
,
y
z
−
m
y
¯
z
¯
{\displaystyle I_{yz}=I_{zy}=I_{G,yz}-m{\bar {y}}{\bar {z}}\,\!}
證明：
慣性張量的平行軸定理 a) 參考右圖 ，讓
(
x
′
,
y
′
,
z
′
)
{\displaystyle (x\,',\ y\,',\ z\,')\,\!}
(
x
,
y
,
z
)
{\displaystyle (x,\ y,\ z)\,\!}
d
m
{\displaystyle dm\,\!}
y
=
y
′
+
y
¯
{\displaystyle y=y\,'+{\bar {y}}\,\!}
z
=
z
′
+
z
¯
{\displaystyle z=z\,'+{\bar {z}}\,\!}
依照慣性張量的慣性矩定義方程式，
I
G
,
x
x
=
∫
y
′
2
+
z
′
2
d
m
{\displaystyle I_{G,xx}=\int \ y\,'\,^{2}+z\,'\,^{2}\ dm\,\!}
I
x
x
=
∫
y
2
+
z
2
d
m
{\displaystyle I_{xx}=\int \ y^{2}+z^{2}\ dm\,\!}
所以，
I
x
x
=
∫
(
y
′
+
y
¯
)
2
+
(
z
′
+
z
¯
)
2
d
m
=
I
G
,
x
x
+
m
(
y
¯
2
+
z
¯
2
)
.
{\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}}\,\!}
相似地，可以求得
I
y
y
{\displaystyle I_{yy}\,\!}
I
z
z
{\displaystyle I_{zz}\,\!}
b) 依照慣性張量的慣性積定義方程式 ，
I
G
,
x
y
=
−
∫
x
′
y
′
d
m
{\displaystyle I_{G,xy}=-\int \ x\,'y\,'\ dm\,\!}
I
x
y
=
−
∫
x
y
d
m
{\displaystyle I_{xy}=-\int \ xy\ dm\,\!}
因為
x
=
x
′
+
x
¯
{\displaystyle x=x\,'+{\bar {x}}\,\!}
y
=
y
′
+
y
¯
{\displaystyle y=y\,'+{\bar {y}}\,\!}
I
x
y
=
−
∫
(
x
′
+
x
¯
)
(
y
′
+
y
¯
)
d
m
=
I
G
,
x
y
−
m
x
¯
y
¯
.
{\displaystyle {\begin{aligned}I_{xy}&=-\int \ (x\,'+{\bar {x}})(y\,'+{\bar {y}})\ dm\\&=I_{G,xy}-m{\bar {x}}{\bar {y}}\ .\\\end{aligned}}\,\!}
相似地，可以求得對於點 O 的其他慣性積方程式。
實心長方體：a)座標系統的原點在質心。b)座標系統的原點在角落。 思考一個實心長方體對於質心 G 的慣性張量，
I
G
=
[
1
12
m
(
w
2
+
h
2
)
0
0
0
1
12
m
(
h
2
+
d
2
)
0
0
0
1
12
m
(
w
2
+
d
2
)
]
{\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}}\,\!}
如圖右，質心 G 的位置是
(
d
2
,
w
2
,
h
2
)
{\displaystyle \left({\frac {d}{2}},\ {\frac {w}{2}},\ {\frac {h}{2}}\right)\,\!}
I
x
x
=
1
12
m
(
w
2
+
h
2
)
+
m
(
(
w
2
)
2
+
(
h
2
)
2
)
{\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)\,\!}
I
y
y
=
1
12
m
(
h
2
+
d
2
)
+
m
(
(
h
2
)
2
+
(
d
2
)
2
)
{\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)\,\!}
I
z
z
=
1
12
m
(
w
2
+
d
2
)
+
m
(
(
w
2
)
2
+
(
d
2
)
2
)
{\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)\,\!}
I
x
y
=
−
m
(
w
2
)
(
d
2
)
=
−
m
w
d
4
{\displaystyle I_{xy}=-m\left({\frac {w}{2}}\right)\left({\frac {d}{2}}\right)=-{\frac {mwd}{4}}\,\!}
I
x
z
=
−
m
(
h
2
)
(
d
2
)
=
−
m
h
d
4
{\displaystyle I_{xz}=-m\left({\frac {h}{2}}\right)\left({\frac {d}{2}}\right)=-{\frac {mhd}{4}}\,\!}
I
y
z
=
−
m
(
w
2
)
(
h
2
)
=
−
m
w
h
4
{\displaystyle I_{yz}=-m\left({\frac {w}{2}}\right)\left({\frac {h}{2}}\right)=-{\frac {mwh}{4}}\,\!}
因此，實心長方體對於點 O 的慣性張量是
I
G
=
[
1
3
m
(
w
2
+
h
2
)
−
1
4
m
w
d
−
1
4
m
h
d
−
1
4
m
w
d
1
3
m
(
h
2
+
d
2
)
−
1
4
m
w
h
−
1
4
m
h
d
−
1
4
m
w
h
1
3
m
(
w
2
+
d
2
)
]
{\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}}\,\!}
Beer, Ferdinand; E. Russell Johnston, Jr., William E. Clausen (2004). Vector Mechanics for Engineers. 7th edition. USA: McGraw-Hill, ISBN 0-07-230492-8 
