Game Development Reference

In-Depth Information

(

)

mb

2

+

c

2

−

mab

−

mac

1

1

1

3

4

4

(

)

2

2

=−

1

mab

1

m a

+

c

−

1

mbc

4

3

4

(

)

−

mac

−

mbc

m a

2

+

b

2

1

1

1

4

4

3

2

a
bc

a
bc

a
bc

−

m

,,

E

−

,,

⊗

,,

.

3

222

222

222

After adding corresponding matrix entries, we have for the inertia tensor of a box

(with side lengths
a
,
b
, and
c
) about its center of mass

(

)

mb

2

+

c

2

0

0

1

12

(

)

2

2

=

0

1

12

ma

+

c

0

.

(

)

0

0

ma

2

+

b

2

1

12

Example 14.7.
Find the inertia tensor of a dome, having constant density
ρ
and

semiaxis lengths
a
,
b
, and
c
as shown in Figure 14.8, in a coordinate system

where center of mass coincides with the origin.

c

a

b

Figure 14.8.
The dome used in Example 14.7.

Solution.
The total mass of the dome is

, and the
z
coordinate of the

m ρπabc

=

2

center of mass can be calculated using the integral

2

2

2

11

−

w

1

−

v

−

w

π

2

2

mC ρabc

=

w du dv dw

=

abc

,

(14.100)

z

4

2

2

2

01

−− −−−

w

1

v

w

where we have made the substitutions

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