Game Development Reference
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and as we said previously, the strain can be described using only its six indepen-
dents components. We can re-arrange this as
∂x
0
0
ε xx
ε yy
ε zz
ε yz
ε zx
ε xy
∂y
0
0
u x
u y
u z
∂z
0
0
= D u e .
=
2 ∂y
2 ∂x
0
2 ∂z
2 ∂y
0
2 ∂z
2 ∂x
0
Replacing the displacement field by its nodal description of Equation (10.3), we
have
ˆ
ε = DH e u e . (10.4)
Hence, we can deduce the displacement-deformation matrix to be B e = DH e ,
which written explicitly is
∂ξ 1
∂x
∂ξ 2
∂x
∂ξ 3
∂x
∂ξ 4
∂x
0
0
0
0
0
0
0
0
∂ξ 1
∂y
∂ξ 2
∂y
∂ξ 3
∂y
∂ξ 4
∂y
0
0
0
0
0
0
0
0
∂ξ 1
∂z
∂ξ 2
∂z
∂ξ 3
∂z
∂ξ 4
∂z
0
0
0
0
0
0
0
0
B e =
.
∂ξ 1
∂y
∂ξ 1
∂x
∂ξ 2
∂y
∂ξ 2
∂x
∂ξ 3
∂y
∂ξ 3
∂x
∂ξ 4
∂y
∂ξ 4
∂x
0
0
0
0
∂ξ 1
∂z
∂ξ 1
∂y
∂ξ 2
∂z
∂ξ 2
∂y
∂ξ 3
∂z
∂ξ 3
∂y
∂ξ 4
∂z
∂ξ 4
∂y
0
0
0
0
∂ξ 1
∂z
∂ξ 1
∂x
∂ξ 2
∂z
∂ξ 2
∂x
∂ξ 3
∂z
∂ξ 3
∂x
∂ξ 4
∂z
∂ξ 4
∂x
0
0
0
0
(10.5)
To compute the partial derivatives of Equation (10.5), we need to further an-
alyze the tetrahedron geometry. Let us remember the geometric definition of a
tetrahedron in terms of its barycentric coordinates and the position p i of its four
nodes:
ξ 1
ξ 2
ξ 3
ξ 4
1
p x
p y
p z
1
1
1
1
p 1 ,x
p 2 ,x
p 3 ,x
p 4 ,x
.
=
(10.6)
p 1 ,y
p 2 ,y
p 3 ,y
p 4 ,y
p 1 ,z
p 2 ,z
p 3 ,z
p 4 ,z
Therefore, the barycentric coordinates can be found by inverting the matrix:
ξ 1
ξ 2
ξ 3
ξ 4
α 1
α 2
α 3
α 4
1
p x
p y
p z
β 1
β 2
β 3
β 4
=
,
(10.7)
γ 1
γ 2
γ 3
γ 4
δ 1
δ 2
δ 3
δ 4
where
1
α 1
α 2
α 3
α 4
1
1
1
1
β 1
β 2
β 3
β 4
p 1 ,x
p 2 ,x
p 3 ,x
p 4 ,x
=
.
(10.8)
γ 1
γ 2
γ 3
γ 4
p 1 ,y
p 2 ,y
p 3 ,y
p 4 ,y
δ 1
δ 2
δ 3
δ 4
p 1 ,z
p 2 ,z
p 3 ,z
p 4 ,z
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