Graphics Reference
In-Depth Information
2
2
2
2
Ê
∂
∂
j
¢+
∂
∂∂
j
∂
∂∂
j
¢+
∂
∂
j
ˆ
1
2
1
1
1
2
a
a
¢
a
a
¢
1
2
1
2
Á
Á
Á
Á
Á
Á
˜
˜
˜
˜
˜
˜
uv
uv
u
v
2
2
2
2
∂
∂
j
¢+
∂
∂∂
j
∂
∂∂
j
¢+
∂
∂
j
2
2
2
2
2
2
(
¢=
J
j
a
a
¢
a
a
¢
.
1
2
1
2
uv
uv
u
v
2
2
2
2
∂
∂
j
¢+
∂
∂∂
j
∂
j
¢+
∂
∂
j
3
3
3
3
a
a
¢
a
a
¢
1
2
1
2
Ë
2
uv
∂∂
uv
2
¯
u
v
The condition on g≤(t) which makes the curve g(t) into a geodesic is that
∂
∂
j
≤
()
(
()
)
=
g
t
•
a
t
0
(15.3a)
u
∂
∂
j
≤
()
(
()
)
=
g
t
•
a
t
0
(15.3b)
v
This would lead to having to solve a second order differential equation for a.
Instead, by introducing a new variable a¢(t) =b(t), one turns the system (15.3)
into a system of first order differential equations, which is the usual preferred
approach. See [PFTV86], for example, for ways to solve such systems of
equations.
Next, assume that we only want a geodesic in the kinematic or static force sense.
In this case, equations (15.3) get replaced by the single equation
g≤
()
()
=
t
•
b
t
0
(15.4)
where
()
=
(
()
)
¥
()
bn
t
g
t
g
t
.
Of course, as was pointed out earlier, such a curve g(t) may not mathematically be a
geodesic since g≤(t) may not be a normal to the surface, but it will trace out a geo-
desic path. Now, assume that the curve a(t) above is an arc-length parameterization
of a curve, so that |a¢(t)|=1. Let us parameterize the unit vectors a¢(t) by the turning
angle q(t). In other words, write
¢
()
=
(
()
()
)
a
t
cos
q
t
, sin
q
t
.
(15.5)
It follows from (15.5) and the chain rule that
¢¢
()
=
()
-
(
()
()
)
a
t
q
t
sin
q
t
, cos
q
t
.
(15.6)
If we replace g≤ in equation (15.4) by the right-hand side of equation (15.2) and also
replace a≤ by the right hand side of equation (15.6), then one can solve for q¢(t). In
fact, it is easy to show that solving (15.4) is equivalent to solving the following system
of equations:
