Graphics Reference
In-Depth Information
Proof.
We can think of k(s), which is the length of the vector T¢(s), as the area of
the rectangle spanned by the orthogonal vectors T(s) and T¢(s), but such an area can
be computed from the cross product of the two vectors. Since the z-component of the
vectors is 0, the formula works out to the determinant as indicated. It is easy to check
that the signs are correct also.
Note 1.
Our parameterizations will typically be assumed to be
regular
. Recall that
“regular” means that the derivative does not vanish and hence avoids various degen-
erate cases that a zero derivative would cause, such as it being in the denominator of
a formula as in the case of curvature.
Note 2.
Because the arc-length parameterization G(s) of a path is unique given a tra-
versal direction, the values of functions defined in terms of it should really be thought
of as associated to the corresponding points on the
path
. Therefore, and this is another
reason for using regular parameterizations, if F(t) is a regular parameterization for the
same curve, then we can think of such values as functions of t because the change of
parameter function between t and s is a one-to-one correspondence. This means that
we may use expressions such as “the principal normal at t” or “the curvature at t” and
write N(t) or k(t), respectively. We must be a little careful though because, if signed
quantities are involved, such as the signed curvature, then the parameter t and the
arc-length parameter s need to traverse the curve in the same direction.
Following up on Note 2, since curves are hardly ever presented via their arc-length
parameterization because that usually involves complicated formulas, functions
defined in terms of arc-length parameterization are not very useful from a computa-
tional point of view. It important therefore that one can compute them with respect
to arbitrary parameterizations.
To begin with, consider a regular parameterization F(t) for a curve and let
G(s) = F(f(s)) be its arc-length parameterization, where t =f(s) and f¢(s) > 0. Now the
equation
()
=¢
()
=¢
()
(
)
¢
()
Ts
G s
F
ff
s
s
(9.5)
together with the facts that T(s) is a unit vector and f¢(s) > 0 imply that
1
¢
()
=
f
s
.
(9.6)
(
()
)
Fs
¢
f
Next, let S(t) = T(f
-1
(t)). Then S¢(t) = T¢(f
-1
(t))f
-1
¢(t), implies that
¢
()
¢
()
.
St
Ft
¢
()
=
Ts
(9.7)
Equations (9.5)-(9.7) lead to the following formulas:
(
()
)
¢
()
¢
()
Fs
Fs
¢
f
f
Ft
Ft
()
=
Ts
=
(9.8)
(
()
)
¢
