DIFFERENTIAL GEOMETRY OF CURVES 81
The unit bi-normal vector Bmay be obtained from Eq. (3.30b) as
B ̇r ̇ ̇r
̇r ̇ ̇r
=
| |
×
×
̇ ̇rijk = – cos – sin + 0atat
r ̇ ̇ ̇r
ijk
= – sin cos ijk
- cos – sin 0
× = sin – cos +^2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
at a t b
at at
ab t ab t a
| | = ( ̇r× ̇ ̇r ab sin ) + (–t^222222 ab cos ) + (t a ) = a a + b
Therefore, B
ijk
sin – cos +
(^22) +
btbta
ab
The normal vector N is given by
N = B×T
Therefore
N
ijk
=^1 ijk
- sin – cos
- sin cos
22 = – (cos + sin + 0 )
ab
bt b ta
at a t b
tt
The curvature is given by Eq. 3.30(c).
κ = | |
| |
=
+
( + )
=
( + )
3.
22
223/2 22
r ̇ ̇ ̇r
̇r
× aa b
ab
a
ab
Therefore, the radius of curvature ρ =
( + ab^22 )
a
.
From Eq. (3.33), the torsion τ is given as
τ =
d
ds
B
d
ds
d
dt
ds
dt
d
dt
d
dt
BB B r
= ⎛ =
⎝
⎞
⎠
d
dt
btbt
ab
Bij
=
cos + sin
(^22) +
d
dt
ab
r
=^22 +
⇒ τ =
cos + sin
(+ )
(^22) (+ ) 22
btbt
ab
b
ab
ij