Computer Aided Engineering Design

194 COMPUTER AIDED ENGINEERING DESIGN

Hence, λ^3 κb = p×Δp=m (say) (6.57)

Therefore, | m | = λ^3 κ (6.58)

Using the operator Δ on (6.57),

λλκΔλλκλκΔ d = (^334 ) = ( ) + =

ds

d

ds

d

ds

bm b

b

⇒ m (6.59)

d

ds

b

= –τn

whereτ is the torsion of the curve of intersection (Chapter 3).

Taking the scalar product with (6.56) results in

λλκ λκτ λκλ

λ

ΔΔ

d

ds

d

ds

⎛ ( (^34) ) – bn n tmp 2 + =
⎝
⎞
⎠
⋅
⎛
⎝
⎜
⎞
⎠
⎟ ⋅ (6.60)
⇒ λ^6 κ^2 τ = –Δm·Δp
From Eqs. (6.58) and (6.60) we can determine the curvature and torsion at any point on the curve of
intersection.
Example 6.8. Determine the torsion and curvature of the curve of intersection between a plane and
a sphere given by
x = 2, x^2 + y^2 + z^2 = 9
The plane intersects the sphere in a circle of radius √5, having its center on the x-axis at (2, 0, 0). Let
fxyz gx
xyz
=^1
2
(^222 + + – 9), = ( – 2), = ∇ ∂ + +
∂
∂
∂
∂
∂
⎛
⎝
⎜
⎞
⎠
ijk⎟
Figure 6.24 Intersection between a sphere and a plane in Example 6.8
z
g = x^2 + y^2 + z^2 – 9 = 0
f = x– 2 = 0
x Y
∇g
∇g
T