Computer Aided Engineering Design

DIFFERENTIAL GEOMETRY OF SURFACES 195

Then

f = (xi + yj + zk) = (x,y,z) = Nf, g = (i + 0j + 0k) = (1, 0, 0) = Ng

Here,Nf and Ng are vectors normal to the sphere f and the plane g, respectively.

∇×∇fgxyz = zy

100

= (0, , – ) = =

ijk

λtp

Sincet is the unit tangent vector to the curve of intersection

( ) ( ) =^222 = (– ) + = (^22 + )

1

λλλtt⋅⇒±yzλ yz^2

It can be observed that at y = 0 on the x = 2 plane, the point on the circle of intersection is given by

(2, 0, √5) and λt = 0i + zj+ 0k = 5j. This shows that the unit tangent t is along the j-direction

(Oy-axis). Using Eqs. (6.55) and (6.57),

pijk = 0 + – and 0zy + – pijk = 0 – –

x

z

y

y

z

ΔΔ≡ ∂ yz

∂

∂

∂

∂

∂

⎛

⎝⎜

⎞

⎠⎟

⇒

Therefore, pp

ijk

= 0 – ii

0– –

×Δλzy = –(^22 + ) = –^2

yz

yz

Now, from equation (5),

λ^3 κb = p×Δp=–λ^2 i⇒λκb = –1i

This shows that the bi-normal b is along the negative x-axis and the curvature κ =^1
( + )

.

22

1

yz^2

The

radius of the circle is √5.
Again, from equation (6.57) ⇒p×Δp = m.

Therefore,Δmijkijk = – {–( + ) + 0 + 0 } = {–(2 – 2 ) + + }
z^22
y
y
z

∂ yz zyyz

∂

∂

∂

⎛

⎝

⎜

⎞

⎠

⎟^00

⇒Δm = (0,0,0)

From equation (6.60),

• Δm·Δp = – (0, 0, 0) · (0, –y,–z) = 0 = λ^6 κ^2 τ

Hence, the torsion of the curve of intersection τ = 0. This is true, because the curve is a circle of radius
√5 lying on the plane x = 2.

Example 6.9. Intersection between a sphere and a cylinder, or a cylinder and another cylinder is
quite common in mechanical design. In some cases, it may be possible to get a parametric representation
of the curve discussed as follows.