180 COMPUTER AIDED ENGINEERING DESIGN
(L–G 11 κn) + (M–G 12 κn)μ = 0⇒
(– ) ( – )
(– ) ( – )1
=0
012 22
11 12MG NG
LG MGnn
nnκκ
κκμ⎡
⎣⎢⎤
⎦⎥⎡
⎣⎢⎤
⎦⎥⎡
⎣⎢⎤
⎦⎥ (6.34)For a non-trivial solution of μ, the determinant of the coefficient matrix must be zero. Therefore
( – ) ( – )
( – ) ( – )= 0
12 22
11 12MG NG
LG MGnn
nnκκ
κκor (GG G 11 22 – 1222 ) κκnn – (GNGL GM 11 + 22 – 2 12 ) + (LNM –^2 ) = 0 (6.35)
The above can be further simplified to find the two optimal values of the normal curvature. Thus
κκnnGN G L GM
GG GLN M
GG G2 11 22 12(^112212)
2
2
(^112212)
- 2
+ – 2+= 0⇒ κκnn^2 – 2HK + = 0
with H
GN G L GM
GG GKLN M
GG G=+ – 2
2( – )and =11 22 12(^112212)
2
2
(^112212)
2
⇒± = ( κn HHK^2 – )
Thus (κκn)max = max = ( + HHK^2 – )
(κκn)min = min = ( –HHK^2 – ) (6.36)
The maximum and minimum normal curvatures (κmax and κmin) at a point on a surface can be
calculated as above. K and H are called the Gaussianandmean curvatures, respectively. It can be
shown that the discriminant (H^2 – K) is either positive or zero. When the discriminant is 0, the surface
point is called an umbilical point for which κn = H. When K=H = 0, the point is aflatorplanar
point. We note that
KH = , =
- max min 2
κκ κκmax min (6.37)
Example 6.3. The parametric equation of a monkey saddle surface (Figure 6.13a) is given by
r(u,v) = ui + vj + (u^3 – 3uv^2 )k = (u,v,u^3 – 3uv^2 )
Compute the Gaussian and mean curvatures.
We can compute
ru = (1, 0, 3u^2 – 3v^2 ), rv = (0, 1, – 6uv),
ruu = (0, 0, 6u), ruv = (0, 0, – 6v),rvv = (0, 0, – 6u)
Therefore, from Eqs. (6.9) and (6.16), we can calculate the coefficients of the first and second
fundamental forms of the monkey saddle as