172 COMPUTER AIDED ENGINEERING DESIGN

The symmetric matrix G is termed as the first fundamental matrix of the surface. In literature, the
components of G matrix are also written as

G 11 = E = ru·ru,G 12 = ru·rv,G 21 = ru·rv,G 22 = rv·rv (6.9a)

G 11 , G 12 and G 22 are called the first fundamental form coefficients. The length of the curve c(t) lying
on the surface is given by

ds G

du

dt

G

du

dt

d

dt

G

d

dt

(^211) dt
2
12 22
2
= ⎛ + 2 + ( )^2
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⎧
⎨
⎩⎪
⎫
⎬
⎭⎪
vv
(6.9b)
This equation can also be expressed as
ds^2 = G 11 du^2 + 2G 12 dudv + G 22 dv^2 (6.9c)
Thefirst fundamental form for a surface is given by
I
G
= Gdu Gd GG G d
1
{( + ) + ( – ) }
11
11 12 2 11 22 1222
⎛
⎝
⎞
⎠
vv(6.9d)
The unit tangent t to the curve is given by
t
rr
rr
rr
G

(, )

• (, )
  (, )


• 
  

  (, )

  
  

  (, )

  
  


• 
  

  ∂ (, )
  ∂
  ∂
  ∂
  ∂
  ∂
  ∂
  ∂
  ∂
  ∂
  ∂
  ∂
  ⎡
  ⎣⎢
  ⎤
  ⎦⎥
  u
  u
  du
  dt
  ud
  dt
  u
  u
  du
  dt
  ud
  dt
  u
  u
  du
  dt
  ud
  dt
  du
  dt
  d
  dt
  du
  d
  vv
  v
  v
  vv
  v
  v
  vv
  v
  v
  v tt
  d
  dt
  v
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  (6.10)
  Thus, for t to exist, G should always be positive definite. For any 2 × 2 matrix M, the condition for
  positive definiteness is that (a) M 11 > 0 and (b) M 11 M 22 – M 12 M 21 > 0, where Mij is the entry in the
  ith row and jth column of M. Now G 11 = ru·ru > 0 and also G 11 G 22 – G 12 G 21 = (ru·ru) (rv·rv)–
  (ru·rv)^2 = (ru×rv)· (ru×rv) > 0 and so G is always positive definite. The length of the curve
  segment in t 0 ≤t≤t 1 can be computed using Eq. (6.8b) as
  sds
  du
  dt
  d
  dt
  du
  dt
  d
  dt
  dt
  t
  t
  t
  t
  = =
  0
  1
  0
  1
  ∫∫
  ⎡
  ⎣⎢
  ⎤
  ⎦⎥
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  v
  v
  G (6.11)
  Ifc(t 1 ) and c(t 2 ) are two curves on the surface r(u,v) that intersect, the angle of intersection θ can be
  computed using
  tt
  rr
  G
  rr
  12
  11
  11
  1
  1
  =^22
  (, )

  
  


• 
  

  (, )
  (, )

  
  


• 
  

  (,)
  ⋅
  ∂
  ∂
  ∂
  ∂
  ⎡
  ⎣⎢
  ⎤
  ⎦⎥
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⋅
  ∂
  ∂
  ∂
  ∂
  u
  u
  du
  dt
  ud
  dt
  du
  dt
  d
  dt
  du
  dt
  d
  dt
  u
  u
  du
  dt
  ud
  dt
  vv
  v
  v
  v
  v
  vv
  v
  v
  = cos
  22
  2
  2
  du
  dt
  d
  dt
  du
  dt
  d
  dt
  v
  v
  ⎡
  ⎣⎢
  ⎤
  ⎦⎥
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  G
  θ (6.12)