184 COMPUTER AIDED ENGINEERING DESIGN
ruu= [–(1 –v)acosu]i–asinu (1 –v)j
ruv= [asinu– 2a/π]i–acosuj
rvv= 0
All we need to show then is M≠ 0. From Eqs. (6.19) and (6.22)
D
xyz
xyz
xyz
au
a
au
au
a
au
aua
au
au b
uu
12 = uuu =
sin –
2
- cos 0
- (1 – ) sin ( ) –
2
(1 – ) cos 0 - cos + –
2 - sin
vvuv
vvv
v
v
v
π
π
π
⎛
⎝
⎞
⎠
= (1 – ) cos sin –
2
- cos (1 – ) sin ( ) +
2
bauau
a
au a u
a
vv
v
ππ
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⎧
⎨
⎩
⎫
⎬
⎭
=
cos sin –
2
cos – cos sin +
2
cos – cos sin
+ cos sin –
2
cos
2
2
2
2
2
2
2
b
auu
a
ua uu
a
ua u u
auu
a
u
ππ
π
vv
v
v
⎧
⎨
⎪⎪
⎩
⎪
⎪
⎫
⎬
⎪⎪
⎭
⎪
⎪
=–
2
cos
2
b
a
u
π
⎧
⎨
⎩
⎫
⎬
⎭
which is zero only when cos u = 0 or when u =^12 π. Since D 12 and hence M≠ 0 for other values of
u, the Gaussian curvature is not zero at all points on the surface. The toothpaste tube, therefore, is
non-developable and cannot be flattened without tearing or stretching.
Some other examples of ruled but non-developable surfaces are those of Plucker polar and hyperbolic
paraboloid surfaces shown in Figure 6.16. The respective equations are
r(u,v) = u cos vi+usinvj + sin nvk (n is an integer ≥ 2)
r(u,v) = ui + vj + uvk
Figure 6.15 A symmetric half of the toothpaste tube