Computer Aided Engineering Design

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