Computer Aided Engineering Design

88 COMPUTER AIDED ENGINEERING DESIGN

Similar treatment can be employed for y(u) and z(u) starting with the cubic form in Eq. (4.3) to obtain
results analogous to Eq. (4.6). The combined result for r(u)≡ [x(u),y(u),z(u)] may be expressed as

r( ) = [ ( ) ( ) ( )] = [ 1]

2–21 1

–3 3 –2 –1

0010

1000

(^32)
+1 1 +1
+1 1 +1
u xuyuzu u u u
xyz
xyz
pq r
pq r
iii
iii
iii
iii
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥

• 
  


• 
  

  or
  r
  P
  P
  T
  T
  ( ) = [ 1] UMG
  2–21 1
  –3 3 –2 –1
  0010
  1000
  (^32) =
  +1
  +1
  uuuu
  i
  i
  i
  i
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⎥
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⎥
  (4.7)
  whereU = [u^3 u^2 u 1] is the row matrix, M is a 4 × 4 square Hermite matrix and G is a 4 × 3 geometric
  matrix containing the end point and slope information Matrices U and M are identical for all Hermite
  cubic segments, but the geometric matrix G is user defined, that is, to alter the curve’s shape, we need
  to alter the entries in the geometric matrix. We may relocate Pi or Pi+1, or alter the end tangents Ti and
  Ti+1, both in magnitude and direction, to effect shape change. Keeping the end points and directions of
  the end tangents the same, we may as well observe the change in shape when altering the magnitudes
  of the end tangents. For a Ferguson segment given by Eq. (4.7), we can compute the first and second
  derivatives by differentiating r(u) with respect to u (Eq. 4.8). This would help in computing differential
  properties like tangents and end curvatures when imposing continuity conditions at the junction points.
  r
  r
  uu PPi i TTi i
  du
  du
  () = uu uu uu uu
  ()
  = (6^22 – 6 ) + (–6 + 6 ) +1 + (3^22 – 4 + 1) + (3 – 2 ) + 1
  r
  r
  uu u PPTTi i i i
  du
  du
  () = uuuu
  ()
  = (12 – 6) + (–12 + 6) + (6 – 4) + (6 – 2)
  2
  2 +1 +1 (4.8)
  The above equations can be written in the matrix form as
  r
  P
  P
  T
  T
  u UM G
  i
  i
  i
  i
  ( ) = [uuuu 1]
  0000
  6–63 3
  –6 6 –4 –2
  0010
  (^32) =
  +1
  +1
  1
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⎥
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⎥
  r
  P
  P
  T
  T
  uu UM G
  i
  i
  i
  i
  ( ) = [uuuu 1]
  0000
  0000
  12 –12 6 6
  –6 6 –4 –2
  (^32) =
  +1
  +1
  2
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⎥
  ⎡
  ⎣
  ⎢
  ⎢
  ⎢
  ⎢
  ⎤
  ⎦
  ⎥
  ⎥
  ⎥
  ⎥
  (4.9)
  Example 4.2. The starting and end points of a planar curve segment are Pi = (–1, 2) and Pi+1 = (8, 5).
  The unit tangent vectors at the ends are ti = (0.94, 0.35) and ti+1 = (0.39, – 0.92) with tangent
  magnitudes as ci = 8.5 and ci+1 = 15.2. Determine the tangent, radius of curvature, normal, and bi-
  normal for a point on the curve at u = 0.5.