98 COMPUTER AIDED ENGINEERING DESIGN
r(u) = (1 – 3u^2 + 2u^3 )Pi + (3u^2 – 2u^3 )Pi+1 + (u – 2u^2 + u^3 )Ti + (–u^2 + u^3 )Ti+1
= (1 – 3u^2 + 2u^3 )Pi + (3u^2 – 2u^3 )Pi+1 + (u – 3u^2 + 2u^3 ) (Pi+1–Pi)
= (1 – u)Pi + (u)Pi+1
which is linear in u and thus is a line segment PiPi+1 for u in [0, 1]. More generally, if Ti =
ci(Pi+1 – Pi) and Ti+1 = ci+1(Pi+1–Pi), then
r(u) = Pi + (Pi+1 – Pi){3u^2 – 2u^3 + ci(u – 2u^2 + u^3 ) + ci+1(–u^2 + u^3 )}
Observe that for ci = ci+1 = 1, we get r(u) = (1 – u)Pi + (u)Pi+1. For ci = ci+1 = c, we have
r(u)= Pi + (Pi+1 – Pi) {3u^2 – 2u^3 + c(u – 3u^2 + 2u^3 )}
=Pi + (Pi+1 – Pi) {(1 – c) (3u^2 – 2u^3 ) + cu}
Ifc = 0, then r(u) = Pi + (Pi+1 – Pi) (3u^2 – 2u^3 ) which is also a line with v = 3u^2 – 2u^3. However, while
the point on r(u) = (1 – u)Pi + (u)Pi+1 moves with a constant speed with respect to u, the point on
r(u) = Pi + (Pi+1 – Pi) (3u^2 – 2u^3 ) moves with a variable speed. Note that for the latter
ru(u) = (– 6u^2 + 6u) (Pi+1 – Pi) and ruu(u) = (–12u + 6) (Pi+1 – Pi)
Thus, the tangents at the start and end points of this line are zero, and the point on the line accelerates
tillu =^12 , when r(u =^12 ) =^12 (Pi+1 + Pi), and then decelerates. For c = –1, it can be determined that
r(u) = Pi + (Pi+1 – Pi)(6u^2 – 4u^3 – u)
Special Cases of Ferguson Curves
- ru(u) = 0 for all u, it can be shown that the curve reduces to a point and r(u) = Pi.
- rr
rr r
rr
rr r
uuu
x
u
x
uu
x
uuu
y
u
y
uu
y
uuu
z
u
z
uu
z
uuu
×≠ 0 and r = 0, then r(u) is a planar curve.
- If ru≠ 0, and ru×ruu = 0, then r(u) is a straight line.
- If ru×ruu≠ 0, and 0,
rr r
rr
rr r
x
u
x
uu
x
uuu
y
u
y
uu
y
uuu
z
u
z
uu
z
uuu
r ≠ , r(u) is a three-dimensional curve.
To generate conics using Ferguson segments, consider
r(u) = (1 – 3u^2 + 2u^3 )Pi + (3u^2 – 2u^3 )Pi+1 + (u – 2u^2 + u^3 )Ti + (–u^2 + u^3 )Ti+1
which, for u =^12 gives
rPPTT() =^1
2
( + ) +^1
8
(^12) i i+1 ( – i i+1)
To determine the geometric matrix G for a conic section, consider the following construction in
Figure 4.9. Let the end tangents at points A (Pi) and B (Pi+1) on a conic section meet at point C(r 2 )
and let D be the mid point of AB, that is,^12 (Pi + Pi+1). Let CD intersect the curve at P. Draw DQ
parallel to AC and PQ parallel to BC.