Computer Aided Engineering Design

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

1. ru(u) = 0 for all u, it can be shown that the curve reduces to a point and r(u) = Pi.


2. 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.

3. If ru≠ 0, and ru×ruu = 0, then r(u) is a straight line.


4. 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.