Computer Aided Engineering Design

DESIGN OF CURVES 99

In similar triangles EBD and CBA, since D is the mid point of AB, the ratio BE/BD = BC/BA =^12
which implies BE =^12 BC=EC. In similar triangles PDQ and CDE,

DP

DC

DQ

DE

PQ

CE

= = = ,α thus DQ = DE =^1 AC QP EC CB

2

, = = –^1

2

αα α α

Now, DP = DQ + QP =^12 α(AC – CB)

and rr DP PPpD = + = (^12 i + i+1) +^12 α[(rP P r2+12 – ) – (i i – )]

Since rrpi = ( = 1/2) = ( + u 21 PPi+1) + (^18 TTi – i+1)

On comparison

Ti = 4α(r 2 – Pi), Ti+1 = 4α(Pi+1 – r 2 )

Therefore, the geometric matrix G for the Ferguson segment of a conic section (except the circle) is
given by

G= [Pi Pi+1 4 α(r 2 – Pi) 4α(Pi+1 – r 2 )]T (4.23)

With the above matrix: (a) if α < 0.5, the curve is an elliptical segment, (b) if α = 0.5, the curve is
a parabolic segment while for (c) 0.5 < α < 1, the curve represents a hyperbolic segment.

Example 4.5. Design a conic with end points Pi = (4, – 8) and Pi+1 = (4, 8) when the end tangents meet
atr 2 = (– 4, 0).
For known α, we can compute the end tangents using Eq. (4.23) as Ti = 4α (–8, 8) and Ti+1 =
4 α (8, 8). The Ferguson’s segment is

r( ) = [ 1]

2–21 1

–3 3 –2 –1

0010

1000

4–8

48

–32 32

32 32

uuuu^32

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

αα

αα

or r(u) = (32 αu^2 – 32αu + 4)i + [(–32 + 64α)u^3 + (48 – 96α)u^2 + 32αu – 8]j = x(u)i + y(u)j

Figure 4.9 Construction for a conic section

1

2 AB

1

2 AB

1

2 BC

1

2 BC

A

B

C

P

E

Q

D

αCD

αDE