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