# Curves and surfaces in computer aided geometric design - Yamaguchi F.

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coefficient of t3 = 2p(Q0 —Q1) + (l— p) (0o + 0i) = O-

Fig. 7.8. Geometrical relation among Q0, Q1} Q0 and Qx to express a conic section curve

7.6 Ã-Conic Curves

349

From this we obtain:

1 -V

Qi-Qo=VL^o + Qi)

2 p

(7.20)

This relation is shown in Fig. 7.8. Letting QT be the intersection of the tangents at points Q0 and Qx, we have:

(7.21)

Qt -Qo^^-Qo 2 p

Qi -QT=^0, 2 p

Solving these equations for Q0 and Q ( and

to 1 to 1"

-3 3 -2 -1

5 t2 t 1] 0 0 1 0

1 0 0 0

" 1 -P 0 0

0 1 -p 0

1 to T to î 1 -p

0 2(1 — 2 p) 0

to ] to Ã

-3 3 -2 -1

= lt: 5 t2 t 1] 0 0 1 0

1 0 0 0

" 1 -P 0 0

0 1 -p 0

— 2(1 — 2p) 0 1-p

0 2(1 — 2 p) 0

0 ‘

0 2 P

0 1 -P

1 -p_ 2 p

.1 -P

1 0 0

0 0 1

2 p 2 p 0

1 -P 1-p

0 2 p 2 P

1-p 1 -p

Qo

Qi

(Qi -Qr)

350

= [>3 t2 t 1]

7. The Rational Polynomial Curves

0 0 0

x 1 -p -2 p 1 -p -2(1-p) 2 p 0 1-p 0 0

-1-2 1- pi — p 0 On rQoi

[t2 t 1] -2 2 0 î "a î Qr

1 î î 1 1 î 0 ò "à 1 _Qi_

showing that this agrees with Eq. (7.5) for a conic section curve.

rQo1

Qr

_Qi_

References (Chap. 7)

32) Forrest, A.R.: “Conic Sections”, Draft of Computer-Aided Geometric Design, Dec. 1978.

Appendix A: Vector Expression of Simple Geometrical Relations

Straight Line Vector Formulas

A straight line which passes through the point P0 = [x0 y0 z0] in the direction of the vector L can be expressed in terms of the parameter t as:

P{t) = P0 + Lt. (A.l)

When L is a unit vector, the distance between points P(t) and P0 along the straight line becomes \P(t) — P0| = \Lt\ = \t\.

The vector formula for the straight line passing through the two points Ë) = 1>î Óî zo] and Pi = L*l Ó1 zi] is, with L = Pl—P0\

P(t) = (l-t)P0 + tP1. (A.2)

Points on the line segment P0PX correspond to values of the parameter t in the range

Perpendicular Bisector of a Line Segment

Let /V be a vector normal to a plane on which lie the two points P0 and Px. Since the vector (P1 — P0)xN is in the direction of the perpendicular bisector,

Fig. A.l

352 Appendix A: Vector Expression of Simple Geometrical Relations

the formula for the perpendicular bisector is found by setting L = (P1—P0)xN and by replacing P0 by (P1—P0)/2 in Eq. (A.l), (Fig. A.l):

P(t) = ^^ + (P1-P0)xNt. (A3)

Length of a Common Perpendicular to Two Non-Parallel Lines

Let Ð0Ðã and PqP[ be two straight lines. Let the intersections of the common perpendicular with these lines be H and H' (Fig. A.2). From Eq. (A.l):

H=P0 + (P1 — P0)t H' = Pl+(Pl-Po)t'.

Therefore:

H-H' = P0-Po + (P1-P0)t-(P[-Po)t'.

Let I be the length of the common perpendicular and è be the unit vector in the direction H— H . Then we have:

1={H-H) è

= (P0-P^)- u + (P1-P0)t ¦ u-(P[-Po)t' ¦ U

= {P0-Pi)-U

è is a unit vector perpendicular to Pt — P0 and P[ — P0:

(P]-P0)x{P[-P’) \(P1-P0)x(P[-P^))-

Appendix A: Vector Expression of Simple Geometrical Relations

Therefore:

_(Ð0-Ð<;)-[(Ë-Ë))õ(^-Ð0)]

/=-

1(Ë-Ë))õ(^-Ðî)1 IPo-PbPi-Po'Pl-Po']*'

\(Pl-Po)x(P[-P^)\

(A.4)

Plane Vector Formulas

A plane is expressed by ax + by + cz + d = 0. Since [a b c]T is a normal vector TV, if P = [x ó z] is a point on the plane, we have:

P-N+d = 0. (A.5)

Formula for a Plane That Passes Through 3 Points not on a Straight Line

Consider the 3 points Q0, Qu Q2 that lie on a plane but not colinear. (Qi — Qo) x(Qi~Qo) is a vector normal to the plane. Therefore, from Eq. (A.5) we have:

P ¦ KQi ~ Qo) x (Q2 - Qo)J = Qo • [(Qi - Qo) x (Q2 - Qo)J ¦

Expanding the right-hand side gives:

right side = Q0 ¦ [(& - Q0) x (Q2 - Q0)]

= Qo' (Qi x Q2 ~ Qi x Qo — Qo x Q2 + Qo x Qo) ~ Qo ¦ (Qi x Q2) = [Qo, Qu Q2]-

Therefore, the formula for the plane that passes through the three points Q0, Qu Qi is:

[Pi Qi~Qoi Q2~Qo]= LQo’ Qu Q{]- (A.6)

The brackets in the numerator indicate the triple scalar product.

354 Appendix A: Vector Expression of Simple Geometrical Relations

Intersection Point of 3 Planes

Consider 3 planes having the formulas:

P-Nt+d^O P-N2 + d2 = 0 P-N3 + d3 = 0.

The intersection point is:

„ dANtxNJ + dziNsxNJ + dsWxfy)

ë1-(ëã2õëó • (A-7)

It is necessary to have • (N2 x N3) ô 0.

A Circle That Passes Through 3 Points

Let us find the circle that passes through the 3 points Q0, Ql9 Q2 (Fig. A.3). The center of the circle is at the intersection of the perpendicular bisector Lx of the line segment Q0Q\ and the perpendicular bisector L2 of the line segment Q0Q2• Assuming that the points Q09 Ql9 Q2 are on a plane in 3-dimensional space, the center Qc of the circle to be found is at the intersection of 3 planes, namely, the plane in which the circle lies, the plane which includes Lx and is perpendicular to the circle plane, and the plane which includes L2 and is perpendicular to the circle plane.

The plane in which the circle lies is expressed by the equation:

(P-Qo) ¦ {(Qi - So) X (É2 - Qo)} =0. (A.8)

The equations of the planes perpendicular to the circle plane and including Lx and L2, respectively:

Fig. A.3

Appendix A: Vector Expression of Simple Geometrical Relations 355

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