Marsh D. Applied Geometry for Computer Graphics and CAD

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5. Curves 127

5.6.5 Converting from Parametric Form to Implicit Form

Conics deﬁned by polynomial coordinate functions are easily converted to im-

plicit form.

Example 5.24

Consider the conic deﬁned by x =2t

− 3t and y = t

+ t − 2. Add scalar

multiples of the equations to eliminate the quadratic terms

x − 2y =



− 3t



− 2



+ t − 2



= −5t +4.

Solving for t in terms of x and y gives t =(−x +2y + 4)/ 5. Substituting for t

in x =2t

− 3t (or alternatively, in y = t

+ t − 2) gives

x =2



−x +2y +4



− 3



−x +2y +4



Expanding and simplifying gives an implicit equation for the conic

− 4xy − 13x +4y

+ y − 14 = 0 .

In general, a conic is parametrized by rational functions and the approach

indicated in Example 5.24 is tedious. A more general method follows from the

following result.

Theorem 5.25

A necessary and suﬃcient condition that two quadratics

+ a

t + a

=0, and (5.12)

+ b

t + b

=0, (5.13)

have a common solution is



− a



=0. (5.14)

Proof

Suppose (5.12) and (5.13) have a common solution. Then b

×(5.12)−a

×(5.13)

yields

− a

)+(a

− a

) t =0. (5.15)

128 Applied Geometry for Computer Graphics and CAD

Similarly, b

×(5.12)−a

×(5.13) yields

((a

− a

)+(a

− a

) t) t =0. (5.16)

Eliminating t from (5.15) and (5.16) gives

− a

)(a

− a

) − (a

− a

)(a

− a

)=0,

and hence (5.14). The proof of the converse is left as an exercise for the reader.

Theorem 5.26

The conic with parametrization

x =

+ a

t + a

+ c

t + c

,y=

+ b

t + b

+ c

t + c

(5.17)

has an implicit equation of the form

x + B

y + C

)

− (A

x + B

y + C

)(A

x + B

y + C

)=0

where the coeﬃcients A

are the signed 2 × 2 minors of the matrix

Q =

⎛

⎝

⎞

⎠

Proof

For all t at which the conic is deﬁned, (5.17) can be multiplied through by the

denominator to give

− c

x +(−c

x + a

) t +(a

− c

x) t

=0, and (5.18)

− c

y +(−c

y + b

) t +(b

− c

y) t

=0. (5.19)

Applying Theorem 5.25 to (5.18) and (5.19) gives the necessary and suﬃcient

condition D

− D

=0where

=(b

− b

) x +(a

− a

) y + a

− a

=(b

− b

) x +(a

− a

) y + a

− a

, and

=(b

− b

) x +(a

− a

) y + a

− a

The proof is now complete since every point (x, y) of the conic satisﬁes D

−

= 0, a quadratic polynomial in x and y of the form

x + B

y + C

)

− (A

x + B

y + C

)(A

x + B

y + C

)=0

where the coeﬃcients A

are the signed minors of the matrix Q.

5. Curves 129

Example 5.27

To determine an implicit equation for the conic

(x(t),y(t)) =



18 − 19t +8t

−2 − 5t +4t

−6+7t +3t

−2 − 5t +4t



Apply Theorem 5.26 to

Q =

⎛

⎝

18 −19 8

−673

−2 −54

⎞

⎠

The required minors are

=(7)(4)− (−5) (3) = 43 ,

= −((−6) (4) − (−2) (3)) = 18 etc.

Alternatively, compute the transpose of the adjugate matrix of Q, to give

⎛

⎝

⎞

⎠

⎛

⎝

43 18 44

36 88 128

−113 −102 12

⎞

⎠

Therefore

x + B

y + C

)

− (A

x + B

y + C

)(A

x + B

y + C

)

=(18x +88y − 102)

− (43x +36y − 113)(44x + 128y + 12) .

Expanding and simplifying gives 784



−2x

− 5xy + x +4y

− 5y +15



=0.

The solution reverses the computation of Example 5.22.

Exercise 5.30

Convert the following conics from parametric to implicit form:

(a)



− 1,t+2



(b)



− 1,t+3



(c)



+ t − 1,t

− 3t +3



(d)



, 2t



(e)



−

+t+1

−2t+1

, −

−2t+1



Theorem 5.26 can be generalized to planar rational curves of any degree (see

[23], [13]):

130 Applied Geometry for Computer Graphics and CAD

Theorem 5.28

Let

(x(t),y(t)) =





i=0



i=0



i=0



i=0



be a rational curve of degree of degree n. Then an implicit form is obtained

from the Bezout resultant



0,0

··· D

0,n

n,0

··· D

n,n



=0,

where D

i,j



k≤min(i,j)

m=i+j−k+1

− c

) x+(a

− a

) y +(a

− a

5.7 Conics in Space

A conic in three-dimensional space is given parametrically by



+ a

t + a

+ d

t + d

+ b

t + b

+ d

t + d

+ c

t + c

+ d

t + d



Any conic in space is contained in a plane. To verify this, suppose every point

of the conic lies in the plane Ax + By + Cz + D =0.Then



+ a

t + a



+ B



+ b

t + b



+ C



+ c

t + c



+ D



+ d

t + d



=0,

that is,

(Aa

+ Bb

+ Cc

+ Dd

) t

+(Aa

+ Bb

+ Cc

+ Dd

) t

+ Aa

+ Bb

+ Cc

+ Dd

=0. (5.20)

Since this holds for all t (in an interval) the coeﬃcients of (5.20) must be

identically zero, implying

+ Bb

+ Cc

+ Dd

=0,

+ Bb

+ Cc

+ Dd

=0, and

+ Bb

+ Cc

+ Dd

=0.

The equations can be interpreted as deﬁning three planes in the three-

dimensional projective space with homogeneous coordinates (A, B, C, D). Thus

5. Curves 131

the coeﬃcients A, B, C, D can be determined using the method for computing

the intersection of three planes given in Section 3.4. This yields

(A, B, C, D)=



or any multiple of the determinant.

A planar representation of the conic can be obtained by applying a view-

plane coordinate mapping. The origin and the X-andY -axis directions in the

derived plane are speciﬁed, and the viewplane coordinate matrix VC is com-

puted. Then VC is applied to (x(t),y(t),z(t)) to give a conic in the speciﬁed

Cartesian coordinate system.

Example 5.29

Consider the conic (x(t),y(t),z(t)) =



3+6t−4t

−1−6t+2t

9−6t

−1−6t+2t

1−3t+t

−1−6t+2t



.Then

(A, B, C, D)=



391−1

60−3 −6

−4 −612



=54e

− 18e

+36e

=(54, −18, 36, 36) .

After dividing (A, B, C, D) through by 18, the conic is found to lie in the plane

3x −y +2z + 2 = 0. This can be veriﬁed by substituting x = x(t),y = y(t),z =

z(t) into the equation of the plane,



3+6t − 4t

−1 − 6t +2t



−



9 − 6t

−1 − 6t +2t





1 − 3t + t

−1 − 6t +2t



+2=0.

Multiplying through by the denominator gives



3+6t − 4t



−



9 − 6t





1 − 3t + t





−1 − 6t +2t



=0.

Expanding the brackets yields that the left-hand side of the equation is identi-

cally zero, and hence the conic lies in the derived plane.

132 Applied Geometry for Computer Graphics and CAD

Let the plane have origin and axes as speciﬁed in Example 4.11. Applying

the viewing coordinate mapping matrix VC to the conic gives



3+6t − 4t

9 − 6t

1 − 3t + t

−1 − 6t +2t



⎛

⎜

⎝

0.385 0.360 −0.333

0.642 0.600 0.111

−0.706 0.960 −0.222

0.449 −1.200 0.778

⎞

⎟

⎠



5.778 + 1.734t − 5.2t

8.64 + 6.48t − 6.48t

−1 − 6t +2t



The planar representation of the conic is



5.778+1.734t−5.2t

−1−6t+2t

8.64+6.48t−6.48t

−1−6t+2t



The vector form for the conic in space is

O +



5.778 + 1.734t − 5.2t

−1 − 6t +2t



X +



8.64 + 6.48t − 6.48t

−1 − 6t +2t



Y ,

where O, X and Y are as given in Example 4.11.

Exercise 5.31

By applying Theorem 5.6, show that the conic

(x(t),y(t),z(t)) =



9+3t − 4t

3+4t +4t

4t − 3t

3+4t +4t

1+t

3+4t +4t



lies in the plane −16x +55y + 273z − 43 = 0.

5.8 Applications of Conics

Example 5.30 (Headlights and Radar)

Parabolas have the special property that rays of light, emanating from a light

source positioned at the focus, are reﬂected in the parabola along parallel lines,

as illustrated in Figure 5.15. This property is used in the design of car headlight

reﬂectors. A reﬂector has the shape of a paraboloid, that is, a surface obtained

by rotating a parabola about its axis of symmetry. If a headlight bulb is posi-

tioned at the focus of the parabola then it produces a beam of light consisting

of the reﬂected parallel rays of light.

The same property is used in the design of radar or satellite dishes. Signals

from a distant point travel along (nearly) parallel rays. Signals which reach a

paraboloid shaped dish are reﬂected along linear paths which pass through the

focus. The satellite receiver is positioned at the focus in order to obtain the

best reception of the signals.

5. Curves 133

Figure 5.15

EXERCISES

5.32. Show that a ray of light emanating from the focus F(0,m)ofthe

parabola y =4mx

is reﬂected parallel to the axis of symmetry, that

is the y-axis, as follows.

(a) Determine the tangent vector/line at a point P(x, y)onthe

parabola.

(b) Determine the angle α between the line FP and the tangent.

and deduce that the reﬂection of the ray is parallel to the y-axis.

5.33. Show that all rays of light emanating from one focus of an ellipse are

reﬂected in lines containing the other focus. The reader may wish to

contemplate an elliptical snooker table with the cue ball positioned

at one focus and a single pocket positioned at the other focus. If the

ball is struck (without spin) with suﬃcient strength then the ball

will hit the elliptical cushion and rebound along a line containing

the pocket.

Example 5.31 (Suspension Bridges)

Suspension bridges are designed so that a cable hanging from two pillars or

towers carries the weight of the bridge uniformly along the cable. The resulting

shape of the cable is a parabola. Suppose the pillars are 1, 410 metres apart (the

span of the Humber Bridge, Hull, UK) and the height of the pillars is h metres.

Let the origin be the lowest point of the parabola, and let the horizontal plane

134 Applied Geometry for Computer Graphics and CAD

of the bridge be the x-axis. Then the parabola is symmetrical about the y-axis

and passes through the points (−705,h), (705,h), (0, 0). Let the parabola be

y = ax

+ bx + c. Then, clearly c =0,and

h = a (−705)

+ b (−705) , and

h = a (705)

+ b (705) .

Thus a = h/ (705)

and b = 0 giving the parabola y =

705

In reality, one must account for the earth’s curvature when modelling large

structures. So in the example of the bridge the distance between the base of

the pillars is 1, 410 metres, but the distance between the tops of the pillars is

greater.

Example 5.32 (Radar)

Discovered as recently as the 1940s, the method known as hyperbolic navi-

gation has had a considerable inﬂuence on sea and air navigation. A receiver

records the radio signals transmitted from two ﬁxed stations. Assuming that

the velocity v of radio energy is constant, the distances travelled by radio en-

ergy are proportional to the time taken. Suppose the times taken to receive

the signals, sent at the same time, from each station are t

and t

. Then the

distance from each station is vt

and vt

. Hence the diﬀerence in distance of

the receiver from the stations is v(t

− t

). The locus of all possible positions

of the receiver relative to the ﬁxed stations is a branch of a hyperbola with the

stations positioned at the foci (see Exercise 5.34). There are two points on the

hyperbola a given distance vt

from the ﬁrst station and it remains to decide

which is the correct location. Commercial hyperbolic navigation systems in-

clude the Decca Navigation System, LORAN, Omega, and Global Positioning

Systems (GPS).

Exercise 5.34

Let F

and F

be ﬁxed points, and let d

and d

be the distances of a

point P from F

and F

respectively. Show that the locus of all points

P such that d

− d

is constant is a hyperbola.

B´ezier Curves I

6.1 Introduction

Two of the most important mathematical representations of curves and sur-

faces used in computer graphics and computer-aided design are the B´ezier and

B-spline forms. The original development of B´ezier curves took place in the au-

tomobile industry during the period 1958–60 by two Frenchmen, Pierre B´ezier

at Renault and Paul de Casteljau at Citr¨oen. The development of B-splines

followed the publication in 1946 of a landmark paper [22] on splines. B-splines

will be discussed in detail in Chapter 8. Further discussion of the historical

development of B´ezier curves may be found in [2], [6] and [10].

B´ezier curves are polynomial curves (see Deﬁnition 5.1) which have a par-

ticular mathematical representation. Their popularity is due to the fact that

they possess a number of mathematical properties which facilitate their manip-

ulation and analysis, and yet no mathematical knowledge is required in order

to use the curves. A B´ezier curve of degree n is speciﬁed by a sequence of n +1

points which are called the control points. The polygon obtained by joining the

control points with line segments in the prescribed order is called the control

polygon.

Control of the shape of a B´ezier curve is facilitated by the fact that the

control polygon reﬂects the basic shape of the curve. In many drawing and

CAD packages the control points of a B´ezier curve may be speciﬁed by clicking

with a mouse at the desired locations within a document window. The control

points are visible on the computer screen, and modiﬁcation of a control point

135

136 Applied Geometry for Computer Graphics and CAD

is executed with a simple click and drag operation of the mouse. As an aid to

design, many packages display the changing curve as a control point is modiﬁed.

6.2 B´ezier Curves of Low Degree

6.2.1 Linear B´ezier Curves

A linear B´ezier curve is a line segment joining two control points b

)

and b

), and parametrized by

(x(t),y(t)) = (1 − t)(p

)+t(p

), for t ∈ [0, 1] ,

so that x(t)=(1− t)p

+ tp

,andy(t)=(1− t)q

+ tq

. Letting B(t)=

(x(t),y(t)), the curve can be written in the vector form

B(t)=(1− t)b

+ tb

. (6.1)

The curve is deﬁned on the interval [0, 1], so the starting point of the curve

is B(0) = b

and the ﬁnishing point is B(1) = b

, that is, the B´ezier curve

interpolates the ﬁrst and last control points.

Example 6.1

The B´ezier form for the linear segment passing through points b

(1, 2) and

(3, 4) is B(t)=(1− t)b

+ tb

=(1− t)(1, 2) + t(3, 4). Hence x(t)=(1−

t)+3t =1+2t and y(t)=2(1− t)+4t =2+2t.

6.2.2 Quadratic B´ezier Curves

Suppose three control points b

), b

), and b

) are speciﬁed.

Then the quadratic B´ezier curve is deﬁned to be

B(t)=(1− t)

)+2(1− t)t(p

)+t

), for t ∈ [0, 1] .

The starting point of the curve is B(0) = b

and the ﬁnishing point is B(1) =

. The curve can be expressed in the parametric form (x(t),y(t)) where

x(t)=(1− t)

+2(1− t)tp

+ t

, and

y(t)=(1− t)

+2(1− t)tq

+ t

The triangle b

obtained by joining the control points with line segments,

in their prescribed order, is called the control polygon.