Conics, Quadrics, and Projective Space

James D Emery

Last Edit 9/3/2015

Contents

1 Introduction 4

2 Overview 7

3 Projective Space and the Conic Sections 15

4 The Cross Ratio 21

5 The Tangent Line 29

6 Computing A Canonical Representation 33

7 Conic Through a Set of Points 44

8 Parametric Conic Arcs, Matrices of Projective Transforma-

tions. 44

9 A Projective Transformation That Takes Four Points To Four

Points 48

10 An Affine Transformation That Approximately Takes Four

Points To Four Points 51

11 The Rational Parametric Arc And The Conic Arc 53

12 Straight Lines In Projective Space 57

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13 Quadric Surfaces 59

14 Quadric Surfaces And Their Matrices 60

15 Transformations 63

16 The Construction Of Solids 67

17 Plane Image 68

18 Monte Carlo Integration 68

19 Stratified sampling 72

20 Point Transformations and Coordinate Transformations 74

21 Cross Product, Axial Vectors 75

22 Angular Velocity 77

23 Angular Momentum And The Inertia Tensor 79

24 Surface Integrals 81

25 Volume Integral 83

26 Polygon Areas 84

27 Perspective Projection 85

28 The Tektronix Hardware Modeler: A Quadric Solid Mod-

eler 92

29 References and bibliography 94

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List of Figures

1 Points of projective 2-space are lines in 3-space. A 3-d lineartransformation is a 2-d projective transformation. A rotationof the cone can project the circle to an ellipse, a parabola, ora hyperbola. In the same way the lines through any figure inthe plane are a projective space representation of the figure.Any sequence of perspective projections of a figure onto a setof planes can be represented as a sequence of linear transfor-mations in 3-space. The plane shown in the figure is knownas the affine plane. . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Let lines PQ and PR be tangent to the ellipse. Tangent pointsQ and R are on the polar of P . Thus the line defined by Qand R is the polar of P . Intersection point S is on both thepolar of P and the polar of P ′. Thus the polar of S is the linethrough P and P ′. . . . . . . . . . . . . . . . . . . . . . . . . 25

3 The parabola x − 2yz = 0 and a point at infinity Q. The linethrough P and Q meets the parabola at P and Q. The polarof a point at infinity is a diameter. The polar of Q is the lineat infinity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 The parabola x−2yz = 0 and a point at infinity Q = (1, 1, 0).The line through P and Q meets the parabola at two finitepoints. The polar of Q is the line x = 1, which is a diame-ter of the parabola. The point at in finity in the diagram isrepresented as an arrow in the direction of the infinity point Q. 30

5 Results of plotting conics with the program pltconic.ftn.(a)Ellipse with equation 10x2+10xy+20y2+3x+−4y−10 = 0,(b)Hyperbola with equation 5x2+10xy−7y2−3x+2y−1 = 0,(c)Parabola with equation x2 + 6xy + 9y2 + 3x − 5y − 1 = 0,(d)and a circle with equation 2x2 + 2y2 + 1

2x − 1

5y − 1 = 0. . 45

6 A conic passing through four points defined by the products oflines passing through the points (1, 1), (2, 2), (2, 1), (3, 3). Thegeneral equation of such a conic is λ(y − x)(y − x + 1) + (1 −λ)(y − 2)(y − 1) = 0 where λ is a parameter. Here λ = 1/2.By letting points come together we can also define conics bytangent lines. . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

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7 (a)Any conic arc can be obtained by mapping this parabola,which is defined by tangent lines AC and BC, and pointD,where A = (0, 0), B = (1, 1), C = (0, 1/2), D = (1/4, 1/2).(b)An example of mapping the parabola to a conic represen-tation of a circle. The intersection of the tangents is at C,which is a point at infinity. . . . . . . . . . . . . . . . . . . . 47

8 A Perspective drawing showing, a building, a railroad track,and the horizon. The vanishing point at the end of the railroadtracks is at infinity in three space, yet its projection is a finitepoint on the two dimensional page. The figure was generatedwith program tracks.ftn. . . . . . . . . . . . . . . . . . . . . 86

9 The intersection of the parabolic cylinder z2 = y and thecone y2 = xz is the union of the twisted cubic curve andthe line along the x axis. This figure was generated with thesolid modeling program qs90.ftn, which was formerly calledQuadric.ftn. The program outputs a bitmap, with pixel val-ues as hex characters. Program bm2ps.c converts it to apostscript file. . . . . . . . . . . . . . . . . . . . . . . . . . . 93

1 Introduction

Quadric solids are solids bounded by quadric surfaces. Quadric solids providea rich collection of sets for constructing models of 3-dimensional objects.A large set of quadric solids may be built by combining primitive quadricsolids, such as spheres, ellipsoids, blocks, cones, wedges, and cylinders. Manymanufactured objects are combinations of quadric shapes. There are 17 typesof quadric surfaces, including such less common surfaces as the parabolichyperboloid.

The world is full of objects built from rectangular blocks, circular cylin-ders, and spheres. These objects are quadric solids. Polyhedra are quitecommon in nature. Minerals and crystals often take a polyhedral form. Someviruses have the shape of a polyhedron. Polyhedra are quadric solids becuasethey can be built from planar half spaces, and planes are quadric surfaces(pairs of planes). Essentially all objects can be approximated, to an ar-bitrary accuracy with polyhedra. More specifically, their surfaces may beapproximated with triangles.

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But a polyhedron is locally flat and has no local curvature. So the planarfaces of a polyhedron can not model a local curved shape exactly. An advan-tage of using quadric surfaces is that they can locally approximate curvatureand shape [O’Neil p.202]. An object can be represented by a relatively smallnumber of primitive quadric solids. If we regard atoms as tiny balls, then in asense, all objects are built from spheres. But Avogadro’s number 6.03× 1023

is very large, so modeling with spheres is somewhat impractical. However, inthe field of scientific visualization we do model with rectangular cells definedby very large numbers of lattice points. When modeling simple manufac-tured parts, we are interested in finding exact models. Quadric solids canserve well in such modeling.

Quadric solids by themselves do not give a full practical set of primitivesfor modeling. For example, surfaces of revolution are ubiquitous. But mostsurfaces of revolution are not quadric surfaces. Further a large number ofquadrics are frequently required to get an acceptable approximation to asurface of revolution. The torus is a surface of revolution that is not aquadric surface. A torus could be approximated by a set of cylinders. Butthe differential geometric properties would not be approximated. Parametricfree form surfaces, such as spline surfaces, are widely used in manufacturing,but are not quadric surfaces.

Computations with quadric solids are simple. The calculation of a normaldirection, at a point on a quadric surface, is accomplished by a matrix multi-plication. The intersection points of a straight line with a primitive quadricsolid require only the solution of a quadratic equation. The transformedequation of a translated or rotated quadric surface is easily found.

With the advent of the computer age, there has been a renaissance ofcalculation. Many well known algorithms and calculating methods, whiletheoretically powerful, were formerly of little interest. They were two bur-densome for the human computer. In the past techniques were emphasizedthat could be carried out easily by hand. But now those calculations, whichwere too lengthy for hand calculation, are proper for computer modeling andcomputer graphics. Frequently the novice in geometric modeling and graph-ics does not use the proper tools. Naturally he uses the calculations taughthim in the schools and colleges, which are often only the small amount of an-alytic geometry found in the calculus course, and deals, for example, with thesimultaneous solution of a system of equations in a fixed coordinate system.A vector approach is superior for many calculations. The vector methods

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tend to be independent of the coordinate system. Those with scientific edu-cation are usually familiar with vectors. But they are frequently unaware ofsome very useful mathematics. This is the mathematics of projective geom-etry, projective space, and specifically the idea of homogeneous coordinates.Projective space is very important in certain advanced areas of mathemat-ics. It is an important aspect of the field of algebraic geometry. But evenamong mathematicians, knowledge of projective and algebraic geometry isnot universal.

Often the course in projective geometry, which is offered in Americanuniversities and colleges, is designed for future teachers of high school math-ematics. This course, or a course called modern geometry, is taught to showthe future teacher of Euclidean geometry, that there are many geometries,and that not all of the geometries are Euclidean. Niether the course onprojective geometry, nor the books designed for it, emphasize the beauti-ful simplifications and generalizations of calculation that come from treatingpoints as belonging to projective space. The few pages discussing projectivespace in Birkhoff and MacLane (Survey of Modern Algebra), are moreilluminating than the numerous pages of the elementary books on projectivegeometry. This is because the connection to linear algebra and vector spacesis not exploited in these books.

The purpose of this report is, first to describe a modeling system whichuses quadric solids as primitive objects, and second to present some ele-mentary ideas of projective geometry which are useful for computing. Theselatter ideas, although elementary, are not readily available. The quadric solidmodeling system is implemented in a computer program. An original ver-sion was called Quadric. A later version is called qs90.FTN. It producesviews of 3-dimensional objects, calculates volumes, calculates moments, andcalculates inertia tensors.

The idea of representing objects as primitive solids was quite fashionablein the 1980’s. Now most solid modeling systems are surface modeling sys-tems. But primitive solid modeling still has some interest. Quadric solidsare obvious candidates for the primitives. There are several papers dealingwith quadric surfaces [Leven].

The modeling system described here represents objects as Boolean com-binations of quadric half space primitives. An image of the model is madeby scanning its surface and calculating an illumination intensity at each sur-face point. The 2-dimensional picture plane is scanned to construct intensity

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values. A surface normal is calculated at the point on the surface, wherethe line through the picture point pierces the object. The intensity value isdetermined from the inner product of the normal at the first pierce pointwith an illumination vector. Since the primitives are quadric solids, this canbe done easily.

A quadric surface is the set of real zeros of a quadratic form in fourdimensions. Such a set of zeros is called a variety in algebraic geometry. Aquadratic form is a polynomial in which each term is of the second degree.The set of points were the form takes nonpositive values is call a lower halfspace. An upper half space is the set where the form takes nonnegativevalues. A quadric form defines two quadric half spaces. Examples of halfspaces are the points inside an ellipsoid, and the points outside two parallelplanes.

It is easy to find the intersection points of a quadric surface with a straightline. The parameters of the intersection points on the line will be the rootsof a quadratic equation. Surface normals are also easily calculated.

The objects are built using the standard set operators, union, intersec-tion and complementation. Some modeling schemes, for example the PADL

system, which was developed at the university of Rochester, uses regular-ized set operators. These operators guarantee that the resulting objects aretopologically closed, and homogeneously three dimensional. Using standardoperators one may get a set that is not closed. Because we have only an ap-proximate model of the surface and do not explicitely calculate the boundary,the use of regularized operations does not seem appropriate. For intersectionpoints which are candidates to be on the boundary, We construct a smallline segment in the direction of the normal that contains the point. If oneend of the segment is inside, and the other outside, the point is a boundarypoint of the object.

2 Overview

We shall start with two-dimensional projective space. This space is easy tovisualize. Projective 2-space is the set of lines in 3-dimensional space whichpass through the origin. Projective 2-space is the set of one dimensional sub-spaces of a 3-dimensional vector space. A plane that does not pass throughthe origin is given a special role. It is called the affine plane. A coordinate

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system is usually selected so that the affine plane is the plane z = 1. Theidea of projective space can be illustrated by considering the intersection ofa cone and a plane. The intersection curve is called a conic section. See Fig.1.

When the cone is rotated, the conic section is transformed to a new conicsection. A circle may go into a hyperbola, a parabola, or an ellipse. Thelines on the cone are points of projective 2-space. The intersection of theselines with the affine plane are the representation of the points in affine space.However the projective points (i.e. lines) and the affine points are to beconsidered the same objects, namely 2-dimensional points. The componentsof any vector which lies in the subspace representing a projective point, arecalled the homogeneous coordinates of the point. Homogeneous coordinatesare determined only up to a scalar multiple. A line in 3-space which is parallelto the affine plane meets it at infinity. The projective point in the directionof this line is called a point at infinity. Tts z coordinate will be zero. Theproperty of being a point at infinity depends on the particular affine planeselected.

In Fig. 1, if after the cone is rotated to produce an ellipse in the plane,we fix the plane to the rotated cone and then rotate the cone back to itsoriginal position, then we see that we have a perspective projection from thecircle on one plane to an ellipse on a second plane. And we see that such aperspective projection is essentially the same as the 3d linear transformation,namely the original rotation of the cone. Again suppose a figure in theplane is translated in the plane with the lines joining the points of the figurefollowing the translation. This is clearly a linear transformation of the linesor more precisely of the vectors in the direction of the lines. The plane figurewould be translated the same if we were to translate the origin or cone vertexwith the translation of the plane figure. So we see that the shifting of thepoint of perspective projection is equivalent to a linear transformation ofthe 3-d space. So suppose we project a figure from one plane to a second,and then from the second figure to a third figure on a third plane from asecond perspective point. Then we see from what we have said before thatthis is equivalent to a set of linear transformations of the fixed 3-d space,namely rotations, translations and possibly a uniform scaling. So the studyof these sequences of perspective projections, which were the original subjectof projective geometry may be studied by considering linear transformationsof one dimensional subspaces of a 3 dimensional vector space, or of a n

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Figure 1: Points of projective 2-space are lines in 3-space. A 3-d lineartransformation is a 2-d projective transformation. A rotation of the cone canproject the circle to an ellipse, a parabola, or a hyperbola. In the same waythe lines through any figure in the plane are a projective space representationof the figure. Any sequence of perspective projections of a figure onto a set ofplanes can be represented as a sequence of linear transformations in 3-space.The plane shown in the figure is known as the affine plane.

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dimension vector space in general. One can illustrate this equivalence easilyin the one dimensional case by drawing sequences of perspective projectionsof points on lines, and then showing how the planes can be lined up parallelby rotations and translations so that there is a single perspective projectionpoint, projecting points onto a stack of parallel planes, which by scaling canbe made to coincide.

Now consider a 2-dimensional line. We may view it as a line lying in theaffine plane. The points of this line in projective space, will be the locus oflines which pass through it and the origin. It is a plane in 3-space. Becauseany two planes with a point in common must coincide or meet in a line, itfollows that any 2-dimensional lines in projctive space meet in a point. Twoparallel lines meet at a point at infinity. Notice that 2-dimensional projectivespace contains more points than affine space, that is it contains the pointsat infinity. In two dimensional space an affine transformation is the sumof a linear transformation and a translation. If the linear transformation isorthogonal (i.e. a rotation) then the affine transformation is a rigid motionor a congruence. Now any nonsingular linear transformation maps the set ofone dimensional subspaces onto itself. So we may consider such a mapping astaking 2-dimensional projective points to projective points. Such a mappingis thus called a projective transformation. These transformations distinguisheuclidean, affine, and projective geometry . This is the thesis of Felix Klein.This is the essence of his famous Erlangen program. He defined a geometryto be the study of the invariants of a group of transformations. Thus theconcept of angle is a property of euclidean space because angle is invariantunder a rigid motion. Parallelism belongs to affine geometry because it ispreserved under an affine transformation. Being a conic section is a propertyof projective geometry, because it is preserved under a projective transfor-mation. Being an ellipse, however, is not a projective property. There is aprojective transformation that takes an ellipse to a different type of conicsection.

We shall now discuss one of the major reasons for using homogeneouscoordinates for calculations. An affine transformation is a special case of aprojective transformation. Since a projective transformation is in essence alinear transformation in a higher dimensional space, it can be representedby a matrix, and the transformation of a point may be accomplished withmatrix multiplication. An affine transformation is represented as a matrixmultiplication. There many advantages to using homogeneous coordinates.

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Quadrics are represented homogeneously as quadratic forms, and can berepresented as symmetric matrices. Perspective views are projections fromprojective 3-space to projective 2-space. A three dimensional vanishing point,that is a point at infinity, may be projected to a finite two dimensionalpoint, i.e. the vanishing point of converging railroad tracks. The intersectionpoint of two lines can be calculated without considering slopes, or withoutconsidering parallelism. That is, the intersection algorithm does not needto consider exceptional cases, such as the case of an infinite slope. Theintersection of two parallel lines, which gives a point at infinity, does notrequire a special case.

The idea of the polar of a point with respect to a quadric plays a large rolein the theory of quadrics. A quadratic form is represented by a symmetricmatrix. There is an associated bilinear form, which is a bilinear functional.This is A function from a vector product space to a scalar field. A realbilinear functional is a mapping from the cartesian product of vector spacesto the real numbers, which is linear in each variable. For a fixed point or polep, the polar is the set of points such that the bilinear form corresponding tothe quadric vanishes on any pair of points consisting of p and a point fromthe set. In two dimensions the polar is a line, in three dimensions it is aplane. If p is on the surface of the quadric, the polar is the tangent plane.This gives the normal vector. If p is at infinity, the polar is a diameter of thequadric.

The cross ratio is an important invariant in projective geometry. It isratio of distances between four collinear points. The cross ratio is used toprove several properties of projective space.

We shall require the intersection of a line and a quadric. A line is definedby any two distinct points, including the case of points at infinity. The in-tersection points are determined by the roots of a quadratic equation. Thereare several cases: (1) There may be two intersection points, (2) There maybe no real intersection points, (3) There may be a tangent point, and (4)The line may be a ruling. A ruling is a straight line that lies entirely in thesurface of the, quadric. For example, bilinear interpolation on a rectangleproduces a ruled quadric surface.

Conic arcs have been used in various applications of geometry. For exam-ple, they are used in the APT program, which generates machine tool paths.A conic arc is simply a portion of a conic section. A projective transforma-tion exists that takes four points to four points. Thus if we desire a given

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conic arc defined by four points, we construct the transformation that takesa given parabola, which is situated so that its points are a function of oneparameter, to the desired arc. Then We have a parametric representation ofthe arc. One advantage is that sines and cosines do not have to be calculatedfor the points of a circular arc.

Surface normals can be considered points at infinity. A vector field isa mapping that assigns a vector to a given point. The surface normal is avector field defined on the surface. But this field is really only a directionfield and a direction is essentially a point at infinity. Thus we may considera surface normal as a point at infinity.

Three dimensional projective space is defined in the same way as 2-dimensional space. It consists of all 1-dimensional subspaces of a 4-dimensionalvector space. The affine plane becomes a 3-dimensional hyperplane. Sincewe can not climb into 4-dimensional space,we can not visualize projectivespace from outside as we did in the 2-dimensional case.

Objects are constructed from quadric half-space primitives. There areonly two bounded primitives, namely the sphere and the ellipsoid. We createbounded objects by suitably combining the primitives. To each set corre-sponds a characteristic function which takes a value of true or false (or 1 and0) on each point of space. Its value at a point x is true if and only if x is inthe set. The mapping which takes a set to its characteristic function, unionto ”and”, intersection to ”or”, and complementation to ”not”, is a Booleanalgebra isomorphism from sets to the algebra of propositional functions. Thelatter algebra is represented in computer programming languages. Thus inour program objects are represented as logical expressions. Actually alge-braic expressions are binary trees [Knuth], so we may consider our object abinary tree with primitives at the terminal nodes and operators at the othernodes.

We determine the characteristic function of a primitive from the cor-responding quadratic form. Then the characteristic function of the objectcomes directly from the definition of the object as a logical expression. Us-ing topology, we may show that a point is on the surface of the object only ifit is on the surface of some primitive, and the normal to the surface at a pointon the surface is equal to the the normal of some primitive. If the surfacepoint lies on more than one primitive, then either the primitives are tangent,or they are not. In the former case they have the same normal, in the latercase the point lies on the intersection curve and hence is on the boundary

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between regions where the normal belongs to one, or to the other primitive.Eeither can be assigned. Usually when a point lies on several primitives, it isnot important how the normal is assigned since the intersection of the prim-itives will be of measure zero and so is negligible in any view or integration.a slight problem is the specification of the primitives.

Although all primitives are specified by symmetric 4 by 4 matrices, itmay be difficult in practice to determine the coefficients from their ordinarycharacterization. The primitives are most easily specified by entities such asradius, center, major axis, and orientation. We know the matrices of certainwell situated primitives. For example, we know the matrix of the sphereof radius one that has its center at the origin. We may transform a simpleprimitive to obtain the matrix of a more complex primitive. Transformationsemployed are rotation, scaling, and translation. For example, by scaling, theunit sphere can be made an ellipse. It then can be rotated and translated.the new matrix of the transformed quadric can be obtained by multiply-ing the original matrix on the left and the right by matrices related to thetransformations.

The motivation for constructing the modeling system was the desire fora system that would compute the volume, center of gravity and the inertiatensor. Since the characteristic function is available, a Monte Carlo integra-tion method is suggested. This technique consists in assigning a probabilitymeasure to a bounded space containing the object, and then obtaining thevalue of the integral as the expectation of some random variable. For ex-ample, the volume integral is defined as the expectation of the characteristicfunction. The expectation is estimated from a random sample of the ran-dom variable. In the case of the volume, the variance of the characteristicfunction is quite high. Thus a large number of samples is required so thatthe sample variance, which is a measure of the error of the estimated inte-gral, is small. Stratified sampling will often decrease the sampling variance.In stratified sampling one breaks up the domain into smaller sub domainsso that the variance is smaller in each subdomain. For example, if a smallblock is chosen completely inside the object then the variance of the charac-teristic function is zero. Likewise the variance of the characteristic functioncompletely outside the function is zero. Only those blocks which containboundary points contribute to the variance. Stratified sampling should de-crease the overall variance and so fewer sample points should be required forthe error to be within a given tolerance. When only one point is sampled

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from each subdomain we have essentially a Riemann sum for the integral.in general, the amount of work required to evaluate an integral numeri-

cally is proportional to the number of function evaluations required in onedimension raised to a power equal to the dimension of the space. This istrue at least when the multiple integral is computed by iterated integration.Hence if 100 function evaluations are required in one dimension, 1,000,000evaluations are required in three dimensions, so it pays to reduce the dimen-sion.

By applying the divergence theorem the integral over a volume can usu-ally be reduced to the integral over a surface; so we need a technique forcalculating surface integrals. To integrate on a surface we must have somemodel of the surface. In this modeling system we do not have such an explicitmodel. We have a multiple surface patch defined by projection. When theobject is projected to two dimensions, to every point in the two dimensionalregion corresponds a set of points where the projection line pierces the object.This defines a multiple valued function.

A manifold is the mathematical generalization of a surface. It is a spacetogether with a set of coordinate systems or surface patches. For a twodimensional manifold, a surface patch is a 1-1 mapping from the manifold toan open set in 2-dimensional euclidian space. A complete set of patches thatcovers the manifold is called an atlas. The projections from several projectionpoints of an object will be a set of patches that completely cover the surfaceof the object. However, when we include all pierce points, these mappings arenot 1-1, so are not legitimate coordinate systems. Still we can do integrationon these multilayer patches, because integration is loosely an adding up ofall infinitesimal surface elements, and we may add the elements in any order.We can add the elements corresponding to pierce points of a given projectionline all at once. There is a difficulty, the set of patches that define a manifoldare not disjoint, so that when we add the results of integrating each patch,we may well be adding a given surface element many times. We need apartition of unity. A partition of unity is a set of weighting functions appliedto each patch so that the sum of the weighted patch integrals is the wholesurface integral [Spivak]. When our surface patches derive from projectionsin the x, y, and z directions a natural partition of unity is the set of squaresof the components of the surface normal vectors. The individual layers ofeach view must be considered as separate patches. So we have reduced ourtask to the calculation of a two dimensional integration over a rectangular

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domain. the integrand will in general be discontinuous and we must choose asuitable integration algorithm. Gaussian quadrature is not suitable becausethe integrand will be discontinuous. The generalized trapezoid method willwork, but may require more calculation and give less accuracy than necessary.

Because the moment of inertia about any axis can be obtained from theinertia tensor, it is economical to compute the full inertia tensor rather thanjust the moments of inertia. There are two different tensors that are calledthe inertia tensor. One definition is given in mathematics books, and anotherin some physics books. In order to clarify what is being calculated, I haveincluded material concerning angular momentum, angular velocity, momentof inertia, and their relation to the inertia tensor. In the next section I willbegin a more detailed treatment of the material.

3 Projective Space and the Conic Sections

Projective geometry is related to the theory of perspective. Perspective wasdeveloped by the Italian painters during the renaissance. To explain per-spective, consider the eye as the origin in a 3-dimensional coordinate system.Let a ray go out from the eye. Suppose this ray impinges on some object ata point. In some sense, the ray represents the point. Let a picture plane beinterposed between the eye and the object. Then the point where the rayintersects the picture plane is the projection of the object point to its pictureimage. Through this process, three dimensional space is projected to twodimensional space. The concern of the artist is to accurately represent theobject as it appears to the eye. The artist is concerned with the appearanceof parallel lines and vanishing points. Vanishing points are the images ofpoints and lines that are infinitely far away. Such points are called pointsat infinity. Spatial points at infinity will be discussed in the section on per-spective. Here we confine ourselves to the two dimensional case. the artistAlbrect Durer, who dabbled a bit in geometry (see Panofsky) ,constructed adevice for recording the perspective image of an object in the picture planeby recording the position on the plane of a chord stretched from a fixed pointto a point on the object. This device appears in one of Durer’s works andmay be familiar to the reader. The idea of projective geometry is to consider,the chord, or more accurately the mathematical ray, as representing a twodimensional point.

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projective space is formally defined as the set of 1-dimensional subspacesof a vector space. it can be illustrated by considering those plane curveswhich are called conic sections. Figure 1 shows a cone with its axis in the zdirection, and with a 90 degree vertex angle.

This cone meets the plane z = 1 in a circle. A point on the surface of thecone satisfies the equation

x2 + y2 − z2 = 0

The circle in the plane has the equation

x2 + y2 − 1 = 0

The first equation can be written as

[

x y z]

1 0 00 1 00 0 −1

xyz

= 0

P T AP = 0

where

P =

xyx

and

A =

1 0 00 1 00 0 −1

The function Q(P ) = P tAP is a quadratic form. The cone is the set

S(Q) = {P : Q(P ) = 0}.

Let L be a nonsingular linear transformation with matrix B.

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Then L maps the set S(Q) to

L(S(Q)) = {P : Q(L−1P ) = 0}

= {P : (B−1P )TA(B−1P ) = 0}= {P : P T (B−1)T A(B−1P ) = 0}= {P : (B−1P )TA(B−1P ) = 0}

= {P : L(Q)P = 0}.where L(Q)(P ) = Q(L−1(p)). L(Q) is a quadratic form with symmetricmatrix

(B−1)T AB−1.

Let L be the transformation of a rotation about the x-axes by an angle a.Let cos(a) = c and sin(a) = s. Then the matrix of L is

B =

1 0 00 c s0 −s c

and B−1 = BT =

1 0 00 c −s0 s c

Then L(Q) has matrix

1 0 00 c0 −s c

1 0 00 1 00 0 −1

1 0 00 c −s0 s c

=

1 0 00 c0 −s c

1 0 00 c −s0 −s −c

=

1 0 00 c2 − s2 −2cs0 −2cs s2 − c2

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When the angle a is 45 degrees, the matrix of L(Q) is

1 0 00 0 −10 −1 0

The equation of the rotated cone is

[

x y z]

1 0 00 0 −10 −1 0

xyz

=

[

x y z]

x−z−y

.

= x2 − 2yz = 0.

The equation of its intersection with the plane z = 1 is

x2 − 2y = 0.

or

y =x2

2.

This is the equation of a parabola.When the angle a is 90 degrees, the equation of the cone becomes

[

x y z]

1 0 00 −1 00 0 1

xyz

= x2 − y2 + z2 = 0.

The equation of its intersection with the plane z = 1 is

y2 − x2 = 1.

This is the equation of a hyperbola. the plane that cuts the cone is calledthe affine plane. for each point in the plane there is a corresponding linewhich passes through the point and through the origin. The set of linesthat pass through the origin is called projective space. Each line is con-sidered a point of the space. Those points in projective space ( i.e. lines in

18

3-dimensional Euclidean space) that are parallel to the affine plane are calledpoints at infinity, because they meet the plane only at infinity. There aremore projective points than affine points. The coordinates of any vector inthe direction of a line which passes through the origin are the homogeneouscoordinates of the projective point represented by the line. Because the coor-dinates of any scalar multiple of the vector are also homogeneous coordinatesof the projective point, homogeneous coordinates are determined only up toa scalar multiple. So (1, 2, 3) and (2, 4, 6) are homogeneous coordinates ofthe same point. A point at infinity has coordinates of the form (x, y, 0). If apoint with coordinates (x, y, z) is not an ideal point ( e.g. a point at infinity),then its affine coordinates are (x/z, y/z).

One of the properties of projective space is that every pair of distinctlines meet in exactly one point. By a line we mean the locus of points thatsatisfy an homogeneous equation of the form

ax + by + cz = 0.

The corresponding affine equation is

ax + by + c = 0.

Then lines which are parallel in the affine plane meet at a point at infinity.For example, the parallel lines with affine equations

ax + by + c = 0.

and

ax + by + d = 0.

meet at the point, which has coordinates (−b, a, 0). Affine rotations andtranslations are represented as linear transformations in projective space. Arotation has matrix

R =

cos(a) sin(a) 0− sin(a) cos(a) 0

0 0 1

19

Thus

cos(a) sin(a) 0 x− sin(a) cos(a) 0 y

0 0 1 1

.

=[

cos(a)x + sin(a)y = − sin(a)x + cos(a)y1]

.

The following calculation represents an affine translation

Tp =

1 0 a x0 1 b y0 0 1 1

=

x + ay + b

1

.

The product of the two matrices above then both rotates and translates.This shows how affine transformations can be calculated using homogeneouscoordinates.

A projective point is a set of all scalar multiples of a vector. It is a onedimensional subspace of a vector space. In general, projective n-space isthe set of 1-dimensional subspaces contained in a vector space of dimensionn + 1. The nonsingular linear transformations on the vector space mapthe set of 1-dimensional subspaces onto themselves. These linear mappingsare called projective transformations when they are applied to projectivespace. Considering the previous example, it is easy to see that an affinetransformation is a special case of a projective transformation.

We now introduce quadratic forms. Let A be a symmetric matrix and letp be a coordinate vector.

p =

xyz

.

Then P TAP is a quadratic form. For example, if

A =

1 2 32 4 53 5 6

.

ThenP TAP = x2 + 4y2 + 6z2 + 4xy + 6xz + 10yz.

20

All quadratic forms can be obtained in this way. Corresponding to eachquadratic form is an associated bilinear form or tensor of rank two. It is afunction of two vector variables and is linear in each. It is given as

B(P, Q) = P TAQ.

A conic section S is the set of zeroes of a quadratic form in 2-dimensionalspace. Thus

S = {vP : P T AP = 0}.Given a point Q, the polar of Q with respect to a quadric represented by amatrix A is the following set.

{P : B(P, Q) = 0}.

Since the equation is linear, the polar of a point is a line in the two dimen-sional case, and a plane in the three dimensional case. Directly from thedefinition it is seen that P is on the polar of Q if and only if Q is on thepolar of P . Properties of the polar can be obtained by using the concept ofthe cross ratio of four collinear points.

4 The Cross Ratio

Given four collinear points in affine space, the cross ratio is defined in termsof the directed distances between the points. Affine space is embedded inprojective space and the cross ratio can be extended to apply to the pointsof this space. It can be defined without the non-projective concept of dis-tance. The importance of cross ratio is that it is invariant under projectivetransformations. Before we define the cross ratio a few remarks on notationare in order. a point may refer to several distinct concepts. a point maybe an n + 1 dimensional vector, or it may be the span of such a vector,that is the 1-dimensional subspace generated by the vector, or it may be thecorresponding affine point which is the intersection of the subspace with theaffine hyperplane. To complicate things further, we must distinguish betweena vector and its corresponding coordinate vector. Its coordinates depend onthe chosen basis of the vector space. To distinguish all of these things wouldresult in too many symbols. So we will use the same symbols for all of the

21

above ideas and hope that context will give the desired meaning. For exam-ple, if we add two points, they are vectors, because addition is not defined forprojective points. Division of vectors has no meaning in general, but if onevector is a multiple of another, then their quotient has an obvious meaning.We shall use this meaning below.

Let A, B, C, and D be four collinear points. In affine space we define thecross ratio as

(AB, CD) =d(C, A)/d(C, B)

(d(D, A)/d(D, B),

where d(X, Y ) is a directed distance from X to Y. Now let A, B, C, and D bevectors representing the points. Then numbers a and b can be found so thataC and bD lie on the straight line connecting A and B. Then there exists at such that

aC = (1 − t)A + tB.

andaC − B = (1 − t)A − (1 − t)B.

Thend(C, A)/d(C, B) = (A − aC)/(B − aC) = −t/(1 − t).

It follows that ifC = c1A + c2B,

thend(C, A)/d(C, B) = −c2/c1

Similarly ifD = d1A + d1B,

thend(D, A)/d(D, B) = −d2/d1.

So the cross ratio is

(AB, CD) =d(C, A)/d(C, B)

(d(D, A)/d(D, B)= (c2/c1)/(d2/d1) = (c2d1)/(c1d2).

This ratio is independent of which representative vectors are chosen for thepoints. For example, if A is replaced by a new vector that is a multiple ofA, then c1 and d1 are multiplied by the same scalar. So the cross ratio is not

22

changed. Thus the cross ratio is defined independently of a distance function.That is it is defined independently of a metric.

Therefore the distance independent projective definition of the cross ratioof points (AB, CD), of the projective points A, B, C, D, is defined in thefollowing way. If A, B, C, D are any vector representations of the projectivepoints, where

C = c1A + c2B,

andD = d1A + d1B,

then the cross ratio (AB, CD) is defined as

(AB, CD) = (c2/c1)/(d2/d1) = (c2d1)/(c1d2).

The preservation of the cross ratio by a projective transformation followsbecause a projective transformation is a linear transformation of the vectorspace of dimension n + 1.Proposition. A projective transformation preserves the cross ratio.Proof. Let A, B, C, and D be representative vectors of four distinct collinearpoints. Suppose

C = c1A + c2B

andD = d1A + d2B.

Then if L is the linear transformation corresponding to the given projectivetransformation we have

L(C) = c1L(A) + c2L(B)

andL(D) = d1L(A) + d2L(B).

So the points L(A), L(B), L(C), and L(D) have the same cross ratio. A setof collinear points is called a range of points. A range of four points is calleda harmonic range when its cross ratio equals -1. The name harmonic rangecan be explained in the following way. Suppose the distance from A to C is1/3, the distance from B to C is -1/6, and the distance from B to D is 1/2.Then the cross ratio is -1, so the points constitute a harmonic range. the

23

distance from A to D is 1,the distance from A to B is 1/2, and the distancefrom A to C is 1/3. Vibrating strings of these lengths produce the first,second, and third harmonic tones.

A point Q is called the harmonic conjugate of P with respect to A andB when (AB, PQ) = −1.Proposition If a line through P meets a quadric at A and B, then theharmonic conjugate Q lies on the polar of P .Proof Suppose P = pA + pB and Q = qA + qB, then pq = −pq, so

qP = qpA + qpB

andpQ = pqA + pqB

Subtracting, we findA = (1/2)(P/p + Q/q).

SimilarlyB = (1/2)(P/p + Q/q).

Let f be the bilinear form defied by the quadric. Since A is on the quadric,we have

0 = f(A, A) = (q/p)f(P, P ) + 2(q/p)f(P, Q) + f(Q, Q).

Similarly0 = (q/p)f(P, P ) + 2(q/p)f(P, Q) + f(Q, Q).

Subtracting and noting that q/p = −q/p, we find that f(P, Q) must be zero.This means that Q is on the polar of PCorollary. If a line through P is tangent to a quadric at Q, then Q is onthe polar of P .Informal Proof. Rotate the line through P slightly, if necessary, so thatit intersects the quadric in two points that are nearly equal. The harmonicconjugate of P is between these two points, and from our discussion above,it is on the polar of P. As the line approaches the tangent line, this pointapproaches Q.Construction. To construct a tangent to a conic from a point P , find theintersection of the conic with the polar of P.

24

R

Q

S

P

P’

Figure 2: Let lines PQ and PR be tangent to the ellipse. Tangent points Qand R are on the polar of P . Thus the line defined by Q and R is the polarof P . Intersection point S is on both the polar of P and the polar of P ′.Thus the polar of S is the line through P and P ′.

25

Corollary. The locus of the harmonic conjugates of p with respect to theintersection points of lines through p with the conic is the polar of P .Example.

In Fig. 2 lines PQ and PR are tangent to the ellipse. We conclude thatQ and R are on the polar of P Thus the line defined by Q and R is the polarof P . It also follows that S is on both the polar of P and the polar of P ′.Thus the polar of S is the line through P and P ′.

A diameter of a quadric is the polar of a point at infinity. thus ”diameter”is an affine idea because ”point at infinity” only makes sense when some affinehyperplane has been identified. We may represent a line through P and Q byP + tQ where the parameter t varies from minus to plus infinity. t equal toinfinity corresponds to the point Q. We may think of this as the line throughP in the direction Q. When a line meets a quadric in two finite intersectionpoints R and S, the line segment from R to S is called the chord determinedby the line. A diameter in the direction Q is the polar of Q where Q is apoint at infinity.

Consider the intersection of the line P + tQ with a quadric. There areseveral cases. Let the bilinear form be f . Then

0 = f(P + tQ, P + tQ)

= f(P, P ) + 2tf(P, Q) + t2f(Q, Q)

There are four cases:Case 1.

P (Q, Q) = 0, f(P, Q) 6= 0.

Q is one intersection point. The parameter of the other is

t = −f(P, P )/(2f(P, Q)).

Case 2.

f(Q, Q) = 0, f(P, Q) = 0, f(P, P ) 6= 0.

The two intersection points coincide. They equal Q.Case 3.

f(Q, Q) = 0, f(P, Q) = 0, f(P, P ) = 0

The line is a ruling. All points lie on the conic.

26

Case 4.

p(Q, Q) 6= 0.

There are two roots, which may be real or complex, and may coincide.Proposition. A diameter in the direction Q bisects all chords in the direc-tion Q.Proof. Since Q is a point at infinity, and the line determines a chord withtwo finite intersection points R and S, from above we must have f(Q, Q)??o0.For otherwise, Q would be an intersection point. Using the quadratic formulafor the two roots, substituting in P + tQ, adding and dividing by two, wefind

(R + S)/2 = P − (f(P, Q)/f(Q, Q))Q.

Then

f((R + S)/2), Q) = f(P, Q) − f(P, Q)f(Q, Q)/f(Q, Q) = 0

So the midpoint of the chord lies on the polar of Q, which is the diameterdetermined by Q.Example. Consider the parabola x − 2yz = 0. Let

p =

xyz

.

Then P TAP is a quadratic form. For example, if

A =

1 2 32 4 53 5 6

.

P =

001

.

and

Q =

010

.

27

P

Q

Figure 3: The parabola x − 2yz = 0 and a point at infinity Q. The linethrough P and Q meets the parabola at P and Q. The polar of a point atinfinity is a diameter. The polar of Q is the line at infinity.

28

The matrix of this conic is

A =

1 0 00 0 −10 −1 0

.

Q does not determine a chord because the line does not meet the conic intwo finite intersection points. In fact Q, which is a point at infinity, lies onthe conic. The polar of Q has line coordinates given by

QA = (0, 0,−1).

The equation of the polar of Q, which is a diameter of the parabola is

−z = 0

This is the line at infinity.Example. Consider the same parabola as above, but let

Q =

110

.

See Fig. 4. Then the polar of Q has equation x − z = 0. Thus the affineequation of the diameter corresponding to Q is x = 1.

5 The Tangent Line

Let P and Q be distinct points. Let P be on the conic.Proposition. P + tQ is tangent to the conic at P if and only if f(P, Q) = 0.Proof. First suppose P + tQ is tangent at P. Then the intersection pointscoincide and they equal P, because P is on the quadric f(P, P ) = 0. But theintersection equation is then

2tf(P, Q) + tf(Q, Q) = 0,

and its roots must be zero. So

f(P, Q) = 0.

29

P

Q

Figure 4: The parabola x−2yz = 0 and a point at infinity Q = (1, 1, 0). Theline through P and Q meets the parabola at two finite points. The polar ofQ is the line x = 1, which is a diameter of the parabola. The point at infinity in the diagram is represented as an arrow in the direction of the infinitypoint Q.

30

Now suppose f(P, Q) = 0. It is given that f(P, P ) = 0. The intersectionequation is

tf(Q, Q) = 0.

If f(Q, Q) 6= 0, then the roots of the equation are zero. So P is a doubleintersection point. If f(Q, Q) = 0, then the line is a ruling. Hence in bothcases the line is the tangent line.Corollary. The tangent line P + tQ at P is the polar of P .Proof. Since the line passes through P , we have f(P, P ) = 0 so P is on thepolar of P . From above f(P, Q) = 0 so Q is on the polar of P. Therefore theline through P and Q is the polar of P.

We have taken the tangent line as the line that intersects the conic attwo coinciding points. With this definition the tangent line does not havepoints on both sides of the conic curve. For consider the set of points wheref(P, P ) < 0, and the set where f(P, P ) > 0. These sets are bounded bythe conic curve. We may consider one of them the inside, and the other, theoutside of the curve. Suppose P and Q are on opposite sides of the conic.then f(P, P )f(Q, Q) < 0. Then the intersection equation of the line throughP and Q has positive discriminant. So the line meets the conic in two realpoints that do not coincide. Hence the line through P and Q is not a tangentline.

If the point Q is on the polar of P , then

f(P, Q) = P TAQ = 0.

Suppose

P T A = (a, b, c),

Q =

xyz

.

Then the first vector is called the line coordinate vector. The equation ofthe polar is

ax + by + cz = 0.

31

The 2-dimensional vector[

ab

]

,

is normal to the line. For if

x1

y1

1

,

and

x2

y2

1

,

1 are on the line, then we have by subtraction

(x − x)a + (y − y)b = 0.

Example. Consider the hyperbola

x2 − y2 = 1.

Let

Q =

110

,

and

P =

001

.

Thenf(q, q) = f(p, q) = 0.

The equation of the tangent line at the ideal point Q is x−y = 0, because

AQ = (1,−1, 0).

32

The tangent line is an asymptote of the hyperbola.The set of centers is the intersection of all diameters. So P is a center

if f(P, Q) = 0 for all ideal points Q. Thus the set of centers is the set of allpoints P such that

AP =

001

.

If A is nonsingular, than the conic has only one center.Example. Consider the conic

(x − 1)2 + y2 = 1.

Its matrix is

A =

1 0 −10 1 0−1 0 0

.

Hence the center is x = 1, y = 0, z = 1.

6 Computing A Canonical Representation

Suppose the conic is given in matrix form as

E = {p : p∗Ap = 0},

where A is a symmetric matrix

A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

.

The computations that will be described here, are realized in programpltconic.ftn.

We shall put E into canonical form by mapping the locus with a rotationthrough an angle θ, followed by a translation by (t1, t2). The combinedtransformation has matrix

33

T =

c −s t1s c t20 0 1

The inverse of T is

R =

c s e−s c f0 0 1

where

e = −(t1c + t2s)

f = t1s − t2c

The mapping of the conic locus is

T (E) = {Tp : p∗Ap = 0}

= {q : (Rq)∗ARq = 0}

= {q : q∗R∗ARq = 0}

= {q : q∗Bq = 0},

where B is the symmetric matrix

R∗AR.

Let the determinant of A be d3. Let the determinant of the upper 2 by 2submatrix be d2. The determinants of R and R∗ are each equal to 1, so thedeterminant of B is also d3. Also for the same reason the transformationpreserves d2. We will show that if d3 = 0, then the conic is degenerate, andthe conic locus is a line, a pair of parallel lines, a pair of intersecting lines,a point, or is empty. If d3 is not zero, then if d2 < 0 it is a hyperbola, if

34

d2 = 0 it is a parabola, and if d2 > 0 and d3 < 0, it is an ellipse. if d2 > 0and d3 > 0 it is the empty set.

The coefficients of B are

b11 = (ca11 − sa12)c − (ca12 − sa22)s

b12 = (ca11 − sa12)s + (ca12 − sa22)c

b13 = (ca11 − sa12)e + (ca12 − sa22)f + ca13 − sa23

b22 = (sa11 + ca12)s + (sa12 + ca22)c

b23 = (sa11 + ca12)e + (sa12 + ca22)f + sa13 + ca23

b33 = (ea11 + fa12 + a13)e + (ea12 + fa22 + a23)f + ea13 + fa23 + a33

= a11e2 + 2a12ef + a22f

2 + 2a13e + 2a23f + a33.

Notice that

b33 = (e, f, 1)A(e, f, 1)∗.

We shall choose θ so that b12 is zero. We have

b12 = (c2 − s2)a12 + sc(a11 − a22) = 0.

That is0 = a12 cos(2θ) + (1/2)(a11 − a22) sin(2θ).

So the rotation angle is determined by

tan(2θ) =2a12

a22 − a11.

Thus2θ = atan2(2a12, a22 − a11).

Let us write

35

b13 = c11e + c12f − g1

b23 = c21e + c22f − g2,

where

c11 = (ca11 − sa12)

c12 = (ca12 − sa22)

c21 = (sa11 + ca12)

c22 = (sa12 + ca22)

g1 = −(ca13 − sa23)

g2 = −(sa13 + ca23)

For purposes of classification, the transformation can be done in two steps,a pure rotation where,

e = 0, f = 0,

followed by a pure translation, where the rotation angle is zero.If the rotation angle is zero, then

c11 = a11

c12 = a12

c21 = a12

c22 = a22)

g1 = −a13

g2 = −a23.

After such a pure rotation, it is easy to see the conic type from theproperties of the coefficients, where the xy cross term is absent.

But here let us return to the single step transformation. We can makeb13 = 0 and b23 = 0, if we can find e and f so that

c11e + c12f = g1

36

c21e + c22f = g2.

We have a linear equation to solve for e and f .The determinant of this system is

d2 = a11a22 − a212.

Because the upper 2 by 2 subdeterminant of R and RT is 1, d2 also equals

b11b22 − b212 = b11b22.

If d2 = 0, then the quadric is neither an ellipse nor a hyperbola.If d2 is not zero, then using the values of e and f that make b13 and b23

zero, wefind that the matrix of the transformed canonical form is

B =

b11 0 00 b22 00 0 b33

.

If d2 = 0, then we can not necessarily choose the translation (e, f) so thatb13 = 0 and b23 = 0. In these cases, either b11 or b22 is zero.

Suppose one of b11 and b22 is zero.In this case we have three equations

b13 = c11e + c12f − g1

b23 = c21e + c22f − g2

b33 = (e, f, 1)A(e, f, 1)∗.

A related set of equations obtained by setting the b coefficients to zero is

0 = c11e + c12f − g1

0 = c21e + c22f − g2

0 = (e, f, 1)A(e, f, 1)∗.

We claim that if b11 is not zero, then the 1st and 3rd equation of theset have a solution. If b22 is not zero, then the 2nd and 3rd equations of

37

the set have a solution. This follows by examining the component transfor-mations, by first applying the rotation, then the translation. Finding thetranslation is equivalent to completing squares. For example, it is equivalentto transforming the terms like

k1x2 + k2x

into terms likek1(x + m1)

2 + m2.

Solving the first or second equation, simultaneously with the third equa-tion is equivalent to finding a finite intersection point

(e, f, 1)

of a line and the conic.This is the intersection of the line with coordinates (c11, c12,−g1), with

the conic A.Suppose b11 is not zero, and b22 = 0.Solving the first and third equation for e and f gives us the canonical

matrix

B =

b11 0 00 0 b23

0 b23 0

.

Suppose b22 is not zero, and b11 = 0. Solving the second and thirdequation for e and f gives us the canonical matrix

B =

0 0 b13

0 b22 0b13 0 0

.

So we have shown that after our transformation, the conics have fourstandard forms. They are:

Form 1:

B =

b11 0 00 b22 00 0 b33

.

38

Form 2:

B =

b11 0 00 0 b23

0 b23 0

.

Form 3:

B =

0 0 b13

0 b22 0b13 0 0

.

Form 4:

B =

0 0 b13

0 0 b23

b13 b23 b33

.

Form 4 is a degenerate case that is just a linear equation.Let us now consider the form 1 case.

Ellipse.

All coefficients b11, b22 and b33, are not zero. The coefficients b11 and b22

have the same sign, which differs from the sign of b22. Then the canonicalequation is

x2

a2+

y2

b2= 1,

where the radii of the ellipse are

a =√

−b33/b11

andb =

√

−b33/b22.

In this case d2 > 0 and d3 < 0.Hyperbola.

All coefficients b11, b22 and b33, are not zero. The coefficients b11 and b22

have different signs. There are two subcases: b11b33 < 0 and b11b33 > 0. Ifb11b33 < 0, then the canonical equation is

39

x2

a2− y2

b2= 1,

wherea =

√

−b33/b11

andb =

√

b33/b22.

If we define the axis of the hyperbola to be the line of symmetry that meetsthe hyperbola at two points, then the axis of the original conic is at angle−θ, where θ is the rotational angle of the canonical transformation.

If b11b33 > 0, then the canonical equation is

y2

a2− x2

b2= 1,

wherea =

√

b33/b11

andb =

√

−b33/b22.

Then the axis of the original conic is at angle −(θ + π/2), where θ is therotational angle of the canonical transformation.

In this case d2 < 0 and d3 < 0.Intersecting lines.

If b11 and b22 have different signs, and b33 = 0, then the conic has equation

|b11|x2 = |b22|y2,

which is equivalent to two equations of a line through the origin

y =α

βx,

andy = −α

βx,

where

40

α =√

|b11|, β =√

|b22|

In this case d2 < 0, and d3 = 0.Parallel lines or no real solution.

If b11b22 = 0 then the equation is

b11x2 = −b33

orb22y

2 = −b33.

This is a pair of parallel lines, a single line , or there is no solution,depending upon signs. In this case d2 = 0 and d3 = 0.Point.

If b11 and b22 have the same signs, and b33 = 0, then the conic has equation

|b11|x2 = −|b22|y2,

whose only solution is

x = 0, y = 0.

In this case d2 > 0, and d3 = 0.No real points.

If b11, b22, and b33 all have the same signs, then the conic has equation

|b11|x2 + |b22|y2 = −|b33|,

which has no real solution.In this case d2 > 0 and d3 > 0 or d3 < 0.

Let us now consider the form 2 case.

B =

b11 0 00 0 b23

0 b23 0

.

Parabola.

41

If b23 is not zero, then form 2 represents a parabola. The equation is

b23y = −b11x2,

4Fy = x2.

The focal distance is

F = − b23

4b11.

In this case d2 = 0, and d3 < 0.Line.

If b23 is zero, form 2 represents the line

x = 0.

In this case d2 = 0, and d3 = 0.Let us now consider the form 3 case.

B =

0 0 b13

0 b22 0b13 0 b33

.

If b13 is not zero, form 3 represents a parabola.The equation is

b13x = −b22y2 − b33

4Fx = y2 + c

which if b13 is not zero is a parabola.

F = − b13

4b22.

In this case d2 = 0, and d3 < 0.Line.

If b13 is zero, form 3 represents the line

y = 0.

In this case d2 = 0, and d3 = 0.Let us now consider the form 4 case.

42

Line.

We have the straight line

2b13x + 2b23y + b33 = 0.

If an ellipse has a center C = (cx, cy), then the original transformation Tmaps it to the origin, which is given in homogeneous coordinates as

(0, 0, 1)∗

Hence R, which is the inverse of T , maps the origin to the original ellipsecenter.

Therefore

cx

cy

1

=

c s e−s c f0 0 1

001

=

ef1

So (e, f) is the center of the original conic, if it is an ellipse, or a hyperbola.If it is a parabola, then (e, f) is the vertex. The center of a parabola is atinfinity.

Alternately, for the case of the ellipse or hyperbola, the first two rows ofmatrix A are homogeneous coordinates of diameter lines (being the polars ofpoints at infinity), so their intersection is the center.

If the matrix A is nonsingular, then the unique center (cx, cy) is thesolution of

A

cx

cy

1

=

001

This follows because the polar of every point at infinity

xy0

is a diameter given by[x, y, 0]A.

So every diameter meets the center if and only if

43

[x, y, 0]A

cx

cy

1

= 0

if and only if

A

cx

cy

1

=

001

7 Conic Through a Set of Points

Five points determine a conic. We can find the equation of a conic throughfour points as a product of lines through the points. Thus if we have pointsp1, p2, p3, p4 let the line through pi and pj be `ij, then

λ`12`34 + (1 − λ)`13`24 = 0

is the equation of a conic through the four points. The parameter λ canbe chosen so that the conic passes through a fifth point. The discriminantcan be examined to determine the type of conic. By letting the points cometogether we can specify tangent constraints.

8 Parametric Conic Arcs, Matrices of Pro-

jective Transformations.

A conic arc is a portion of a conic curve. A conic arc can be expressed inparametric form. This can be done so that each coordinate is a rationalfunction of a single variable. To construct such a function, we map a specialparabola to the required arc using a projective transformation.

We shall describe the mapping in a later section. First we need to provethe following proposition.Proposition. If L and M represent the same projective transformation,then L is a scalar multiple of M .

44

(a) (b)

(c) (d)

Figure 5: Results of plotting conics with the program pltconic.ftn.(a)Ellipse with equation 10x2+10xy+20y2+3x+−4y−10 = 0, (b)Hyperbolawith equation 5x2 + 10xy − 7y2 − 3x + 2y − 1 = 0, (c)Parabola with equa-tion x2 + 6xy + 9y2 + 3x − 5y − 1 = 0, (d)and a circle with equation2x2 + 2y2 + 1

2x − 1

5y − 1 = 0.

45

Figure 6: A conic passing through four points defined by the products of linespassing through the points (1, 1), (2, 2), (2, 1), (3, 3). The general equation ofsuch a conic is λ(y − x)(y − x + 1) + (1 − λ)(y − 2)(y − 1) = 0 where λ isa parameter. Here λ = 1/2. By letting points come together we can alsodefine conics by tangent lines.

46

(a)

(b)

A’ B’

D’

C’

A

C

D

B

Figure 7: (a)Any conic arc can be obtained by mapping this parabola, whichis defined by tangent lines AC and BC, and point D,where A = (0, 0), B =(1, 1), C = (0, 1/2), D = (1/4, 1/2). (b)An example of mapping the parabolato a conic representation of a circle. The intersection of the tangents is atC, which is a point at infinity.

47

Proof. We shall write [v] for the subspace spanned by the vector V . Let Vand U be linearly independent. Let P be the given projective transformation.Then

P ([U ]) = [L(U)] + [M(U)].

So there is a number a so that L(U) = aM(U). Similarly, there is anumber b, so that L(V ) = bM(V ). Then

P ([U + V ]) = [L(U) + L(V )]

= [M(aU) + M(bV )]

= P ([aU + bV ]).

But a projective transformation is one to one. Therefore there is a c, so that

U + V = c(aU + bV ).

Then by the linear independence of U and V , we have

ca = 1 = cb.

It follows that a = b and L = aM.

9 A Projective Transformation That Takes

Four Points To Four Points

Now we shall show that there is a projective transformation mapping fourpoints in the plane to four points in the plane.Proposition. Let A, B, C, and D be distinct points in the plane, no threeof which are collinear. Let A′, B′, C ′, and D′, also be distinct points in theplane, no three of which are collinear. Then there is a unique projectivetransformation taking the first four points to the second four points.Proof. Consider the points as coordinate vectors. Then there are numbersa, b, c so that

D = aA + bB + cC.

This is true because A, B, and C are a basis of the vector space. Similarlythere are numbers a’, b’, c’ so that

48

D′ = a′A + b′B + c′C.

Define a”, b”, c” by

a” = a′/a, b” = b′/b,

andc” = c′/c.

None of a, b, or c are zero, because no three of the given points are collinear.Define a matrix M as a product of three by three matrices, formed from thepoint column vectors. Let

M =[

a′′A′ b′′B′ c′′C ′] [

A B C]−1

.

Then the projective transformation defined by M maps the first threeunprimed points to the first three primed points. Also D is mapped to D’,as we show here.

MD = aMA + bMB + cMC

= aa”A′ + bb”B′ + cc”C ′

= a′A′ + b′B′ + c′C ′ = D′.

It remains to show that the projective transformation defined by M isunique. We have

MA = a”A′, MB = b”B′, MC = c”C ′,

andMD = D′.

Suppose there is a matrix N and numbers aNbN , cN , and dN , so that

NA = aNA′, NB = bNB′, NC = cNC ′,

andND = dND′.

49

Then ND is given by

ND = N(aA + bB + cC)

= aaNA′ + bbNB′ + ccNC ′.

And ND is also given byND = dND′.

Equating these two expressions for ND, and dividing by dN , we get

D′ = (aaN/dN)A′ + (bbN/dN)B′ + (ccN/dN)C ′.

ButD′ = a′A′ + b′B′ + c′C ′,

and because A′, B′, C ′ are linearly independent, we must have

aaN/dN = a′,

bbN/dN = b′,

andccN/dN = c′.

Thena = dN(a′/aN) = dNa”,

b = dNb”,

andc = dNc”.

So matrix N is a scalar multiple of matrix M ,

N = dNM.

Therefore N and M represent the same projective transformation.Here is a MatLab script to create this projective transformation:

50

% prjtrans.m, A projective transformation taking quadrilateral a,b,c,d

%to quadrilateral ap,bp,cp,dp

%Reference: quadric.pdf "Conics, Quadrics, and Projective Space"

%

a=[20;20;1]

b=[60;30;1]

c=[70;60;1]

d=[25;50;1]

m1=[a’;b’;c’]

m1=m1’

m1i=inv(m1)

e=m1i*d

ap=[1;1;1]

bp=[100;1;1]

cp=[100;100;1]

dp=[1;100;1]

m2=[ap’;bp’;cp’]

m2=m2’

m2i=inv(m2)

f=m2i*dp

g1=f(1)/e(1)

g2=f(2)/e(2)

g3=f(3)/e(3)

ap2=g1*ap

bp2=g2*bp

cp2=g3*cp

m3=[ap2’;bp2’;cp2’]

m3=m3’

p=m3*m1i

pa=p*a

pb=p*b

pc=p*c

pd=p*d

pa=(1/pa(3))*pa

pb=(1/pb(3))*pb

pc=(1/pc(3))*pc

pd=(1/pd(3))*pd

10 An Affine Transformation That Approxi-

mately Takes Four Points To Four Points

Let the affine transformation acting on a point with coordinates x and y begiven by

x′ = m11x + m12y + m13,

y′ = m21x + m22y + m23.

51

Four unprimed points are to be mapped to four primed points. Thus we getthe following eight equations in the six matrix coefficients mij ,

m11xk + m12yk + m13 = x′k

m21xk + m22yk + m23 = y′k

for k = 1, .., 4. Let C be a the six dimensional coefficient vector,

C =

m11

m12

m13

m21

m22

m23

The linear system becomesAC = B,

where

A =

x1 y1 1 0 0 00 0 0 x1 y1 1x2 y2 1 0 0 00 0 0 x2 y2 1.. .. .. .. .. ..x4 y4 1 0 0 00 0 0 x4 y4 1

,

and

B

x′1

y′1

x′2

y′2

...x′

4

y′4

.

The six by six normal equation for C is

AT AC = AT B.

52

11 The Rational Parametric Arc And The

Conic Arc

We can now construct the rational parametric arc. First we shall constructthe special case of the conic arc. The parabola with equation x− y2 = 0 andwith matrix

0 0 1/20 −1 0

1/2 0 0

,

is to be mapped to the required conic arc, (see Fig. 5). If we take t = y,then

tt2

1

is a point on the parabola. Points A, B, and D lie on the parabola and aregiven as

A =

001

B =

111

D =

1/41/21

C is the intersection point of the tangent lines at A and B. The tangent lineat B has equation

[

x y 1]

0 0 1/2 10 −1 0 1

1/2 0 0 1

53

= (1/2)x− y + 1/2 = 0.

So

c =

01/31

.

For future reference note that

D = aA + bB + cC,

where a = 1/4, b = 1/4, and c = 1/2.It takes five points to determine a conic. Here we do have five pieces of in-

formation, namely three points which lie on the conic, and two tangents. Thisparabola is the set {P : f(P, P ) = 0}. A projective transformation M mapsit to the conic {P ′ : f ′(P ′, P ′) = 0}, where f ′(P ′, P ′) = f(M?(P ′), M?(P ′)).By the construction above, let M be the projective transformation that takesthe four points defining the parabola to the four primed points that definingsome conic arc. From the definition of f ′, we see that

f ′(C ′, A′) = f(C, A) = 0.

andf ′(C ′, B′) = f(C, B) = 0.

So C ′ is the intersection of the tangents to the conic arc at A′ and B′. Aparametric equation of the arc from A′ to B′ is given by

x′

y′

z′

= M

t2

t1

,

for 0 ≤ t ≤ 1.Example. Let the arc be a half circle.

A′ =

−101

B′ =

101

54

D′ =

011

.

For this example we see by inspection that

C ′ =

001

That is, the intersection of the tangents at A′ and B′, is the point atinfinity in the direction of the y-axis. In general we may find the polars of A′

and B′ and then calculate the cross product of their line coordinates to getC ′. We need the coordinates a′, b′, c′ of D′ with respect to the basis A′, B′, C ′.We have

a′

b′

c′

=[

A′ B′ C ′]−1

D′

=

−1/2 0 1/21/2 0 1/20 1 0

011

=

1/21/21

.

We know a, b, c from above, so

a” = a′/a = (1/2)/(1/4) = 2, b” = b′/b = 2, c” = c′/c = 2.

NowM =

[

a”A′ b”B′ c”C] [

A B C]−1

= 2

−1 1 00 0 11 1 1

0 1 00 1 1/21 1 1

−1

= 2

−1 1 00 0 11 1 1

0 −2 11 0 0−2 2 0

=

0 4 −2−4 4 04 −4 2

.

55

Finally, we get

x′

y′

x′

= M

t2

t1

.

So in the affine plane we have

x′ =2t − 1

2t2 − 2t + 1,

and

y′ =2t − 2t

2t − 2t + 1.

There appears to be much calculation here. It is not as bad as it ap-pears. There is a Fortran conic arc subroutine that is called conarc inlibrary emerylib.ftn. The subroutine shows that the calculation can bequite simple.

Here is a listing:

c+ conarc parametric conic arc

subroutine conarc(r0,r1,r2,p,u,x,y)

implicit real*8 (a-h,o-z)

c parameters

c r0-start point

c r1-arc is tangent to the line through r0 and r1, and

c tangent to the line through r1 and r2

c r2-end point

c p-value controls fullness of arc: if near 1 arc

c is a sharp hyperbola, if 1/2 it is a parabola,

c if near zero it is a flat ellipse. p must be

c greater than zero and less than 1.

c u-parameter, the arc is parameterized on the unit interval

c x,y-point on the arc at parameter u.

c Notice that if p = 1/2, this is the second degree Bezier curve.

dimension r0(*),r1(*),r2(*)

w0=1-p

w2=1-p

w1=p

b0=w0*(1-u)**2

56

b1=2*w1*u*(1-u)

b2=w2*u*u

d=b0+b1+b2

x=(r0(1)*b0+r1(1)*b1+r2(1)*b2)/d

y=(r0(2)*b0+r1(2)*b1+r2(2)*b2)/d

return

end

A reference for the equations that are used in the subroutine is Faux andPratt.

It is advantageous to calculate the rational parametric form when pointsof the arc must be displayed or plotted. When we have a rational parametricform we can easily compute the x and y coordinates of any point.

Some early books on aircraft construction have material on conic arcs.One such book is: Aircraft Analytic Geometry, by J. J. Apalategui andL. J. Adams, McGraw-Hill, 1944. Many of the early aircraft shapes wereconstructed using composite conic arcs.

A classic method for finding a parameterization of an algebraic curve isgiven in books on Algebraic Geometry. This method uses a set of lines thatare defined by a single parameter. Each line is intersected with the algebraiccurve, giving each point on the curve as a function of the line parameter.Frequently one uses the set of lines through the origin: y = tx where t is theparameter. As an example one can get a simple rational parameterization ofthe circle

(x − 1)2 + y2 = 1,

using this method.

12 Straight Lines In Projective Space

The equation of a line is of the form

ax + by + cz = 0

The coordinates of the line

abc

57

are perpendicular to the coordinates of the point

xyz

If A and B are coordinate vectors of two lines then the cross product isthe coordinate vector of the intersection point because it is perpendicular toboth. In the same way if X and Y are coordinate vectors of two points thentheir cross product is the coordinate vector of the line containing both. Thisis an example of duality in projective space.Example. Find the intersection point of the two lines with equations

x + y = 0,

and2x + y = 1.

The cross product is

2−5−1

So the affine coordinates of the intersection point are (−2, 5).Example. Find the line through the points

111

and

321

The cross product is

−12−1

58

So the equation is

−x + 2y = 1.

Example. Find the line passing through (1, 1) and making an angle of 30degrees with the x-axis. The line passes through the point at infinity givenby

cos(30)sin(30)

0

=

√3/2

1/20

,

and the point

111

.

Taking the cross product we find the line to be

x −√

3y = 1 −√

3.

13 Quadric Surfaces

Quadric surfaces are the analogue of conic sections in three dimensionalspace. This space is constructed in a four dimensional vector space so thereare four coordinates for each point. Again corresponding to each quadric isa bilinear form f . The quadric surface is the set of points P where

f(P, P ) = 0.

The properties and propositions that have been treated for 2-space are es-sentially the same in 3-space.

The equation for the intersection of a line P + tQ is still

0 = f(P + tQ, P + tQ) = f(P, P ) + 2tf(P, Q) + tf(Q, Q),

and the various cases are the same as for the conics. If P is a point then

(P, Q) = 0.

59

Is the equation of a plane, where (P, Q) is the inner product of P and Q. Qis the coordinate vector of the plane. There is a duality between points andplanes. For example, three points determine a plane. Dually three planesdetermine a point. In two dimensions, the duality was between points andlines. The first three coordinates of the plane give a vector normal to theplane in affine space, where the affine hyperplane has equation w = 1. Thehomogeneous coordinates of a point are written as x, y, z, w. If P is a point,then the set of points Q, where

f(P, Q) = 0,

is the polar of P . In 3-space, the polar is a plane. If A is the symmetricmatrix of the quadric

f(P, Q) = PAQ = QAP = f(Q, P ) = (Q, AP ).

If P is on the quadric surface then the polar of P is the tangent plane. Thiscan be seen by examining the equations of intersection of lines in the tangentplane passing through P . The equation of the tangent plane is then

(Q, AP ) = 0,

and AP is the coordinate vector of this plane. The first three coordinatesof AP are the coordinates of a normal vector. A diametrical plane in thedirection Q, where Q is a point at infinity, is the polar of Q. The diametricalplane is the analogue in 3-space of the diameter in 2-space. A center is apoint in the intersection of all diametrical planes. points on a quadric surfaceare called singular points if they are centers and regular points otherwise. Anexample of a singular point is the apex of a cone. The statement above thatthe polar of a point on the surface is the tangent plane must be qualified.It holds only for regular points. obviously the apex of a cone has no welldefined tangent plane.

As in the two dimensional case, a diametrical plane in the direction Qbisects all chords in the direction Q. The two dimensional proof holds.

14 Quadric Surfaces And Their Matrices

There are 17 quadric surfaces (see Olmstead). some are complex and someare degenerate. The names, equations and matrices of some of the realnondegenerate quadric surfaces are:

60

(1) Sphere:x2 + y2 + z2 − w2 = 0.

1 0 0 00 1 0 00 0 1 00 0 0 −1

(2) Hyperbolic paraboloid:

x2 − y2 − 2wz = 0.

1 0 0 00 −1 0 00 0 0 −10 0 −1 0

(3) Elliptic paraboloid:x2 + y2 − 2zw = 0.

1 0 0 00 1 0 00 0 0 −10 0 −1 0

(4) Parabolic cylinder:x2 − 2yz = 0.

1 0 0 00 0 −1 00 −1 0 00 0 0 0

(5) Hyperbolic cylinder:x2 − y2 − w2 = 0.

1 0 0 00 −1 0 00 0 0 00 0 0 −1

(6) Intersecting planes:x2 − y2 = 0.

61

1 0 0 00 −1 0 00 0 0 00 0 0 0

(7) Parallel planes:x2 − w2 = 0.

1 0 0 00 0 0 00 0 0 00 0 0 −1

(8) Circular cylinder:x2 + y2 − w2 = 0.

1 0 0 00 1 0 00 0 0 00 0 0 −1

(9) Hyperboloid of two sheets:

x2 + y2 − z2 + w2 = 0.

1 0 0 00 1 0 00 0 −1 00 0 0 1

(10) Cone:x2 + y2 − z2 = 0.

1 0 0 00 1 0 00 0 −1 00 0 0 1

(11) hyperboloid of one sheet:

x2 + y2 − z2 − w2 = 0.

1 0 0 00 1 0 00 0 −1 00 0 0 −1

62

15 Transformations

A one to one linear transformation maps a quadric surface to a new quadricsurface. Suppose

S = {P : f(P, P ) ≤ 0}then

LS = {LP : f(P, P ) ≤ 0}= {Q : f(L−1Q, L−1Q) ≤ 0}= {Q : (L−1Q, AL−1Q) ≤ 0}

= {Q : (Q, L−1T AL−1Q) <= 0}= {Q : f ∗(Q, Q) ≤ 0},

where f ∗ is a bilinear form with matrix

L−1T AL−1 = A′

The matrices for the scaling, translating and rotation transformations aregiven here.

(1) Scaling matrix:

a 0 0 00 b 0 00 0 c 00 0 0 1

Inverse of scaling matrix:

1/a 0 0 00 1/b 0 00 0 1/c 00 0 0 1

(2) Affine rotation: Let c = cos(θ) and s = sin(θ). The sense of therotations about the unit coordinate vectors is given by the right hand rule.With the thumb of the right hand extended in the direction of the vector,positive angle is the direction in which the fingers curl while grasping thevector.

63

Rotation about the x-axis:

1 0 0 00 c −s 00 s c 00 0 0 1

Rotation about the y-axis:

c 0 s 00 1 0 0−s 0 c 00 0 0 1

Rotation about the z-axis:

c −s 0 0s c 0 00 0 1 00 0 0 1

Now if R is a rotation matrix, then

R−1(θ) = R(−θ).

An orthogonal matrix R has the property RT = R−1, so the matrix of thetransformed quadric surface is

A′ = R(θ)AR(−θ).

(3)Affine translation: The matrix

1 0 0 a0 1 0 b0 0 0 c0 0 0 1

translates by

abc

.

64

The inverse of T is

T−1 =

1 0 0 −a0 1 0 −b0 0 0 −c0 0 0 1

.

The transformed matrix is

A′ =

1 0 0 00 1 0 00 0 1 0−a −b −c 1

A

1 0 0 −a0 1 0 −b0 0 1 −c0 0 0 −1

.

Example. Given a sphere with matrix

1 0 0 00 1 0 00 0 1 00 0 0 −1

Then the scaling transformation matrix

2 0 0 00 1 0 00 0 1 00 0 0 1

Maps the sphere to an ellipse with matrix

1/2 0 0 00 1 0 00 0 1 00 0 0 1

1 0 0 00 1 0 00 0 1 00 0 0 −1

1/2 0 0 00 1 0 00 0 1 00 0 0 1

=

1/2 0 0 00 1 0 00 0 1 00 0 0 1

1/2 0 0 00 1 0 00 0 1 00 0 0 −1

65

=

1/4 0 0 00 1 0 00 0 1 00 0 0 −1

.

The translation matrix T shifts it by 2

1 0 0 −20 1 0 00 0 1 00 0 0 1

The matrix of the shifted conic becomes

(T−1)T (S−1)T AS−1T−1

=

1 0 0 00 1 0 00 0 1 02 0 0 1

1/4 0 0 00 1 0 00 0 1 00 0 0 −1

1 0 0 20 1 0 00 0 1 00 0 0 1

=

1 0 0 00 1 0 00 0 1 02 0 0 1

1/4 0 0 1/20 1 0 00 0 1 00 0 0 −1

=

1/4 0 0 1/20 1 0 00 0 1 0

1/2 0 0 0

.

The equation of the shifted ellipsoid is

x2/4 + y2 + z2 + 1/2xw + 1/2wx = 0.

Suppose we rotate about the z-axis by angle 45 degrees. The rotation matrixis

Rz(45) =

√2/2 −

√2/2 0 0√

2/2√

2/2 0 00 0 1 00 0 0 1

.

66

The matrix of the quadric becomes

Rz(45)(T−1)T (S−1)T AS−1T−1Rz(−45).

This is

√2/2 −

√2/2 0 0√

2/2√

2/2 0 00 0 1 00 0 0 1

1/4 0 0 1/20 1 0 00 0 1 0

1/2 0 0 0

√2/2

√2/2 0 0√

2/2√

2/2 0 00 0 1 00 0 0 1

=

√2/2 −

√2/2 0 0√

2/2√

2/2 0 00 0 1 00 0 0 1

√2/8

√2/8 0 1/2

−√

2/2√

2/2 0 00 0 1 0√2/4

√2/4 0 0

.

The equation of the shifted rotated quadric is

(5/8)x2 + (5/8)y2 − (3/4)xy + z2 + xw√

2/2 + yw√

2/2 = 0.

16 The Construction Of Solids

A primitive solid is the set of points ”inside” a quadric surface. If f is thebilinear form of the quadric surface the primitive solid is

{P : f(P, P ) ≤ 0}.

A solid object is constructed by combining the primitive solids using theset operators union, intersection and complementation. The characteristicfunction, CS corresponding to the set S, is a logic valued function mappingpoints to a value true or false. The value of CS(P ) is true if P ∈ S, otherwiseit is false. By definition P ∈ A ∪ B, if and only if, P ∈ A or P ∈ B. So thecharacteristic function of A ∪ B is

C(A ∪ B) = C(A) ∨ C(B),

where ∨ is the logical ”or” operator. Similarly

C(A ∩ B) = C(A) ∧ C(B),

67

where ∧ is the logical ”and” operator, and

C(A)c = ¬C(A),

¬ is the negation operator. Objects are defined as logic valued functionswhose domain is 3-space. These functions may be represented with logicalvariables of programming languages.

17 Plane Image

A shaded image is produced by finding the intersection of the object anda line that passes through the projection point. Let L be such a line. Itsintersection with each primitive surface is calculated. For each intersectionpoint there is a surface normal. A small line segment of fixed length δl, in thenormal direction with center at the intersection point, determines when thepoint is on the surface of the object. It is on the object when one endpointis inside and the other is outside the object. This is determined by thecharacteristic function.

One or more light sources illuminate the object. The inner product ofthese light source vectors with the surface normal give an intensity functiondefining the shading. The point on the surface is projected to 2-space withthe calculated shading value. The projection point may be at infinity.

18 Monte Carlo Integration

An integral on a three dimensional set can be evaluated by enclosing the setin a rectangular box, and then taking a uniformly distributed random samplein the box. Suppose the integral

∫

Sf(x)dm

is to be evaluated. Let x be a point. Let B be a rectangular hexahedronenclosing S. Let m be the volume measure (Lebesgue measure). Define aprobability measure on the set B by

p(S) = m(S)/m(B).

68

Let χS be the characteristic function of the set S. If x ∈ S then χS(x) is1, otherwise it is 0. Let g be the product of f and χS. Then g is a randomvariable defined on the enclosing set B. Its expectation is

E(g) =∫

Bgdp =

∫

Sfdp =

1

B

∫

Sfdm.

∫

Sfdm = m(B)E(g)

Let x1, ...., xn be a sample of independent uniformly distributed random vari-ables. Then g(x1), ..., g(xn) is also independent. Let g be the sample mean

g = (g(x1) + ..... + g(xn))/n.

The expectation of the sample mean is the mean of the original distribution,

E(g) = E(g).

The variance is

V (g) = E((g − E(g))2) = E(g2) − E(g)2 = V (g)/n.

By the central limit theorem, the random variable

(g − E(g)/√

V (g) = (g − E(g))/√

V (g)/n,

has approximately, for sufficiently large n, the standard normal distribution.Now

1√2π

∫ 1.96

−1.96exp(−x2/2)dx ≈ .95.

SoP (|g − E(g)| ≤ 1.96

√

V (g)/n ≈ .95.

This gives a 95 percent confidence interval for the estimated value of themean. Multiplying by m(B), we get a confidence interval for the integral.

P (|m(B)g −∫

Sfdm| ≤ 1.96m(B)

√

V (g)/n ≈ .95.

We are interested in the specific quantities:(1) Volume; f = 1

69

(2) Moments;f = xi, i = 1, 2, 3(3) Inertia tensor: f = xixj , i, j = 1, 2, 3

Example. let S be a sphere of radius a. Let B be the cube of side 2a,centered at the origin. Then m(B) = 8a3, and m(S) = (4/3)πa3 . Let f = 1and g = χS. Then the expectation of g is

E(g) =∫

Sg(x)dp =

∫

Sdp = p(S) =

(4/3)πa3

8a3=

π

6.

The variance of g is

v(g) =∫

g2dp − E(g)2 = E(g)(1 − E(g)).

Let us choose the sample size n, so that the relative error of the estimatedvolume is less than 1/100. The error is

1.96m(B)√

E(g)(1 − E(g))/n.

We can get a relative error by dividing by the volume m(B)E(g). We get

1.96√

(1 − E(g))/(E(g)n)) < .01.

Then1/E(g) − 1 < (.01/1.96)2n,

and son > 1962(6π − 1) = 34953

This is a large number of samples. We must decide how to choose the blockB. In the case of volume, the error is

1.96m(B)√

[m(S)/m(B)][1 − m(S)/m(B))]/n

1.96m(B)√

m(S)[m(B) − m(S)/n.

We must minimize a function of the form

h(x) =√

c(x − c),

70

where x ≥ c. The derivative is

dh

dx=

c

2√

c(x − c),

so the function is increasing. The minimum occurs at x = c. We shouldchoose the box B to be as small as possible.

Usually we do not know the variance V (g), but we can use the samplevariance as an estimate. The sample variance is defined as

S2 =

∑nk=1(g(Pk) − g)2

n

=

∑nk=1 g(Pk)

2

n− g2.

An unbiased estimator of the variance of g is (p. 165 Brunk)

n

n − 1S2.

So∫

Sfdm = m(B)g + 1.96m(B)

√

S2

n − 1.

Example. We shall calculate the x1 moment of the sphere of the previousexample. We have f = x1 and

E(g) =∫

Bgdp =

1

8a3

∫

Sx1dm = 0.

By symmetry the variance is

1

8a3

∫

Sx2dm =

π

8a3

∫ a

−ax2

1(a2 − x2

1)dx

=πa2

30.

So the error is

(1.96)a4 sqrtπ√30n

.

71

In this case we can’t divide by the moment to get a relative error, becausethe moment is zero. But a moment is the product of a volume and a length,so a relative error results from dividing by a4 .Example. We shall calculate the inertia tensor components for the sphere.When i 6= j, by symmetry,

∫

Sxixjdm = 0.

When i = j we have∫

x2i dm =

∫ a

−ax2

i (a2 − x2

i )dx

= π

[

x3i a

2

3− x5

i

5

]a

−a

= 2πa5(1/3 − /15)

=4

15πa5.

19 Stratified sampling

It may be possible to decrease the error by doing stratified sampling. Wedivide the domain B into m subdomains. We introduce a probability measurepi on each Bi by

pi(Si) = m(Si)/m(Bi).

Then ∫

S∩Bi

fdm = m(Bi)∫

Bi

gidpi = E(gi)m(Bi),

where gi is the product of f and the characteristic function χ(Si) of Si. Then

∫

Sfdm =

m∑

i=1

∫

S∩Bi

m(Bi)gidpi.

Let hi = m(Bi)gi. Then, defining

h =m

∑

i=1

hi,

72

we have∫

Sfdm = E(h) =

m∑

i=1

E(hi).

Also

v(h) =m

∑

i=1

v(hi).

Let hi1, ....., him, be an independent sample of random variables having thedistribution of hi . We take k samples from each strata. Let b2

i = v(hi) and

B2 = v(h11) + .... + v(hmk

= k(b21 + ... + b2

m).

Let ai = E(hi) and A = k(a1 + ... + am). Consider

((h11 − a1) + (h12 − a2) + ... + (hmk − ak))/B

=h11 + ... + hmk − k(a1 + .... + am)

√

k(b21 + ... + b2

m

=(h11 + ... + hmk)/k − ∫

fdm√

(b21 + ... + b2

m)/k.

Let

yk =h11 + ... + hmk

k,

and

v(k) =b21 + ... + b2

m

k.

The Lindeberg condition is satisfied (p.455 Eisen). As k → ∞ the followingprobability approaches a normal probability. That is,

p((yk −∫

fdm)/v(k) ≤ x) →√

(1/2π)∫ x

−∞exp(−t2/2)dt.

It follows thatp(|yk −

∫

fdm| ≤ 1.96v(k)) ≈ .95.

yk is a better estimate of the integral than m(B)g, provided

v(k) < m(B)√

v(g)/(mk)

73

Note that there are mk samples. Suppose that either v(gi) = v(g) or elsev(g1) = 0. Then

b2i = v(hi) = m(Bi)

2v(gi) =

{

m(Bi)2v(g)

0

Suppose j of the b2i are not zero, and suppose the strata are equal, so that

m(Bi) = m(B)/m. Then

v(k) =

√

b21 + .... + b2

k√k

=√

(j(m(B)2/m2)v(g)/k

=√

(j/m)m(B)√

v(g)/(mk)

≤ m(B)√

v(g)/(mk).

Many strata will be completely inside or completely outside the object, andwill have zero variance. If

j < m,

then the stratified variable yk should be a better estimate of the integral thanthe unstratified variable.

20 Point Transformations and Coordinate Trans-

formations

Let u1, u2, u3, and u′1, u

′2, u

′3, be bases of a vector space and let c and c′ be

corresponding coordinate maps. That is, c maps a vector to its coordinatevector with respect to the unprimed basis. Let A be the change of coordinatematrix so that

c′(P ) = Ac(P )andc(P ) = Ac′(P )

Let T be a linear transformation defined by

T (u) = u′

74

Thenc(P ) = c′(TP ) = Ac(TP )

Thusc(TP ) = Ac(P )

Andc′(TP ) = c(P ) = Ac′(P )

Hence the matrix of T is A in both bases. In summation we have shown thatwhen an unprimed basis is transformed to a primed basis, then the matrix ofthe coordinate transformation from the unprimed to the primed coordinatesystem is the inverse of the matrix of the point transformation.

21 Cross Product, Axial Vectors

The cross product of two vectors in a Euclidean space is defined by its com-ponents with respect to an orthogonal coordinate system by

a × b =

u1 u2 u3

a1 a2 a3

b1 b2 b3

=

a1 a2 a3

b1 b2 b3

u1 u2 u3

= εijkaibjuk,

where εijk is the permutation symbol and u1, u2, u3, are unit coordinate vec-tors. If ijk is an even permutation, the permutation symbol is 1, if ijk is odd,the permutation symbol is -1, if ijk are not distinct, then the permutationsymbol is 0. The kth component of a × b is

ck = εijkaibj .

Let there be a new primed coordinate system. We have

εijk

∂xi

∂x′p∂xj

∂x′q∂xk

∂x′r = Det(J)εpqr.

The Jacobian is

J =

∂x1

∂x′1

∂x2

∂x′1

∂x3

∂x′1

∂x1

∂x′2

∂x2

∂x′2

∂x3

∂x′3

∂x1

∂x′3

∂x2

∂x′3

∂x3

∂x′3

75

[Lass, page 8]. The definition of a× b depends on the coordinate system.But, if the coordinate systems are related by a proper orthogonal transfor-mation, rather than by a general transformation, then the determinant ofthe Jacobian is 1, that is

Det(J) = 1.

Then

εijk

∂xi

∂x′p∂xj

∂x′q∂xk

∂x′r = εpqr.

So the permutation symbol transforms as a tensor. The components of a× b,are obtained by contracting the product of a tensor and two vectors, andso the cross product is a Cartesian vector. A second order tensor, whosecomponents obey

cij = −cji,

is called antisymmetric. An antisymmetric tensor defines a Cartesian vector.The components of this vector are

cij = −(1/2)εijkcij .

These are the components of a Cartesian vector, because the vector is thecontraction of a Cartesian tensor and a general tensor. We find that

c =

c32

−c31

c21

.

A vector such as c is called an axial vector. Let dr be the rth component ofthe cross product c × x. Then

dr = εpqr(−(1/2)εijpcij)xq.

We claim thatεpqr(−(1/2)εijpcij) = crq.

We have

−(1/2)εpqrεpijcij

=3

∑

p=1

(−1/2)εpqrεpijcij

76

=∑

p 6=q,p 6=r

(−1/2)εpqrεpijcij

=∑

p 6=q,p 6=r

[(−1/2)εpqrεpijcij − (−1/2)εpqrεpqrcrq]

=∑

p 6=q,p 6=r

[(−1/2)cqr + (1/2)crq = crq.

Sodr = crqx

q.

If L is the linear transformation which has components crq, then the valueto which a vector x is mapped, is given by the cross product

L(x) = c × x.

where c is the axial vector defined by the antisymmetric tensor with compo-nents ck. These ideas are applied to the concept of angular velocity.

22 Angular Velocity

Consider a rigid body, with a point at the origin fixed. Let P be a point inthe body and let u1, u2, u3 and u′

1, u′2, u

′3 be two bases of a vector space. The

unprimed basis is fixed in space. The primed basis is fixed in the body. Then

u′i = R(ui),

where R is an orthogonal transformation, which varies with time. Then thecoordinates of P are related by

x′i = aijxj .

where (aij) is the inverse of the matrix (aij) of R. By differentiating, andnoting that the components of P are constant in the primed system, we findthat (Lass page 46)

dxi

dt= wijxj ,

for

wij = −aik

dakj

dt.

77

By differentiating the norm of P , which is invariant, we find that (wij) isskew symmetric, that is

wij = −wji.

So (wij) can be represented as an axial vector. We have

c(w) =

w32

−w31

w21

=

w32

w13

w21

and

v =dc(P )

dt= c(w) × c(P ).

w is the angular velocity vector. It transforms like a vector under an orthog-onal transformation.Example. Let rotation transformation R have matrix

(aij) =

c(t2) −s(t2) 0s(t2) c(t2) 0

0 0 1

.

So

(aij) = (aij)T =

c(t2) s(t2) 0−s(t2) c(t2) 0

0 0 1

,

and

daij

dt= 2t

−s(t2) c(t2) 0−c(t2) −s(t2) 0

0 0 0

.

Then

(wij) = −2t

c(t2) −s(t2) 0s(t2) c(t2) 0

0 0 1

−s(t2) c(t2) 0−c(t2) −s(t2) 0

0 0 0

= −2t

0 1 0−1 0 00 0 0

.

78

So

c(w) =

002t

.

Suppose P = u′1. Let c(P ) be the coordinate vector of P in the unprimed

system and c′(P ) the coordinate vector in the primed system. Then

c(P ) = (aij)c′(P )

= (aij)

100

=

c(t2)s(t2)

0

.

Now

dc(P )

dt= 2t

−s(t2)c(t2)

0

.

This is the velocity computed with matrices. The velocity computed withthe cross product is

c(w) × c(P ) =

002t

×

c(t2)s(t2)

0

= 2t

−s(t2)c(t2)

0

.

We get the same result with either method.

23 Angular Momentum And The Inertia Ten-

sor

The angular momentum vector is defined as (Lass p.115)

L =∫

r × vdm,

79

where r is the position, v the velocity and m the mass. The kth componentis

Lk =∫

εijkxixjdm.

ThendLk

dt=

∫

εijkxixjdm +

∫

εijkxixjdm

=∫

εijkxixjdm.

The second term vanishes because the cross product of parallel vectors iszero. The torque vector T is

∫

r × df =∫

r × d2r

dt2dm.

The kth component of the torque is

Tk =∫

εijkxixjdm.

Thus dL/dt = T . Becausexj = wj

pxp,

we haveLk =

∫

εijkxiwj

pxpdm = εijk

∫

xixpdm

= εijkwjpI

ip,

where I is the inertia tensor. The components of the inertia tensor are

I ip =∫

xixpdm.

The moment of inertia about an arbitrary axis can be found from theinertia tensor. Given a straight line through the origin, let p be a function,whose value at a point is the distance from the point to the line. The momentof inertia about the line is then defined as (McConnel p.233)

∫

p2dm.

Suppose u is a unit vector in the direction of the line. Let t be the trace.Then

p2 = xixi − (r · u)2

80

= δijxixj − (δijx

iuj)2.

So the moment of inertia is

= δijIij − δijδkpu

jupI ik

= t(I) − I ikuiuk

= t(I)δikuiuk − I ikuiuk

= (t(I)δik − I ik)uiuk

= J ikuiuk.

The tensor J has components

J ik = t(I)δik − I ik.

In some books (e.g. Goldstein), J , rather than I, is called the inertia tensor.The explicit components of J are:

J11 =∫

[(x2)2 + (x3)2]dm,

J22 =∫

[(x1)2 + (x3)2]dm,

J33 =∫

[(x1)2 + (x2)2]dm.

When i 6= j,

J ij = −∫

[(xi)2 + (xj)2]dm.

If f is the bilinear form corresponding to the symmetric matrix J , thenthe moment of inertia about a unit vector u is f(u, u).

24 Surface Integrals

A surface area element dσ goes to ds = dσ cos(θ), when it is projected to aplane, where θ is the angle between the surface normal of the plane, and theprojection direction. If u is a unit vector in the projection direction, then

dσ = ds/(n · u).

81

A general surface integral is∫

f(x, n)dσ.

f is a function of position x and normal n. Suppose p is the projectionfunction. Then

∫

Af(x, n)dσ =

∫

p(A)f(p−1(y), n(p−1(y))(ds/(n · u).

Now suppose A1, ...., An covers the surface, and p is the plane projection fromA1 to 2-space. Suppose there are functions gi defined on the surface, whereeach gi vanishes outside of Ai. And suppose that each gi has values between0 and 1, and suppose that the gi sum to 1. That is

n∑

i=1

gi(x) = 1.

Such a set of functions is called a partition of unity, subordinate to the coverA1, ..., An. Then we have

f =n

∑

i=1

gi(x)f(x) = 1.

The surface integral is

∫

Af(x, n)dσ =

n∑

i=1

∫

p(Ai)gi(p

−1(y), n(p−1(y))f(p−1(y), n(p−1(y))(ds/(n · u).

When pi is the projection in the xi direction, let gi be the function whosevalue at a point is the square of the ith component of the unit normal vectorto the surface. In general pi is a many to one function and when we defineAik to be the kth layer of the surface, which is pierced by the projecting ray,the functions

fi : Aik → R,

are surface patches, and the

gi : Aik → R,

constitute a partition of unity.

82

25 Volume Integral

A volume integral can be changed to a surface integral by using the divergencetheorem. Define a vector field

H =r

3=

1

3

xyz

.

Let B be a set and ∂B its bounding surface. Then the volume of B is∫

BdV =

∫

B∇ · HdV

=∫

∂BH · ndS

=1

3

∫

∂B(xnx + yny + znz)dS

Note that the integrand is the distance from the origin to the plane containingthe area element dS. In particular for any plane area, say A, the volume ofthe pyramid, with vertex at the origin and base A, is

V =hA

3,

where h = n · R, is the height of the pyramid. The volume of a polyhedronconsisting of k polygons, is therefore

V =k

∑

i=1

Ai

Pi · ni

3,

where Ai is the area of the ith polygon, Pi is some point on the polygon, andni is the outward normal to the polygon. The divergence theorem can alsobe used to compute moments. For example, applying the divergence theoremto the vector field

G =1

2

x2

11

,

gives a surface calculation of the x moment.

mx =1

2

∫

∂Bx2nxdS.

83

26 Polygon Areas

We shall calculate polygon areas. Let a closed curve γ have position vectorr(t), 0 ≤ t ≤ 1. The area enclosed by the curve is

A =1

2

∮

γr × dr.

This of course is a vector. It is the magnetic moment of a circuit withunit current. When the curve lies in a plane, the magnitude of this integralis equal to the area enclosed by the curve. This is obvious from the definitionof the cross product when the plane passes through the origin. When theplane does not pass through the origin, the vector r can be written as a sumof a constant vector normal to the plane, and a vector that lies in the plane.The constant vector is everywhere normal to the line element, and so doesnot make a contribution to the integral. This case reduces to the former case.

As an aside, this calculation can be used to find the inner region enclosedby a plane curve. In the case that the curve lies in the xy plane, the vectorintegral will be in the z direction. If the component is positive then theinner region is to the left. This is obvious if the curve is the unit circlewith the counterclockwise orientation. For a general proof, note that a curvewhich bounds an inner region can be continuously deformed to a unit circlein such a way that the area never vanishes. Thus the sign of the area mustbe maintained.

If the sign is negative then the inner region lies to the right. The curvedirection is defined by the increasing parameter value.

Returning to the polygon area problem, the integral over a line segmentis

A =r1 × r2

2,

where r1 and r2 are the starting and ending points. This is true because theline element has the same direction as both r2 − r1 and r − r1. The crossproduct of parallel vectors is zero, so the integral is

1

2r1 ×

∫

dr

=1

2r1 × (r2 − r1)

84

=1

2r1 × r2.

Thus the area of a closed polygon Ai with m sides and m vertices equals themagnitude of

1

2

m∑

j=1

rj × rj+1,

whererm+1 = r1.

This sum is clearly normal to the polygon, since each term is. A unit normalis obtained by dividing by the area Ai.

The number of cross products that need to be calculated can be reduced.For example for a triangle with m = 3 the sum of the cross products is

r1 × r2 + r2 × r3 + r3 × r1

= r2 × r3 − r2 × r2 − r1 × r3 + r1 × r2

= (r2 − r1) × (r3 − r2)

For another example we may decompose a hexagon into 4 triangles and writethe area as

1

2[

3∑

i=1

(ri+1 − ri) × (ri+2 − ri+1) + (r3 − r1) × (r5 − r3)].

27 Perspective Projection

Let Q be a point in real projective 3 Space (RP3). Q has component vector

Q = (Q1, Q2, Q3, Q4) = (x, y, z, w),

where the components are the homogeneous coordinates of the point.Consider a plane with coordinate vector A. The equation of the plane is

given as an inner product,

(A, Q) = A1Q1 + A2Q2 + A3Q3 + A4Q4 = 0.

85

Figure 8: A Perspective drawing showing, a building, a railroad track, andthe horizon. The vanishing point at the end of the railroad tracks is at infinityin three space, yet its projection is a finite point on the two dimensional page.The figure was generated with program tracks.ftn.

86

Let P be a projection point not in the plane. Let L(Q) be the point ofintersection of the plane with the line

Q + tP.

Then(A, Q + tP ) = 0.

Solving for t, we find

L(Q) = Q − (A, Q)

(A, P )P.

L is a linear projection operator, i.e. L ∗ L = L. Using a rigid motion trans-formation, we may transform the plane to the xy plane, and the projectionpoint to the positive z axis. Hence we consider the special case where theprojection plane has equation

z = 0,

and the projection point P is on the positive z axis. Then

A = (0, 0, 1, 0)T ,

andP = (0, 0, p3, p4)

T .

In this special case of projection into the xy plane,we write

L = Lxy.

Then(x′, y′, w′)T = LxyQ = (x, y, z, w)T − z

p3(0, 0, p3, p4)

T .

= (x, y, 0, w − zp4

p3

)T .

Lxy =

1 0 0 00 1 0 00 0 −p4/p3 1

=

1 0 0 00 1 0 00 0 −1/d 1

,

where d is the distance from the projection plane to the projection point.Given a projection point P and a center point C, let T1 be an affine trans-formation translating C to the origin. Let P have elevation angle θ, and

87

azimuth angle φ. The azimuth angle gives the direction in the xy plane mea-sured from the positive x axis. The azimuth is the rotation angle, about thez axis, that the xz plane would have to be rotated, so that it would includethe projection direction. For example, the azimuth of the positive x axis iszero, and the azimuth of the positive y axis is π/2 . The elevation is theangle from the xy plane to the projection direction. We rotate the image ofthe projection point into the yz plane with a rotation T2 about the z-axisby angle −φ − π/2. We rotate the image of the projection point up to thez-axis, with a rotation T3 about the positive x axis, by an angle −(π/2− θ).The complete perspective transformation matrix is

L = LxyT3T2T1,

where

T1 =

1 0 0 −Cx

0 1 0 −Cy

0 0 1 −Cz

0 0 0 1

,

T2 =

c(φ + π/2) s(φ + π/2) 0 0−s(φ + π/2) c(φ + π/2) 0 0

0 0 1 00 0 0 1

=

−s(φ) c(φ) 0 0−c(φ) −s(φ) 0 0

0 0 1 00 0 0 1

,

and

T3 =

1 0 0 00 c(θ − π/2) −s(θ − π/2) 00 s(θ − π/2) c(θ − π/2) 00 0 0 1

=

1 0 0 00 s(θ) c(θ 00 −c(θ) s(θ) 00 0 0 1

.

88

Example. Isometric projection. The picture plane has coordinate vector

R =

1110

The projection point is at infinity,

P =

1110

Hence d = ∞, and so

Lxy =

1 0 0 00 1 0 00 0 −1/d 1

=

1 0 0 00 1 0 00 0 0 1

The azimuth angle is φ = π/4, and

sin(φ) =

√2

2,

cos(φ) =

√2

2.

The tangent of the elevation angle is

tan(θ) =1√2.

Then

sin(θ) =

√3

3

cos(θ) =

√6

3

T1 is the identity, because the center of projection is already at the origin.

T1 = I.

89

T2 =

−√

22

√2

20 0

−√

22

−√

22

0 00 0 1 00 0 0 1

,

T3 =

1 0 0 0

0√

33

√6

30

0 −√

63

√3

30

0 0 0 1

.

Then

T3T2T1 =

−√

22

√2

20 0

−√

66

−√

66

√6

30√

126

√126

√3

30

0 0 0 1

Then

L = LxyT3T2T1 =

−√

22

√2

20 0

−√

66

−√

66

√6

30

0 0 0 1

A projective transformation matrix is defined only up to a scalar multiple.Hence we can factor out

√6/6, to get

L =

−√

3√

3 0 0−1 −1 2 0

0 0 0√

6

.

The images of the unit points are,

L

1001

=

−√

3−10√6

,

L

0101

=

√3

−10√6

,

90

and

L

0011

=

020√6

.

The image of the coordinate axes are as shown in the Figure. The lengthsare foreshortened by the factor 1/

√6. Conventional isometric drawing, true

lengths are laid off along the coordinate axes, so that the matrix is

−√

3/2√

3/2 0 0−1/2 −1/2 1 0

0 0 0 1

.

In perspective drawing, the horizon is defined as the intersection of the xy-plane with the plane at infinity

w = 0.

A horizon point can not be reached in 3-space, but we may ”see” its pro-jection. A line at infinity can project to a finite line in the picture plane.Similarly the point at infinity where two parallel lines meet, may project intoa finite point in 2-dimensional space. Such a point is called a vanishing point.For example, the point

Q =

1100

is the point at infinity, where parallel 45 degree lines in the plane meet atinfinity. The vanishing point is L(Q) in the picture plane. The horizon inthe picture plane is determined by two vanishing points, corresponding totwo sets of parallel lines in the xy-plane. When the projection point P isat infinity, as in isometric projection, parallel lines in 3-space are mapped toparallel lines in 2-space so there are no vanishing points.

If the line through P and Q is parallel to the picture plane, then theprojection of Q will be a point at infinity in the picture plane.

The following information is needed to construct the perspective trans-formation: (1)The center of the projection plane

(Cx, Cy, Cz)T .

91

. (2)The direction of the projection point from the center, which is specifiedby the azimuth angle φ, and the elevation angle θ. (3)The reciprocal of thedistance from the projection point to the picture plane

1

d,

which is zero for a projection point at infinity.Note that the usual projection from an infinite point on the z axis is

specified by elevation θ = π/2, azimuth φ = −π/2, and reciprocal distance1/d = 0. This gives Lxy.

28 The Tektronix Hardware Modeler: A Quadric

Solid Modeler

In conjunction with the Portland CAM-I (Computer Aided ManufacturingInternational) meeting, held August 3-7 1987, there was a tour of Tektronixlaboratories, where A CSG (Constructive Solid Geometry) system usingquadric halfspaces, with almost real time response, say three seconds foran average model, was given its first public viewing. This system consistedof both software and hardware.

The method:Space is divided recursively using an octree technique. Each block is

tested against each half space to determine whether it is inside, outside oron. If a given block is completely inside or completely outside the model it isdiscarded and not subdivided. A Taylor series expansion is used to classifya block with respect to a halfspace. When a block is sufficiently small andmeets the model surface it is considered a point on the surface and a normalis computed. It is mapped to the viewing plane and an intensity is computed.The voxels are searched from front to back and once a voxel is found to be”on” the object, its image on the screen, which is a rectangle is reservedand no voxels behind it need be searched. This algorithm is implementedin hardware and is very fast. The algorithm is very much faster than raytracing. It is claimed to be of order less than n, so that the larger n is (n=number of half space primitives) the greater its advantage.

Also seen at the labs was a discrete simulation system using Smalltalk,a sophisticated computer algebra system using the symbolic mathematical

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Z

-Y

X

Figure 9: The intersection of the parabolic cylinder z2 = y and the coney2 = xz is the union of the twisted cubic curve and the line along the x axis.This figure was generated with the solid modeling program qs90.ftn, whichwas formerly called Quadric.ftn. The program outputs a bitmap, with pixelvalues as hex characters. Program bm2ps.c converts it to a postscript file.

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system Reduce as basis, a 3d terminal using a quarter wave plate, and a fastswitching polarizing liquid crystal display (required 3d glasses).

Arnie Karush, Tektronix GMP (CAM-I Geometric Modeling Program)representative, was formerly in charge of the Tektronix 4029 (polygon proces-sor) development project. He says that the 4029 has advantages over theSilicon-Graphics terminal in certain cases. The 4029 must be driven by anexternal computer while the Silicon-Graphics can function alone.

29 References and bibliography

[1]Aslander Louis And Mackenzie Robert, Introduction To Differential

Manifolds, Dover N.y 1977.

[2]Barnhill R E, Riesenfeld R F (Editors), Computer Aided Geometric

Design, Academic Press 1974.

[3]Birkhoff Garrett And Maclane Saunders, A Survey Of Modern Alge-

bra Revised Edition, Macmillan 1953.

[4]Brunk H D, Introduction To Mathematical Statistics, Blaisdell 2ndEd. 1965.

[5]Courant R, Robbins H, What Is Mathematics?, Oxford UniversityPress 1947.

[6]Eisen Martin, Introduction To Mathematical Probability Theory,Prentice-hall Englewood Cliffs N.j. 1969.

[7]Faux Ivor D. And Pratt Michel J., Computational Geometry For De-

sign And Manufacture, Halsted New York 1979.

[8]Goldstein Herbert, Classical Mechanics, Addison-wesley, 1950.

[9]Halmos Paul R., Introduction To Hilbert Space, Chelsea New York2nd Ed. 1957.

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[10]Hodge W V D, Pedoe D, Methods Of Algebraic Geometry, Cam-bridge U. Press 1968.

[11]Knuth Donald, The Art Of Computer Programming, Vol. 1 Addison-wesley 2nd Ed.1973..

[12]Lass Harry, Elements Of Pure And Applied Mathematics, Mcgraw-hill 1957..

[13]Leven Joshua, A Parametric Algorithm For Drawing Pictures Of

Solid Objects Composed Of Quadric Surfaces, Comm. Acm V. 19 No.10 P555 Oct 1976.

[14]Mcconnell Albert J, Applications Of Tensor Calculus, Dover 1931.

[15]Olmsted John M H, Solid Analytic Geometry, Appleton-century-crofts New York 1947..

[16]O’neill Barrett, Elementary Differential Geometry, Academic PressNew York 1966.

[17]Panofsky Erwin, Durer As A Mathematician, In The World Of Math-ematics, Vol. 1 James Newman Ed., Simon And Schuster New York 1956.

[18]Requicha A A G, Voelker H B, Constructive Solid Geometry, Tech-nical Memorandum 25,, Production Automation Project The University Of,Rochester Rochester N. Y.1977.

[19]Sasaki Tateaki, Multidimensional Monte Carlo Integration Based

On Factorized Approximation Functions, Siam Jour. Num. AnalysisVol. 15 No. 5 Oct. 1978.

[20]Semple J G, Kneebone G T, Algebraic Projective Geometry, Oxford1952.

[21]Smart E H, A First Course In Projective Geometry, Macmillan1913.

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[22]Spivak Michael, Calculus On Manifolds, W.a Benjamin, New York1965.

[23]Struik Dirk J, Analytic And Projective Geometry, Addison-wesleyCambridge Mass. 1953.

[24]Willmore T J, An Introduction To Differential Geometry, OxfordU. Press 1959.

[25]Yakowitz S, Krimmel J E , Szidarouszky F, Weighted Monte Carlo

Integration, Siam Jour. Num. Anal. Vol. 15, No. 6 Dec. 1978.

[26] Emery James D, Projective space, Quadric surfaces, Conics And

Rational Curves, rational.tex, rational.pdf, 2001.

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