Conics, Quadrics, and Projective Space - STEM2

Conics, Quadrics, and Projective Space
James D Emery
Last Edit 10/22/2012
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 Transformations.
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
1

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 Modeler
92
29 References and bibliography 94
2

List of Figures
1 Points of projective 2-space are lines in 3-space. A 3-d linear
transformation is a 2-d projective transformation. A rotation
of the cone can project the circle to an ellipse, a parabola, or
ahyperbola. Inthesamewaythe lines through any figure in
the plane are a projective space representation of the figure.
Any sequence of perspective projections of a figure onto a set
of planes can be represented as a sequence of linear transformations
in 3-space. The plane shown in the figure is known
as the affine plane. . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Let lines PQand PRbe tangent to the ellipse. Tangent points
Q and R are on the polar of P . Thus the line defined by Q
and R is the polar of P . Intersection point S is on both the
polar of P and the polar of P ′ .ThusthepolarofS is the line
through P and P ′ . . . . . . . . . . . . . . . . . . . . . . . . . 25
3 The parabola x − 2yz = 0 and a point at infinity Q. The line
through P and Q meets the parabola at P and Q. Thepolar
of a point at infinity is a diameter. The polar of Q is the line
at 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 finite
points. The polar of Q is the line x = 1, which is a diameter
of the parabola. The point at in finity in the diagram is
represented 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 10x 2 +10xy+20y 2 +3x+−4y−10 = 0,
(b)Hyperbola with equation 5x 2 +10xy −7y 2 −3x+2y −1 =0,
(c)Parabola with equation x 2 +6xy +9y 2 +3x − 5y − 1=0,
(d)and a circle with equation 2x 2 +2y 2 + 1x − 1 y − 1=0. . 45
2 5
6 A conic passing through four points defined by the products of
lines passing through the points (1, 1), (2, 2), (2, 1), (3, 3). The
general 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 by
tangent lines. . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3

7 (a)Any conic arc can be obtained by mapping this parabola,
which is 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 parabola to a conic representation
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 railroad
tracks is at infinity in three space, yet its projection is a finite
point on the two dimensional page. The figure was generated
with program tracks.ftn. . . . . . . . . . . . . . . . . . . . . 86
9 The intersection of the parabolic cylinder z 2 = y and the cone
y 2 = 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 Quadric. . . . . . . . . . . . . . . . . . . . 93
1 Introduction
Quadric solids are solids bounded by quadric surfaces. Quadric solids provide
a rich collection of sets for constructing models of 3-dimensional objects.
A large set of quadric solids may be built by combining primitive quadric
solids, such as spheres, ellipsoids, blocks, cones, wedges, and cylinders. Many
manufactured objects are combinations of quadric shapes. There are 17 types
of quadric surfaces, including such less common surfaces as the parabolic
hyperboloid.
The world is full of objects built from rectangular blocks, circular cylinders,
and spheres. These objects are quadric solids. Polyhedra are quite
common in nature. Minerals and crystals often take a polyhedral form. Some
viruses have the shape of a polyhedron. Polyhedra are quadric solids becuase
they can be built from planar half spaces, and planes are quadric surfaces
(pairs of planes). Essentially all objects can be approximated, to an arbitrary
accuracy with polyhedra. More specifically, their surfaces may be
approximated with triangles.
But a polyhedron is locally flat and has no local curvature. So the planar
faces of a polyhedron can not model a local curved shape exactly. An advantage
of using quadric surfaces is that they can locally approximate curvature
4

and shape [O’Neil p.202]. An object can be represented by a relatively small
number of primitive quadric solids. If we regard atoms as tiny balls, then in a
sense, all objects are built from spheres. But Avogadro’s number 6.03 × 10 23
is very large, so modeling with spheres is somewhat impractical. However, in
the field of scientific visualization we do model with rectangular cells defined
by very large numbers of lattice points. When modeling simple manufactured
parts, we are interested in finding exact models. Quadric solids can
serve well in such modeling.
Quadric solids by themselves do not give a full practical set of primitives
for modeling. For example, surfaces of revolution are ubiquitous. But most
surfaces of revolution are not quadric surfaces. Further a large number of
quadrics are frequently required to get an acceptable approximation to a
surface of revolution. The torus is a surface of revolution that is not a
quadric surface. A torus could be approximated by a set of cylinders. But
the differential geometric properties would not be approximated. Parametric
free 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 normal
direction, at a point on a quadric surface, is accomplished by a matrix multiplication.
The intersection points of a straight line with a primitive quadric
solid require only the solution of a quadratic equation. The transformed
equation of a translated or rotated quadric surface is easily found.
With the advent of the computer age, there has been a renaissance of
calculation. Many well known algorithms and calculating methods, while
theoretically powerful, were formerly of little interest. They were two burdensome
for the human computer. In the past techniques were emphasized
that could be carried out easily by hand. But now those calculations, which
were too lengthy for hand calculation, are proper for computer modeling and
computer graphics. Frequently the novice in geometric modeling and graphics
does not use the proper tools. Naturally he uses the calculations taught
him in the schools and colleges, which are often only the small amount of analytic
geometry found in the calculus course, and deals, for example, with the
simultaneous solution of a system of equations in a fixed coordinate system.
A vector approach is superior for many calculations. The vector methods
tend to be independent of the coordinate system. Those with scientific education
are usually familiar with vectors. But they are frequently unaware of
some very useful mathematics. This is the mathematics of projective geom-
5

etry, projective space, and specifically the idea of homogeneous coordinates.
Projective space is very important in certain advanced areas of mathematics.
It is an important aspect of the field of algebraic geometry. But even
among mathematicians, knowledge of projective and algebraic geometry is
not universal.
Often the course in projective geometry, which is offered in American
universities and colleges, is designed for future teachers of high school mathematics.
This course, or a course called modern geometry, is taught to show
the future teacher of Euclidean geometry, that there are many geometries,
and that not all of the geometries are Euclidean. Niether the course on
projective geometry, nor the books designed for it, emphasize the beautiful
simplifications and generalizations of calculation that come from treating
points as belonging to projective space. The few pages discussing projective
space in Birkhoff and MacLane (Survey of Modern Algebra), are more
illuminating than the numerous pages of the elementary books on projective
geometry. This is because the connection to linear algebra and vector spaces
is not exploited in these books.
The purpose of this report is, first to describe a modeling system which
uses quadric solids as primitive objects, and second to present some elementary
ideas of projective geometry which are useful for computing. These
latter ideas, although elementary, are not readily available. The quadric solid
modeling system is implemented in a computer program. An original version
was called Quadric. A later version is called QS90.FTN. It produces
views of 3-dimensional objects, calculates volumes, calculates moments, and
calculates inertia tensors.
The idea of representing objects as primitive solids was quite fashionable
in the 1980’s. Now most solid modeling systems are surface modeling systems.
But primitive solid modeling still has some interest. Quadric solids
are obvious candidates for the primitives. There are several papers dealing
with quadric surfaces [Leven].
The modeling system described here represents objects as Boolean combinations
of quadric half space primitives. An image of the model is made
by scanning its surface and calculating an illumination intensity at each surface
point. The 2-dimensional picture plane is scanned to construct intensity
values. A surface normal is calculated at the point on the surface, where
the line through the picture point pierces the object. The intensity value is
determined from the inner product of the normal at the first pierce point
6

with an illumination vector. Since the primitives are quadric solids, this can
be done easily.
A quadric surface is the set of real zeros of a quadratic form in four
dimensions. Such a set of zeros is called a variety in algebraic geometry. A
quadratic 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 half
space. An upper half space is the set where the form takes nonnegative
values. A quadric form defines two quadric half spaces. Examples of half
spaces are the points inside an ellipsoid, and the points outside two parallel
planes.
It is easy to find the intersection points of a quadric surface with a straight
line. The parameters of the intersection points on the line will be the roots
of a quadratic equation. Surface normals are also easily calculated.
The objects are built using the standard set operators, union, intersection
and complementation. Some modeling schemes, for example the PADL
system, which was developed at the university of Rochester, uses regularized
set operators. These operators guarantee that the resulting objects are
topologically closed, and homogeneously three dimensional. Using standard
operators one may get a set that is not closed. Because we have only an approximate
model of the surface and do not explicitely calculate the boundary,
the use of regularized operations does not seem appropriate. For intersection
points which are candidates to be on the boundary, We construct a small
line segment in the direction of the normal that contains the point. If one
end of the segment is inside, and the other outside, the point is a boundary
point of the object.
2 Overview
We shall start with two-dimensional projective space. This space is easy to
visualize. Projective 2-space is the set of lines in 3-dimensional space which
pass through the origin. Projective 2-space is the set of one dimensional subspaces
of a 3-dimensional vector space. A plane that does not pass through
the origin is given a special role. It is called the affine plane. A coordinate
system is usually selected so that the affine plane is the plane z =1. The
idea of projective space can be illustrated by considering the intersection of
a cone and a plane. The intersection curve is called a conic section. See Fig.
7

1.
When the cone is rotated, the conic section is transformed to a new conic
section. A circle may go into a hyperbola, a parabola, or an ellipse. The
lines on the cone are points of projective 2-space. The intersection of these
lines 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 be
considered the same objects, namely 2-dimensional points. The components
of any vector which lies in the subspace representing a projective point, are
called the homogeneous coordinates of the point. Homogeneous coordinates
are determined only up to a scalar multiple. A line in 3-space which is parallel
to the affine plane meets it at infinity. The projective point in the direction
of this line is called a point at infinity. Tts z coordinate will be zero. The
property of being a point at infinity depends on the particular affine plane
selected.
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 its
original position, then we see that we have a perspective projection from the
circle on one plane to an ellipse on a second plane. And we see that such a
perspective projection is essentially the same as the 3d linear transformation,
namely the original rotation of the cone. Again suppose a figure in the
plane is translated in the plane with the lines joining the points of the figure
following the translation. This is clearly a linear transformation of the lines
or more precisely of the vectors in the direction of the lines. The plane figure
would be translated the same if we were to translate the origin or cone vertex
with the translation of the plane figure. So we see that the shifting of the
point of perspective projection is equivalent to a linear transformation of
the 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 a
second perspective point. Then we see from what we have said before that
this is equivalent to a set of linear transformations of the fixed 3-d space,
namely rotations, translations and possibly a uniform scaling. So the study
of these sequences of perspective projections, which were the original subject
of projective geometry may be studied by considering linear transformations
of one dimensional subspaces of a 3 dimensional vector space, or of a n
dimension vector space in general. One can illustrate this equivalence easily
in the one dimensional case by drawing sequences of perspective projections
of points on lines, and then showing how the planes can be lined up parallel
8

Figure 1: Points of projective 2-space are lines in 3-space. A 3-d linear
transformation is a 2-d projective transformation. A rotation of the cone can
project the circle to an ellipse, a parabola, or a hyperbola. In the same way
the lines through any figure in the plane are a projective space representation
of the figure. Any sequence of perspective projections of a figure onto a set of
planes can be represented as a sequence of linear transformations in 3-space.
The plane shown in the figure is known as the affine plane.
9

y rotations and translations so that there is a single perspective projection
point, projecting points onto a stack of parallel planes, which by scaling can
be made to coincide.
Now consider a 2-dimensional line. We may view it as a line lying in the
affine plane. The points of this line in projective space, will be the locus of
lines which pass through it and the origin. It is a plane in 3-space. Because
any two planes with a point in common must coincide or meet in a line, it
follows that any 2-dimensional lines in projctive space meet in a point. Two
parallel lines meet at a point at infinity. Notice that 2-dimensional projective
space contains more points than affine space, that is it contains the points
at infinity. In two dimensional space an affine transformation is the sum
of a linear transformation and a translation. If the linear transformation is
orthogonal (i.e. a rotation) then the affine transformation is a rigid motion
or a congruence. Now any nonsingular linear transformation maps the set of
one dimensional subspaces onto itself. So we may consider such a mapping as
taking 2-dimensional projective points to projective points. Such a mapping
is thus called a projective transformation. These transformations distinguish
euclidean, affine, and projective geometry . This is the thesis of Felix Klein.
This is the essence of his famous Erlangen program. He defined a geometry
to be the study of the invariants of a group of transformations. Thus the
concept of angle is a property of euclidean space because angle is invariant
under a rigid motion. Parallelism belongs to affine geometry because it is
preserved under an affine transformation. Being a conic section is a property
of projective geometry, because it is preserved under a projective transformation.
Being an ellipse, however, is not a projective property. There is a
projective transformation that takes an ellipse to a different type of conic
section.
We shall now discuss one of the major reasons for using homogeneous
coordinates for calculations. An affine transformation is a special case of a
projective transformation. Since a projective transformation is in essence a
linear transformation in a higher dimensional space, it can be represented
by a matrix, and the transformation of a point may be accomplished with
matrix multiplication. An affine transformation is represented as a matrix
multiplication. There many advantages to using homogeneous coordinates.
Quadrics are represented homogeneously as quadratic forms, and can be
represented as symmetric matrices. Perspective views are projections from
projective 3-space to projective 2-space. A three dimensional vanishing point,
10

that is a point at infinity, may be projected to a finite two dimensional
point, i.e. the vanishing point of converging railroad tracks. The intersection
point of two lines can be calculated without considering slopes, or without
considering parallelism. That is, the intersection algorithm does not need
to consider exceptional cases, such as the case of an infinite slope. The
intersection of two parallel lines, which gives a point at infinity, does not
require a special case.
The idea of the polar of a point with respect to a quadric plays a large role
in the theory of quadrics. A quadratic form is represented by a symmetric
matrix. 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 real
bilinear functional is a mapping from the cartesian product of vector spaces
to the real numbers, which is linear in each variable. For a fixed point or pole
p, the polar is the set of points such that the bilinear form corresponding to
the quadric vanishes on any pair of points consisting of p and a point from
the set. In two dimensions the polar is a line, in three dimensions it is a
plane. 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 the
quadric.
The cross ratio is an important invariant in projective geometry. It is
ratio of distances between four collinear points. The cross ratio is used to
prove several properties of projective space.
We shall require the intersection of a line and a quadric. A line is defined
by any two distinct points, including the case of points at infinity. The intersection
points are determined by the roots of a quadratic equation. There
are several cases: (1) There may be two intersection points, (2) There may
be 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 the
surface of the, quadric. For example, bilinear interpolation on a rectangle
produces a ruled quadric surface.
Conic arcs have been used in various applications of geometry. For example,
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 transformation
exists that takes four points to four points. Thus if we desire a given
conic arc defined by four points, we construct the transformation that takes
a given parabola, which is situated so that its points are a function of one
parameter, to the desired arc. Then We have a parametric representation of
11

the arc. One advantage is that sines and cosines do not have to be calculated
for the points of a circular arc.
Surface normals can be considered points at infinity. A vector field is
a mapping that assigns a vector to a given point. The surface normal is a
vector field defined on the surface. But this field is really only a direction
field and a direction is essentially a point at infinity. Thus we may consider
a 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-dimensional
vector space. The affine plane becomes a 3-dimensional hyperplane. Since
we can not climb into 4-dimensional space,we can not visualize projective
space from outside as we did in the 2-dimensional case.
Objects are constructed from quadric half-space primitives. There are
only two bounded primitives, namely the sphere and the ellipsoid. We create
bounded objects by suitably combining the primitives. To each set corresponds
a characteristic function which takes a value of true or false (or 1 and
0) on each point of space. Its value at a point x is true if and only if x is in
the set. The mapping which takes a set to its characteristic function, union
to ”and”, intersection to ”or”, and complementation to ”not”, is a Boolean
algebra isomorphism from sets to the algebra of propositional functions. The
latter algebra is represented in computer programming languages. Thus in
our program objects are represented as logical expressions. Actually algebraic
expressions are binary trees [Knuth], so we may consider our object a
binary tree with primitives at the terminal nodes and operators at the other
nodes.
We determine the characteristic function of a primitive from the corresponding
quadratic form. Then the characteristic function of the object
comes directly from the definition of the object as a logical expression. Using
topology, we may show that a point is on the surface of the object only if
it is on the surface of some primitive, and the normal to the surface at a point
on the surface is equal to the the normal of some primitive. If the surface
point 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 later
case the point lies on the intersection curve and hence is on the boundary
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 is
not important how the normal is assigned since the intersection of the prim-
12

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, it
may be difficult in practice to determine the coefficients from their ordinary
characterization. The primitives are most easily specified by entities such as
radius, center, major axis, and orientation. We know the matrices of certain
well situated primitives. For example, we know the matrix of the sphere
of radius one that has its center at the origin. We may transform a simple
primitive to obtain the matrix of a more complex primitive. Transformations
employed are rotation, scaling, and translation. For example, by scaling, the
unit sphere can be made an ellipse. It then can be rotated and translated.
the new matrix of the transformed quadric can be obtained by multiplying
the original matrix on the left and the right by matrices related to the
transformations.
The motivation for constructing the modeling system was the desire for
a system that would compute the volume, center of gravity and the inertia
tensor. Since the characteristic function is available, a Monte Carlo integration
method is suggested. This technique consists in assigning a probability
measure to a bounded space containing the object, and then obtaining the
value of the integral as the expectation of some random variable. For example,
the volume integral is defined as the expectation of the characteristic
function. The expectation is estimated from a random sample of the random
variable. In the case of the volume, the variance of the characteristic
function is quite high. Thus a large number of samples is required so that
the sample variance, which is a measure of the error of the estimated integral,
is small. Stratified sampling will often decrease the sampling variance.
In stratified sampling one breaks up the domain into smaller sub domains
so that the variance is smaller in each subdomain. For example, if a small
block is chosen completely inside the object then the variance of the characteristic
function is zero. Likewise the variance of the characteristic function
completely outside the function is zero. Only those blocks which contain
boundary points contribute to the variance. Stratified sampling should decrease
the overall variance and so fewer sample points should be required for
the error to be within a given tolerance. When only one point is sampled
from each subdomain we have essentially a Riemann sum for the integral.
in general, the amount of work required to evaluate an integral numerically
is proportional to the number of function evaluations required in one
13

dimension raised to a power equal to the dimension of the space. This is
true at least when the multiple integral is computed by iterated integration.
Hence if 100 function evaluations are required in one dimension, 1,000,000
evaluations are required in three dimensions, so it pays to reduce the dimension.
By applying the divergence theorem the integral over a volume can usually
be reduced to the integral over a surface; so we need a technique for
calculating surface integrals. To integrate on a surface we must have some
model of the surface. In this modeling system we do not have such an explicit
model. We have a multiple surface patch defined by projection. When the
object is projected to two dimensions, to every point in the two dimensional
region 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 space
together with a set of coordinate systems or surface patches. For a two
dimensional manifold, a surface patch is a 1-1 mapping from the manifold to
an open set in 2-dimensional euclidian space. A complete set of patches that
covers the manifold is called an atlas. The projections from several projection
points of an object will be a set of patches that completely cover the surface
of the object. However, when we include all pierce points, these mappings are
not 1-1, so are not legitimate coordinate systems. Still we can do integration
on these multilayer patches, because integration is loosely an adding up of
all infinitesimal surface elements, and we may add the elements in any order.
We can add the elements corresponding to pierce points of a given projection
line all at once. There is a difficulty, the set of patches that define a manifold
are 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 a
partition of unity. A partition of unity is a set of weighting functions applied
to each patch so that the sum of the weighted patch integrals is the whole
surface integral [Spivak]. When our surface patches derive from projections
in the x, y, andz directions a natural partition of unity is the set of squares
of the components of the surface normal vectors. The individual layers of
each view must be considered as separate patches. So we have reduced our
task to the calculation of a two dimensional integration over a rectangular
domain. the integrand will in general be discontinuous and we must choose a
suitable integration algorithm. Gaussian quadrature is not suitable because
the integrand will be discontinuous. The generalized trapezoid method will
14

work, but may require more calculation and give less accuracy than necessary.
Because the moment of inertia about any axis can be obtained from the
inertia tensor, it is economical to compute the full inertia tensor rather than
just the moments of inertia. There are two different tensors that are called
the inertia tensor. One definition is given in mathematics books, and another
in some physics books. In order to clarify what is being calculated, I have
included material concerning angular momentum, angular velocity, moment
of inertia, and their relation to the inertia tensor. In the next section I will
begin a more detailed treatment of the material.
3 Projective Space and the Conic Sections
Projective geometry is related to the theory of perspective. Perspective was
developed by the Italian painters during the renaissance. To explain perspective,
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 at
a point. In some sense, the ray represents the point. Let a picture plane be
interposed between the eye and the object. Then the point where the ray
intersects the picture plane is the projection of the object point to its picture
image. Through this process, three dimensional space is projected to two
dimensional space. The concern of the artist is to accurately represent the
object as it appears to the eye. The artist is concerned with the appearance
of parallel lines and vanishing points. Vanishing points are the images of
points and lines that are infinitely far away. Such points are called points
at infinity. Spatial points at infinity will be discussed in the section on perspective.
Here we confine ourselves to the two dimensional case. the artist
Albrect Durer, who dabbled a bit in geometry (see Panofsky) ,constructed a
device for recording the perspective image of an object in the picture plane
by recording the position on the plane of a chord stretched from a fixed point
to a point on the object. This device appears in one of Durer’s works and
may be familiar to the reader. The idea of projective geometry is to consider,
the chord, or more accurately the mathematical ray, as representing a two
dimensional point.
projective space is formally defined as the set of 1-dimensional subspaces
of a vector space. it can be illustrated by considering those plane curves
which are called conic sections. Figure 1 shows a cone with its axis in the z
15

direction, and with a 90 degree vertex angle.
This cone meets the plane z = 1 in a circle. A point on the surface of the
cone satisfies the equation
x 2 + y 2 − z 2 =0
Thecircleintheplanehastheequation
x 2 + y 2 − 1=0
The first equation can be written as
⎡
[ ]
x y z
⎢
⎣
1 0 0
0 1 0
0 0 −1
⎤ ⎡
⎥ ⎢
⎦ ⎣
x
y
z
⎤
⎥
⎦ =0
P T AP =0
and
where
A =
P =
⎡
⎢
⎣
⎡
⎢
⎣
x
y
x
⎤
⎥
⎦
1 0 0
0 1 0
0 0 −1
⎤
⎥
⎦
The function Q(P )=P t AP 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.
Then L maps the set S(Q) to
L(S(Q)) = {P : Q(L −1 P )=0}
= {P :(B −1 P ) T A(B −1 P )=0}
16

= {P : P T (B −1 ) T A(B −1 P )=0}
= {P :(B −1 P ) T A(B −1 P )=0}
= {P : L(Q)P =0}.
where L(Q)(P ) = Q(L −1 (p)). L(Q) is a quadratic form with symmetric
matrix
(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 0
0 c s
0 −s c
⎤
⎥
⎦
and B −1 = B T =
⎡
⎢
⎣
1 0 0
0 c −s
0 s c
⎤
⎥
⎦
Then L(Q) has matrix
⎡
⎢
⎣
1 0 0
0 c
0 −s c
=
⎡
⎢
⎣
⎤ ⎡
⎥ ⎢
⎦ ⎣
1 0 0
0 c
0 −s c
=
1 0 0
0 1 0
0 0 −1
⎤ ⎡
⎥ ⎢
⎦ ⎣
1 0 0
0 c −s
0 s c
⎤ ⎡
1 0 0
⎥ ⎢
⎦ ⎣ 0 c −s
0 −s −c
⎤
1 0 0
0 c 2 − s 2 ⎥
−2cs ⎦
0 −2cs s 2 − c 2
When the angle a is 45 degrees, the matrix of L(Q) is
⎡
⎢
⎣
⎡
⎢
⎣
1 0 0
0 0 −1
0 −1 0
17
⎤
⎥
⎦
⎤
⎥
⎦
⎤
⎥
⎦

The equation of the rotated cone is
⎡
[ ]
x y z
⎢
⎣
1 0 0
0 0 −1
0 −1 0
⎤ ⎡
⎥ ⎢
⎦ ⎣
x
y
z
⎤
⎡
⎥
⎦ = [ x y z ] ⎢
⎣
= x 2 − 2yz =0.
The equation of its intersection with the plane z =1is
or
x 2 − 2y =0.
x
−z
−y
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 0
0 −1 0
0 0 1
⎤ ⎡
⎥ ⎢
⎦ ⎣
= x 2 − y 2 + z 2 =0.
The equation of its intersection with the plane z =1is
y 2 − x 2 =1.
This is the equation of a hyperbola. the plane that cuts the cone is called
the affine plane. for each point in the plane there is a corresponding line
which passes through the point and through the origin. The set of lines
that pass through the origin is called projective space. Each line is considered
a point of the space. Those points in projective space ( i.e. lines in
3-dimensional Euclidean space) that are parallel to the affine plane are called
points at infinity, because they meet the plane only at infinity. There are
more projective points than affine points. The coordinates of any vector in
the direction of a line which passes through the origin are the homogeneous
coordinates of the projective point represented by the line. Because the coordinates
of any scalar multiple of the vector are also homogeneous coordinates
18
x
y
z
⎤
⎥
⎦
⎤
⎥
⎦ .

of the projective point, homogeneous coordinates are determined only up to
a scalar multiple. So (1, 2, 3) and (2, 4, 6) are homogeneous coordinates of
the same point. A point at infinity has coordinates of the form (x, y, 0). If a
point 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 distinct
lines meet in exactly one point. By a line we mean the locus of points that
satisfy 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
and
ax + by + c =0.
ax + by + d =0.
meet at the point, which has coordinates (−b, a, 0). Affine rotations and
translations are represented as linear transformations in projective space. A
rotation has matrix
R =
⎡
⎢
⎣
cos(a) sin(a) 0
− sin(a) cos(a) 0
0 0 1
⎤
⎥
⎦
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 ] .
19

The following calculation represents an affine translation
⎡
⎤ ⎡ ⎤
1 0 a x x + a
⎢
⎥ ⎢ ⎥
Tp = ⎣ 0 1 b y ⎦ = ⎣ y + b ⎦ .
0 0 1 1 1
The product of the two matrices above then both rotates and translates.
This shows how affine transformations can be calculated using homogeneous
coordinates.
A projective point is a set of all scalar multiples of a vector. It is a one
dimensional subspace of a vector space. In general, projective n-space is
the set of 1-dimensional subspaces contained in a vector space of dimension
n + 1. The nonsingular linear transformations on the vector space map
the set of 1-dimensional subspaces onto themselves. These linear mappings
are called projective transformations when they are applied to projective
space. Considering the previous example, it is easy to see that an affine
transformation is a special case of a projective transformation.
We now introduce quadratic forms. Let A be a symmetric matrix and let
p be a coordinate vector.
p =
⎡
⎢
⎣
x
y
z
⎤
⎥
⎦ .
Then P T AP is a quadratic form. For example, if
⎡ ⎤
1 2 3
⎢ ⎥
A = ⎣ 2 4 5 ⎦ .
3 5 6
Then
P T AP = x 2 +4y 2 +6z 2 +4xy +6xz +10yz.
All quadratic forms can be obtained in this way. Corresponding to each
quadratic form is an associated bilinear form or tensor of rank two. It is a
function of two vector variables and is linear in each. It is given as
B(P, Q) =P T AQ.
20

A conic section S is the set of zeroes of a quadratic form in 2-dimensional
space. Thus
S = {vP : P T AP =0}.
Given a point Q, thepolarofQ with respect to a quadric represented by a
matrix 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 dimensional
case, and a plane in the three dimensional case. Directly from the
definition it is seen that P is on the polar of Q if and only if Q is on the
polar of P . Properties of the polar can be obtained by using the concept of
the cross ratio of four collinear points.
4 The Cross Ratio
Given four collinear points in affine space, the cross ratio is defined in terms
of the directed distances between the points. Affine space is embedded in
projective space and the cross ratio can be extended to apply to the points
of this space. It can be defined without the non-projective concept of distance.
The importance of cross ratio is that it is invariant under projective
transformations. Before we define the cross ratio a few remarks on notation
are in order. a point may refer to several distinct concepts. a point may
be 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 the
corresponding affine point which is the intersection of the subspace with the
affine hyperplane. To complicate things further, we must distinguish between
a vector and its corresponding coordinate vector. Its coordinates depend on
the chosen basis of the vector space. To distinguish all of these things would
result in too many symbols. So we will use the same symbols for all of the
above ideas and hope that context will give the desired meaning. For example,
if we add two points, they are vectors, because addition is not defined for
projective points. Division of vectors has no meaning in general, but if one
vector is a multiple of another, then their quotient has an obvious meaning.
We shall use this meaning below.
21

Let A, B, C, and D be four collinear points. In affine space we define the
cross ratio as
d(C, A)/d(C, B)
(AB, CD) =
(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 be
vectors representing the points. Then numbers a and b can be found so that
aC and bD lie on the straight line connecting A and B. Then there exists a
t such that
aC =(1− t)A + tB.
and
Then
It follows that if
then
then
Similarly if
So the cross ratio is
aC − B =(1− t)A − (1 − t)B.
d(C, A)/d(C, B) =(A − aC)/(B − aC) =−t/(1 − t).
(AB, CD) =
C = c 1 A + c 2 B,
d(C, A)/d(C, B) =−c 2 /c 1
D = d 1 A + d 1 B,
d(D, A)/d(D, B) =−d 2 /d 1 .
d(C, A)/d(C, B)
(d(D, A)/d(D, B) =(c 2/c 1 )/(d 2 /d 1 )=(c 2 d 1 )/(c 1 d 2 ).
This ratio is independent of which representative vectors are chosen for the
points. For example, if A is replaced by a new vector that is a multiple of
A, thenc 1 and d 1 are multiplied by the same scalar. So the cross ratio is not
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 ratio
of points (AB, CD), of the projective points A, B, C, D, is defined in the
22

following way. If A, B, C, D are any vector representations of the projective
points, where
C = c 1 A + c 2 B,
and
D = d 1 A + d 1 B,
then the cross ratio (AB, CD) is defined as
(AB, CD) =(c 2 /c 1 )/(d 2 /d 1 )=(c 2 d 1 )/(c 1 d 2 ).
The preservation of the cross ratio by a projective transformation follows
because a projective transformation is a linear transformation of the vector
space 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 collinear
points. Suppose
C = c 1 A + c 2 B
and
D = d 1 A + d 2 B.
Then if L is the linear transformation corresponding to the given projective
transformation we have
L(C) =c 1 L(A)+c 2 L(B)
and
L(D) =d 1 L(A)+d 2 L(B).
So the points L(A),L(B),L(C), and L(D) have the same cross ratio. A set
of collinear points is called a range of points. A range of four points is called
a harmonic range when its cross ratio equals -1. The name harmonic range
can be explained in the following way. Suppose the distance from A to C is
1/3, 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
distance from A to D is 1,the distance from A to B is 1/2, and the distance
from A to C is 1/3. Vibrating strings of these lengths produce the first,
second, and third harmonic tones.
ApointQ is called the harmonic conjugate of P with respect to A and
B when (AB, P Q) =−1.
23

Proposition If a line through P meets a quadric at A and B, then the
harmonic conjugate Q lies on the polar of P .
Proof Suppose P = pA + pB and Q = qA + qB, thenpq = −pq, so
and
Subtracting, we find
Similarly
qP = qpA + qpB
pQ = pqA + pqB
A =(1/2)(P/p + Q/q).
B =(1/2)(P/p + Q/q).
Let f be the bilinear form defied by the quadric. Since A is on the quadric,
we have
Similarly
0=f(A, A) =(q/p)f(P, P)+2(q/p)f(P, Q)+f(Q, Q).
0=(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 P
Corollary. If a line through P is tangent to a quadric at Q, thenQ is on
the polar of P .
Informal Proof. Rotate the line through P slightly, if necessary, so that
it intersects the quadric in two points that are nearly equal. The harmonic
conjugate 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 point
approaches Q.
Construction. To construct a tangent to a conic from a point P , find the
intersection of the conic with the polar of P.
Corollary. The locus of the harmonic conjugates of p with respect to the
intersection 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 that
Q and R are on the polar of P Thus the line defined by Q and R is the polar
24

P
Q
S
P’
R
Figure 2: Let lines PQ and PR be tangent to the ellipse. Tangent points Q
and R are on the polar of P . Thus the line defined by Q and R is the polar
of 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

of 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 affine
hyperplane has been identified. We may represent a line through P and Q by
P + tQ where the parameter t varies from minus to plus infinity. t equal to
infinity corresponds to the point Q. Wemaythinkofthisasthelinethrough
P in the direction Q. When a line meets a quadric in two finite intersection
points R and S, the line segment from R to S is called the chord determined
by the line. A diameter in the direction Q is the polar of Q where Q is a
point at infinity.
Consider the intersection of the line P + tQ with a quadric. There are
several cases. Let the bilinear form be f. Then
There are four cases:
Case 1.
0=f(P + tQ, P + tQ)
= f(P, P)+2tf(P, Q)+t 2 f(Q, Q)
P (Q, Q) =0,f(P, Q) ≠0.
Q is one intersection point. The parameter of the other is
Case 2.
t = −f(P, P)/(2f(P, Q)).
f(Q, Q) =0,f(P, Q) =0,f(P, P) ≠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.
Case 4.
p(Q, Q) ≠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 direction
Q.
26

Proof. Since Q is a point at infinity, and the line determines a chord with
two 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 formula
for the two roots, substituting in P + tQ, adding and dividing by two, we
find
(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 diameter
determined by Q.
Example. Consider the parabola x − 2yz =0. Let
p =
⎡
⎢
⎣
x
y
z
⎤
⎥
⎦ .
Then P T AP is a quadratic form. For example, if
⎡ ⎤
1 2 3
⎢ ⎥
A = ⎣ 2 4 5 ⎦ .
3 5 6
and
Thematrixofthisconicis
A =
P =
Q =
⎡
⎢
⎣
⎡
⎢
⎣
⎡
⎢
⎣
0
0
1
0
1
0
⎤
⎥
⎦ .
⎤
⎥
⎦ .
1 0 0
0 0 −1
0 −1 0
⎤
⎥
⎦ .
27

Q
P
Figure 3: The parabola x − 2yz = 0 and a point at infinity Q. The line
through P and Q meets the parabola at P and Q. The polar of a point at
infinity is a diameter. The polar of Q is the line at infinity.
28

Q does not determine a chord because the line does not meet the conic in
two finite intersection points. In fact Q, which is a point at infinity, lies on
the 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 =
⎡
⎢
⎣
See Fig. 4. Then the polar of Q has equation x − z =0. Thus the affine
equation of the diameter corresponding to Q is x =1.
1
1
0
⎤
⎥
⎦ .
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 points
coincide and they equal P, because P is on the quadric f(P, P) =0. But the
intersection equation is then
and its roots must be zero. So
2tf(P, Q)+tf(Q, Q) =0,
f(P, Q) =0.
Now suppose f(P, Q) =0. It is given that f(P, P) =0. The intersection
equation is
tf(Q, Q) =0.
29

Q
P
Figure 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 finite points. The polar of
Q is the line x = 1, which is a diameter of the parabola. The point at in
finity in the diagram is represented as an arrow in the direction of the infinity
point Q.
30

If f(Q, Q) ≠ 0, then the roots of the equation are zero. So P is a double
intersection point. If f(Q, Q) = 0, then the line is a ruling. Hence in both
cases 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 ,wehavef(P, P) =0soP is on the
polar of P .Fromabovef(P, Q) =0soQ is on the polar of P. Therefore the
line through P and Q is the polar of P.
We have taken the tangent line as the line that intersects the conic at
two coinciding points. With this definition the tangent line does not have
points on both sides of the conic curve. For consider the set of points where
f(P, P) < 0, and the set where f(P, P) > 0. These sets are bounded by
the conic curve. We may consider one of them the inside, and the other, the
outside 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 through
P and Q has positive discriminant. So the line meets the conic in two real
points that do not coincide. Hence the line through P and Q is not a tangent
line.
If the point Q is on the polar of P ,then
Suppose
f(P, Q) =P T AQ =0.
P T A =(a, b, c),
Q =
⎡
⎢
⎣
Then the first vector is called the line coordinate vector. The equation of
the polar is
x
y
z
⎤
⎥
⎦ .
ax + by + cz =0.
The 2-dimensional vector
[
a
b
]
,
31

and
is normal to the line. For if
⎡
⎢
⎣
⎡
⎢
⎣
x 1
y 1
1
x 2
y 2
1
⎤
⎥
⎦ ,
⎤
⎥
⎦ ,
1 are on the line, then we have by subtraction
(x − x)a +(y − y)b =0.
Example. Consider the hyperbola
x 2 − y 2 =1.
Let
and
Q =
P =
⎡
⎢
⎣
⎡
⎢
⎣
1
1
0
0
0
1
⎤
⎥
⎦ ,
⎤
⎥
⎦ .
Then
f(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).
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 all
points P such that
32

AP =
⎡
⎢
⎣
0
0
1
⎤
⎥
⎦ .
If A is nonsingular, than the conic has only one center.
Example. Consider the conic
(x − 1) 2 + y 2 =1.
Itsmatrixis
A =
⎡
⎢
⎣
1 0 −1
0 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
where A is a symmetric matrix
E = {p : p ∗ Ap =0},
⎡
⎤
a 11 a 12 a 13
⎢
⎥
A = ⎣ a 21 a 22 a 23 ⎦ .
a 31 a 32 a 33
The computations that will be described here, are realized in program
pltconic.ftn.
We shall put E into canonical form by mapping the locus with a rotation
through an angle θ, followed by a translation by (t 1 ,t 2 ). The combined
transformation has matrix
T =
⎡
⎢
⎣
c −s t 1
s c t 2
0 0 1
33
⎤
⎥
⎦

The inverse of T is
R =
⎡
⎢
⎣
c s e
−s c f
0 0 1
⎤
⎥
⎦
where
e = −(t 1 c + t 2 s)
f = t 1 s − t 2 c
The mapping of the conic locus is
T (E) ={Tp : p ∗ Ap =0}
= {q :(Rq) ∗ ARq =0}
= {q : q ∗ R ∗ ARq =0}
where B is the symmetric matrix
= {q : q ∗ Bq =0},
R ∗ AR.
Let the determinant of A be d 3 . Let the determinant of the upper 2 by 2
submatrix be d 2 . The determinants of R and R ∗ are each equal to 1, so the
determinant of B is also d 3 . Also for the same reason the transformation
preserves d 2 . We will show that if d 3 = 0, then the conic is degenerate, and
the conic locus is a line, a pair of parallel lines, a pair of intersecting lines,
apoint,orisempty. Ifd 3 is not zero, then if d 2 < 0 it is a hyperbola, if
d 2 = 0 it is a parabola, and if d 2 > 0andd 3 < 0, it is an ellipse. if d 2 > 0
and d 3 > 0 it is the empty set.
The coefficients of B are
b 11 =(ca 11 − sa 12 )c − (ca 12 − sa 22 )s
34

12 =(ca 11 − sa 12 )s +(ca 12 − sa 22 )c
b 13 =(ca 11 − sa 12 )e +(ca 12 − sa 22 )f + ca 13 − sa 23
b 22 =(sa 11 + ca 12 )s +(sa 12 + ca 22 )c
b 23 =(sa 11 + ca 12 )e +(sa 12 + ca 22 )f + sa 13 + ca 23
b 33 =(ea 11 + fa 12 + a 13 )e +(ea 12 + fa 22 + a 23 )f + ea 13 + fa 23 + a 33
Notice that
= a 11 e 2 +2a 12 ef + a 22 f 2 +2a 13 e +2a 23 f + a 33 .
b 33 =(e, f, 1)A(e, f, 1) ∗ .
We shall choose θ so that b 12 is zero. We have
b 12 =(c 2 − s 2 )a 12 + sc(a 11 − a 22 )=0.
That is
0=a 12 cos(2θ)+(1/2)(a 11 − a 22 )sin(2θ).
So the rotation angle is determined by
tan(2θ) = 2a 12
a 22 − a 11
.
Thus
Let us write
2θ = atan2(2a 12 ,a 22 − a 11 ).
b 13 = c 11 e + c 12 f − g 1
where
b 23 = c 21 e + c 22 f − g 2 ,
35

c 11 =(ca 11 − sa 12 )
c 12 =(ca 12 − sa 22 )
c 21 =(sa 11 + ca 12 )
c 22 =(sa 12 + ca 22 )
g 1 = −(ca 13 − sa 23 )
g 2 = −(sa 13 + ca 23 )
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
c 11 = a 11
c 12 = a 12
c 21 = a 12
c 22 = a 22 )
g 1 = −a 13
g 2 = −a 23 .
After such a pure rotation, it is easy to see the conic type from the
properties of the coefficients, where the xy cross term is absent.
But here let us return to the single step transformation. We can make
b 13 =0andb 23 = 0, if we can find e and f so that
c 11 e + c 12 f = g 1
c 21 e + c 22 f = g 2 .
We have a linear equation to solve for e and f.
The determinant of this system is
36

d 2 = a 11 a 22 − a 2 12 .
Because the upper 2 by 2 subdeterminant of R and R T is 1, d 2 also equals
b 11 b 22 − b 2 12 = b 11b 22 .
If d 2 = 0, then the quadric is neither an ellipse nor a hyperbola.
If d 2 is not zero, then using the values of e and f that make b 13 and b 23
zero, we
find that the matrix of the transformed canonical form is
⎡
⎤
b 11 0 0
⎢
⎥
B = ⎣ 0 b 22 0 ⎦ .
0 0 b 33
If d 2 = 0, then we can not necessarily choose the translation (e, f) sothat
b 13 =0andb 23 = 0. In these cases, either b 11 or b 22 is zero.
Suppose one of b 11 and b 22 is zero.
In this case we have three equations
b 13 = c 11 e + c 12 f − g 1
b 23 = c 21 e + c 22 f − g 2
b 33 =(e, f, 1)A(e, f, 1) ∗ .
A related set of equations obtained by setting the b coefficients to zero is
0=c 11 e + c 12 f − g 1
0=c 21 e + c 22 f − g 2
0=(e, f, 1)A(e, f, 1) ∗ .
We claim that if b 11 is not zero, then the 1st and 3rd equation of the
set have a solution. If b 22 is not zero, then the 2nd and 3rd equations of
the set have a solution. This follows by examining the component transformations,
by first applying the rotation, then the translation. Finding the
37

translation is equivalent to completing squares. For example, it is equivalent
to transforming the terms like
into terms like
k 1 x 2 + k 2 x
k 1 (x + m 1 ) 2 + m 2 .
Solving the first or second equation, simultaneously with the third equation
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 (c 11 ,c 12 , −g 1 ), with
the conic A.
Suppose b 11 is not zero, and b 22 =0.
Solving the first and third equation for e and f gives us the canonical
matrix
⎡
⎢
B = ⎣
b 11 0 0
0 0 b 23
0 b 23 0
Suppose b 22 is not zero, and b 11 = 0. Solving the second and third
equation for e and f gives us the canonical matrix
B =
⎡
⎢
⎣
0 0 b 13
0 b 22 0
b 13 0 0
⎤
⎥
⎦ .
⎤
⎥
⎦ .
So we have shown that after our transformation, the conics have four
standard forms. They are:
Form 1:
⎡
Form 2:
B =
⎢
⎣
⎤
b 11 0 0
⎥
0 b 22 0 ⎦ .
0 0 b 33
38

⎡
⎢
B = ⎣
b 11 0 0
0 0 b 23
0 b 23 0
⎤
⎥
⎦ .
Form 3:
Form 4:
B =
⎡
⎢
⎣
0 0 b 13
0 b 22 0
b 13 0 0
⎤
⎥
⎦ .
⎡
⎤
0 0 b 13
⎢
⎥
B = ⎣ 0 0 b 23 ⎦ .
b 13 b 23 b 33
Form 4 is a degenerate case that is just a linear equation.
Let us now consider the form 1 case.
Ellipse.
All coefficients b 11 ,b 22 and b 33 , are not zero. The coefficients b 11 and b 22
have the same sign, which differs from the sign of b 22 . Then the canonical
equation is
and
x 2
a + y2
2 b =1, 2
where the radii of the ellipse are
a =
b =
√
−b 33 /b 11
√
−b 33 /b 22 .
In this case d 2 > 0andd 3 < 0.
Hyperbola.
All coefficients b 11 ,b 22 and b 33 , are not zero. The coefficients b 11 and b 22
have different signs. There are two subcases: b 11 b 33 < 0andb 11 b 33 > 0. If
b 11 b 33 < 0, then the canonical equation is
39

and
where
x 2
a 2 − y2
b 2 =1,
a =
b =
√
−b 33 /b 11
√
b 33 /b 22 .
If we define the axis of the hyperbola to be the line of symmetry that meets
the hyperbola at two points, then the axis of the original conic is at angle
−θ, whereθ is the rotational angle of the canonical transformation.
If b 11 b 33 > 0, then the canonical equation is
and
where
y 2
a 2 − x2
b 2 =1,
a =
b =
√
b 33 /b 11
√
−b 33 /b 22 .
Then the axis of the original conic is at angle −(θ + π/2), where θ is the
rotational angle of the canonical transformation.
In this case d 2 < 0andd 3 < 0.
Intersecting lines.
If b 11 and b 22 have different signs, and b 33 = 0, then the conic has equation
|b 11 |x 2 = |b 22 |y 2 ,
which is equivalent to two equations of a line through the origin
y = α β x,
and
y = − α β x,
where
40

In this case d 2 < 0, and d 3 =0.
Parallel lines or no real solution.
If b 11 b 22 = 0 then the equation is
√ √
α = |b 11 |,β = |b 22 |
or
b 11 x 2 = −b 33
b 22 y 2 = −b 33 .
This is a pair of parallel lines, a single line , or there is no solution,
depending upon signs. In this case d 2 =0andd 3 =0.
Point.
If b 11 and b 22 have the same signs, and b 33 = 0, then the conic has equation
whose only solution is
|b 11 |x 2 = −|b 22 |y 2 ,
x =0,y =0.
In this case d 2 > 0, and d 3 =0.
No real points.
If b 11 , b 22 ,andb 33 all have the same signs, then the conic has equation
|b 11 |x 2 + |b 22 |y 2 = −|b 33 |,
which has no real solution.
In this case d 2 > 0andd 3 > 0ord 3 < 0.
Let us now consider the form 2 case.
Parabola.
⎡
⎢
B = ⎣
b 11 0 0
0 0 b 23
0 b 23 0
⎤
⎥
⎦ .
41

If b 23 is not zero, then form 2 represents a parabola. The equation is
b 23 y = −b 11 x 2 ,
4Fy = x 2 .
The focal distance is
F = − b 23
.
4b 11
In this case d 2 =0,andd 3 < 0.
Line.
If b 23 is zero, form 2 represents the line
In this case d 2 =0,andd 3 =0.
Let us now consider the form 3 case.
x =0.
⎡
⎤
0 0 b 13
⎢
⎥
B = ⎣ 0 b 22 0 ⎦ .
b 13 0 b 33
If b 13 is not zero, form 3 represents a parabola.
The equation is
b 13 x = −b 22 y 2 − b 33
which if b 13 is not zero is a parabola.
4Fx = y 2 + c
F = − b 13
4b 22
.
In this case d 2 =0,andd 3 < 0.
Line.
If b 13 is zero, form 3 represents the line
In this case d 2 =0,andd 3 =0.
Let us now consider the form 4 case.
y =0.
42

Line.
We have the straight line
2b 13 x +2b23y + b 33 =0.
If an ellipse has a center C =(c x ,c y ), then the original transformation T
maps it to the origin, which is given in homogeneous coordinates as
(0, 0, 1) ∗
Hence R, whichistheinverseofT , maps the origin to the original ellipse
center.
Therefore ⎡ ⎤ ⎡
⎤ ⎡ ⎤ ⎡ ⎤
⎢
⎣
c x
c y
1
⎥
⎦ =
⎢
⎣
c s e
−s c f
0 0 1
⎥ ⎢
⎦ ⎣
0
0
1
⎥
⎦ =
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 at
infinity.
Alternately, for the case of the ellipse or hyperbola, the first two rows of
matrix A are homogeneous coordinates of diameter lines (being the polars of
points at infinity), so their intersection is the center.
If the matrix A is nonsingular, then the unique center (c x ,c y )isthe
solution of
⎡ ⎤ ⎡ ⎤
⎢
A ⎣
c x
c y
1
⎥
⎦ =
This follows because the polar of every point at infinity
⎡ ⎤
x
⎢ ⎥
⎣ y ⎦
0
is a diameter given by
⎢
⎣
[x, y, 0]A.
So every diameter meets the center if and only if
0
0
1
⎥
⎦
⎢
⎣
e
f
1
⎥
⎦
43

⎡
⎢
[x, y, 0]A ⎣
c x
c y
1
⎤
⎥
⎦ =0
if and only if
⎡
⎢
A ⎣
c x
c y
1
⎤
⎥
⎦ =
⎡
⎢
⎣
0
0
1
⎤
⎥
⎦
7 Conic Through a Set of Points
Five points determine a conic. We can find the equation of a conic through
four points as a product of lines through the points. Thus if we have points
p 1 ,p 2 ,p 3 ,p 4 let the line through p i and p j be l ij ,then
λl 12 l 34 +(1− λ)l 13 l 24 =0
is the equation of a conic through the four points. The parameter λ can
be chosen so that the conic passes through a fifth point. The discriminant
can be examined to determine the type of conic. By letting the points come
together we can specify tangent constraints.
8 Parametric Conic Arcs, Matrices of Projective
Transformations.
A conic arc is a portion of a conic curve. A conic arc can be expressed in
parametric form. This can be done so that each coordinate is a rational
function of a single variable. To construct such a function, we map a special
parabola to the required arc using a projective transformation.
We shall describe the mapping in a later section. First we need to prove
the 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 10x 2 +10xy+20y 2 +3x+−4y−10 = 0, (b)Hyperbola
with equation 5x 2 +10xy − 7y 2 − 3x +2y − 1 = 0, (c)Parabola with equation
x 2 +6xy +9y 2 +3x − 5y − 1 = 0, (d)and a circle with equation
2x 2 +2y 2 + 1x − 1y − 1=0. 2 5
45

Figure 6: A conic passing through four points defined by the products of lines
passing through the points (1, 1), (2, 2), (2, 1), (3, 3). The general 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 by tangent lines.
46

B
C
D
A
(a)
D’
C’
A’ B’
(b)
Figure 7: (a)Any conic arc can be obtained by mapping this parabola, which
is defined by tangent lines AC and BC, andpointD,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 representation of a circle. The intersection of the tangents is at
C, which is a point at infinity.
47

Proof. We shall write [v] for the subspace spanned by the vector V .LetV
and 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 a
number b, sothatL(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, sothat
U + V = c(aU + bV ).
Then by the linear independence of U and V ,wehave
It follows that a = b and L = aM.
ca =1=cb.
9 A Projective Transformation That Takes
Four Points To Four Points
Now we shall show that there is a projective transformation mapping four
points in the plane to four points in the plane.
Proposition. Let A, B, C, and D be distinct points in the plane, no three
of which are collinear. Let A ′ ,B ′ ,C ′ , and D ′ , also be distinct points in the
plane, no three of which are collinear. Then there is a unique projective
transformation taking the first four points to the second four points.
Proof. Consider the points as coordinate vectors. Then there are numbers
a, b, c so that
D = aA + bB + cC.
This is true because A, B, andC are a basis of the vector space. Similarly
there are numbers a’, b’, c’ sothat
48

D ′ = a ′ A + b ′ B + c ′ C.
Define a”, b”, c” by
and
a” =a ′ /a, b” =b ′ /b,
c” =c ′ /c.
None of a, b, orc 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 the
point column vectors. Let
M = [ a ′′ A ′ b ′′ B ′ c ′′ C ′ ][
A B C
] −1
.
Then the projective transformation defined by M maps the first three
unprimed 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 is
unique. We have
and
MA = a”A ′ ,MB = b”B ′ ,MC = c”C ′ ,
MD = D ′ .
Suppose there is a matrix N and numbers a N b N ,c N , and d N ,sothat
NA = a N A ′ ,NB = b N B ′ ,NC = c N C ′ ,
and
ND = d N D ′ .
49

Then ND is given by
And ND is also given by
ND = N(aA + bB + cC)
= aa N A ′ + bb N B ′ + cc N C ′ .
ND = d N D ′ .
Equating these two expressions for ND, and dividing by d N ,weget
But
D ′ =(aa N /d N )A ′ +(bb N /d N )B ′ +(cc N /d N )C ′ .
D ′ = a ′ A ′ + b ′ B ′ + c ′ C ′ ,
and because A ′ ,B ′ ,C ′ are linearly independent, we must have
and
and
Then
aa N /d N = a ′ ,
bb N /d N = b ′ ,
cc N /d N = c ′ .
a = d N (a ′ /a N )=d N a”,
b = d N b”,
c = d N c”.
So matrix N is a scalar multiple of matrix M,
N = d N M.
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 Approximately
Takes Four Points To Four Points
Let the affine transformation acting on a point with coordinates x and y be
given by
x ′ = m 11 x + m 12 y + m 13 ,
y ′ = m 21 x + m 22 y + m 23 .
51

Four unprimed points are to be mapped to four primed points. Thus we get
the following eight equations in the six matrix coefficients m ij ,
m 11 x k + m 12 y k + m 13 = x ′ k
m 21 x k + m 22 y k + m 23 = y ′ k
for k =1, .., 4. Let C be a the six dimensional coefficient vector,
The linear system becomes
where
and
⎡
A =
⎢
⎣
⎡
C =
⎢
⎣
⎤
m 11
m 12
m 13
m 21
⎥
m 22 ⎦
m 23
AC = B,
x 1 y 1 1 0 0 0
0 0 0 x 1 y 1 1
x 2 y 2 1 0 0 0
0 0 0 x 2 y 2 1
.. .. .. .. .. ..
x 4 y 4 1 0 0 0
0 0 0 x 4 y 4 1
⎡
B
⎢
⎣
x ′ 1
y ′ 1
x ′ 2
y ′ 2
...
x ′ 4
y ′ 4
The six by six normal equation for C is
⎤
.
⎥
⎦
A T AC = A T B.
⎤
,
⎥
⎦
52

11 The Rational Parametric Arc And The
Conic Arc
We can now construct the rational parametric arc. First we shall construct
the special case of the conic arc. The parabola with equation x − y 2 =0and
with matrix
⎡
⎤
⎢
⎣
0 0 1/2
0 −1 0
1/2 0 0
is to be mapped to the required conic arc, (see Fig. 5). If we take t = y,
then
⎡ ⎤
t
⎢
⎣ t 2 ⎥
⎦
1
⎥
⎦ ,
is a point on the parabola. Points A, B, andD lie on the parabola and are
given as
A =
B =
D =
⎡
⎢
⎣
⎡
⎢
⎣
⎡
⎢
⎣
0
0
1
1
1
1
⎤
⎥
⎦
1/4
1/2
1
C is the intersection point of the tangent lines at A and B. The tangent line
at B has equation
⎡
[ ]
x y 1
⎢
⎣
⎤
⎥
⎦
⎤
⎥
⎦
0 0 1/2 1
0 −1 0 1
1/2 0 0 1
⎤
⎥
⎦
53

So
=(1/2)x − y +1/2 =0.
c =
For future reference note that
⎡
⎢
⎣
0
1/3
1
⎤
⎥
⎦ .
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 information,
namely three points which lie on the conic, and two tangents. This
parabola is the set {P : f(P, P) =0}. A projective transformation M maps
it to the conic {P ′ : f ′ (P ′ ,P ′ )=0}, wheref ′ (P ′ ,P ′ )=f(M (P ′ ),M (P ′ )).
By the construction above, let M be the projective transformation that takes
the four points defining the parabola to the four primed points that defining
some conic arc. From the definition of f ′ ,weseethat
f ′ (C ′ ,A ′ )=f(C, A) =0.
and
f ′ (C ′ ,B ′ )=f(C, B) =0.
So C ′ is the intersection of the tangents to the conic arc at A ′ and B ′ . A
parametric equation of the arc from A ′ to B ′ is given by
⎡
x ′ ⎤ ⎡
t 2 ⎤
⎢
⎣ y ′ ⎥ ⎢ ⎥
⎦ = M ⎣ t ⎦ ,
z ′ 1
for 0 ≤ t ≤ 1.
Example. Letthearcbeahalfcircle.
⎡
A ′ =
B ′ =
⎢
⎣
⎡
⎢
⎣
54
−1
0
1
1
0
1
⎤
⎥
⎦
⎤
⎥
⎦

D ′ =
⎡
⎢
⎣
0
1
1
⎤
⎥
⎦ .
For this example we see by inspection that
⎡ ⎤
0
C ′ ⎢ ⎥
= ⎣ 0 ⎦
1
That is, the intersection of the tangents at A ′ and B ′ ,isthepointat
infinity 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 get
C ′ . 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 1/2
1/2 0 1/2
0 1 0
We know a, b, c from above, so
⎤ ⎡
⎥ ⎢
⎦ ⎣
0
1
1
⎤
⎥
⎦ =
⎡
⎢
⎣
1/2
1/2
1
⎤
⎥
⎦ .
a” =a ′ /a =(1/2)/(1/4) = 2,b”=b ′ /b =2,c”=c ′ /c =2.
Now
M = [ a”A ′ b”B ′ c”C ][ A B C ] −1
⎡
⎢
=2⎣
⎡
⎢
=2⎣
−1 1 0
0 0 1
1 1 1
−1 1 0
0 0 1
1 1 1
=
⎡
⎢
⎣
⎤ ⎡
⎥ ⎢
⎦ ⎣
⎤ ⎡
⎥ ⎢
⎦ ⎣
0 4 −2
−4 4 0
4 −4 2
0 1 0
0 1 1/2
1 1 1
⎤−1
⎥
⎦
0 −2 1
1 0 0
−2 2 0
⎤
⎥
⎦ .
⎤
⎥
⎦
55

Finally, we get
⎡
⎢
⎣
So in the affine plane we have
x ′ ⎤
y ′ ⎥
x ′
x ′ =
⎡
⎢
⎦ = M ⎣
t 2 t
1
⎤
⎥
⎦ .
2t − 1
2t 2 − 2t +1 ,
and
y ′ 2t − 2t
=
2t − 2t +1 .
There appears to be much calculation here. It is not as bad as it appears.
There is a Fortran conic arc subroutine that is called conarc in
library emerylib.ftn. The subroutine shows that the calculation can be
quite 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

1=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 and
Pratt.
It is advantageous to calculate the rational parametric form when points
of the arc must be displayed or plotted. When we have a rational parametric
form 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, byJ.J.Apalateguiand
L. J. Adams, McGraw-Hill, 1944. Many of the early aircraft shapes were
constructed using composite conic arcs.
A classic method for finding a parameterization of an algebraic curve is
given in books on Algebraic Geometry. This method uses a set of lines that
are defined by a single parameter. Each line is intersected with the algebraic
curve, 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 the
parameter. As an example one can get a simple rational parameterization of
the circle
(x − 1) 2 + y 2 =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 ⎡ ⎤
a
⎢ ⎥
⎣ b ⎦
c
57

are perpendicular to the coordinates of the point
⎡ ⎤
x
⎢ ⎥
⎣ y ⎦
z
If A and B are coordinate vectors of two lines then the cross product is
the coordinate vector of the intersection point because it is perpendicular to
both. In the same way if X and Y are coordinate vectors of two points then
their cross product is the coordinate vector of the line containing both. This
is an example of duality in projective space.
Example. Find the intersection point of the two lines with equations
x + y =0,
and
The cross product is
2x + y =1.
⎡
⎢
⎣
2
−5
−1
⎤
⎥
⎦
So the affine coordinates of the intersection point are (−2, 5).
Example. Find the line through the points
⎡
⎢
⎣
1
1
1
⎤
⎥
⎦
and
The cross product is
⎡
⎢
⎣
⎡
⎢
⎣
3
2
1
−1
2
−1
⎤
⎥
⎦
⎤
⎥
⎦
58

So the equation is
−x +2y =1.
Example. Find the line passing through (1, 1) and making an angle of 30
degrees with the x-axis. The line passes through the point at infinity given
by
and the point
⎡
⎢
⎣
cos(30)
sin(30)
0
⎡
⎢
⎣
⎤
⎥
⎦ =
1
1
1
⎤
⎡
⎢
⎣
⎥
⎦ .
3/2
1/2
0
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 dimensional
space. This space is constructed in a four dimensional vector space so there
are four coordinates for each point. Again corresponding to each quadric is
a 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 essentially
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. Q
is the coordinate vector of the plane. There is a duality between points and
planes. For example, three points determine a plane. Dually three planes
determine a point. In two dimensions, the duality was between points and
lines. The first three coordinates of the plane give a vector normal to the
plane in affine space, where the affine hyperplane has equation w =1. The
homogeneous coordinates of a point are written as x, y, z, w. IfP 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 symmetric
matrix 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. This
can be seen by examining the equations of intersection of lines in the tangent
plane 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 coordinates
of AP are the coordinates of a normal vector. A diametrical plane in the
direction Q, whereQ is a point at infinity, is the polar of Q. The diametrical
plane is the analogue in 3-space of the diameter in 2-space. A center is a
point in the intersection of all diametrical planes. points on a quadric surface
are called singular points if they are centers and regular points otherwise. An
example of a singular point is the apex of a cone. The statement above that
the 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 well
defined tangent plane.
As in the two dimensional case, a diametrical plane in the direction Q
bisects 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 some
are degenerate. The names, equations and matrices of some of the real
nondegenerate quadric surfaces are:
60

(1) Sphere:
(2) Hyperbolic paraboloid:
x 2 + y 2 + z 2 − w 2 =0.
⎡
⎤
1 0 0 0
0 1 0 0
⎢
⎥
⎣ 0 0 1 0 ⎦
0 0 0 −1
(3) Elliptic paraboloid:
(4) Parabolic cylinder:
(5) Hyperbolic cylinder:
(6) Intersecting planes:
⎡
⎢
⎣
x 2 − y 2 − 2wz =0.
1 0 0 0
0 −1 0 0
0 0 0 −1
0 0 −1 0
⎤
⎥
⎦
x 2 + y 2 − 2zw =0.
⎡
⎤
1 0 0 0
0 1 0 0
⎢
⎥
⎣ 0 0 0 −1 ⎦
0 0 −1 0
⎡
⎢
⎣
x 2 − 2yz =0.
1 0 0 0
0 0 −1 0
0 −1 0 0
0 0 0 0
⎤
⎥
⎦
x 2 − y 2 − w 2 =0.
⎡
⎤
1 0 0 0
0 −1 0 0
⎢
⎥
⎣ 0 0 0 0 ⎦
0 0 0 −1
x 2 − y 2 =0.
61

(7) Parallel planes:
(8) Circular cylinder:
⎡
1 0 0 0
0 −1 0 0
⎢
⎣ 0 0 0 0
0 0 0 0
⎤
⎥
⎦
x 2 − w 2 =0.
⎡
⎤
1 0 0 0
0 0 0 0
⎢
⎥
⎣ 0 0 0 0 ⎦
0 0 0 −1
x 2 + y 2 − w 2 =0.
⎡
⎤
1 0 0 0
0 1 0 0
⎢
⎥
⎣ 0 0 0 0 ⎦
0 0 0 −1
(9) Hyperboloid of two sheets:
x 2 + y 2 − z 2 + w 2 =0.
⎡
⎤
1 0 0 0
0 1 0 0
⎢
⎥
⎣ 0 0 −1 0 ⎦
0 0 0 1
(10) Cone:
x 2 + y 2 − z 2 =0.
⎡
⎤
1 0 0 0
0 1 0 0
⎢
⎥
⎣ 0 0 −1 0 ⎦
0 0 0 1
(11) hyperboloid of one sheet:
x 2 + y 2 − z 2 − w 2 =0.
⎡
⎤
1 0 0 0
0 1 0 0
⎢
⎥
⎣ 0 0 −1 0 ⎦
0 0 0 −1
62

15 Transformations
A one to one linear transformation maps a quadric surface to a new quadric
surface. Suppose
S = {P : f(P, P) ≤ 0}
then
LS = {LP : f(P, P) ≤ 0}
= {Q : f(L −1 Q, L −1 Q) ≤ 0}
= {Q :(L −1 Q, AL −1 Q) ≤ 0}
= {Q :(Q, L −1T AL −1 Q)

Rotation about the x-axis:
⎡
⎢
⎣
1 0 0 0
0 c −s 0
0 s c 0
0 0 0 1
⎤
⎥
⎦
Rotation about the y-axis:
⎡
⎢
⎣
c 0 s 0
0 1 0 0
−s 0 c 0
0 0 0 1
⎤
⎥
⎦
Rotation about the z-axis:
⎡
⎢
⎣
c −s 0 0
s c 0 0
0 0 1 0
0 0 0 1
Now if R is a rotation matrix, then
⎤
⎥
⎦
R −1 (θ) =R(−θ).
An orthogonal matrix R has the property R T
transformed quadric surface is
= R −1 , so the matrix of the
A ′ = R(θ)AR(−θ).
(3)Affine translation: The matrix
⎡
⎤
1 0 0 a
0 1 0 b
⎢
⎥
⎣ 0 0 0 c ⎦
0 0 0 1
translates by
⎡
⎢
⎣
a
b
c
⎤
⎥
⎦ .
64

The inverse of T is
⎡
T −1 = ⎢
⎣
1 0 0 −a
0 1 0 −b
0 0 0 −c
0 0 0 1
⎤
⎥
⎦ .
The transformed matrix is
⎡
A ′ = ⎢
⎣
1 0 0 0
0 1 0 0
0 0 1 0
−a −b −c 1
Example. Given a sphere with matrix
⎡
⎢
⎣
⎤ ⎡
⎥
⎦ A ⎢
⎣
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
Then the scaling transformation matrix
⎡
⎢
⎣
2 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Maps the sphere to an ellipse with matrix
⎡
⎢
⎣
1/2 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
⎡
= ⎢
⎣
⎤ ⎡
⎥ ⎢
⎦ ⎣
1/2 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
⎤ ⎡
⎥ ⎢
⎦ ⎣
1 0 0 −a
0 1 0 −b
0 0 1 −c
0 0 0 −1
⎤
⎥
⎦
⎤
⎥
⎦
⎤ ⎡
⎥ ⎢
⎦ ⎣
⎤
⎥
⎦ .
1/2 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1/2 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
⎤
⎥
⎦
⎤
⎥
⎦
65

⎡
= ⎢
⎣
The translation matrix T shifts it by 2
⎡
⎢
⎣
1/4 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
1 0 0 −2
0 1 0 0
0 0 1 0
0 0 0 1
The matrix of the shifted conic becomes
⎤
⎥
⎦
⎤
⎥
⎦ .
(T −1 ) T (S −1 ) T AS −1 T −1
⎡
⎤ ⎡
⎤ ⎡
⎤
1 0 0 0 1/4 0 0 0 1 0 0 2
0 1 0 0
0 1 0 0
0 1 0 0
= ⎢
⎥ ⎢
⎥ ⎢
⎥
⎣ 0 0 1 0 ⎦ ⎣ 0 0 1 0 ⎦ ⎣ 0 0 1 0 ⎦
2 0 0 1 0 0 0 −1 0 0 0 1
⎡
⎤ ⎡
⎤
1 0 0 0 1/4 0 0 1/2
0 1 0 0
0 1 0 0
= ⎢
⎥ ⎢
⎥
⎣ 0 0 1 0 ⎦ ⎣ 0 0 1 0 ⎦
2 0 0 1 0 0 0 −1
⎡
⎤
1/4 0 0 1/2
0 1 0 0
= ⎢
⎥
⎣ 0 0 1 0 ⎦ .
1/2 0 0 0
The equation of the shifted ellipsoid is
x 2 /4+y 2 + z 2 +1/2xw +1/2wx =0.
Suppose we rotate about the z-axis by angle 45 degrees. The rotation matrix
is
⎡ √ √ ⎤
√ 2/2 −
√ 2/2 0 0
R z (45) =
2/2 2/2 0 0
⎢
⎣ 0 0 1 0
⎥ . ⎦
0 0 0 1
66

The matrix of the quadric becomes
This is
⎡ √ √ ⎤ ⎡
√ 2/2 −
√ 2/2 0 0
2/2 2/2 0 0
⎢
⎣ 0 0 1 0
⎥ ⎢
⎦ ⎣
0 0 0 1
⎡
=
⎢
⎣
R z (45)(T −1 ) T (S −1 ) T AS −1 T −1 R z (−45).
√ √
√ 2/2 −
√ 2/2 0 0
2/2 2/2 0 0
0 0 1 0
0 0 0 1
1/4 0 0 1/2
0 1 0 0
0 0 1 0
1/2 0 0 0
⎤ ⎡
⎥ ⎢
⎦ ⎣
The equation of the shifted rotated quadric is
⎤ ⎡
⎥
⎦ ⎢
⎣
√ √
√ 2/2
√ 2/2 0 0
2/2 2/2 0 0
0 0 1 0
0 0 0 1
√ √ ⎤
2/8 2/8 0 1/2
− √ 2/2 √ 2/2 0 0
√
0
√
0 1 0 ⎥
2/4 2/4 0 0
(5/8)x 2 +(5/8)y 2 − (3/4)xy + z 2 + 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 the
bilinear 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 the
set operators union, intersection and complementation. The characteristic
function, C S corresponding to the set S, is a logic valued function mapping
points to a value true or false. The value of C S (P ) is true if P ∈ S, otherwise
it is false. By definition P ∈ A ∪ B, if and only if, P ∈ A or P ∈ B. Sothe
characteristic 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 functions
whose domain is 3-space. These functions may be represented with logical
variables of programming languages.
17 Plane Image
A shaded image is produced by finding the intersection of the object and
a line that passes through the projection point. Let L be such a line. Its
intersection with each primitive surface is calculated. For each intersection
point there is a surface normal. A small line segment of fixed length δl, inthe
normal direction with center at the intersection point, determines when the
point is on the surface of the object. It is on the object when one endpoint
is inside and the other is outside the object. This is determined by the
characteristic function.
One or more light sources illuminate the object. The inner product of
these light source vectors with the surface normal give an intensity function
defining the shading. The point on the surface is projected to 2-space with
the 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 set
in a rectangular box, and then taking a uniformly distributed random sample
in the box. Suppose the integral
∫
f(x)dm
S
is to be evaluated. Let x be a point. Let B be a rectangular hexahedron
enclosing S. Let m be the volume measure (Lebesgue measure). Define a
probability measure on the set B by
p(S) =m(S)/m(B).
68

Let χ S be the characteristic function of the set S. Ifx ∈ S then χ S (x) is
1, otherwise it is 0. Let g be the product of f and χ S .Theng is a random
variable defined on the enclosing set B. Its expectation is
∫ ∫
E(g) = gdp = fdp = 1 ∫
fdm.
B
S B S
∫
fdm = m(B)E(g)
S
Let x 1 , ...., x n be a sample of independent uniformly distributed random variables.
Then g(x 1 ), ..., g(x n ) is also independent. Let ḡ bethesamplemean
ḡ =(g(x 1 )+..... + g(x n ))/n.
The expectation of the sample mean is the mean of the original distribution,
The variance is
E(ḡ) =E(g).
V (ḡ) =E((ḡ − E(g)) 2 )=E(ḡ 2 ) − E(ḡ) 2 = V (g)/n.
By the central limit theorem, the random variable
(ḡ − E(ḡ)/
√
V (g) =(ḡ − E(g))/
√
V (g)/n,
has approximately, for sufficiently large n, the standard normal distribution.
Now
∫
1 1.96
√ exp(−x 2 /2)dx ≈ .95.
2π
−1.96
So
√
P (|ḡ − E(g)| ≤1.96 V (g)/n ≈ .95.
This gives a 95 percent confidence interval for the estimated value of the
mean. Multiplying by m(B), we get a confidence interval for the integral.
∫
P (|m(B)ḡ −
We are interested in the specific quantities:
(1) Volume; f =1
S
√
fdm|≤1.96m(B) V (g)/n ≈ .95.
69

(2) Moments;f = x i ,i=1, 2, 3
(3) Inertia tensor: f = x i x j ,i,j =1, 2, 3
Example. let S be a sphere of radius a. Let B bethecubeofside2a,
centered at the origin. Then m(B) =8a 3 ,andm(S) =(4/3)πa 3 .Letf =1
and g = χ S . Then the expectation of g is
∫
∫
E(g) = g(x)dp = dp = p(S) =
S
S
The variance of g is
∫
v(g) =
(4/3)πa 3
8a 3 = π 6 .
g 2 dp − E(g) 2 = E(g)(1 − E(g)).
Let us choose the sample size n, so that the relative error of the estimated
volume 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)) 196 2 (6π − 1) = 34953
This is a large number of samples. We must decide how to choose the block
B. 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 should
choose the box B to be as small as possible.
Usually we do not know the variance V (g), but we can use the sample
variance as an estimate. The sample variance is defined as
S 2 =
∑ nk=1
(g(P k ) − ḡ) 2
n
∑ nk=1
g(P k ) 2
=
− ḡ 2 .
n
An unbiased estimator of the variance of g is (p. 165 Brunk)
n
n − 1 S2 .
So
√
∫
S
2
fdm = m(B)g +1.96m(B)
S
n − 1 .
Example. We shall calculate the x 1 moment of the sphere of the previous
example. We have f = x 1 and
∫
E(g) = gdp = 1 x
B 8a
∫S
3 1 dm =0.
By symmetry the variance is
1
8a 3 ∫S
x 2 dm =
π ∫ a
x 2
8a 3 1 (a2 − x 2 1 )dx
−a
So the error is
= πa2
30 .
(1.96)a 4 sqrtπ √
30n
.
71

In this case we can’t divide by the moment to get a relative error, because
the moment is zero. But a moment is the product of a volume and a length,
so a relative error results from dividing by a 4 .
Example. We shall calculate the inertia tensor components for the sphere.
When i ≠ j, by symmetry,
∫
x i x j dm =0.
When i = j we have
∫
S
x 2 i dm = ∫ a
= π
−a
[ x
3
i a 2
3
x 2 i (a2 − x 2 i )dx
− x5 i
5
] a
−a
=2πa 5 (1/3 − /15)
= 4 15 πa5 .
19 Stratified sampling
It may be possible to decrease the error by doing stratified sampling. We
divide the domain B into m subdomains. We introduce a probability measure
p i on each B i by
p i (S i )=m(S i )/m(B i ).
Then
∫
∫
fdm = m(B i ) g i dp i = E(g i )m(B i ),
S∩B i B i
where g i is the product of f and the characteristic function χ(S i )ofS i .Then
∫
S
fdm =
Let h i = m(B i )g i . Then, defining
m∑
∫
i=1
S∩B i
m(B i )g i dp i .
m∑
h = h i ,
i=1
72

we have
∫
m∑
fdm = E(h) = E(h i ).
S
i=1
Also
m∑
v(h) = v(h i ).
i=1
Let h i1 , ....., h im , be an independent sample of random variables having the
distribution of h i .Wetakek samples from each strata. Let b 2 i = v(h i)and
B 2 = v(h 11 )+.... + v(h mk
= k(b 2 1 + ... + b2 m ).
Let a i = E(h i )andA = k(a 1 + ... + a m ). Consider
((h 11 − a 1 )+(h 12 − a 2 )+... +(h mk − a k ))/B
= h 11 + ... + h mk − k(a 1 + .... + a m )
√
k(b 2 1 + ... + b 2 m
= (h 11 + ... + h mk )/k − ∫ fdm
√
.
(b 2 1 + ... + b 2 m)/k
Let
y k = h 11 + ... + h mk
,
k
and
v(k) = b2 1 + ... + b 2 m
.
k
The Lindeberg condition is satisfied (p.455 Eisen). As k →∞the following
probability approaches a normal probability. That is,
∫
√ ∫ x
p((y k − fdm)/v(k) ≤ x) → (1/2π) exp(−t 2 /2)dt.
It follows that
−∞
∫
p(|y k − fdm|≤1.96v(k)) ≈ .95.
y k is a better estimate of the integral than m(B)ḡ, provided
√
v(k)

Note that there are mk samples. Suppose that either v(g i )=v(g) orelse
v(g 1 )=0. Then
{
b 2 i = v(h i )=m(B i ) 2 m(Bi )
v(g i )=
2 v(g)
0
Suppose j of the b 2 i are not zero, and suppose the strata are equal, so that
m(B i )=m(B)/m. Then
√
v(k) =
b 2 1 + .... + b 2 k
√
k
√
= (j(m(B) 2 /m 2 )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, and
will have zero variance. If
j

Then
Thus
c(P )=c ′ (TP)=Ac(TP)
c(TP)=Ac(P )
And
c ′ (TP)=c(P )=Ac ′ (P )
Hence the matrix of T is A in both bases. In summation we have shown that
when an unprimed basis is transformed to a primed basis, then the matrix of
the coordinate transformation from the unprimed to the primed coordinate
system 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 components
with respect to an orthogonal coordinate system by
⎡
⎤ ⎡
⎤
u 1 u 2 u 3 a 1 a 2 a 3
⎢
⎥ ⎢
⎥
a × b = ⎣ a 1 a 2 a 3 ⎦ = ⎣ b 1 b 2 b 3 ⎦
b 1 b 2 b 3 u 1 u 2 u 3
= ɛ ijk a i b j u k ,
where ɛ ijk is the permutation symbol and u 1 ,u 2 ,u 3 , are unit coordinate vectors.
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 permutation
symbol is 0. The kth component of a × b is
c k = ɛ ijk a i b j .
Let there be a new primed coordinate system. We have
The Jacobian is
ɛ ijk
∂x i
∂x ′p ∂x j
∂x ′q ∂x k
∂x ′r =Det(J)ɛ pqr.
J =
⎡
⎢
⎣
∂x 1
∂x ′1 ∂x 2
∂x ′1 ∂x 3
∂x ′1
∂x 1
∂x ′2 ∂x 2
∂x ′2 ∂x 3
∂x ′3
∂x 1
∂x ′3 ∂x 2
∂x ′3 ∂x 3
∂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 transformation,
rather than by a general transformation, then the determinant of
the Jacobian is 1, that is
Det(J) =1.
Then
∂x i ∂x j ∂x k
ɛ ijk
∂x ′p ∂x ′q ∂x = ɛ pqr.
′r
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, and
so the cross product is a Cartesian vector. A second order tensor, whose
components obey
c ij = −c ji ,
is called antisymmetric. An antisymmetric tensor defines a Cartesian vector.
The components of this vector are
c ij = −(1/2)ɛ ijk c ij .
These are the components of a Cartesian vector, because the vector is the
contraction of a Cartesian tensor and a general tensor. We find that
⎡ ⎤
c 32
⎢ ⎥
c = ⎣ −c 31 ⎦ .
c 21
A vector such as c is called an axial vector. Let d r be the rth component of
the cross product c × x. Then
d r = ɛ pqr (−(1/2)ɛ ijp c ij )x q .
We claim that
We have
ɛ pqr (−(1/2)ɛ ijp c ij )=c rq .
−(1/2)ɛ pqr ɛ pij c ij
3∑
= (−1/2)ɛ pqr ɛ pij c ij
p=1
76

So
= ∑
p≠q,p≠r
= ∑
p≠q,p≠r
= ∑
p≠q,p≠r
(−1/2)ɛ pqr ɛ pij c ij
[(−1/2)ɛ pqr ɛ pij c ij − (−1/2)ɛ pqr ɛ pqr c rq ]
[(−1/2)c qr +(1/2)c rq = c rq .
d r = c rq x q .
If L is the linear transformation which has components c rq , then the value
to 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 components
c k . 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 in
the body and let u 1 ,u 2 ,u 3 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(u i ),
where R is an orthogonal transformation, which varies with time. Then the
coordinates of P are related by
x ′ i = a ij x j .
where (a ij ) is the inverse of the matrix (ā ij )ofR. By differentiating, and
noting that the components of P are constant in the primed system, we find
that (Lass page 46)
dx i
dt = w ijx j ,
for
w ij = −ā ik
da kj
dt .
77

By differentiating the norm of P , which is invariant, we find that (w ij )is
skew symmetric, that is
w ij = −w ji .
So (w ij ) can be represented as an axial vector. We have
c(w) =
⎡
⎢
⎣
w 32
−w 31
w 21
⎤
⎥
⎦ =
⎡
⎢
⎣
⎤
w 32
⎥
w 13 ⎦
w 21
and
v = dc(P ) = c(w) × c(P ).
dt
w is the angular velocity vector. It transforms like a vector under an orthogonal
transformation.
Example. Let rotation transformation R have matrix
⎡
c(t 2 ) −s(t 2 ⎤
) 0
⎢
(ā ij )= ⎣ s(t 2 ) c(t 2 ⎥
) 0 ⎦ .
0 0 1
and
So
Then
(a ij )=(ā ij ) T =
da ij
dt
⎡
⎢
(w ij )=−2t ⎣
⎡
⎢
=2t ⎣
⎡
⎢
⎣
c(t 2 ) −s(t 2 ) 0
s(t 2 ) c(t 2 ) 0
0 0 1
c(t 2 ) s(t 2 ) 0
−s(t 2 ) c(t 2 ) 0
0 0 1
−s(t 2 ) c(t 2 ) 0
−c(t 2 ) −s(t 2 ) 0
0 0 0
⎡
⎢
= −2t ⎣
⎤ ⎡
⎥ ⎢
⎦ ⎣
0 1 0
−1 0 0
0 0 0
⎤
⎥
⎦ .
⎤
⎥
⎦ ,
−s(t 2 ) c(t 2 ) 0
−c(t 2 ) −s(t 2 ) 0
0 0 0
⎤
⎥
⎦ .
⎤
⎥
⎦
78

So
c(w) =
⎡
⎢
⎣
Suppose P = u ′ 1 .Letc(P ) be the coordinate vector of P in the unprimed
system and c ′ (P ) the coordinate vector in the primed system. Then
Now
0
0
2t
⎤
⎥
⎦ .
c(P )=(ā ij )c ′ (P )
⎡ ⎤
1
⎢ ⎥
=(ā ij ) ⎣ 0 ⎦
0
⎡
c(t 2 ⎤
)
⎢
= ⎣ s(t 2 ⎥
) ⎦ .
0
dc(P )
dt
⎡
⎢
=2t ⎣
−s(t 2 )
c(t 2 )
0
⎤
⎥
⎦ .
This is the velocity computed with matrices. The velocity computed with
the cross product is
⎡ ⎤ ⎡
0 c(t 2 ⎤
)
⎢ ⎥ ⎢
c(w) × c(P )= ⎣ 0 ⎦ × ⎣ s(t 2 ⎥
) ⎦
2t 0
⎡
⎢
=2t ⎣
−s(t 2 )
c(t 2 )
0
We get the same result with either method.
⎤
⎥
⎦ .
23 Angular Momentum And The Inertia Tensor
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 component
is
∫
L k = ɛ ijk x i ẋ j dm.
Then
dL k
dt
∫
=
∫
ɛ ijk x i ẍ j dm + ɛ ijk ẋ i ẋ j dm
∫
= ɛ ijk x i ẍ j dm.
The second term vanishes because the cross product of parallel vectors is
zero. The torque vector T is
∫
∫
r × df =
r × d2 r
dt 2 dm.
The kth component of the torque is
∫
T k = ɛ ijk x i ẍ j dm.
Thus dL/dt = T . Because
we have
∫
L k =
ẋ j = w j p xp ,
ɛ ijk x i w j px p dm = ɛ ijk
∫
x i x p dm
= ɛ ijk wp j Iip ,
where I is the inertia tensor. The components of the inertia tensor are
∫
I ip = x i x p dm.
The moment of inertia about an arbitrary axis can be found from the
inertia 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 moment
of inertia about the line is then defined as (McConnel p.233)
∫
p 2 dm.
Suppose u is a unit vector in the direction of the line. Let t be the trace.
Then
p 2 = x i x i − (r · u) 2
80

= δ ij x i x j − (δ ij x i u j ) 2 .
So the moment of inertia is
= δ ij I ij − δ ij δ kp u j u p I ik
= t(I) − I ik u i u k
= t(I)δ ik u i u k − I ik u i u k
=(t(I)δ ik − I ik )u i u k
= J ik u i u k .
The tensor J has components
J ik = t(I)δ ik − I ik .
In some books (e.g. Goldstein), J, rather than I, is called the inertia tensor.
The explicit components of J are:
∫
J 11 = [(x 2 ) 2 +(x 3 ) 2 ]dm,
∫
J 22 =
∫
J 33 =
[(x 1 ) 2 +(x 3 ) 2 ]dm,
[(x 1 ) 2 +(x 2 ) 2 ]dm.
When i ≠ j,
∫
J ij = − [(x i ) 2 +(x j ) 2 ]dm.
If f is the bilinear form corresponding to the symmetric matrix J, then
the 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 a
plane, where θ is the angle between the surface normal of the plane, and the
projection 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 projection
function. Then
∫
∫
f(x, n)dσ = f(p −1 (y),n(p −1 (y))(ds/(n · u).
A
p(A)
Now suppose A 1 , ...., A n covers the surface, and p is the plane projection from
A 1 to 2-space. Suppose there are functions g i defined on the surface, where
each g i vanishes outside of A i . And suppose that each g i has values between
0 and 1, and suppose that the g i sum to 1. That is
n∑
g i (x) =1.
i=1
Such a set of functions is called a partition of unity, subordinate to the cover
A 1 , ..., A n .Thenwehave
The surface integral is
∫
n∑
∫
f(x, n)dσ =
A
i=1
p(A i )
n∑
f = g i (x)f(x) =1.
i=1
g i (p −1 (y),n(p −1 (y))f(p −1 (y),n(p −1 (y))(ds/(n · u).
When p i is the projection in the x i direction, let g i be the function whose
value at a point is the square of the ith component of the unit normal vector
to the surface. In general p i is a many to one function and when we define
A ik to be the kth layer of the surface, which is pierced by the projecting ray,
the functions
f i : A ik → R,
are surface patches, and the
constitute a partition of unity.
g i : A ik → R,
82

25 Volume Integral
A volume integral can be changed to a surface integral by using the divergence
theorem. Define a vector field
⎡ ⎤
H = r 3 = 1 x
⎢ ⎥
⎣ y ⎦ .
3
z
Let B be a set and ∂B its bounding surface. Then the volume of B is
∫ ∫
dV = ∇·HdV
B B
∫
= H · ndS
∂B
= 1 ∫
(xn x + yn y + zn z )dS
3 ∂B
Note that the integrand is the distance from the origin to the plane containing
the area element dS. In particular for any plane area, say A, thevolumeof
the 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 polyhedron
consisting of k polygons, is therefore
V =
k∑
i=1
P i · n i
A i ,
3
where A i is the area of the ith polygon, P i is some point on the polygon, and
n i is the outward normal to the polygon. The divergence theorem can also
be used to compute moments. For example, applying the divergence theorem
to the vector field
⎡ ⎤
G = 1 2
⎢
⎣
x 2
1
1
⎥
⎦ ,
gives a surface calculation of the x moment.
m x = 1 ∫
x 2 n x dS.
2 ∂B
83

26 Polygon Areas
We shall calculate polygon areas. Let a closed curve γ have position vector
r(t), 0 ≤ t ≤ 1. The area enclosed by the curve is
A = 1 ∮
r × dr.
2 γ
This of course is a vector. It is the magnetic moment of a circuit with
unit current. When the curve lies in a plane, the magnitude of this integral
is equal to the area enclosed by the curve. This is obvious from the definition
of the cross product when the plane passes through the origin. When the
plane does not pass through the origin, the vector r can be written as a sum
of 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 does
not 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 enclosed
by a plane curve. In the case that the curve lies in the xy plane, the vector
integral will be in the z direction. If the component is positive then the
inner region is to the left. This is obvious if the curve is the unit circle
with the counterclockwise orientation. For a general proof, note that a curve
which bounds an inner region can be continuously deformed to a unit circle
in such a way that the area never vanishes. Thus the sign of the area must
be maintained.
If the sign is negative then the inner region lies to the right. The curve
direction is defined by the increasing parameter value.
Returning to the polygon area problem, the integral over a line segment
is
A = r 1 × r 2
,
2
where r 1 and r 2 are the starting and ending points. This is true because the
line element has the same direction as both r 2 − r 1 and r − r 1 . The cross
product of parallel vectors is zero, so the integral is
∫
1
2 r 1 × dr
= 1 2 r 1 × (r 2 − r 1 )
84

= 1 2 r 1 × r 2 .
Thus the area of a closed polygon A i with m sides and m vertices equals the
magnitude of
1
m∑
r j × r j+1 ,
2
j=1
where
r m+1 = r 1 .
This sum is clearly normal to the polygon, since each term is. A unit normal
is obtained by dividing by the area A i .
The number of cross products that need to be calculated can be reduced.
Forexampleforatrianglewithm = 3 the sum of the cross products is
r 1 × r 2 + r 2 × r 3 + r 3 × r 1
= r 2 × r 3 − r 2 × r 2 − r 1 × r 3 + r 1 × r 2
=(r 2 − r 1 ) × (r 3 − r 2 )
For another example we may decompose a hexagon into 4 triangles and write
the area as
1
3∑
2 [ (r i+1 − r i ) × (r i+2 − r i+1 )+(r 3 − r 1 ) × (r 5 − r 3 )].
i=1
27 Perspective Projection
Let Q be a point in real projective 3 Space (RP3). Q has component vector
Q =(Q 1 ,Q 2 ,Q 3 ,Q 4 )=(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) =A 1 Q 1 + A 2 Q 2 + A 3 Q 3 + A 4 Q 4 =0.
85

Figure 8: A Perspective drawing showing, a building, a railroad track, and
the horizon. The vanishing point at the end of the railroad tracks is at infinity
in 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.
intersection of the plane with the line
Let L(Q) bethepointof
Then
Q + tP.
(A, Q + tP )=0.
Solving for t, we find
(A, Q)
L(Q) =Q −
(A, P ) P.
L is a linear projection operator, i.e. L ∗ L = L. Using a rigid motion transformation,
we may transform the plane to the xy plane, and the projection
point to the positive z axis. Hence we consider the special case where the
projection plane has equation
z =0,
and the projection point P is on the positive z axis. Then
and
A =(0, 0, 1, 0) T ,
P =(0, 0,p 3 ,p 4 ) T .
In this special case of projection into the xy plane,we write
L = L xy .
Then
(x ′ ,y ′ ,w ′ ) T = L xy Q =(x, y, z, w) T − z p 3
(0, 0,p 3 ,p 4 ) T .
L xy =
⎡
⎢
⎣
=(x, y, 0,w− z p 4
p 3
) T .
1 0 0 0
0 1 0 0
0 0 −p 4 /p 3 1
⎤
⎥
⎦ =
⎡
⎢
⎣
1 0 0 0
0 1 0 0
0 0 −1/d 1
where d is the distance from the projection plane to the projection point.
GivenaprojectionpointP and a center point C, letT 1 be an affine transformation
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 measured
from the positive x axis. The azimuth is the rotation angle, about the
z axis, that the xz plane would have to be rotated, so that it would include
the projection direction. For example, the azimuth of the positive x axis is
zero, and the azimuth of the positive y axis is π/2 . The elevation is the
angle from the xy plane to the projection direction. We rotate the image of
the projection point into the yz plane with a rotation T 2 about the z-axis
by angle −φ − π/2. We rotate the image of the projection point up to the
z-axis, with a rotation T 3 about the positive x axis, by an angle −(π/2 − θ).
The complete perspective transformation matrix is
L = L xy T 3 T 2 T 1 ,
where
and
⎡
T 2 = ⎢
⎣
⎡
T 3 = ⎢
⎣
⎡
T 1 = ⎢
⎣
1 0 0 −C x
0 1 0 −C y
0 0 1 −C z
0 0 0 1
⎤
⎥
⎦ ,
c(φ + π/2) s(φ + π/2) 0 0
−s(φ + π/2) c(φ + π/2) 0 0
0 0 1 0
0 0 0 1
⎡
= ⎢
⎣
−s(φ) c(φ) 0 0
−c(φ) −s(φ) 0 0
0 0 1 0
0 0 0 1
⎤
⎥
⎦ ,
1 0 0 0
0 c(θ − π/2) −s(θ − π/2) 0
0 s(θ − π/2) c(θ − π/2) 0
0 0 0 1
⎡
= ⎢
⎣
1 0 0 0
0 s(θ) c(θ 0
0 −c(θ) s(θ) 0
0 0 0 1
⎤
⎥
⎦ .
⎤
⎥
⎦
⎤
⎥
⎦
88

Example. Isometric projection. The picture plane has coordinate vector
⎡ ⎤
1
1
R = ⎢ ⎥
⎣ 1 ⎦
0
The projection point is at infinity,
Hence d = ∞, andso
L xy =
⎡
⎢
⎣
⎡
P = ⎢
⎣
1 0 0 0
0 1 0 0
0 0 −1/d 1
1
1
1
0
⎤
⎤
⎥
⎦
⎥
⎦ =
The azimuth angle is φ = π/4, and
√
2
sin(φ) =
2 ,
√
2
cos(φ) =
2 .
The tangent of the elevation angle is
tan(θ) = 1 √
2
.
⎡
⎢
⎣
1 0 0 0
0 1 0 0
0 0 0 1
Then
√
3
sin(θ) =
3
√
6
cos(θ) =
3
T 1 is the identity, because the center of projection is already at the origin.
T 1 = I.
89
⎤
⎥
⎦

Then
Then
⎡
− √ √ ⎤
2 2
0 0
2 2 T 2 =
− √ 2
− √ 2
0 0
2 2 ⎢
⎥
⎣ 0 0 1 0
, ⎦
0 0 0 1
⎡
⎤
1 0 0 0
√ √ T 3 =
0 3 6
0
3 3
⎢
⎣ 0 − √ √ 6 3 ⎥
0
.
3 3 ⎦
0 0 0 1
⎡
T 3 T 2 T 1 =
⎢
⎣
− √ √
2 2
2
− √ 6
− √ 6
√ 6 √ 6
12 12
6 6
2
0 0
√
6
0
√3 3
0
3
0 0 0 1
⎡
− √ √ ⎤
2 2
0 0
2 2
L = L xy T 3 T 2 T 1 = ⎢
⎣ − √ 6
− √ √ 6 6
0
⎥
6 6 3 ⎦
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,
⎡ ⎤ ⎡
1
0
L ⎢ ⎥
⎣ 0 ⎦ = ⎢
⎣
1
− √ 3
−1
0
√
6
⎤
⎥ , ⎦
⎡
L ⎢
⎣
0
1
0
1
⎤ ⎡
⎥
⎦ = ⎢
⎣
90
√ ⎤
3
−1
√
0 ⎥
6
⎦ ,

and
⎡
L ⎢
⎣
0
0
1
1
⎤ ⎡
⎥
⎦ = ⎢
⎣
⎤
0
2
⎥
√
0 ⎦ .
6
The image of the coordinate axes are as shown in the Figure. The lengths
are 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 xyplane
with the plane at infinity
w =0.
A horizon point can not be reached in 3-space, but we may ”see” its projection.
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 into
a finite point in 2-dimensional space. Such a point is called a vanishing point.
For example, the point
⎡ ⎤
1
1
Q = ⎢ ⎥
⎣ 0 ⎦
0
is the point at infinity, where parallel 45 degree lines in the plane meet at
infinity. The vanishing point is L(Q) in the picture plane. The horizon in
the picture plane is determined by two vanishing points, corresponding to
two sets of parallel lines in the xy-plane. When the projection point P is
at infinity, as in isometric projection, parallel lines in 3-space are mapped to
parallel lines in 2-space so there are no vanishing points.
If the line through P and Q is parallel to the picture plane, then the
projection of Q will be a point at infinity in the picture plane.
The following information is needed to construct the perspective transformation:
(1)The center of the projection plane
(C x ,C y ,C z ) T .
91
⎤
⎥
⎦ .

. (2)The direction of the projection point from the center, which is specified
by the azimuth angle φ, and the elevation angle θ. (3)The reciprocal of the
distance 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 distance
1/d =0. ThisgivesL xy .
28 The Tektronix Hardware Modeler: A Quadric
Solid Modeler
In conjunction with the Portland CAM-I (Computer Aided Manufacturing
International) meeting, held August 3-7 1987, there was a tour of Tektronix
laboratories, where A CSG (Constructive Solid Geometry) system using
quadric halfspaces, with almost real time response, say three seconds for
an average model, was given its first public viewing. This system consisted
of 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 or
on. If a given block is completely inside or completely outside the model it is
discarded and not subdivided. A Taylor series expansion is used to classify
a block with respect to a halfspace. When a block is sufficiently small and
meets the model surface it is considered a point on the surface and a normal
is 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 reserved
and no voxels behind it need be searched. This algorithm is implemented
in hardware and is very fast. The algorithm is very much faster than ray
tracing. 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
92

Z
-Y
X
Figure 9: The intersection of the parabolic cylinder z 2 = y and the cone
y 2 = 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 Quadric.
93

system Reduce as basis, a 3d terminal using a quarter wave plate, and a fast
switching 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 processor)
development project. He says that the 4029 has advantages over the
Silicon-Graphics terminal in certain cases. The 4029 must be driven by an
external computer while the Silicon-Graphics can function alone.
29 References and bibliography
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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 Algebra
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[4]Brunk H D, Introduction To Mathematical Statistics, Blaisdell 2nd
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[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 Design
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[9]Halmos Paul R., Introduction To Hilbert Space, ChelseaNewYork
2nd Ed. 1957.
94

[10]Hodge W V D, Pedoe D, Methods Of Algebraic Geometry, Cambridge
U. Press 1968.
[11]Knuth Donald, The Art Of Computer Programming, Vol. 1 Addisonwesley
2nd Ed.1973..
[12]Lass Harry, Elements Of Pure And Applied Mathematics, Mcgrawhill
1957..
[13]Leven Joshua, A Parametric Algorithm For Drawing Pictures Of
Solid Objects Composed Of Quadric Surfaces, Comm. Acm V. 19 No.
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[14]Mcconnell Albert J, Applications Of Tensor Calculus, Dover 1931.
[15]Olmsted John M H, Solid Analytic Geometry, Appleton-centurycrofts
New York 1947..
[16]O’neill Barrett, Elementary Differential Geometry, AcademicPress
New York 1966.
[17]Panofsky Erwin, Durer As A Mathematician,InTheWorldOfMathematics,
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[18]Requicha A A G, Voelker H B, Constructive Solid Geometry, Technical
Memorandum 25,, Production Automation Project The University Of,
Rochester Rochester N. Y.1977.
[19]Sasaki Tateaki, Multidimensional Monte Carlo Integration Based
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95

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