Mathematics_and_Art-Marc_Frantz.pdf

Draft, May, 1997

MATHEMATICS AND ART

Marc Frantz

Department of Mathematical Sciences

IUPUI

402 North Blackford Street

Indianapolis, Indiana 46202-3216

mfrantz@math.iupui.edu

n f

This project was supported, in part,

by the

National Science Foundation

Opinions expressed are those of the authors

and not necessarily those of the Foundation

Linear Perspective

1. Basic Results

Figure 1(a).

Figure 1(b).

Take a good look at the images in Figure 1. Figure 1(a) is a detail from an illumi-

nated manuscript called the Tr es Riches Heures, painted by the Limbourg brothers, Paul,

Hermann and Jean, in the fteenth century. Figure 1(b) is a detail from an eighteenth

century oil painting by the Italian artist Canaletto (Giovanni Antonio Canal), entitled

View of the Grand Canal towards the Pallazzo Contarini dagli Scrigni. Of all the things

one might notice about these works, we would like to focus on one speci c aspect, namely

the illusion of depth in three-dimensional space, which is far more e ective and believable

in the painting on the right. Given that the artists in either case were considered very

competent in their time, it would appear that something happened in Europe between

the fteenth century and the eighteenth century that added a powerful new tool to the

artist’s toolbox. The \something" was the Renaissance, a period of vigorous and creative

activity in the arts and sciences. The powerful new tool was a mathematical technique

called \linear perspective," or just \perspective."

An illusion of depth is not necessarily the most important aspect|or even a necessary

aspect|of a successful painting. But whenever the e ect has been needed, it has been

found that a knowledge of mathematics is an indispensable tool in achieving it. In fact, this

is why we have chosen to compare two paintings from Europe, for nowhere was the e ect

of perspective upon art more dramatic. Moreover, the use of mathematical perspective is

perhaps more important today than ever before, if one considers modern creative media

such as lmmaking, computer graphics, and virtual reality that have been made possible

by advances in technology.

Figure 2. Computer-aided perspective helps bring to life a

landscape with dinosaurs in Jurassic Park .

The idea behind the concept of \linear perspective" is illustrated in Figure 3. We

imagine a viewer using only one eye, with the eye located at E(0; 0; d) in a three di-

mensional coordinate system, looking in the direction of the positive z-axis. The point

P (x; y; z) on the vase in the gure is visible to the viewer, and light re ected from that

point travels in a straight line to the viewer’s eye, piercing the \picture plane" z =0at

the point P 0 (x 0 ;y 0 ;0). We think of the picture plane as the surface of a canvas on which

we wish to paint a realistic picture of the vase; thus the point P 0 on the canvas should

be painted as a small dot of the same color as the light re ected from the point P on the

vase, so that the viewer will see the same thing from that direction that he or she would

see if the vase were actually there.

EP= á x,y,z+d ñ

EP'= á x',y',d ñ

P ( x,y,z )

P' ( x',y', 0)

eye of viewer

at E (0,0, –d )

picture plane z= 0

Figure 3. P 0

is the perspective image of P .

The point P 0 is called the perspective image (or sometimes the central projection )

of P . If we identify the vase with the set of all visible points on its surface, then the

perspective image of the vase (represented by the small vase image in the gure) is the set

of perspective images of those visible points.

We only need the two coordinates ( x 0 ;y 0 ) to locate a point in the picture plane. From

now on a pair ( x 0 ;y 0 ) of primed coordinates will indicate a point in the picture plane (the

coordinates of this point in space would be ( x 0 ;y 0 ;0)). To determine x 0 and y 0

in Figure 3,

thx 0 ;y 0 ;di=hx; y; z + di:

Equating components and solving for u, v, and t,weget

x 0 =x=t; y 0 = y=t;

t =(z+d)=d:

Substituting for t in the expressions for x 0 and y 0 ,weget

x 0 = dx

z + d

and y 0 =

z + d :

(1)

Thus, if we know the coordinates ( x; y; z) of a point on the vase|or any other object|

and if we know the distance d from the viewer’s eye to the picture plane, we can determine

the coordinates ( x 0 ;y 0 ) of the corresponding perspective image in the picture plane using

the formulas in (1).

Example 1. Assume we have the setup in Figure 3, with the viewer 3 units from the

picture plane. If P (2; 4; 5) is a point on an object we wish to paint, nd the picture plane

coordinates coordinates ( x 0 ;y 0 ) of the perspective image of P .

Solution. We have d =3,x=2,y= 4, and z = 5. Thus, by (1),

x 0 =

(5) + (3) = 6

8 = 3

and y 0 =

(5) + (3) = 12

8 = 3

2 :

Notice in Figure 3 that we have put the vase on the opposite side of the picture

plane from the viewer; that is, the z-coordinates of all the points on the vase are positive.

Although it is possible to do the same kind of perspective treatment when the object

is between the viewer and the picture plane (in which case the image gets larger than

the object), we shall assume from now on that any object we wish to make a picture of

consists of points whose z-coordinates are positive. Thus, in a sense, the image will always

be smaller than the object. (We say \in a sense" because we have not been clear about the

meaning of \smaller." To be more precise, if P 1 and P 2 are two points on the object, then

notice that the vectors !

EP 0 = hx 0 ;y 0 ;di and EP = hx; y; z + di are parallel, and hence

there is a positive number t such that

(3)(2)

(3)(4)

the distance between their corresponding perspective images P 1 and P 2 will be smaller

than the distance between P 1 and P 2 ; Exercise # gives you some hints and asks you to

prove this.)

Of course, we would like to do problems that are more interesting than Example 1, but

on the other hand, we don’t want to knock ourselves out doing lots of tedious computations.

To obtain some nice shortcuts that will help us do interesting drawings, we will derive

important drawing rules from (1). To start with, we would like to say that the perspective

image of a straight line is also a straight line. Unfortunately, that’s not quite true. For

example, a line that goes through the viewer’s eye, such as the line through E and P in

Figure 3, would be represented by a single point in the picture plane (which makes sense,

because a thin pencil lead, seen end-on, looks like a dot, not a line segment). However, the

following theorem gives us some help in this regard. For convenience, we stick to our rule

that any object we might want to depict consists of points with positive z-coordinates.

Theorem 1. If S is a subset of a straight line L in space (S might be a line segment, a

point, a ray, etc.), and if all the points of S have positive z-coordinates, then the perspective

image S 0 of S is a subset of a straight line Ax 0 + By 0 = C in the picture plane.

Proof. Let us assume that L is given parametrically by

x = x 0 + at; y = y 0 + bt; z = z 0 + ct;

where 1 <t<1and x 0 ;a;y 0 ;b;z 0 ;c are constants. Now let P be any point of S.

Then P =(x 0 +at; y 0 + bt; z 0 + ct) for some value of t, and by (1), the perspective image

P 0 =(x 0 ;y 0 ;0) of P satis es

x 0 =

d(x 0 + at)

(z 0 + ct)+d

and y 0 =

(z 0 + ct)+d :

(Since z = z 0 + ct is positive, we are not dividing by zero.) If we de ne the constants A,

B, and C by

A = bz 0 + bd cy 0 ; B=cx 0 ad az 0 ; C=d(bx 0 ay 0 );

it is straightforward (but tedious) to check that Ax 0 + By 0 = C. Since the point P of S

was arbitrary, the theorem is proved.

Although it’s unfortunate that we had to say something technical instead of \The

perspective image of a straight line is also a straight line," there are some useful drawing

tricks rel ated t o Theorem 1. One useful related fact is that the perspective image of a line

segment P 1 P 2 is not only a subset of a straight line, it is (not surprisingly) either a point

or a line segment P 1 P 2 , where P 1 and P 2 are the corresponding perspective images of P 1

and P 2 (Figure 4).

d(y 0 + bt)

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