Mathematics and Art (pdf document) free file download at fliiby.com

"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 c by Marc Frantz 1997. All rights reserved. IO NA L SC I This project was supported, in part, E NC N AT nf O UN N National Science Foundation Opinions expressed are those of the authors and not necessarily those of the Foundation by the E F DA O TI 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 illuminated manuscript called the Tres 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 specic aspect, namely the illusion of depth in three-dimensional space, which is far more eective 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 eect 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 eect 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. 1 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 dimensional 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 = 0 at the point P 0 (x0 ; y0 ; 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=〈x,y,z+d〉 EP'=〈x',y',d〉 y P(x,y,z) P'(x',y',0) eye of viewer at E(0,0,–d) z x picture plane z=0 Figure 3. P 0 is the perspective image of P . 2 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 (x0 ; y0 ) to locate a point in the picture plane. From now on a pair (x0 ; y0 ) of primed coordinates will indicate a point in the picture plane (the coordinates of this point in space would be (x0 ; y0 ; 0)). To determine x0 and y0 in Figure 3, ! ! notice that the vectors EP 0 = hx0 ; y0 ; di and EP = hx; y; z + di are parallel, and hence there is a positive number t such that thx0 ; y0 ; di = hx; y; z + di: Equating components and solving for u, v, and t, we get x0 = x=t; x0 = z dx d + y0 = y=t; and t = (z + d)=d: y0 = z dy d : + (1) Substituting for t in the expressions for x0 and y0 , we get 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 (x0 ; y0 ) of the corresponding perspective image in the picture plane using the formulas in (1). picture plane. If P (2; 4; 5) is a point on an object we wish to paint, nd the picture plane coordinates coordinates (x0 ; y0 ) of the perspective image of P . Example 1. Assume we have the setup in Figure 3, with the viewer 3 units from the Solution. We have d = 3, x = 2, y = 4, and z = 5. Thus, by (1), 6 3 (3)(2) x0 = (5) + (3) = 8 = 4 and 3 (3)(4) y0 = (5) + (3) = 12 = 2 : 8 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 P1 and P2 are two points on the object, then 3 the distance between their corresponding perspective images P10 and P20 will be smaller than the distance between P1 and P2 ; 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 Ax0 + By0 = C in the picture plane. Proof. Let us assume that L is given parametrically by x = x0 + at; y = y0 + bt; z = z0 + ct; where 1 < t < 1 and x0 ; a; y0 ; b; z0 ; c are constants. Now let P be any point of S . Then P = (x0 + at; y0 + bt; z0 + ct) for some value of t, and by (1), the perspective image P 0 = (x0 ; y0 ; 0) of P satises d + x0 = (z (x0 ct)at)d + + 0 and + y0 = (zd(y0 ct)bt) d : + + 0 (Since z = z0 + ct is positive, we are not dividing by zero.) If we dene the constants A, B , and C by A = bz0 + bd cy0 ; B = cx0 ad az0 ; C = d(bx0 ay0 ); it is straightforward (but tedious) to check that Ax0 + By0 = 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 related to Theorem 1. One useful related fact is that the perspective image of a line segment P1 P2 is not only a subset of a straight line, it is (not surprisingly) either a point or a line segment P10 P20 , where P10 and P20 are the corresponding perspective images of P1 and P2 (Figure 4). 4 y P2 P'2 E z x P'1 P1 picture plane Figure 4. The perspective image of P1 P2 is P10 P20 . Instead of proving this fact, let's use it to do Example 2. The visible corners of a cube in space are M (4; 2; 3); N (6; 2; 3); P (4; 4; 3); Q(6; 4; 3); R(4; 2; 5); S (6; 2; 5); T (4; 4; 5): If the viewer is located at E (0; 0; 3), make a line drawing to show the cube the way the viewer sees it. Solution. Our rst step is to use (1) to determine the x0 y0 -coordinates of the corresponding 3 (3)(4) x0 = (5) + (3) = 12 = 2 8 and 12 3 (3)( 4) y0 = (5) + (3) = 8 = 2 : perspective images M 0 , N 0 , P 0 , Q0 , R0 , S 0 , T 0 . Since the z -coordinate of E is 3, we have d = 3, and we can, for example, compute the x0 y0 -coordinates of T 0 as follows: Proceeding in this fashion, we can specify each perspective image with an ordered pair (x0 ; y0 ) and get the points M 0 (2; 1); N 0 (3; 1); P 0 (2; 2); Q0 (3; 2); 0 3; 3 ; R 2 4 0 9; 3 ; S 4 4 0 3; 3 : T 2 2 Now the edges of the cube are line segments in space, and by our comment above, their perspective images are line segments in the picture plane. Thus, to make a line drawing of the cube, we plot the primed points that represent the corners, and connect appropriate corners with line segments (Figure 5). 5 y' 1 -3 -2 -1 1 R' 2 S' M' T' P' 3 x' -1 -2 N' Q' Figure 5. A cube in perspective. Although Figure 5 is not exactly a masterpiece, we can make some interesting observations about it. First however, test your grasp of the geometry involved by answering these questions: (i) There is one corner point U (x; y; z ) of the cube which is not visible. What are its coordinates? Deduce them by thinking about the coordinates of M , N , P , Q, R, S , and T . (ii) In Figure 5, use a pencil to indicate where you think the perspective image U 0 of U would be if the cube were transparent. What are its picture plane coordinates (x0 ; y0 )? (iii) Check your answers to questions (i) and (ii) for consistency by applying (1) to your coordinates for U to determine x0 and y0 . Now let's make some observations about Figure 5. We know that the object depicted there is a cube, and hence angles such as 6 QPM and 6 TPM are in reality right angles. However, if we look at the corresponding angles 6 Q0 P 0 M 0 and 6 T 0 P 0 M 0 in Figure 5, we see that 6 Q0 P 0 M 0 is a right angle, while 6 T 0 P 0 M 0 is not. Similarly, each face of the cube is in reality a 22 square, but the only visible face of the cube whose image is a square is the face MNPQ, whose image is the 1 1 square M 0 N 0 P 0 Q0 in Figure 5. If you answered Questions (i) and (ii) correctly, you found that the hidden corner U had coordinates (6; 4; 5) and its image U 0 had picture plane coordinates (9=4; 3=2). Thus the image R0 S 0 T 0 U 0 of the hidden \back face" RSTU is also a square, and its dimensions are 3=4 3=4. In other words, the two faces RSTU and MNPQ of the cube have perspective images R0 S 0 T 0 U 0 and M 0 N 0 P 0 Q0 that are miniature, but undistorted, versions of the original square faces. It also happens that these two faces of the cube are parallel to the picture plane z = 0. This can be checked by noticing that the points R, S , T , U all have a z -coordinate of 5, and the points M , N , P , Q all have a z -coordinate of 3. Thus R, S , T , U belong to the plane z = 5, while M , N , P , Q belong to the plane z = 3. 6 This state of aairs is not an accident; in fact, it reflects a more general principle which artists have taken advantage of for hundreds of years, namely, that any shape which lies in a plane parallel to the picture plane will have a perspective image that is an exact, undistorted miniature of the original (Figure 6). Such shapes are relatively easy to draw, because they don't have to be \distorted" mathematically to give an illusion of depth. P1 surface parallel to picture plane picture plane y' P'1 P2 CAT FOOD P'2 CAT FOOD x' perspective image viewer Figure 6. Since the cat food label is parallel to the picture plane, its perspective image is undistorted. You probably have a pretty good idea of what we mean by an \undistorted miniature" by looking at Figure 6, but we should be precise about the idea, since it turns out to be very convenient for artists. As a rst step in making things more precise, we present Theorem 2: while reading it, think of S as the set of points on the cat food label in Figure 6, and think of P1 and P2 as any two points on the label. plane, and let the viewer be located at E (0; 0; d) with d > 0. Then there exists a constant , with 0 < < 1, such that for any two points P1 (x1 ; y1 ; z0 ) and P2 (x2 ; y2 ; z0 ) in S , their 0 0 corresponding perspective images P10 (x01 ; y1 ; 0) and P20 (x02 ; y2 ; 0) are exactly times as far ! ! apart as P1 and P2 , and the vector P10 P20 is parallel to the vector P1 P2 . In other words, ! ! P10 P20 = P1 P2 . 7 Theorem 2. Let S be a set of points in a plane z = z (z > 0) parallel to the picture 0 0 Proof. From (1) we can compute 0 0 x01 = z dx1 d ; y1 = z dy1 d ; x02 = z dx2 d ; y2 = z dy2 d : + + + + 0 0 0 0 If we dene the constant by = d=(z0 + d), then 0 < < 1 since d and z0 are positive constants, and we can rewrite the above equations as 0 0 x01 = x1 ; y1 = y1 ; x02 = x2 ; y2 = y2 : Thus ! ! 0 0 P10 P20 = hx02 x01 ; y2 y1 ; 0i = hx2 x1 ; y2 y1 ; 0i = hx2 x1 ; y2 y1 ; z0 z0 i = P1 P2 : To grasp the meaning of Theorem 2, suppose that the viewer is 3 units from the picture plane (that is, let d = 3), and consider an object, say, an equilateral triangle with sides of length 2, that lies in the plane z = 4. Then the number in Theorem 2 is given by = d=(z0 + d) = 3=(4 + 3) = 3=7. Theorem 2 says that the images of any two points in the plane z = 4 will be 3=7 as far apart as the actual points; thus the image of each side of the equilateral triangle will be a line segment of length (3=7) 2 = 6=7. It follows that the image of the equilateral triangle is also an equilateral triangle, and the angles are preserved (their images are still 60 angles). It's not hard to prove in general that any angle in a plane z = z0 will have an image with the same degree measure (Exercise #). What we have said so far means that \shapes are preserved." But what about orientation? That is, is it possible that the image of an equilateral triangle is an upside down version of the original triangle, or tilted at some strange angle? The answer is no, because the ! vector P1 P2 connecting any two points P1 , P2 (such as two vertices of the triangle) in the ! ! plane z = 4 will have a perspective image P10 P20 = (3=7)P1 P2 that is parallel to it, so the image is not rotated in any way. This non-distortion eect can be seen in Figure 7. In this picture, we are looking at a picture plane that is vertical with respect to the (locally flat) earth. Several buildings on the right side of the painting have walls that are parallel to the picture plane, and hence their images, as well as the images of their doors and windows, are undistorted. As we mentioned earlier, this makes it easier for the artist to render these parts of the painting, and indeed many paintings and drawings of buildings feature walls that are parallel to the picture plane. However, convenience is not the only reason for drawing buildings in this way. Remember that most paintings are done on a rectangular canvas, and hence objects in the painting that have a rectangular shape will help organize the composition and keep it in harmony with its border. (Several lines that parallel the edges of the canvas have been indicated in the gure.) Even if the face of a building is not parallel to the picture plane, the edges of the buildings|which are 8 vertical lines in space|will each lie in some plane that is parallel to the picture plane, and hence by Theorem 2 the images of the lines will also be vertical; this adds still more lines that are parallel to edges of the canvas, such as the two vertical lines indicated at the left side of the picture. One compositional use of such lines is illustrated by the vertical edge of the tall building that can just barely be seen at the far right edge of the canvas. As the viewer's eye scans the rooftops from left to right, the eye tends to be led o the edge of the painting. This vertical line serves to \stop the eye" and bring the viewer's attention back to the center of the painting. This little trick is a direct application of the ! ! ! rule P10 P20 = P1 P2 derived in Theorem 2: The vector P1 P2 is the actual vertical edge of ! the building, and the line segment in the painting is the vector P10 P20 , which is also vertical, and times smaller. Figure 7. A larger view of Canaletto's painting in Figure 1a. Horizontal and vertical lines help organize the composition and relate the objects in the painting to the border of the canvas. Besides the non-distortion phenomenon, there is another important idea we can grasp by looking at the cube from Figure 5, which has been reproduced in Figure 8 without all the labels. Take a look at the cube we drew. Does it really look like a cube? If you're looking at it from a typical reading distance, it probably seems too elongated in the direction of 9 your line of sight|more like the shape of dumpster or a cedar chest, than a die or a sugar cube. y' 1 -3 -2 -1 1 2 3 x' -1 -2 Figure 8. The drawing from Figure 5. At a natural viewing distance (as opposed to the correct 3 units) the cube looks too long in the direction of the line of sight. y' 1 -5 -4 -3 -2 -1 1 2 3 4 5 x' -1 -2 -3 Figure 9. The same cube and picture plane, with the correct viewing distance increased to a more natural 21 units. There's a good reason{or perhaps we should say a bad reason|for this. Remember that the correct viewing distance from the picture plane is d = 3 units. If you hold a scrap of paper up to the x0 -axis, mark o 3 units, and use the paper to place your eye 3 units from the origin, your nose will almost touch the page! At this distance you won't be able to focus properly, but if you glance o to the side towards the image, you should be able to notice that it now looks like a (blurry) cube. Thus, even though our drawing is mathematically correct, it's not a good drawing, because during the planning stage we failed to take into account the actual nished size of the drawing and the natural distance 10 at which a viewer would want to look at the picture. If we use the same cube and picture plane, but change the viewing distance to a more reasonable value of 21 units (that is, leave everything the same as in Example 2, except move the viewer back to E (0; 0; 21)) we get the drawing in Figure 9, which looks much more cube-like from a normal viewing distance. Now you may argue that the only thing wrong with Figure 8 is that the drawing is too small. After all, if it were so big that the units were, say, in feet, then the viewing distance would be 3 feet, which is far enough away that one could at least focus on the image. To see if this approach works, we have made an enlarged detail of Figure 8 in Figure 10. 1 2 3 -1 -2 Figure 10. Enlarged detail of Figure 8. Of course, we couldn't make it big enough so that the viewing distance is 3 feet, but you should be able to focus on the image with your eye 3 units from the origin. At this point, however, a new problem becomes evident: the image of the box is so far from the center of view (the origin) that it is awkward to look that far o to the side. In fact, no matter how much we enlarge the picture, anyone viewing from 3 units away from the origin (measured perpendicularly from the page) would have to look o to the side at more than a 45 angle to see the lower right corner of the box. (Prove it!) It is gen..."

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