Chapter 4 Drawing a Square and Inverse Problems

Chapter 4

Drawing a square and inverse problems Math 4520, Fall 2017

4.1 Artistic deﬁnitions

Let us return to our artist painting some object in the picture plane. There is more to the picture than incidence properties such as Desargues’ Theorem. In fact, if the artist draws accurately what is seen, things such as Desargues’ Theorem will take care of themselves. On the other hand, the picture is a very distorted representation of the object. How distorted can it be? What are some properties that we can measure in the picture that will give us information about the object and how it was painted? We make a few definitions that every artist uses whether he knows it or not. The artist’s eye, or center of projection, is called the station point. All the points in the object plane are projected along a straight line through the station point into the picture plane. The line from the station point perpendicular to the picture plane is called the line of sight. The unique point on the line of light in the picture plane is called the center of vision. Note that the center of vision is the nearest point on the picture plane to the station point. Normally one tries to to keep the center of vision near the center of the actual physical painting. The distance between the station point and the center of vision is called the viewing distance. See Figure 4.1 It is usually desirable to keep the viewing distance within some reasonable range. For me 14 inches seems about right. It is not necessary to be ultra exact about these quantities, but if they are grossly out of line, the picture will probably look “wrong”. On the other hand, an artist might deliberately draw with a very short viewing distance in order to enhance the three-dimensional effect. For example, later we will see that when the station point is at infinity, a cube will have two possibilities as to which face is in front. This reversal effect can be disturbing if it is unintended. Recall that the points at infinity in the object plane can be projected onto ordinary points (sometimes called finite points) in the picture plane. We call the projections of these ideal points, the horizon in the picture plane. Note that if the object plane is perpendicular to the picture plane, then the center of vision lies on the horizon. If two lines are parallel in the object plane, then they project onto lines that intersect on the horizon in the object plane. This point of intersection will be called the vanishing point

1 R. Connelly Math 452, Spring 2002

CLASSICAL GEOMETRIES

5. Drawing a square 5.1 Artistic definitions Let us return to our artist pajnting some object in the picture plane. There is more to the picture than incidence properties such as Desargues' Theorem. In fact, if the artist dra\vs accurately what is seen, things such as Desargues'Theorem will take care of themselves. On the other hand, the picture is a very distorted representation of the object. How distorted can it be? What are some properties that we can measure in the picture that will give us information about the object and how it was painted? We make a few definitions that every artist uses whether he knows it or not. The line from the station point perpendicular to the picture plane is called the line of sight. The unique point on the line of light in the picture plane is called the center of vision. Note that the center of vision is the nearest point on the picture plane to the station point. Normally one tries to to keep the center of vision near the the center of the 2CHAPTERcenter 4. of DRAWINGthe actual physical A SQUARE pajnting.. ANDThe distance INVERSE between PROBLEMSMATH the station point and 4520,the FALL 2017 center of vision is called the called the viewing distance. SeeFigure 5.1.1

Center of Horizon line ... Station vision

point ! Vanishing point Viewing for the parallel distance , lines

-.-4' ~

J L Picture plane Object plane Figure 5.1.1 Figure 4.1 It is usually desirable to keep the viewing distance within some reasonable range. For me 14 inches seemsabout right. It is not necessaryto be ultra exact about these quantities, but if they are grossly out of line, the picture will probably look "wrong" . for the parallelRecall lines. that the In points fact, at any infinity single in linethe object (not plane the line can be at projected inﬁnity) onto has ordinary an ideal point on points (sometimes called finite point..) in the picture plane. We call the projections of it, on thethese line ideal at inﬁnity. points, the The horizon projection in the ofpicture this plane. ideal pointNote that will if be the regarded object plane as theis vanishing point of thatperpendicular single line.to the Explicitly,picture plane, the then line the throughcenter of vision the stationlies on the point horizon. parallel to the given line (in the object plane say) intersects the object plane at the vanishing point. 1

4.2 One point perspective

What happens if one of the sides of a square is parallel to the intersection of the picture plane and the object plane? This is called one point perspective. For example, the projection of a square grid in one point perspective is shown in Figure 4.2. The center of vision is then the vanishing point of the side of the square not parallel to the horizon line. The viewing distance is then the distance from the center of vision to either of the 45◦ diagonal vanishing points. DRAWING A SQUARE 7

Vanishing point and Center of vision Horizon 45° Vanishing point

One point perspective

Figure 5.5.1 Figure 4.2 line. The vie\ving distance is then the distance from the center of vision to either of the 45 o diagonal vanishing points. ,\Then both vanishing points of the sides of the projection of the square are finite points, When boththen ,ve vanishing say that the points square of ( or the thesides square of grid the it projectiongenerates) is ofin thetwo point square perspective. are ﬁnite points, then we sayFor that instance, the the square ske\\' (orsquare the in squarethe grid gridof Figure it generates)5.5.1 is shown is generating in two point its own perspective grid . For in two point perspective in Figure 5.5.2. DRAWING A SQUARE 7

Vanishing point and Center of vision Horizon 45° Vanishing point

One point perspective

Figure 5.5.1

line. The vie\ving distance is then the distance from the center of vision to either of the 4.2. ONE45 o diagonal POINT vanishing PERSPECTIVE points. 3 ,\Then both vanishing points of the sides of the projection of the square are finite points, instance,then the ,ve skewsay that square the square in the ( or grid the square of Figure grid 4.2it generates) is shown is generatingin two point perspective. its own grid in two point perspectiveFor instance, intheFigure ske\\' square 4.3. in the grid of Figure 5.5.1 is shown generating its own grid in two point perspective in Figure 5.5.2.

Figure 4.3 4CHAPTER 4. DRAWING A SQUARE AND INVERSE PROBLEMSMATH 4520, FALL 2017 4.3 Projecting a square and locating the center of vision 2 CLASSICAL GEOMETRIES If two lines are parallel in the object plane, then they project onto lines that intersect on the horizon in the object plane. This point of intersection will be called the vanishing point for the parallel lines. In fact, any single line (not the line at infinity) has an ideal Suppose we havepoint a squareon it, on the in linethe at infinity. object The projection plane, of this not ideal necessarily point will be regarded in one as the point perspective. We will describe avanishing construction point of that single in thisline. Explicitly, object the plane line through that the station will point help parallel us locateto the center of vision in the picture.the given Imagine line (in the object we areplane lookingsay) intersects down the object on plane the at the whole vanishing picture point. drawing process

from a long distance5.2 Projecting away, aa bird’ssquare and eye locating view. the Then center theor vision picture plane is viewed edge-on and we see the square withSuppose nowe pave distortion. a square in the Weobject assumeplane. We will that describe the a construction picture inplane this and the object object plane that will help us locate the center of vision in the picture. Imagine we are plane are perpendicularlooking down. (This on the meanswhole picture that drawing the process linesfrom in a long both distance planes, away, a perpendicularbird's eye to the line of intersection, areview. perpendicular Then the picture plane to is eachviewed edge-on other.) and we Let seethe L squarebe withthe no line distortion. of intersection of the ,\re assume that the picture plane and the object plane are perpendicular. (This means object plane and thethat the picture lines in both plane. planes, Figure perpendicular 4.4 to andthe line Figure of intersection, 4.5 showare perpendicular a construction for a line perpendicular to theto each horizon other. I Let line. L be the line of intersection of the object plane and the picture plane.

L L

Start with the given square. Draw Line 1 parallel to L through one corner of the square.

L Draw Line 2 parallel to the diagonal of the square starting at the other end of Line 1

Figure 4.4

It is easy to check (use the similar triangles indicated in Figure 4.5, as well the corresponding equal angles) that Line 4 is perpendicular to Line L. Note that the construction never uses the ability to draw perpendicular lines explicitly. Line 4 turns out to be perpendicular to Line L, but its construction uses only the ability to draw parallel lines. We now mimic this construction in the picture plane as shown in Figure 4.6 and 4.7. We simply remember that parallel lines in the object plane project to lines in the picture plane that intersect at their vanishing point at inﬁnity. All the vanishing points are on the horizon line, which is parallel to the Line L. Since Line 4 is perpendicular to the line L in the object plane, the vanishing point of the corresponding Line 4 on the horizon in the picture plane will be the center of vision. This is because Line 4 in the object plane is parallel to the Line from the station point to the center of vision, perpendicular to the picture plane. See Figure 4.7. 4.3. PROJECTING A SQUARE ANDDRAWING LOCATING A SQUARE THE CENTER OF VISION3 5

DRAWING A SQUARE 3 Une 4 ,1

.. Une 4 ,1

L Draw Line 4 through the other end of Line 3 and the far comer of the square.

It is easy to check ( use theL similar triangles indicated above, as well the corresponding equal angles) that LineDraw 4 is perpendicularLine 4 through to Line the L. other end of Note that the aboveLine construction 3 and the never far comeruses the of ability the square.to draw perpendicular lines. Line 4 turns out to be perpendicular to Line L, but its construction does not use the ability to draw a perpendicular line. We did use the ability to draw parallel lines, however. It is easyWe to now check mimic ( usethis construction the similar in triangles the picture indicated plane. We above,simply remember as well the that correspondingparallel equal angles)lines in thatthe object Line plane 4 is project perpendicular to linesFigure in theto pictureLine 4.5 L. plane that intersect at their vanishing Note pointthat atthe infinity. above All constructionthe vanishing pointsnever are uses on the the horizon ability line, to whichdraw is perpendicular parallel to the lines. Line 4 turnsLine L. out to be perpendicular to Line L, but its construction does not use the ability to draw a perpendicular line. We did use the ability to draw parallel lines, however. We now mimic this construction in the picture plane. We simply remember that parallel lines in the object plane project to lines in the picture plane that intersect at their vanishing point at infinity. All the vanishing points are on the horizon line, which is parallel to the Line L. L L Start with the projected squarewhich Draw Line 1, as before, parallel to the has its vanishing points for both pairs of Line L. opposite sides on the horizon line. The vanishing point for the diagonal also lies on the horizon line.

L L Start with the projected squarewhich Draw Line 1, as before, parallel to the has its vanishing points for both pairs of Line L. opposite sides on the horizon line. The 'V L L vanishing point for the diagonal also lies Draw Line 2 from the vanishing point Draw Line 3 from the endpoint of Line on the horizonof the diagonal line. through the endpoint of 2 in the square to the vanishing point of Line 1. the other side.

Figure 4.6

'V Thus we nowL have a simple construction thatL allows us to locate the center of vision, assuming thatDraw we haveLine 2 afrom picture the vanishing of a square point in aDraw perpendicular Line 3 from the object endpoint plane. of Line of the diagonal through the endpoint of 2 in the square to the vanishing point of Line 1. the other side. 6CHAPTER 4. DRAWING A SQUARE AND INVERSE PROBLEMSMATH 4520, FALL 2017 4 CLASSICAL GEOMETRIES

L Draw Line 4 through the end of Line 3 and the far corner of the projected square.

Since Line 4 is perpendicular to theFigure line L in 4.7 the object plane, the vanishing point of the corresponding Line 4 on the horizon in the picture plane will be the center of vision. This is because Line 4 in the object plane is parallel to the Line from the station point to the center of vision, perpendicular to the picture plane. Thus we now have a simple construction that allov.'s us to locate the center of vision, assuming that we have a picture 4.4 Determiningof a square in a perpendicular the object viewing plane. distance

5.3 Determining the viewing distance We now take up the problem of finding how far away from the picture plane we should stand to view a squareWe now properly. take up the In problem other of words, finding wehow wantfar away to from find the the picture viewing plane we distance. should We make stand to ,.iew a square properly. In other words, we want to find the viewing distance. We the same assumptionsmake the same as assumptions in Section as in 4.3. Section The 5.2. objectThe object plane plane is is perpendicular perpendicular to the to the picture plane and wepicture have plane a squareand we have in the a square object in the plane object projected plane projected into into the the picture picture plane. plane, as seen in Figure 4.8 Here again we project the whole configuration into the object plane so that the picture plane is projected onto a line. It is viewed edge-on. In Figure 5.3.1 we show this parallel Here againprojection we project perpendicular the wholeto the object configuration plane. (This intois projection the object from the plane point at so infinity that the picture plane is projectedon the line onto perpendicular a line. toIt the is object viewed plane. edge-on.) In Figure 4.8 we show this parallel From Figure 5.3.1 it is clear that if one draws the lines from the station point parallel projection perpendicularto the sides of theto square the in object the object plane. plane, (This then these is projection two lines are from perpendicular the point in at infinity on the lineEuclidean perpendicular 3-space toand thewell objectas i~ Figure plane.) 5.2. So the station point and the two vanishing points form a right triangle with the line of sight as the altitude. The viewing distance is From Figurethe length 4.8 of it this is clearaltitude. that Note if that one the draws vanishing the points lines and from the thestation station point lie point on a parallel to the adjacentplane sides parallel of the to the square object inplane. the object plane, then these two lines are perpendicular in Euclidean 3-spaceSo if one wishes and wellto find as the in proper Figure point 5.2. to view So a the square, station one can point combine and the thebottom two vanishing half of Figure 5.3.1 and the construction above to construct the center of vision and the points form"folded a right down" triangle flap that with determines the line the of viewing sight distance. as the altitude.Since the triangle The in viewing Figure 5.3.1 distance is the length of thisis a altitude.right triangle, Note it lies that on a circle the whose vanishing center is points the midpoint and between the station the two point vanishing lie on a plane parallel to thepoints object of the plane.sides of the projected square. SeeFigure 5.3.2. . H one wishes a more analytic description of the viewing distance, let dl and d2 be the So if onedistances wishes from to findthe center the properof vision pointto the two to viewvanishing a square, points for one the cantwo sides combine of the the bottom half of Figureprojected 4.8 and square. the constructionThen the viewing above distance to constructd is the geometric the center mean of of dl vision and d2. and In the “folded down” flap that determines the viewing distance. Since the triangle in 4.8 is a right triangle, it lies on a circle whose center is the midpoint between the two vanishing points of the sides of the projected square.

If one wishes a more analytic description of the viewing distance, let d1 and d2 be the distances from the center of vision to the two vanishing points for the two sides of the projected square. Then the viewing distance d is the geometric mean of d1 and d2. (This is a good exercise in Euclidean geometry.) In other words,

p d = d1d2 . 4.5. DETERMINING THE SQUARE 7 DRAWING A SQUARE 5

Figure 5.3.2 Figure 4.8

4.5 Determining the square

Notice that the vanishing point for the 45◦ diagonal is between the two vanishing points for the two sides. Also the line from the 45◦ vanishing point to the corner of the square above the horizon line in Figure 4.8, when folded out actually makes an angle of 45◦. This fact can be used to run the construction another way. Suppose that the vanishing points for the sides of the square are known (or chosen), and the center of vision is known. Then the corner of the right triangle above the horizon lies on the circle as before. This corner also lies on the 6 CLASSICAL GEOMETRIES

other words, d = .;d;ii; .

This is a ,vell-know result in Euclidean geometry (and a good exercise).

5.4 Determining the square 8CHAPTER 4. DRAWING A SQUARE AND INVERSE PROBLEMSMATH 4520, FALL 2017 line through the center of vision perpendicular to the horizon line. The intersection of this line and the semi-circle determines the corner. Then the right angle can be bisected. Where this bisecting line intersects the horizon line will be the vanishing point for the 45◦ diagonal of the projection of the square. Once all three vanishing points for the square are known (or constructed), then the projection of the square can be readily drawn. See Figure 4.9, where the order of construction is indicated. Notice that this construction uses more than just the incidence structure in that at a crucial point we need to bisect an angle. Implicitly a compass is used.

, , ... , ".. ".- , ~

Figure 5.4.1 Figure 4.9

5.5 One point perspective and two point p'erspective What happens if one of the sides of the square is parallel to the intersection of the picture plane and the object plane? This is called one point per-,pective.For exanlple, the 4.6 Exercisesprojection of a square grid in one point perspective is shown in Figure 5.5.1. The center of vision is then the vanishing point of the side of the square not parallel to the horizon In problem 1, you will need to copy some pictures that have good examples of one or two point perspective onto a tranparancy. (Be careful to get the correct sort of transparancy for this. The wrong type will melt in the copying machine.) You can use some of the sketches in Geometer’s Sketchpad (or you can make your own, you can use pencil and paper, or use whatever software you are comfortable with) to locate the center of vision and the viewing distance. ~ 1. Using Geometer’s Sketchpad, Cinderella (or other methods, if you like) to locate the center of vision and the viewing distance for your original square and the drawings in the handouts and other pictures that you can find. Try to find 4 or 5 drawings for this exercise. Hand in a copy of the transparancy and the final display as you have positioned the points for the calculation.

1 00 00 2. Suppose that a standard 8 2 by 11 piece of paper has a picture of a square with the vanishing points of both sides on the paper. If the paper is held as this paper is read, long side up, what is the largest the viewing distance can be? Is this a comfortable distance for you? 3. Suppose one has a quadrilateral in the picture plane below the horizon with all internal angles less than 180◦ Using the construction outlined in Sections 4.3 and 4.4 , one ﬁnds a station point in 3-space. When the quadrilateral is projected back into object plane, do we get a square? In other words, do all reasonable quadrilaterals in the picture plane come from a square in the object plane? (Hint: Use the 45◦ diagonal line.) 8 CLASSICAL GEOMETRIES

Exercises:

1. Locate the center of vision and the viewing distance for your original square and the drawings in the handouts. Supposethat a standard 8! " by 11" piece of paper has a picture of a square with 2. the vanishing points of both sides on the paper. H the paper is held as this paper is read, long side up, what is the largest the viewing distance can be? Is this a comfortable distance for you? 3. Describe a procedure for drawing a square, where one or both vanishing points are off the paper in the picture plane. Draw such a square with viewing distance of about 16" with center of vision in the center of the paper. Thy not to mark up your desk. You can allow yourself to draw parallels without a compass. 4. Suppose one has a quadrilateral in the picture plane below the horizon with all internal angles less than 180° Using the construction outlined in Sections 5.2 and 5.3, one finds a station point in 3-space. When the quadrilateral is projected back into object plane, do we get a square? In other words, do all reasonable 4.6.quadrilaterals EXERCISES in the picture plane come from a square in the object plane? (Hint:9 Use the 45° diagonal line.) 5. 4.Prove Prove the statementstatement claimed claimed earlier earlier that that the altitudethe altitude of a right of trianglea right istriangle the geometric is the geometricmean of the mean two of segments the two of segments the hypotenuse.of the hypotenuse.

d, ~

Figure 5.E.l Figure 4.10

6. Justify the claim in Section 5.6 about the viewing distancefor one point perspective. 7. Finish the dra\\'ings in the handouts where the center of vision, the horizon line, 5.and Justify the thetwo claim vanishing in Section points 4.2 are about given. the viewingDraw a distancegrid in two forone point point perspective. perspective.