These are three completely different things. Access is designed to be a database thatâs what it does best. It front-ends itself which means that you donât need anything external in order to create a GUI that can be used by typical users. Draw the axes first. Then loop over X coordinates drawing lines between points on a function. Private Sub FormLoad Dim i As Integer Dim x As Single Dim y As Single Picture1.Scale (-10, 10)-(10, -10) ' Draw X axis.
Abstract. We present some mathematical ideas that occur in art and computergraphics. We touch upon the geometry of similar triangles, rigid motions in three space,perspective transformations, and projective geometry. We discuss computations behind renderingobjects in perspective. We then describe vanishing points, answer how to measure distance in areceding direction in a perspective drawing and why a circle in three space becomes an ellipse whendrawn in perspective. We use MAPLE generated computations and graphics to illustrate the ideas. ``I freely confess that I never had a taste for study or researcheither in physics or geometry except in so far as they could serve as a means of arriving at somesort of knowledge of the proximate causes...for the good and convenience of life, in maintaininghealth, in the practice of some art...having observed that a good part of the arts is based ongeometry, among others the cutting of stone in architecture, that of sundials, and that ofperspective in particular.´´
Artists now make amazing images using computers (e.g., visit PIXAR.). But taking into account the complexity of all thephysical laws of light requires extensive computation. We'll focus on mathematical concepts andderive the basic formulas that are at the heart of all rendering machines. We shall concentrate onthe geometry of drawing objects, which can be described by points in space. We'll calculate theperspective transformations that locate the points on the drawing. Then we'll use MAPLE'scapability to draw polygons and lines to show the effect of transformation. We'll apply this toshow some mathematical / artistic concepts. (This note is a slightly expanded version of lecturenotes [TA].)
A mathematical theory of perspective drawing could only be developed when the Renaissance freedpainters to depict nature in a way closer to what they observed [IW]. The biographer Vasari(1511-74) says that the Florentine architect Filipo Brunelleschi (1337-1446) studied Greekgeometry, developed a theory of perspective and undertook painting just to apply his geometry [KM].The first treatise, Della pittura(1435) by Leone Battista Alberti (1404-72) furnished mostof the rules. Our diagram of the perspective view of the circle occurs inhis text. A complete mathematical treatment De prospectiva pingendi (1478) was given by theItalian fresco painter Piero della Francesca (1410-1492). Leonardo da Vinci (1452-1519)incorporated geometry in his painting and wrote a now lost text on perspective Tratto dellapittura. Albrecht Dürer (1471-1528) also wrote a text on the practice of geometryUnderweysung der Messung mid dem Zyrkel und Rychtscheyd (1525) which was important inpassing on to the Germans the Italian knowledge of perspective drawing. In it, Dürer invented several drawing machines to teach perspective. Alberti was first to ask iftwo drawing screens are interposed between the viewer and the object, and the object is projectedonto both resulting in two different pictures of the same scene, what properties do the twopictures have in common [KM2].
Alberti's Problem: What do the two projections have in common?
This question prompted the development of a new subject, projective geometry whoseexponent was Girard Desargues (1591-1661). Desargues studied perspective geometry from a syntheticpoint of view, meaning he built up the geometry from axioms about points, lines and planes. Asampling is given in the section on projective geometry. There we addressthe question why the perspective image of a circle necessarily the ellipse. It can also be answeredusing analytic geometry methods, such as in our chapter on analytic geometry,where first, points and lines are reduced to equations. A modern deductive footing for perspectivedrawing was given later by Brook Taylor (1685-1731) and J. H. Lambert (1728-77). Acompeting point of view has held by mathematicians such as René Decartes (1596-1650), Pierrede Fermat (1601-1665) and Julius Plücker (1801-1868) who studied these question algebraically.Their work spurred the development of algebraic geometry. Mathematical issues and historyare more completely covered in [BP], [DH], [FP], [M], [OW], [PD], [SD] and [WC].
Some current popular drawing texts such as Edwards' [EB] de-emphasize the analytical approach infavor of an intuitive sighting method for perspective. However, complicated drawing situationsrequire more analysis [EB], [EM]. If a student wishes to pursue linear perspective in the historyand art see, e.g. [CA]. There are a number of web sites worth perusing, [DJ], [EM], [MS] asare sites discussing perspective and the internet [KJ], [XX]. The computer science of graphics isdiscussed in [BP], [FP], [PP]. Many of our illustrations were generated by the MAPLE symbolicalgebra, graphics and computation system. Our code in the MAPLE programming language is availableat http://www.math.utah.edu/~treiberg/HS.htmlDifferential geometric graphic applications of MAPLE are described in Oprea [OJ] and Rovenski [RV]and of MATHEMATICA in Gray [GA]. Other graphics packages to render differential geometric objects, e.g., are Richard Palais' 3D-Filmstripor Konrad Polthier's JavaView.
The perspective transformations that describe how a point in three space is mapped to thedrawing plane can be simply explained using elementary geometry. We begin by setting upcoordinates. A projection involves two coordinate systems. A point in the coordinate system of anobject to be drawn is given by X=(x,y,z) and the corresponding in the imaging system (onthe drawing plane) is P=(u,v). If we use the standard right handed
Projecting an object to the drawing plane.
system, then x and y correspond to width and depth and zcorresponds to height. On the drawing plane, we let u be the horizontal variable andv the vertical.
We can measure the distances between pairs of points in the usual way using the Euclideanmetric. If
X1 = (x1, y1, z1) andP1 = (u1, v1) and so on, then
dist(X1, X2) = {(x1 - x2)2 +(y1 - y2)2 + (z1 - z2)2}1/2,
dist(P1, P2) = {(u1 - u2)2 + (v1 - v2)2}1/2
The projection from X to P is called a parallelprojection if all sets of parallel lines in the object are mapped to parallel lines on thedrawing. Such a mapping is given by an affine transformation, which is of the form
= f(X) = T + AX
where T is a fixed vector in the planeand A is a 3 x 2 constant matrix. Parallel projection has thefurther property that ratios are preserved. That is if X1,X2, X3 and X4 are collinearpoints in the object, then the ratio of distances is preserved under parallel projection
Of course denominators are assumed to be nonzero.
To illustrate, let's begin with an object in three space, say a simplified house. It consists ofthe points [0,0,0], [0,0,3], [3.5,0,5], [7,0,3],[7,0,0], [0,9,3], [0,9,0], [7,9,3], [7,9,0],[3.5,9,5] which define eight corners of a box and two gable points and[3,9.1,0], [5,9.1,0], [5,9.1,8], [3,9.1,8],[3,10.2,0], [5,10.2,0], [5,10.2,8], [5,10.2,8] whichdefine the chimney.
MAPLE generated 3d plot of house.
The most frequent parallel projections are called elevations, oblique projectionsand isometric projections. The elevations are just the front, top and side views of theobject. Thus the projections are given by the functions
Ffront(x,y,z)=(x,z), Fside(x,y,z)=(y,z); Ftop(x,y,z)=(x,y)
Applied to the house object, we get three views.
MAPLE generated elevations.
In oblique projection, which is also called Cavalier projection, the front view isundistorted, but the sections of the object corresponding to y = y0 constantare drawn up and to the right depending on how far back y0 is. If we writew=(2-1/2, 2-1/2)T, the unit vector inthe plane at 45°, then we may write
The vectors have been written in column form to here to facilitate matrix multiplication, butwe'll not fuss about whether a vector is a row or column and use both forms interchangeably. Notethat since w is a unit vector, lengths in the y directions are mappedto equal lengths along the 45° line in the drawing. Indeed, puttingc = 2-1/2,
dist((x0,y1,z0),(x0,y0,z0)) = |y1 - y0|; Of course horizontal and vertical lines also preserve measurement. There is nothing specialabout 45° except that it is common. One can find drafting paper that hashorizontal, vertical and 45° lines used to make oblique projections. Any unitvector w will do as well.
dist(F(x0,y1,z0),F(x0,y0,z0)) = dist((x0 + c y1, z0 + cy1), (x0 + c y0,z0 + c y0)) ={(c y1 - c y0)2 + (c y1 - cy0)2}1/2 = |y1 - y0|.
Let R denote a rotation in the plane which moves points (x,y) about theorigin an angle h0. If a vector is written in polar coordinates (x,y)=(rcos h,r sin h) where r={x2 + y2}1/2 is thedistance of the point to the origin and h is the angle from the positivex-axis, dist((0,0),(x,y)), and h is the angle, then usingtrigonometric identities the rotated point
R(x,y) = (rcos(h0 + h), r sin(h0 + h)) has the same distance from the origin and a new angleh+h0. Bob Palais has a nice way of seeing this [PR http://www.math.utah.edu/~cherk/ccli/bob/Rotation/Rotation.html] and we present a modification of his approach. Without knowing that sines andcosines are involved, it's possible to write down the rotation transformation just knowing whatvectors are rotated into. For example, if we are given a vector (a,b) and wish to rotateit into the vector (0,r), then we know thatr={a2+b2}1/2 is the length of the vector, and the angleof rotation has to be h0 where sin h0=a/r and cosh0 = b/r. Thus
= (r cos h0 cosh - r sin h0 sin h, r sin h0 cos h + r cos h0 sin h) = (x cos h0 - y sin h0, x sin h0 + y cosh0)
Hence R(a,b)=(0,r) as desired.
A variant of oblique projection is called military projection. In this case thehorizontal sections are isometrically drawn so that the floor plans are not distorted and theverticals are drawn at an angle. The military projection is given by rotation in thex-y plane and a vertical translation an amount z. Thus
Fmilitary(x,y,z) = ( c x - s y, s x + c y + z).
Wechose h0=-53.1° so that c=0.6 and s=-0.8. Notethat the floorplan is drawn rotated but without distortion.
MAPLE generated military projection.
The isometric projections are that class or parallel projections for which a round sphereprojects to a round circle. The most common case is when measurements along the x-axisare plotted at 30°, those along the y axis at+150° and the vertical axis. Thus ifw1=(31/2/2,1/2),w2=(-31/2/2,1/2) and w3=(0,1)are unit vectors at -30°, +30° and 90°(vertical), the isometric projection is
MAPLE generated standard isometric projection.
The general parallel projection is obtained by applying a general affine transformation of theform F(X) = AX + T. If we choose T=0 and
and measure the depth according to GH(x,y,z) = x + 2y - z and then the picture isdistorted so that none of the directions measure actual length. As mentioned before, parallel linesand proportions are preserved.
MAPLE generated projection by the affine function F(X) = AX + T.
We now describe the perspective transformation. It is the composition of a rigid motion followedby the perspective transformation that reduces distant objects. The rigid motion moves the objectin front of the drawing plane in such a way that the eye point ep=(xe,ye, ze) is moved to the origin, so that the vector from eyepoint tocenterpoint cp=(xc, yc,zc) toward which the eye islooking is moved to the positive y-axis and so that the vertical line through thecenterpoint is drawn vertical. We shall accomplish the rigid motion by first translating the objectto move the eyepoint to the origin using
T(x,y,z)=(x - xe,y - ye, z - ze).
Let the new vector eye to center be thedisplacement dp:=T(cp). Then we rotate the object around the origin. Every rotation isthe composition of a rotation around the z-axis by an angle h, around the newx-axis by an angle k and around the y-axis by an anglel. The three angles h, k, k are called the Eulerangles. We only need the first two rotations, and we can compute the cosines and sines involvedusing only the eyepoint and centerpoint coordinates. First we rotate dp around thez-axis so that (dp1, dp2) moves to (0,r1) where
r1 = {dp12 +dp22}1/2. Letting s1 =dp1/r1, c1 = dp2/r1 thenthe rotation is as before R( x, y, z) = (c1 x - s1 y,s1 x + c1 y, z).
Let rdp= R(dp)=(0, rdp2,rdp3) be the rotated dp. The length of rdp is the same asthe length of dp which is r2 = {dp12 +dp22 + dp32}1/2 ={rdp22 + rdp32}1/2. The secondrotation takes rdp to (0, r2, 0). Setting c2 =rdp2/r2, s2 = rdp3/r2, therotation around the x-axis becomes
S( x, y, z) = (x,c2 y + s2 z, - s2 y + c2 z).
Thecomposite SRT(X) = S(R(T(X))) is the desired rigid motion
The perpendicular projection is the front view or (x,z) part of the rotated object
Fperp.(x,y,z)= [ c1(x-xe) -s1(y - ye), -s1 s2(x - xe) - c1s2(y - ye) + c2(z - ze)]
and the depthis computed by GPpP(x, y, z)= s1 c2(x - xe) + c1c2(y - ye) + s2(z - ze). For example, taking theeyepoint ep=(11.0,-15.0,2.0) and centerpoint cp=(3.5,5.0,3.0) projects thehouse so:
MAPLE generated orthogonal projection.
Because light reflecting off the object travels in straight lines, the object point is seen onthe drawing plane at the point where the line from the eyepoint to the object point intersects thedrawing plane. The perspective transformation is simply to deduce the coordinates (u,v)on the drawing plane, which is a distance d from the origin, from the pointX=(x,y,z) using triangles. The triangles (0,0):(0,d):(u,d) and(0,0):(0,y):(x,y) in the x-y-plane and the triangles(0,0):(d,0):(d,v) and (0,0):(y,0):(y,z) in they-z-plane are similar. It follows that
Similar triangles used in computing perspective projection.
We have been using d=1 from which the perspective transformation may be calculated.
This is just the x-z-coordinates of the perpendicular transformationdivided by the depth (y-coordinate.) Using the same eyepoint and centerpoint as for theperpendicular transformation, we plot the house by perspective transformation.
MAPLE generated perspective projection.
Perspective transformations have the property that parallel lines on the object are mapped topencils of lines passing through a fixed point in the drawing plane. To see this, note that eachline in the rotated object lies in the plane passing through the line and through the eyepoint.This plane intersects the drawing plane in a line hence the image of a line in space is a line inthe drawing. Any parallel lines in the object are parallel to the drawing plane or not. If thelines are parallel to the drawing plane (the y-coordinates on the line are constant)then the division by the depth (the y coordinate of the rotated object) is division byconstant. Thus the formula reduces to a constant multiple of the numerator which is an affinetransformation that maps parallel lines to parallel lines. If the parallel lines are not parallelwith the drawing plane, then their image on the drawing plane passes through a fixed point, calledthe vanishing point. The easiest way to see this is to consider a pair of points on twoparallel lines that travel together away from the drawing plane. Imagine that a wire of fixedlength connects the points. Because the pair can get farther and farther from the drawing planewithout letting go the wire, their perspective images get closer and closer in the drawing sincethe denominators are getting large whereas the difference in their (x,z) directions arebounded. Imagine the cliché of two rails of a track converging at infinity.
For general choices of the eyepoint and centerpoint, the parallel lines originally in thex, y and z-axis directions are not rotated to a position parallelto the drawing plane. Thus these three directions each have their own vanishing points. This iscalled three-point perspective. The three points may not so easily seen since they may notbe within the cone of vision that limits the width of our view. To illustrate one and two pointperspective we change our eye and center points to guarantee some parallel lines parallel to thedrawing plane.
Let us consider specific choices of eyepoint and centerpoint for which some of the objects axesare parallel to the drawing plane. Let the eyepoint ep=[6.0,-15.0,2.0] and thecenterpoint cp=[6.0,5.0,2.0]. Because dp=[0,20,0] no rotation is necessary.The x and z-axes are parallel to the y=1 plane. The perpendicularprojection is just the front elevation and the perspective view has one vanishing pointcorresponding to the y-axis direction. The vanishing point is indicated (it is theposition of the centerpoint.)
MAPLE generated front elevation and one point perspective projection. Two-point perspective.
Let the eyepoint ep=[16.0,-15.0,2.0] and the centerpoint cp=[6.0,5.0,2.0].This time dp=[-10,20,0] so that the only rotation is about the z axis. Thez-axis is parallel to the y=1 plane. The perpendicular projection is now acorner elevation and the perspective view has two vanishing points corresponding to thex- and y-axis directions. The centerpoint is indicated.
MAPLE generated orthogonal and two-point perspective projection.
Another pair of views come by taking the eyepoint ep=[-6.0,5.0,9.0] and thecenterpoint cp=[6.0,5.0,2.0]. This time the horizontal lines are parallel to the drawingplane but the vertical and receding lines are not. Therefore the vanishing points correspond to thevertical and receding directions.
Another MAPLE generated orthogonal and two-point perspective projection.
The maximum view that the eye can take in is a cone of about 30° about itsaxis (the cone of vision.) It is possible for the computer to plot points outside the coneof vision, but such a drawing has a distortion like a fisheye camera photo. Thus usually bothvanishing points aren't visible in the same scene, as in this computer-generated view of a cubewith parallel lines.
MAPLE generated perspective view of unit cube showing vanishing points.
How do we locate the vanishing points in the drawing? The vanishing points for thex-axis and y axis parallels are always on the horizon line. If dis the distance from eye to drawing, then the two vanishing points in the drawing forx-axis and y-axis lines are on lines which meet at the eyepoint at90°. This is easiest to see by imagining the top view.
MAPLE generated perspective view and construction of vanishing points from topview.
The drawing plane is a distance d from the eyepoint E. The rays emanatingfrom the eyepoint at right angles parallel to the y and x-axes are the linesegments EA and EB. A is the u-coordinate of they-axis vanishing point V1 and B is the u-coordinate ofthe x-axis vanishing point V2. The v-coordinates arev=0 which corresponds to the eyelevel and horizon line. A circle whose center is on thedrawing line and passes through the eyepoint intersects the drawings line at two points, sayA and B for which AEB is a right angle. This is the geometric factthat a diameter AB subtends an angle 90° from any point Eon the arc AEB.
MAPLE generated perspective and top view of vanishing points and theirconstruction.
How do we measure distances in the receding direction? The idea is to figure out sets ofparallel lines which transfer measurements along the baseline, a line parallel to the drawingplane, to the receding line. The projective transformation may scale but not distort distancesalong the baseline. To see how this works, consider the top view of a 3 x 3square.
MAPLE generated parallel sets of measuring lines.
The baseline is the line af. The baseline has equally spaced points a,b, c, o, d, e, f in order. Thespacing is the same as along the square o, c', b'a'and od', e', f'. The square has been rotated an anglefoX. The parallels to oX and the parallels to oY are along the twosides of the square. Their perspective images converge to two vanishing points. The other two setsof lines are called measuring lines. One family are the parallels oP, dd',ee', ff' measure the oX side of the square and the other set ofparallels oQ, cc', bb', aa' measure the oYside of the square. This is what it looks like in perspective.
MAPLE generated top and perspective views showing parallel measuring lines,vanishing and measuring points.
Because thelines connect equally spaced points, the triangles fof' andaoa' are isosceles. This means that if the line oW is chosen so that the angle foW bisectsthe angle foX, then the lines oP and oW are perpendicular and the angle Similarly, the triangle aoa' is isosceles so the angle But since the total angle of a triangle is and since they are supplementary, It follows that so These angles may be easily constructed on the circle.
MAPLE generated top view for constructing vanishing points and measuring points.
As before, we locate the eyepoint E and centerpoint O on the drawing andlet line EF be parallel to AB. The sides of the box from the previous diagramare along the rays EA and EB so that the vanishing points in the drawing arelocated at A and B. Since a line intersects parallel lines so that oppositeangles are equal, Draw a circular arc with center B and radiusBE until it meets the drawing plane line AB at M1. EBM1is a similar triangle to fof' so Thus M1 is the point where the eye views the first familyof measuring parallels; thus M1 is the vanishing point for this set of parallels.Similarly, so that if one draws acircle with center A and radius AE then this circle intersects the pictureplane line at M2. Now the angle so that the point M2 is the vanishing point for the second family of measuringlines. Now we can use the measuring lines to mark off equispaced points on the perspectivelyreceding lines.
MAPLE generated measuring lines viewed in perspective and their constructionviewed from the top.
We mark off equally spaced points a-f on the baseline as before. The linesoV1 and oV2 correspond to the bottom edges of the box. Moving up one unitfrom o gives the upper corner of the box and the rays to V1 and V2give the upper front edges. Now, the first measuring family was chosen so that the intersectionswith the right front edge were points spaced the same distance apart as on the baseline. Thus,where the lines dM1, eM1 and fM1 intersect oV1 are theequally spaced points d', e'f' in the perspective drawing.Similarly, the measuring family of parallel lines for the left side of the box have a vanishingpoint at M2. The intersection of aM2, bM2 and cM2 withoV2 correspond to the equally spaced points a', b', c'on the line oV2. The rest of the box is constructed by extending the vertical lines upfrom a'-f'. If one is using two-point perspective, these are truly verticalin the u-v-plane. Otherwise you have to use the vanishing point correspondingto the vertical family (which is probably way below the picture.)
MAPLE generated measuring lines viewed in perspective and their constructionviewed from the top.
One is taught in drawing class, that circular objects in three-dimensional Euclidean Space aredrawn in perspective as ellipses. The usual construction is to draw a square around the circle, andthen project the perspective view of the square by finding its edges using the vanishing points andmeasuring points, the center by drawing the diagonals, and then sketching the projected circle bydrawing it tangent to the projected square. A beginner will sometimes make the mistake of trying tomake the tangency points the same as the endpoints of the axes of the ellipse, but they are not thesame as seen in the p. 17 figure. But why is the image exactly the ellipse and not some otherclosed curve?
We shall answer this question by figuring out the equation of the image of the circle on theperspective drawing. We'll be using the methods of analytic geometry, where curves are representedby equations. Thus we shall describe a circle in three space by describing it as the locus ofpoints satisfying certain equations. We then compute the corresponding perspective locus in termsof the Cartesian coordinates of the drawing plane. Finally, after some simplification, we will beable to recognize the curve as an ellipse.
The conic sections in the plane are given as the locus, that is the set of all points(u,v) in E2 which satisfy a quadratic equation of the form
(1.)
au2 + 2buv + cv2 + eu + fv + g = 0,
where a, b, c, d, e, f,g are constants. This can be deduced from the geometric description of the conic sectionas the intersection in three space of a plane with a right circular cone. All possible conicsections arise this way including degenerate ones such as lines and points and the empty set. Forexample if a=b=c=0 then
e u + f v + g = 0
isthe equation of a line and if a=c=1, b=0, d=-2u0,e=-2v0, g=-u02-v02then
a u2 + c v2 + e u + f v + g =(u-u0)2 + (v-v0)2 = 0
is satisfied onlyby one point (u,v)=(u0,v0) whereas
u2 + v2 + 1 = 0
has no real solution at all. Onthe other hand if the discriminant
D = a c - b2
is negative, then the conic is a hyperbola, if D=0 the conic is a parabola and ifD is positive the conic is an ellipse. The easiest to see are the canonical conic curves given bythe formulae
Of course if a=b the ellipse is a circle.
Now let's see what a projective transformation looks like analytically. For simplicity, weassume that the set is located in front of the observer (all points of the circle satisfyy0.) Then the horizontal and vertical coordinates of the drawing plane (points whichsatisfy y=1) are
(2.)
where (x, y, z) runs through all points of the original set. Now suppose that weconsider a circle in space with center (x0, y0, z0) andradius r and which lies on a plane not parallel to the drawing plane. By a rotationaround the y-axis, we may arrange that the intersection line of the circle plane and thedrawing plane is horizontal. In other words, the equation of the plane through the center of thecircle sloping away from the drawing plane with slope m is given by
(3.)
z - z0 = m (y - y0).
To beable to see the circle, we require that the eyepoint (0, 0, 0) is not on the plane ofthe circle, which means z0 does not equal m y0. Thecircle also lies on the sphere of radius r centered at (x0,y0, z0), which has the equation
(4.)
(x - x0)2 + (y -y0)2 + (z - z0)2 = r2.
Thecircle is the collection of points satisfying both (3.) and (4.) These are projected using (2.) tothe drawing plane. By substituting (3.) into (4.),
(5.)
(x - x0)2 + (1 + m2)(y -y0)2 = r2.
We are trying to see how these equationsrelate u to v. Using (2.), we substitute in the equations (3.) and (5.)
v y - z0 = m (y - y0)
so
Substituting into equation (5.) and multiplying by (v - m)2 yields
[u(z0 - m y0) - v x0 + mx0]2 + (1 + m2)(z0 - v y0)2 =r2(v - m)2.
Multiplying out and collecting factors of u2, uv, ...yields
(z0 - my0)2u2 + 2 x0(z0 - m y0)u v +[x02 + (1 + m2)y02 -r2]v2 +
Thus(u,v) satisfy a quadratic equation in the plane. The discriminant is D= (z0 - m y0)2[x02 + (1 +m2)y02 - r2] - (z0 - my0)2x02
= (z0 - my0)2[(1 + m2)y02 - r2].
Since the eyepoint is not on the plane of the circle z0 - m y0 > 0. Since the circle is in front of the y=0 plane, the point(x0, 0, z0 + m y0) which is both in the y=0plane and on the circle plane is can't be on the circle, in fact it is farther from the center thanany point of the circle, hence
(1 +m2)y02 - r2 0.
Thus D > 0 and the locus is an ellipse.
Perspective view of the circle
Here is a diagram from Alberti's treatise. The square that surrounds the circle projects to atrapezoid. The circle itself projects to an ellipse which is tangent to all four sides of thetrapezoid. Observe that the left and right endpoints of the axes of the ellipse where the ellipseis widest occur below the tangency points. But be careful when drawing the ellipse which is notcentered on the eyepoint to centerpoint line!
The original impetus to projective geometry came from perspective drawing. Alberti's textbookDella Pittura (1435) formulated new questions that tempted mathematicians to study newquestions beyond those addressed by the Greeks. If two artists make perspective drawings of thesame object, their drawings will not be the same, for example because different parts of the objectwill be closer to each of the the two artists. But what properties of the drawings remain the same?(Diagram of Alberti's question.)
The perspective projection, which takes points X of the object which are in threespace and plots them as points P on the drawing plane. Let us write this
P=f(X).
It has the property that points are mapped to points andlines to lines. However, parallel lines in three space which are not parallel to the drawing planemust be drawn to converge at their vanishing points. Thus the correspondence between lines andpoints in three space and lines and points on the drawing isn't perfect. Thus ifL1 and L2 are parallel lines in three space thenf(L1) and f(L1) are lines which intersect atV their vanishing point. However L1 and L2don't intersect at any point. In the diagram, lines AB, CD and E'V' areparallel. Their projections A'B', C'D' intersect at a point V' which iscalled the vanishing point since it has no corresponding point in three space.
The solution was proposed by Girard Desargues (1591-1661) a self educated man who worked as anarchitect after leaving the army. His opus with the ponderous name, Broullion project d`une atteinteaux événemens des renconteres du cône avec un plan, (1693) which describesprojective methods in geometry went unnoticed. Jean-Victor Poncelet (1788-1867), an engineer inNapoleon's army reworked the theory in Traité des proprietiés projectives desFigures(1822) while a prisoner of war in Russia in 1813 [KF]. This elevated Desargues work in projectivegeometry to one of the success stories of synthetic geometry, whose merits versus analytic geometrywere being debated at the time. We sketch two theorems from projective geometry. For a more rigorous treatment, the reader shouldconsult any of a number of texts, such as O'Hara & Ward [OW] or Wylie [WC].
To complete the correspondence, Desargues introduced ideal points, called points atinfinity one for each set of parallel lines. The points at infinity don't contradict anyaxioms. They function as a convenience since now every pair of lines intersects at one point, thecase of parallel lines does not have to be treated as an exceptional case. The following is nowcalled Desargues' Theorem of Homologous Triangles.
Theorem. Supposethere is a point O and triangles ABC andA'B'C' in the plane or three space. If they are projectively related from thepoint O, that is, the triples {O, A, A'}, {O, B,B'} and {O, C, C'} are all collinear. Then the points ofintersections of the corresponding sides AB and A'B',AC and A'C' and BC andB'C' (or their prolongations) are collinear. Conversely, if the three pairs ofcorresponding sides meet in three points which lie on one straight line, then the lines joiningcorresponding vertices meet at one point (are projectively related.)
The proof is easier for the case that the triangles are not coplanar. See Dörrie [HD] orMeserve [MB] for proofs.
Diagram of Desargues' Theorem of Homologous Triangles.
To see how we may use projective geometry directly to argue that the perspective image of acircle is an ellipse, we use a theorem due to Blaise Pascal (1623-1662). Pascal, who was urged toinvestigate the relationship between projectivities and conics by Desargues, published his Essaisur les Coniques when he was sixteen. Although he didn't prove the converse part, the theoremis known as Pascal's Hexagon Theorem.
Theorem. Let a hexagon beinscribed in a (nonsingular point-) conic. Then the three points of intersection of pairs ofopposite sides are collinear. Conversely, if the opposite sides of a hexagon, (of which no threevertices lie on a straight line) intersect on a straight line, the six vertices lie on anon-singular point-conic.
Diagram for Pascal's Hexagon Theorem.
Pascal's Theorem may be used to deduce that the perspective image of a circle is an ellipse.Thus if c is the circle and f(c) is its image in a perspective drawingrelative to the eyepoint O, then we have to show that if any six points A, B, C, D,E, F are chosen on f(c) so that no three of them lie on a straight line then thepairs of opposite sides intersect in collinear points. Then by the converse of Pascal's Theorem,the six points lie on a nonsingular point-conic. But since five points determine a conic, the sixthpoint which may be any general point of f(c) must be on the same on the conic. Itfollows that no matter which six points are chosen, they lie on the same conic, thusf(C) is (part of) a single point-conic. One argues that f(c) is bounded andnondegenerate thus can only be the ellipse. But the six points are in perspective correspondence topoints A', B', C', D', E', F' on c which is a circle, hence a point-conic.Therefore, by Pascal's Theorem, the pairs of opposite sides (A'B' and E'D'),(B'C' and F'E'), and (C'D' and A'F') intersect atpoints P', Q', R' respectively, which are collinear in the plane of c. Theirperspective images P, Q, R in the plane of F(c) must also be collinear sincethe perspective image of a line not containing O is a line. Moreover the planesOA'B', OE'D' contain the edges AB, ED, resp., since they are perspective toeach other, and thus the planes intersect along the line OP'P. In other words, the pointP is the intersection of the edges AB and ED. SimilarlyQ is the intersection of the edges BC and FE and R isthe intersection of the edges CD and BF. Thus P, Q, R arecollinear and we are done.
An analytic version is in the previous section Analytic Treatment of thePerspective View of a Circle.
Without going very deep into computer science complications, we explain something about themathematics behind computer drawing. Computer science issues are treated, e.g. in [PP]. Oneof the ways that a computer renders three-dimensional object is to build up the image from littleconstituent pieces. The object is regarded as a collection of polygons. The visual position of eachlittle piece is computed and the polygons are drawn one polygon at a time. The computer screen isgiven a Cartesian (horizontal and vertical axis) coordinate system and the polygon is drawnspecifying the position of each P=(u,v) of its vertices. For example, as the threetriangles red (0,0),(4,-1),(1,.5), green (4,2),(1,-.5),(4,1) andblue(4,-2),(3,1.5),(3,-1.5) are drawn, each one covers the previous ones.
If for example, we wish to draw the front elevation of the an object in space consisting of thethree triangles [(0,1,0),(4,0,-1),(1,1,.5)], [(4,1,2),(1,0,-.5),(4,1,1)] and[(4,1,-2),(3,0,1.5),(3,1,-1.5)] viewed toward the +y-axis, we have to drawthe triangles as before, because the projection is given by
F(x, y,z )=(x, z).
This would result in an incorrectpicture because one of tips of each triangle is closer to the viewer than some one of the othertriangles. e.g., the base of the first triangle (0,1,0),(1,1,.5) aty=1 is farther from the viewer than the remaining vertex (4,0,-1) aty=0. Another source of error would be if polygons in the object actually intersected. Tocorrectly render the front elevation, the triangles have to be subdivided further into parts andthe parts in front have to be drawn on top of parts in back.
The most fundamental way to depict depth in a picture is overlapping closer objects over fartherones. In general it is quite involved to decide if some part of the object can be seen or not. Thesimple way to deal with this is to draw all polygons of the object back to front. Some of thepolygons which are in back of the object eventually get completely covered up. This is called thepainter's algorithm. The way it works in our MAPLE program, first we compute the distancesof each point to the eye. Then a typical distance is given for each polygon, which in our case isthe distance to the nearest point. Then the polygons are sorted according to their typicaldistances, and are rendered back to front. Our program does not try to account for complicatedoverlaps or intersections so will sometimes render objects incorrectly. To illustrate the painter'salgorithm, suppose we render a cube. The faces are drawn back to front, depending on the distanceof each side to the viewer. In the example, the sky being farthest is drawn first, followed by theearth, the back face, the base, the sides, the top and finally the front, eventually covering upall but two sides.
Another way of rendering a three-dimensional object is called ray tracing. In raytracing, the computer follows light rays back from the eye to a point on the object from where itfigures out how intense the light is and what its color is by following back the rays whichilluminate that point. This can continue for several stages. At each stage the computation accountsfor surface properties like shine and color and body properties such as refractive index andtransmittivity. An example of ray tracing is a rendering of the same house made by the programPOV-RayTM (Persistence of VisionTM Ray-Tracer Version 3.1.) We havespecified that the body be made of gray glass and be positioned on a chessboard.
To get a sense of what is state of the art in ray-tracing, visit Steven Parker's website Interactive Ray Tracing -- MPEG demo at theScientific Computing and Imaging Institute http://www.sci.utah.edu/index.html in the Graphics andVisualization group http://www.cs.utah.edu/research/areas/graphics/ in theSchool of Computing at the University of Utah.
Last updated: 07 / 24 / 01
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