Exercise on Perspective Squares

\begin{document} \maketitle \secton{Introduction} It is surprising that every convex quadrilateral $ABCD$ in a painter's canvas is a square in perspective from at least one position of the painter's eye relative to the canvas. The problem to be solved here is to locate such viewpoints. \subsection{Using KSEG} Renaissance painters, who invented perspective, obviously didn't have a computer drawing program. They use rule and compass. So the following recipe is equally valid for ruler and compass as it is in KSEG. But for this assignment, you submit one .seg file with the solution. \section{Exercise} Choose four initial points in the plane representing the painter's canvas. Be sure they form a convex quadrilateral. Let the vertices be $ABCD$ in counterclockwise order. Next, construct the horizon line $ h = ((AB)(CD))((BC)(DA)) $. You may need to wiggle the given points so that the two vanishing points $(AB)(CD)$ and $(BC)(DA)$ are in the visible part of your canvas. If you rotate the quadrilateral, you can get $h$ to be really horizontal, and these two vanishing points to be the $LVP$ and the $RVP$ as in the notes. This much alone establishes your quadrilateral to be a rectangle in perspective from every viewpoint. Why is that? You need to understand the theory of lesson P3 to follow the next construction. On the horizon, construct a semicircle with diameter the segment between the left and right vanishing points of the rectangle in perspective. One of the two diagonals of the quadrilateral, when extended, crosses the horizon at the diagonal vanishing point, $DVP$. In the semicircle it is necessary to construct a Thales triangle so that the bisector of the right angle crosses the horizon at the $DVP$. But how to do that? One solution is trial and error. Construct any Thales triangle, bisect the right angle at $T$, and wiggle $T$ along the semicircle until the construction is correct. This will lead to a correct picture, but not to a correct solution. Why not? With ruler and compass, once you have drawn something, you can't change it. With KSEG you can. We call this the "wiggle test". Your construction must persist under changing the initial conditions. So I will now give you the construction that works. In a subsequent lesson (P6) you'll find out why. \subsection{Hint on locating the Thales triangle} Instead of a semicircle, draw the entire circle. Drop a radius perpendicular to the diameter to the point at the very bottom of the circle. Call it $S$ for "south pole". Next, draw $(SD)$ through $S$ and the diagonal vanishing point. Where $(SD)$ crosses the upper semicircle, $T$, is where to put the right angle of the Thales triangle. Verify this experimentally by showing that $(SD)$ really does bisect the right angle at $T$. Finally, by our theory, the eye-points in the canvas lies along the altitude of the Thales' triangle you have constructed, but still inside the circle with diameter from the left to the right vanishing point. \section{Extra Credit Exercise on finding the focal distance.} What remains to be determined is just how far the painter's eye must be away from the canvas. This is the focal distance. For this you need to understand the Horizon-Zenith construction. \subsection{Hint} Look at the figure above. It is a detail from a figure in the lesson. In it you see the horizon line, the eyepoint, called $C$ here. You do not see the Thales triangle itself. But the point $B$ is the base of the altitude of your Thales' triangle. And the distance from $B$ to $A$ in the picture is the length of the altitude. Therefore, the focal distance is represented by the segment from $A$ to $C$. To understand why this must be true requires some serious spatial imagination, and I do not require you to supply it on this first exercise. A labelled .seg file what holds together under wiggling is suffcient. \end{document}