Is the line created by the minor axis of an ellipse concurrent to the lines running to the opposite vanishing point?

Question

My questions needs more context than what can fit into the title so let me elaborate. In pretty much all the art textbooks I am reading on linear perspective they state that the correct way of placing an ellipse is to imagine the minor axis of an ellipse as an axel of a wheel running to the opposite vanishing point. Here are examples for the various types of perspective the textbooks mention.

One point perspective:

In one point perspective only one set of parallel lines of a cuboid are concurrent, the other two sets are parallel (one set parallel to the horizon, the other perpendicular to it). If we plot an ellipses inside a one-point square the minor axis' should all run vertical (to the non-concurrent vanishing point). We find that only ellipses whose major axis are perpendicular to the vanishing point (A`) have this property.

Two point perspective:

In two point perspective two of the sets of parallel lines of a cuboid are concurrent (meeting at R3 & S3 in the image) and the other set are not concurrent (parallel). I tried placing an ellipse such that the perspective center points of the sides of the quadrilateral are the tangent points of the ellipse but the minor axis does not seem concurrent with the lines running to the opposite concurrence (vanishing point). It should be stated that the two quadrilateral and their concurrences are a mirror reflection of each other, but this is not always the case in two point perspective.

If I adjust the size of the bounding quadrilateral I can make the minor axis concurrent with the lines running to the opposite vanishing point... but what if I want an ellipse placed inside a different sized quadrilateral?

Am I missing something in my plotting of ellipses or are the authors of these textbooks mistaken?

Answer 1

You are absolutely right; the authors of these textbooks are mistaken, and your diagrams show why.

I happen to have watched a fair number of instruction videos on perspective drawing, and to my frustration, I haven't yet found a single one that addresses this mistake, even from the most meticulous and formal instructors.

This is somewhat understandable. Broadly speaking, artists establish rules based on aesthetics, intuition, and approximation; whereas mathematicians remain skeptical without proof. These are two valid mindsets for two different fields, and your question belongs to the overlap of the two. The danger of an artist learning the technical craft of perspective is that they might use rules of thumb indiscriminately without fully understanding when they apply. The danger of a mathematician learning to draw is that they might break out a straightedge and compass without considering whether that will actually contribute to a successful composition. It would be nice to find a happy middle ground between these two camps, but ellipses turn out to be surprisingly complex. A perfect geometric construction exists, but just isn't worth it in the the context of a larger drawing, so it's natural to look for a simpler way.

Note that the rule actually does work for isometric drawings. In other words, if all parallel lines remain parallel (as opposed to converging to a vanishing point), then it holds. There are other cases where it holds, as you allude to in the comments. Most importantly, the rule will be reasonably close when the drawing is within the cone of vision. In contrast, in your first example, the leftmost circle is outside the cone of vision and therefore quite distorted.

In summary, the "rule" should be considered a good approximation most of the time, but the more distortion there is, the more likely you should abandon it and use your instincts (or a straightedge and compass).

Answer 2

Minor/major axes, foci, etc. are "metric objects" that have a meaning in the framework of metric geometry, therefore are outside the scope of projective geometry which (in particular) doesn't preserve neither the lengths nor the ratio of lengths (but the cross-ratio). As a consequence, it is not astonishing that you find such discrepancies...

Besides, are you aware of the fact that a figure like your first figure can be easily obtained by programmation in a convenient language. For example, this figure:

Fig. 1. Figure generated by a program written in Matlab (see bottom of this answer) The vanishing points are materialized by red dots. We see on this figure that we cannot infer anything on the orientations of the major/minor axis of the ellipses...

Here is another figure, similar to the first one you have given with trapezoids instead of general quadrilaterals:

Fig. 2: Figure obtained by changing $A$ into $[5+0.i,1+2i,0+0i]$ and changing denominator $(x+y+t)$ into $(y+t)$ in function $Z$. Only a single vanishing point at finite distance (the other one being at infinity).

Explanation: everything is based on the common (and simple) expression of all projective transformations (in particular those giving a perspective):

$$X=\dfrac{ax+by+c}{gx+hy+k}, \ \ Y=\dfrac{dx+ey+f}{gx+hy+k}, \tag{1}$$

(please note the common denominator) or equivalently:

$$Z=\dfrac{Ax+By+Ct}{gx+hy+kt}\tag{2}$$

$$\text{with} \ Z=X+iY, A=a+id, B=b+ie, C=c+if,$$

with $t=1$ for ordinary points and $t=0$ for points at infinity.

(using complex numbers for plane geometry is very handy).

Remark:

Here, in the plane with coordinates $(x,y)$, I have first described a chain of 6 tangent circles (+ circumscribed squares) to which I have applied a certain projective transform of the type given by (2).

Program:

clear all;close all;hold on
A=[2+i,1+1i,0+0i]; % could be (almost) any set of comp. numbers
Z=@(x,y,t)((A(1)*x+A(2)*y+A(3)*t)./(x+y+t)); % see formula (2)
s=5; % shift
a=0:0.01:2*pi; % parameter
for k=1:4;
   % ellipses as images of circles by "Z transform":
   plot(Z(cos(a)+2*k,sin(a)+s,1)); 
   % quadrilaterals as images of squares: 
   plot(Z([-1,1,1,-1,-1]+2*k,[-1,-1,1,1,-1]+s,1)); 
end;
% vanishing points in the 2 directions, with $t=0$:
plot(Z(1,0,0),'or','markerfacecolor','r'); 
plot(Z(0,1,0),'or','markerfacecolor','r');