The shape of parabola in a 1-point perspective drawing with the vanishing point at the parabola's point at infinity

Question

Edit: After a little bit of careful construction, I've worked out that the map I should have been using instead is $$\phi\left(x,y\right)=\left(\frac{x}{ay+1},\frac{ay}{ay+1}\right)$$ where $a\in(0,\infty)$ determines the spacing between horizontal lines as well as the eccentricity of the ellipse. Choosing an $a$ value also determines to the "height" of the observer/camera/eye, and smoothly varying $a$ corresponds to moving the "camera" closer to or father away from the plane.

This would be the appropriate map for modeling 1-point perspective.

More generally, if we desire $(p,q)$ to be the coordinates of our vanishing point, then

$$\Phi(x,y)=\left(p+\frac{x}{ay+1},q-1+\frac{ay}{ay+1}\right)$$ for any $a>0$

See Constructing a map that accurately mimics 1-point perspective for a derivation of this result.

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It is common to see introductory explanations of projective geometry that claim the following:

• if you take a parabola in the plane and create a 1-point perspective drawing of it such that the vanishing point is at the parabola's point at infinity, then the parabola will appear as an ellipse in the perspective drawing $\tag1$

However, if you continue plotting the points $(n,n^2)$ for integers $n$ as I've started to in this picture, you will get a teardrop-like shape given by $x^2=(1-y)^2\log_{b}\left(1-y\right)$ for some $b\in(0,1)$, and not an ellipse.

As can be weakly verified by looking at my desmos simulation, the map $\Phi:\mathbb{R^3} \rightarrow \mathbb{R^2}$ given by $$\Phi(x,y,t)=(xt^y,1-t^y)$$

takes the point at infinity in the direction of the positive $y$-axis to $(0,1)$, and for $t\in(0,1)$ the map continuously deforms the $xy$-plane in such a way that exactly mimics looking down onto the $xy$-plane from a vantage point that continuously increases in height above the $xy$-plane as $t$ increases.

However, there is no $t \in (0,1)$ such that the curve parametrized in $x$ by $\Phi(x,x^2,t)=(xt^{x^2},1-t^{x^2})$ is an ellipse.

Does this prove that (1) is false and should be replaced with the following statement?

• If you take a parabola and create a 1-point perspective drawing of it such that the vanishing point is at the parabola's point at infinity, then the parabola will appear as a teardrop-shaped curve in the perspective drawing $\tag2$

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If instead of $(0,1)$ we desire $(p,q)$ to be the image of the point at infinity in the direction of the positive $y$-axis, then the map $\Phi$ can be modified to $\Phi(x,y,t)=(p+xt^y,q-t^y)$

Answer 1

Instead of the parabola, I used your function to plot the projections of the two diagonal lines $y=x$ and $y=-x$. The result is below (the diagonal lines are purple) for $t = 0.677$.

In other words, the purple curve is what your formula $\Phi(x,y,t)=(xt^y,1-t^y)$ did to two straight lines that cross at right angles. These two perpendicular lines appear to meet at infinity.

Your so-called "one-point perspective" is really weird. Who draws like this? (That's not a rhetorical question. Let's see a reference.)

Your examples drawn from other sources are actually counterexamples of your claim. Here is the perspective grid you used in the question. I have added lines corresponding to the lines $y=x$ and $y=-x$. These two diagonal lines are perfectly straight, passing through the corners of every square in the grid.

Here is the example of a one-point perspective grid you cited in a comment. Again I have drawn two perfectly straight diagonal lines through the corners of two diagonal sequences of squares.

This is, in fact, a well-known technique for drawing one-point perspective. It is documented by DrawingAcademy.com, for example.

Conclusion: in standard one-point perspective, when you draw a picture of a square grid in perspective (as, for example, a floor of a room covered with square tiles), the diagonals of diagonally adjacent "tiles" in your grid form straight lines.

The diagonals in your grid are not straight, therefore your grid is not standard one-point perspective.

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But the idea that you have a function $\Phi(x,y,t)$ that plots points in a "world" $x,y$ plane onto points on an "image" plane, using $t$ as a parameter to control the perspective, is a good idea. As you would think, for the kind of perspective you're trying to draw -- a square grid such all the "constant $y$" lines in the grid plot onto horizontal lines on the image and the point at infinity is plotted at $(0,1)$ -- it is also a good idea for $\Phi$ to have a formula of the form

$$ \Phi(x,y,t) = (x f(y,t), 1 - f(y, t)). $$

You need something equivalent to this in order to ensure that the images of lines parallel to the $y$ axis in the $x,y$ plane are straight lines that converge at the vanishing point. The only defect in the formula proposed in the question is that $f(y,t)\stackrel?=t^y$ lets the horizontal lines approach the horizon too quickly.

As a model of one-point perspective, suppose that the image plane is perpendicular to a horizontal $x,y$ plane and that you, the artist, are viewing the scene from the point $E$ at a distance $t$ from the image plane. The figure below is what this setup might look like to a person standing far away to your right.

We suppose that $E$ is at a distance $1$ above the $x,y$ plane so that the point $H$ (the vanishing point in the image) is also distance $1$ above the $x,y$ plane. We set up coordinates on the image plane so that $C = (0,0)$ and $H = (1,0).$

Now if you look in the direction of the point $S$ in the $x,y$ plane, at a distance $y$ beyond the image plane, your line of sight passes through the point $P$ on the image plane, so that's where you want to plot the image of $S$. Using the facts that the legs of triangle $\triangle ADE$ are in the ratio $1:t+y$ and that triangles $\triangle ADE$ and $\triangle ACP$ are similar, we can work out that the distance $CP$ is

$$ \frac{y}{t+y} = 1 - \frac{t}{t+y}. $$

So let $f(t,y) = \dfrac{t}{t+y}.$ Then

$$ \Phi(x,y,t) = \left(x \frac{t}{t+y}, 1 - \frac{t}{t+y}\right) $$

The results are shown in this Desmos graph.

Note that this formula for $\Phi$ is equivalent to the one you worked out independently; just substitute $1/t$ in place of $a$ in your formula and simplify. So you can take this last part of the answer as verification of your later result.

Answer 2

Here's a geometric argument for why a parabola and an ellipse are projectively equivalent. Think of the cone $z^2=x^2+y^2$. If I slice it with the plane $z=1$ (this will be one choice of viewing plane), I get the circle $x^2+y^2=1$. Now let's slice with the plane $H_t$ given by the equation $-(\sin t)y+(\cos t)z=1$ (as pictured below).

If I rotate the usual coordinates in $\Bbb R^3$ so that $x=x'$ and $H_t$ is given by the equation $z'=1$, then with a bit of algebra, you can check that the equation of the slice of the cone with $H_t$ has the equation $$(x')^2 + (\cos 2t)(y')^2 - (2\sin2t) y' = \cos 2t.$$ When $t=0$, we have our circle. When $0<t<\pi/4$, we have an ellipse, and when $t=\pi/4$, we have the parabola $y'=(x')^2$.

Here's the traditional algebraic approach to the projective geometry. We work with homogeneous coordinates $(X,Y,Z)$ on $\Bbb R^3-\{0\}$ with the understanding that a point of the projective plane represents a line through the origin in $\Bbb R^3$. Thus, any two points $(X,Y,Z)$ and $c(X,Y,Z)$ with $c\ne 0$ represent the same point in the projective plane. We see that the equation $XZ=Y^2$ gives the parabola $x=y^2$ when we set $x=X/Z$ and $y=Y/Z$. On the other hand the equation $Z^2-X^2=Y^2$ gives the circle $x^2+y^2=1$ when we set $x=X/Z$ and $y=Y/Z$. (You can similarly get ellipses with different eccentricities.) But the point is that we can make a linear change of coordinates in $X,Z$ that turns $XZ$ into $Z^2-X^2$. If you understand why $xy=1$ and $x^2-y^2=1$ are both hyperbolas in the Euclidean plane, then you can easily work this out.)

I'm not going to address the mathematics of "perspective" at this point, but I will add that the whole point of projective geometry is to include points at infinity, and the parabola has a point at infinity (which is the point you're not seeing when you complain that your vertical line doesn't intersect the parabola twice), whereas the standard picture of the ellipse has all its points in the finite part of the plane.