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Problem

Provide the set of all possible maps $\Phi(x,y)$ such that the image of the usual square coordinate grid under $\Phi$ is a 1-point perspective grid with vanishing point at $(p,q)$.

Note that such a map can be used to demonstrate the projective equivalence of all conics.

Solution

Here is what I believe to be a set of conditions which, when satisfied, are necessary and sufficient for accurately mimicking 1-point perspective in the way we desire our map to:

  1. the image of any vertical line is an element of a pencil of lines through $(p,q)$
  2. the image of any vertical line has an accumulation point at $(p,q)$
  3. the image of any horizontal line is an horizontal line
  4. the image of any point on the line $y=q-1$ is the same as the preimage
  5. the image of any diagonal line is a diagonal line whose slope has the same sign

Let $\Phi(x,y)=\left(f(x,y),g(x,y)\right)$

WLOG, let $(p,q)=(0,1)$ (we can perform the necessary translation later)

To satisfy conditions 1 and 4 simultaneously, we require that $g(x,y)=1-\frac{1}{x}f(x,y)$

Thus, $\Phi(x,y)=(x(1-g(x,y)),g(x,y))$

To satisfy condition 4, we also require that $\Phi(x,0)=(x(1-g(x,0)),g(x,0))=(x,0)$ $\forall x\in\mathbb{R}$

Since this implies that $g(x,0)=0$ $\forall x\in\mathbb{R}$, we can conclude that $g_x(x,y)=0$ and thus $\Phi(x,y)=(x(1-h(y)),h(y))$ for some univariate function $h$

To satisfy condition 5, we require $\Phi(x,x)$ to be such that $h(x)=ax(1-h(x))$ for some $a>0$

Thus, if $a>0$ then $h(x)$ can be any function of the form $\frac{ax}{ax+1}$.

Thus, we finally have $\Phi(x,y)=(\frac{x}{1+ay},\frac{ay}{1+ay})$

After a translation by $(p,q-1)$, we finally have our answer:

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

Here is a desmos simulation to verify this result

Simon M
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