Curvilinear perspective

Horizon and vanishing points

The system uses both curved perspective lines and an array of straight converging ones to approximate the image on the retina of the eye, which is itself spherical, more accurately than the traditional linear perspective, which only uses straight lines but is very distorted at the edges.

It uses either four, five or more vanishing points:

• In five-point (fisheye) perspective: Four vanishing points are placed around in a circle, they are named N, W, S, E, plus one vanishing point in the center of the circle.
• Four, or infinite-point perspective is the one that (arguably) most approximates the perspective of the human eye, while at the same time being effective for making impossible spaces, while five point is the curvilinear equivalent of one point perspective, so is four point the equivalent of two point perspective.

This technique can, like two-point perspective, use a vertical line as a horizon line, creating both a worms and birds eye view at the same time. It uses four or more points equally spaced along a horizon line, all vertical lines are made perpendicular to the horizon line, while orthogonals are created using a compass set on a line made at a 90-degree angle through each of the four vanishing points.

Geometric relationship

Distances a and c between the viewer and the wall are greater than the b distance, so adopting the principle that when an object is a greater distance from the observer, it becomes smaller, the wall is reduced and thus appears distorted at the edges.

Mathematics

If a point has the 3D Cartesian coordinates (x,y,z):

${\displaystyle P_{3\mathrm{D}}=(x,y,z)}$

Denoting distance from the point to the origin by d = √x² + y² + z²,

then the transformation of the point to a curvilinear reference system of radius R is

${\displaystyle P_{2\mathrm{D}}=\left(\frac{xR}{d},\frac{yR}{d}\right)}$

(if d = 0, then the point is at the origin, which means its projection is undefined)

This is derived by first projecting the 3D point onto a sphere with radius R that centers on the origin, so that we obtain an image of the point that has coordinates

${\displaystyle P_{\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}}=(x,y,z)\ast \left(\frac{R}{d}\right)}$

Then, we do a parallel projection that is parallel with the z-axis to project the point on the sphere onto the paper at z = R, thus obtaining

${\displaystyle P_{\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{e}}=\left(\frac{xR}{d},\frac{yR}{d},R\right)}$

Since we are not concerned with the fact that the paper is resting on the z = R plane, we ignore the z-coordinate of the image point, thus obtaining

${\displaystyle P_{2\mathrm{D}}=\left(\frac{xR}{d},\frac{yR}{d}\right)=R\ast \left(\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}}\right)}$

Since changing ${\displaystyle R}$ only amounts to a scaling, it is usually defined to be unity, simplifying the formula further to:

${\displaystyle P_{2\mathrm{D}}=\left(\frac{x}{d},\frac{y}{d}\right)=\left(\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}}\right)}$

A line that does not pass through the origin is projected to a great circle on the sphere, which is further projected to an ellipse on the plane. The ellipse has the property that its long axis is a diameter of the "bounding circle".

Examples