## Geometry of Stereo 3D Projection

This page was created in response to several requests
for details on how the hypercube applet projects
a point from 3-space onto the 2-dimensional surface of the monitor.

**
Figure: projection from an arbitrary 3D point (x1,y1,z1) to a 2D point on the monitor
(looking down from above the monitor).
**

### Notation:

Points in 3-dimensional space are represented by triples of numbers or symbols enclosed
in parentheses, e.g. "(x1,y1,z1)".
The first number or symbol represents the x-coordinate, the second
represents the y-coordinate, the third represents the z-coordinate.
### Features of the Figure:

- The monitor: the center of the monitor surface is the "origin" of the coordinate system.
The monitor surface lies in the x-y plane (all points on the monitor surface
have a z-coordinate of zero).
- A brown hexagon: a 3D figure to be projected onto the monitor.
- A person's right eye, located at (d,0,z0). The separation between the right
eye and the left eye (not shown) is 2d.
The eyes are at a distance z0 from the monitor and vertically
centered (the y-coordinate is zero).
- A blue ray of light which originates at a point on the monitor surface (x2,y2,0),
passes through a vertex (x1,y1,z1) of the hexagon, and enters the right eye.

### Overview of the Math:

The 2 points (x1,y1,z1) and (d,0,z0) determine a straight line. We need to find the point
(x2,y2,0) where that line touches the plane of the monitor. The first step is to get a
formula for the line. Then we find the point on the line where the z-coordinate is zero.
### The Math

It is easy to verify that the following is a "parametric" representation of the line:
x = t * x1 + (1 - t) * d
y = t * y1 + (1 - t) * 0
z = t * z1 + (1 - t) * z0

where the asterisk (*) represents multiplication.
The "parameter" is "t". By varying "t", you can make (x,y,z) equal any point on the line.
It is easy to see that if t = 1, parametric representation gives (x,y,z) = (x1,y1,z1).
If t = 0, the parametric representation gives (x,y,z) = (d,0,z0).
Thus we have verified that the line passes through the necessary two points.

Now we need to find out which value of "t" gives a z-coordinate of zero (i.e. a point in
the plane of the monitor). So we set z=0 in the 3rd equation:

0 = t * z1 + (1 - t) * z0

Solve for "t":
t = z0 / (z0 - z1)

Plug this value of "t" into the equations for "x" and "y" to get the intersection
(x2,y2,0) of the line with the screen.
x2 = (z0 * x1 - z1 * d) / (z0 - z1)
y2 = (z0 * y1) / (z0 - z1)

This is the desired result. For the left eye, replace "d" with "-d".

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