Perspective projection matrix


A VERY OLD STUPID JOKE:

Idiot-A: Hey!!! Look!!! An aeroplane!!!

Idiot-B: I wonder who has the patience to paint such a huge plane!!!

Idiot-C: When the aeroplane goes far into the sky, it becomes so small. And then, any painter can easily paint that tiny aeroplane.

Consider a group of boxes arranged on the floor


The nearer boxes appear bigger.
And the farther boxes appear smaller.

Perspective projection

In computer graphics, we often face the situation to transform a view-frustum into a cuboid shape. The motive behind this transformation is to achieve the realistic effect of diminishing the size of the graphics objects as they move away from the camera.


The above volume should be transformed into...

And, as you have guessed it, this transformation is done through the matrix multiplication. Before getting deeper into the problem, let's observe the transformation a little closer.

Inputs:
1. width of the viewport
2. height of the viewport
3. near plane distance from the camera
4. far plane distance from the camera
5. θ - the field of view

Output:
A 4x4 matrix

We need a 4x4 matrix, that can transform

the frustum EFGHPQRS into...
the cuboid E'F'G'H'P'Q'R'S'

Formal representation:

Lets consider the points A and B on the camera's direction
A is a point on the near plane with the coordinates A(0, 0, near).
B is a point on the far plane with the coordinates B(0, 0, far).

Recall from the previous page that, when we add a 4th component, we get the following 4D coordinates...
A(0, 0, near, 1) and B(0, 0, far, 1)

Now, consider the cuboid E'F'G'H'P'Q'R'S' in which...
On X-axis (parallel to E'F'), left = -1 and right = 1
On Y-axis (parallel to E'H'), top = 1 and bottom = -1
On Z-axis (parallel to E'P'), front = 0 and back = 1

Then, we have A'(0, 0, 0) and B'(0, 0, 1)

If these are to be projected at a unit distance from the camera(or origin), we get the following 4D coordinates...
(0, 0, 0, 1) and (0, 0, 1, 1)

Recall from the previous page that a point(x, y, z, w) is same as (x/w, y/w, z/w, 1)
So, we can have...
A'(0, 0, 0, near) and B'(0, 0, far, far)

Solving for the matrix:

The 4x4 matrix should transform the vectors A into A' and B into B'

Lets Assume the matrix is given by...

M = m11 m12 m13 m14
m21 m22 m23 m24
m31 m32 m33 m34
m41 m42 m43 m44


M x A = A'

m11 m12 m13 m14 X 0 =0
m21 m22 m23 m24 0 0
m31 m32 m33 m34 near 0
m41 m42 m43 m44 1 near


That gives us...

near.m13+m14 = 0
near.m23+m24 0
near.m33+m34 0
near.m43+m44 near


Similarly, if we take M x B = B' then we get...

far.m13+m14 = 0
far.m23+m24 0
far.m33+m34 far
far.m43+m44 far


Solving the above simultaneous equations, we get...

m13 = 0
m14 = 0
m23 = 0
m24 = 0
m44 = 0
m43 = 1
m33 = far / (far - near)
m34 = near.far / (near - far)

Now, our matrix looks like...

M = m11 m12 0 0
m21 m22 0 0
m31 m32 far / (far - near) near.far / (near - far)
m41 m42 1 0


Cool. We have already found half the values of the matrix. For the remaining half, here we go...
(Go and have a cup of coffee if you need a break)

Now, consider the point C which is the mid point of GH


When viewed from the side angle, it looks like this...


With little trigonometric workout, we can deduct...
C(0, near.tan(θ/2), near, 1)

And it's counterpart on the cuboid would be a 4D version of (0, 1, 0)
C'(0, near, 0, near)

M x C = C'

m11 m12 m13 m14 X 0 =0
m21 m22 m23 m24 near.tan(θ/2) near
m31 m32 m33 m34 near 0
m41 m42 m43 m44 1 near


Substituting the already found values will give us...

m11 m12 0 0 X 0 =0
m21 m22 0 0 near.tan(θ/2) near
m31 m32 far / (far - near) near.far / (near - far) near 0
m41 m42 1 0 1 near


That give us...

m12.near.tan(θ/2) = 0
m22.near.tan(θ/2) near
m32.near.tan(θ/2) 0
m42.near.tan(θ/2) + near near


And a little algebraic workout will give us...

m12 = 0
m22 = cot(θ/2)
m32 = 0
m42 = 0

Now, our matrix looks like...

M = m11 0 0 0
m21 cot(θ/2) 0 0
m31 0 far / (far - near) near.far / (near - far)
m41 0 1 0


Cool. We are almost there. We just need one more column in our matrix.

Now, consider the point D, which is the mid-point of GF

Recall that A is the center of the Rectangle EFGH
And C is the mid-point of GH
From the image, we can say that AD / AC = width / height

If we represent the width / height value as 'aspect' ratio, we get...
D(aspect.near.tan(θ/2), 0, near, 1)

And its counterpart on the cuboid would be a 4D version of (1, 0, 0)
D'(near, 0, 0, near)


If we consider M x D = D' then we get...

m11 0 0 0 X aspect.near.tan(θ/2) =near
m21 cot(θ/2) 0 0 0 0
m31 0 far / (far - near) near.far / (near - far) near 0
m41 0 1 0 1 near


That give us...

m11.aspect.near.tan(θ/2) = near
m21.aspect.near.tan(θ/2) 0
m31.aspect.near.tan(θ/2) 0
m41.aspect.near.tan(θ/2) + near near


And a little algebraic workout will give us...

m11 = cot(θ/2) / aspect
m21 = 0
m31 = 0
m41 = 0


Finally, our matrix looks like...

M = cot(θ/2) / aspect 0 0 0
0 cot(θ/2) 0 0
0 0 far / (far - near) near.far / (near - far)
0 0 1 0


EXERCISE:

1) This derivation was based on the assumption that the near plane corresponds to the Z-axis value of front = 0 on the cuboid. Find the projection matrix if the near plane corresponds to Z-axis value of front = -1 on the cuboid.

2) This derivation was also based on the left handed coordinate system, where front has lesser value than the back. Find the projection matrix for the right handed coordinate system, where the Z-axis values are front = 1 and back = -1 on the cuboid.