perspective projection transformation matrix

Question

In one text, the derivation for perspective projection goes like this:

If $x',y',z'$ represent any point along the projection line,and $x_{prp},y_{prp},z_{prp}$ are the projection reference point, then:

$$x'=x-(x-x_{prp})u$$
$$y'=y-(y-y_{prp})u$$ $$z'=z-(z-z_{prp})u$$ where $0\le u \le 1$

Now, finding the value of u from the third equation above and replacing in the first and second equations, we get at $z_vp$:

$$ x_p=x'= x.\frac{z_{vp}-z_{prp}}{z-z_prp}+x_{prp}.\frac{z-z_{vp}}{z-z_{prp}}$$ $$ y_p=y'= y.\frac{z_{vp}-z_{prp}}{z-z_prp}+y_{prp}.\frac{z-z_{vp}}{z-z_{prp}}$$

and there is no projection transformation matrix given.

In the second text, the derivation is using similar triangles:

comparing the similar triangles we get: $$x_p=\frac{-x_c}{\frac{z_c}{d}}$$ $$y_p=\frac{y_c}{\frac{z_c}{d}}$$

and the projection transformation for homogeneous coordinates x,y,z is given by: $$ \begin{bmatrix} -1&&0&&0&&0\\ 0&&1&&0&&0\\ 0&&0&&1&&0\\ 0&&0&&\frac{1}{d}&&0 \end{bmatrix} $$

so, my questions are:

• How do I find the projection matrix of coordinates as found from derivation 1?
• Both are given in the perspective projection area of these different texts, are both transformations the same? If not, how are they different?

Answer 1

Given your equations (which i just assume are correct): $$ \begin{align} x_p &= x \frac{z_{vp}-z_{prp}}{z-z_{prp}}+x_{prp} \frac{z-z_{vp}}{z-z_{prp}}\\ y_p &= y \frac{z_{vp}-z_{prp}}{z-z_{prp}}+y_{prp} \frac{z-z_{vp}}{z-z_{prp}} \end{align}, $$

if we take $(x_{prp}, y_{prp}, z_{prp}) = (0, 0, 0)$, then we have

$$ \begin{align} x_p &= x\frac{z_{vp}}{z}\\ y_p &= y\frac{z_{vp}}{z} \end{align} $$

Assuming the projection plane is at distance $d$ from the origin, and, by your drawing, that the projection plane is on the negative side of the Z axis, we have $z_{vp} = -d$, which leads to

$$ \begin{align} x_p &= - x\frac{d}{z}\\ y_p &= - y\frac{d}{z} \end{align} $$

Now we can follow the same reasoning (homogeneous coordinates) as in my answer to your other question (using column notation) to write

$$ \begin{align} \begin{bmatrix} x\\y\\z\\ \frac{-z}{d} \end{bmatrix} = \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&\frac{-1}{d}&0 \end{bmatrix} \begin{bmatrix} x\\y\\z\\1 \end{bmatrix} \end{align} $$

which after the division by the last coordinate will give the projected point. Notice that because we are writing it in column notation, the perspective matrix is the transpose of the matrix in your other question. Also, the negative sign is because the camera reference frame is a right-handed one.

As for the second textbook that you are reading, I cannot understand why you would have a negative sign only in the x axis without knowing more about the camera reference frame. By the diagram you posted, there should not be a negative sign.