[Math] 3D Perspective projection

I have this following question to answer, however I am not sure how I should combine my calculation into one final answer.

Suppose the Centre of Projection in a viewing space is at an offset
$(0, 0, -5)$ from $(0,0,0)$, and the view plane is the $UV$ plane containing
$c$. Find the transform matrix for the perspective projection, and give
the projected Word Coordinate point $(10,-20,-10)$ on the view plane.

So this is the transformation matrix: with $1/d=0.2$. ($1/5$)
$$
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1/d & 1
\end{pmatrix}
$$
Because the view plane is placed at $z=0$ we use the following similar triangles:
$$
x' = x/(z/d+1), \quad
y' = y/(z/d+1), \quad
z'=z/(z/d+1), \quad
w=z/d
$$
My question is, what values do I use for the $x$,$y$,$z$. If using the project world coordinates $x'$, $y'$, $z'$ will be undefined due to the $0$. Once I have the values for $x'$, $y'$, $z'$ do I multiple them by the transformation matrix?
Thanks for any help

Best Answer

We want to map $P = (x,y,z)^\top$ to $P'=(x',y',z')^\top$.

All rays go through $C = (0,0,-5)^\top = (0,0,-d)^\top$ and hit the plane $z = 0$.

(Large version here and here)

We have the line with intersection $$ (0,0,-d)^\top + t ((x, y, z)^\top - (0,0,-d)^\top) = (x', y', 0)^\top \iff \\ (tx,ty,t(z+d) - d)^\top = (x', y', 0)^\top $$

so we need $$ t(z+d) -d = 0 \iff t = d/(z+d) $$ This leads to \begin{align} P' &= (x', y', z')^\top \\ &= (x', y', 0)^\top \\ &= \left( \frac{d}{z + d} x, \frac{d}{z + d} y, \frac{d}{z + d} (z+d) - d \right)^\top \\ &= \left( \frac{d}{z + d} x, \frac{d}{z + d} y, 0 \right)^\top \quad (*) \end{align}

So far we are in agreement regarding $x'$ and $y'$. We have difference in $z'$, which should be $$ z' = \frac{1}{z/d + 1} (z+d) - d = \frac{d}{z + d} (z+d) - d = 0 $$ and $w'$ will be different as well, see below.

Using homogeneous coordinates we can write the transformation $(*)$ as $$ p' = T p \iff \\ \begin{pmatrix} x' \\ y' \\ z' \\ w' \end{pmatrix} = \begin{pmatrix} d & 0 & 0 & 0 \\ 0 & d & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & d \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} \quad (**) $$ we get a homogeneous image vector $$ p' = \left( d x, d y, 0, z + d \right)^\top $$ which can be normalized to $$ p' = \left( \frac{d}{z + d} x, \frac{d}{z + d} y, 0, 1 \right)^\top $$

Finally one can apply the above transformation $(**)$ to $p = ( 10, -20, -10, 1)^\top$.

This gives $p' = (50, -100, 0, -5)^\top$ which normalizes to $p' = (-10, 20, 0, 1)^\top$ or $P'=(-10,20,0)^\top$, where the $x'$ value agrees with the 2D image view shown above.

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[Math] Finding a 3D transformation matrix based on the 2D coordinates

I would guess that you're supposed to use "homogeneous coordinates", and that the "3D" matrix you're supposed to compute is the matrix of a projective transformation. When we use homogeneous coordinates, a coordinate vector for a point in a plane has three components rather than two. This may seem unintuitive or unappealing, but there are advantages to homogeneous coordinates: 1) "points at infinity" can be assigned coordinates as easily as ordinary points ; 2) every projective transformation is given by multiplication by a $3 \times 3$ matrix. (There are other advantages too; mathematicians have discovered that homogeneous coordinates are a natural and elegant way to assign coordinates to points in a projective plane.)

Suppose we have a projective transformation that is represented by a matrix $H \in \mathbb R^{3 \times 3}$, so that point with coordinate vector $x$ maps to a point with coordinate vector $Hx$. If we have four points $x_1,\ldots,x_4$ in "general position" and we know these four points map to $y_1,\ldots, y_4$, then we can use this information to compute $H$.

We cannot simply say that $y_i = H x_i \, \,$, because one quirky thing about homogeneous coordinates is that if $y$ is a coordinate vector for a point in a projective plane, then any nonzero scalar multiple of $y$ is also a coordinate vector for the same point. This might seem weird but it's something we must get used to.

The most we can say is that $y_i$ is a (nonzero) scalar multiple of $H x_i$. An equivalent way to say the same thing is \begin{equation} y_i \times H x_i = 0 \end{equation} where $\times$ denotes the cross product.

Notice that this equation is linear in the components of $H$ ! For each $i$, we have three linear equations for the components of $H$. However, it turns out that one of those three equations is redundant, so each $i$ really only gives us two equations. All together, we have eight linear equations.

$H$ is found by finding a nonzero solution of this system of eight equations. Note that this only determines $H$ up to a scalar multiple. However, that's to be expected, because in homogeneous coordinates, any nonzero scalar multiple of $H$ represents the same projective transformation as $H$.

[Math] A controlled trapezoid transformation with perspective projecton

Grabbed a pencil, piece of paper, and Maple and solved it by myself.

var eqRoot:Number = -Math.sqrt(newBottomWidth*newBottomWidth*oldHeight*oldHeight - newHeight*newHeight*oldWidth*oldWidth);
var focalLength:Number = Math.abs(eqRoot/(newBottomWidth - oldWidth));
var angle:Number = Math.atan2(eqRoot/(newBottomWidth*oldHeight), newHeight*oldWidth/(newBottomWidth*oldHeight));
var angleDeg:Number = angle*180/Math.PI;

Best Answer

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