The First Ratio

Let's begin by calculating the Euclidean distance between points A and B. I'll start with the 2D homogeneous coordinates of each point, which I will name as follows:

$\begin{array}{l}\text{A}=\left[\begin{array}{ccc}{A}_x& A_y& A_w\end{array}\right]\\ \text{B}=\left[\begin{array}{ccc}{B}_x& B_y& B_w\end{array}\right]\end{array}$

To calculate a Euclidean distance, we must turn these into the “real” coordinates by dividing out the homogeneous w coordinate giving the two 2D Euclidean points:

$\left[\begin{array}{cc}\frac{A_x}{A_w}& \frac{A_y}{A_w}\end{array}\right]\text{and}\left[\begin{array}{cc}\frac{B_x}{B_w}& \frac{B_y}{B_w}\end{array}\right]$

Then subtract giving

$\left[\frac{A_x}{A_w}-\frac{B_x}{B_w}\frac{A_y}{A_w}-\frac{B_y}{B_w}\right]$

The desired distance is the length of this 2D vector. The fact that this quantity doesn't treat all three of the x, y, and w components symmetrically is a hint that Euclidean distance is not a meaningful concept in projective geometry.

Anyway, if we bash ahead to get the Euclidean length of the above vector, we would need to calculate the square root of the sum of the squares of these two components—wotta pain. Instead, let's pretend for a minute that the line m is horizontal and points A … D line up horizontally. In that case, the length of the vector is just the x component of the vector. In other words,

$\frac{A_x}{A_w}-\frac{B_x}{B_w}=\left|\text{AB}\right|$

In general, though, the line will be at some angle θ. In this case, the x component will equal the length times the cosine of the tilt angle, or

$\frac{A_x}{A_w}-\frac{B_x}{B_w}=\left|\text{AB}\right|\cos \theta$

For the collinear segment BD, tipped by the same angle θ the x component would be

$\frac{B_x}{B_w}-\frac{D_x}{D_w}=\left|\text{BD}\right|\cos \vartheta$

Here's where the first ratio will come in to simplify life. The top half of our desired cross ratio is the ratio between these segment AB and segment BD. And, by similarity, the ratio of the x components is the same as the ratio of the lengths. Taking the ratio of the above two equations, we find

$\frac{\frac{A_x}{A_w}-\frac{B_x}{B_w}}{\frac{B_x}{B_w}-\frac{D_x}{D_w}}=\frac{\left|\text{AB}\right|\cos \vartheta }{\left|\text{BD}\right|\cos \vartheta }=\frac{\left|\text{AB}\right|}{\left|\text{BD}\right|}$

The angle dependency cancels. Yay! We can calculate the ratio of lengths without any squaring and square rooting.

The Second Ratio

Now let's look at the second ratio in the “ratio of ratios” game. Remember that what we are really interested in is

$\chi =\frac{\left|\text{AB}\right|/\left|\text{BD}\right|}{\left|\text{AC}\right|/\left|\text{CD}\right|}$

A little algebra on the AB/BD ratio turns it into

$\frac{\left|\text{AB}\right|}{\left|\text{BD}\right|}=\frac{D_w}{A_w}\left(\frac{A_x{B}_w-A_w{B}_x}{B_x{D}_w-B_w{D}_x}\right)$

And by going through the same schtick, but everywhere changing B to C, we can see that the second of our two ratios is

$\frac{\left|\text{AC}\right|}{\left|\text{CD}\right|}=\frac{D_w}{A_w}\left(\frac{A_x{C}_w-A_w{C}_x}{C_x{D}_w-C_w{D}_x}\right)$

Now we can see why the ratio of ratios is something interesting homogeneously—it lets us cancel the ugly w component ratios. The whole cross ratio then boils down to

$\chi =\frac{\left|\text{AB}\right|/\left|\text{BD}\right|}{\left|\text{AC}\right|/\left|\text{CD}\right|}=\frac{A_x{B}_w-A_w{B}_x}{B_x{D}_w-B_w{D}_x}\frac{C_x{D}_w-C_w{D}_x}{A_x{C}_w-A_w{C}_x}$

This is much more symmetric in x, y, and w, and closer to the sort of thing we expect we can make homogeneously constant.

Constancy with Changing m

Now we can ask why this ratio is independent of the position of the line m. The root cause must have something to do with the fact that we generate the points A, B, C, and D from the lines p, q, r, and s that have the special relationship of all intersecting at the same point E.

First, then, here's a brief reminder of the relationship between homogeneous points and lines. You can calculate the point at the intersection of two lines by taking the cross product of the line vectors. This means that the point at the intersection of the four lines is any of the six quantities:

$\begin{array}{l}\text{E}=c_1\left(\text{p}\times \text{q}\right)=c_2\left(\text{q}\times \text{r}\right)=c_3\left(\text{r}\times \text{s}\right)\\ =c_4\left(\text{p}\times \text{r}\right)=c_5\left(\text{q}\times \text{s}\right)=c_3\left(\text{p}\times \text{s}\right)\end{array}$

The inclusion of the constants c_i recognizes the fact that the cross products can represent the same intersection point even though there might be a homogeneous scale factor applied to all three of the x, y, and w components.

After some fooling around, I have found that, for our purposes, the neatest algebraic way to use this relationship is to express the lines q and r in terms of p and s. I'll write this as

$\begin{array}{l}\text{q}=q_{\alpha }\text{p}+q_{\beta }\text{s}\\ \text{r}=r_{\alpha }\text{p}+r_{\beta }\text{s}\end{array}$

You can think of the pairs $\left[\begin{array}{cc}{q}_{\alpha }& q_{\beta }\end{array}\right]$ and $\left[\begin{array}{cc}{r}_{\alpha }& r_{\beta }\end{array}\right]$ as ID homogeneous coordinates for the collection of lines passing through E, the intersection of p and s. Any point passing through this intersection has an $\left[\begin{array}{cc}\alpha & \beta \end{array}\right]$ pair that describes it. Any nonzero multiple of an $\left[\begin{array}{cc}\alpha & \beta \end{array}\right]$ pair represents the same line though.

Now, let's relate this to our points. Remember, the cross product is how we intersect lines, so we have

$\begin{array}{l}\text{A}=\text{p}\times \text{m}\\ \text{B}=\text{q}\times \text{m}\\ \text{C}=\text{r}\times \text{m}\\ \text{D}=\text{s}\times \text{m}\end{array}$

Write the interior point B in terms of the outside lines and points:

$\begin{array}{l}\text{B}=\text{q}\times \text{m}\text{=}\left(q_{\alpha }\text{p+}{q}_{\beta }\text{s}\right)\times \text{m}\\ =q_{\alpha }\left(\text{p}\times \text{m}\right)+q_{\beta }\left(\text{s}\times \text{m}\right)\\ =q_{\alpha }\text{A+}{q}_{\beta }\text{D}\end{array}$

We can then write the top half of the cross ratio as

$\begin{array}{l}\left|\text{AB}\right|/\left|\text{BD}\right|=\frac{A_w}{D_w}\frac{A_x{B}_w-A_w{B}_x}{B_x{D}_w-B_w{D}_x}\\ =\frac{A_w}{D_w}\frac{A_x\left(q_{\alpha }{A}_w+q_{\beta }{D}_w\right)-A_w\left(q_{\alpha }{A}_x+q_{\beta }{D}_x\right)}{\left(q_{\alpha }{A}_x+q_{\beta }{D}_x\right)D_w-\left(q_{\alpha }{A}_w+q_{\beta }{D}_w\right)D_x}\\ =\frac{A_w}{D_w}\frac{q_{\beta }{A}_x{D}_w-q_{\beta }{A}_w{D}_x}{q_{\alpha }{A}_x{D}_w-q_{\alpha }{A}_w{D}_x}\\ =\frac{A_w}{D_w}\frac{q_{\beta }}{q_{\alpha }}\end{array}$

Again, run through this derivation with C instead of B, and you get

$\left|\text{AC}\right|/\left|\text{CD}\right|=\frac{A_w}{D_w}\frac{r_{\beta }}{r_{\alpha }}$

The net cross ratio is now

$\chi =\frac{\left|\text{AB}\right|/\left|\text{BD}\right|}{\left|\text{AC}\right|/\left|\text{CD}\right|}=\frac{q_{\beta }/q_{\alpha }}{r_{\beta }/r_{\alpha }}$

The interesting thing about this is that the whole dependence on the location of m has disappeared, as well as the dependence on which coordinate (here we used x) we chose to use as the measure of distance ratios. That is, it is only dependant on the orientations of the lines q and r relative to P and s. Any line ${\text{m}}^{\prime }$ intersecting these will generate the same cross ratio. The cross ratio is thus a property of the locations of the four original lines, rather than the extra line m.

As another way of looking at this, start with the equations in (12.1) and cross them with P and s:

$\begin{array}{l}\text{q}\times \text{p}={\mathbf{q}}_{\alpha }\text{p}\times \text{p}+q_{\beta }\text{s}\times \text{p}=q_{\beta }\text{s}\times \text{p}\\ \text{q}\times \text{s}=q_{\alpha }\text{p}\times \text{s}+q_{\beta }\text{s}\times \text{s}=q_{\beta }\text{p}\times \text{s}\\ \text{r}\times \text{p}=r_{\alpha }\text{p}\times \text{p}+q_{\beta }\text{s}\times \text{p}=q_{\beta }\text{s}\times \text{p}\\ \text{r}\times \text{s}=r_{\alpha }\text{p}\times \text{s}+q_{\beta }\text{s}\times \text{s}=q_{\beta }\text{p}\times \text{s}\end{array}$

These are four vector equations. Turn them into scalars by dotting them with m:

$\begin{array}{l}\text{m}\cdot \text{q}\times \text{p}=q_{\beta }\text{m}\cdot \text{s}\times \text{p}\\ \text{m}\cdot \text{q}\times \text{s}=q_{\beta }\text{m}\cdot \text{p}\times \text{s}\\ \text{m}\cdot \text{r}\times \text{p}=q_{\beta }\text{m}\cdot \text{s}\times \text{p}\\ \text{m}\cdot \text{r}\times \text{s}=q_{\beta }\text{m}\cdot \text{p}\times \text{s}\end{array}$

And now you can write the cross product in the following form:

$x=\frac{q_{\beta }{r}_{\alpha }}{q_{\alpha }{r}_{\beta }}=\frac{\left(\text{m}\cdot \text{q}\times \text{p}\right)\left(\text{m}\cdot \text{r}\times \text{s}\right)}{\left(\text{m}\cdot \text{q}\times \text{s}\right)\left(\text{m}\cdot \text{r}\times \text{p}\right)}$

Here's another key fact: χ contains only the ratios of the alpha/beta pairs, so it is also independent of arbitrary homogeneous scalings of the vector for q and r. For example, if we replaced q with $\gamma$q (geometrically the same line), it means that we are scaling both $q_{\alpha }$ and $q_{\beta }$ by $\gamma$, which will again cancel out in the formula for χ

Constancy under Perspective

Now let's see why the cross ratio is unchanged when we transform the four lines via an arbitrary transformation matrix. If we have a transformation T that changes our outer two lines as

$\begin{array}{c}{\text{p}}^{\mathbf{\prime }}\text{=}\text{Tp}\\ {\text{s}}^{\mathbf{\prime }}\text{=}\text{Ts}\end{array}$

the inner lines will be

$\begin{array}{l}{\text{q}}^{\mathbf{\prime }}\text{=}\text{Tq}\text{=}\text{T}\mathbf{\left(}{\mathbf{q}}_{\mathbf{\alpha }}\text{p}\text{+}{\text{q}}_{\mathbf{\beta }}\text{s}\mathbf{\right)}\mathbf{=}{\mathbf{q}}_{\mathbf{\alpha }}\text{Tp}\text{+}{\mathbf{q}}_{\mathbf{\beta }}\text{Ts}\\ \mathbf{=}{\mathbf{q}}_{\mathbf{\alpha }}{\text{p}}^{\prime }\text{+}{\mathbf{q}}_{\mathbf{\beta }}{\text{s}}^{\mathbf{\prime }}\end{array}$

And similarly

${\text{r}}^{\mathbf{\prime }}\text{=}{\mathbf{r}}_{\mathbf{\alpha }}{\mathbf{p}}^{\mathbf{\prime }}\mathbf{+}{\mathbf{r}}_{\mathbf{\beta }}{\text{s}}^{\mathbf{\prime }}$

Even if we try to disguise line ${\text{q}}^{\mathbf{\prime }}$ or ${\text{r}}^{\mathbf{\prime }}$ by multiplying by some homogeneous scale, we get the same value of cross ratio—the ratio of ratios:

$\chi =\frac{q_{\beta }/q_{\alpha }}{r_{\beta }/r_{\alpha }}$

A particularly interesting special case of this occurs when we take the outside lines p and s as the x and y axes, and transform by a simple nonuniform scale factor. Numerically, this would be

$\begin{array}{l}\text{p}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\text{q}=\left[\begin{array}{c}{q}_{\alpha }\\ q_{\beta }\\ 0\end{array}\right],\text{r}=\left[\begin{array}{c}{r}_{\alpha }\\ r_{\beta }\\ 0\end{array}\right]\text{s}=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\\ \text{T}=\left[\begin{array}{ccc}{f}_x& 0& 0\\ 0& f_y& 0\\ 0& 0& 1\end{array}\right]\end{array}$

From this we get

${\text{p}}^{\prime }=\left[\begin{array}{c}{f}_x\\ 0\\ 0\end{array}\right],{\text{q}}^{\prime }=\left[\begin{array}{c}{f}_x{q}_{\alpha }\\ f_y{q}_{\beta }\\ 0\end{array}\right],{\text{r}}^{\prime }=\left[\begin{array}{c}{f}_x{r}_{\alpha }\\ f_y{r}_{\beta }\\ 0\end{array}\right]{\text{s}}^{\prime }=\left[\begin{array}{c}0\\ f_y\\ 0\end{array}\right]$

The alpha and beta components of q and r have changed to

$\begin{array}{l}{q^{\prime }}_{\alpha }=f_x{q}_{\alpha },{q^{\prime }}_{\beta }=f_y{q}_{\beta }\\ {r^{\prime }}_{\alpha }=f_x{r}_{\alpha },{r^{\prime }}_{\beta }=f_y{r}_{\beta }\end{array}$

The lines p and s haven't moved, but the lines q and r both have. And the ratios of the $\left[\alpha ,\beta \right]$ values of the lines q and r are different:

$\begin{array}{l}{q^{\prime }}_{\beta }/{q^{\prime }}_{\alpha }=\frac{f_y}{f_x}(q_{\beta }/q_{\alpha })\\ {r^{\prime }}_{\beta }/{r^{\prime }}_{\alpha }=\frac{f_y}{f_x}(r_{\beta }/r_{\alpha })\end{array}$

But sure enough, the ratio of ratios remains the same:

$\chi =\frac{{q^{\prime }}_{\beta }/{q^{\prime }}_{\alpha }}{{r^{\prime }}_{\beta }/{r^{\prime }}_{\alpha }}=\frac{q_{\beta }/q_{\alpha }}{r_{\beta }/r_{\alpha }}$

In other words, even though a transformation might not move the outside lines p and s, it will move the inside lines q and r in such a manner that the change in the ratio $\left(q_{\beta }/q_{\alpha }\right)$ is matched by the same change in the ratio $\left(r_{\beta }/r_{\alpha }\right)$

Summary

The cross ratio is as much a property of the four mutually intersecting lines p, q, r, and s as it is of the four collinear points A, B, C, and D. Any line m that you throw across the lines will generate four points with the same cross ratio.

You can, in turn, take any four collinear points A, B, C, and D, and throw various collections of four mutually intersecting lines through them. Each of these line collections will have the same cross ratio. Finally, you can project any of these figures perspectively and also get an unchanged cross ratio.

The true homogeneous nature of the cross ratio can best be seen by writing it as

$\chi =\frac{q_{\beta }{r}_{\alpha }}{q_{\alpha }{r}_{\beta }}$

and reviewing the effect of an arbitrary homogeneous scaling on each of the four lines. Remember that

$\begin{array}{l}\text{q}=q_{\alpha }\text{p +}{q}_{\beta }\text{s}\\ \text{r}=r_{\alpha }\text{p +}{r}_{\beta }\text{s}\end{array}$

■ Scaling only q will scale both q_α and q_β but this cancels out.

$\chi =\frac{\begin{array}{|c|}\hline q_{\beta }\\ \hline\end{array}{r}_a}{\begin{array}{|c|}\hline q_a\\ \hline\end{array}{r}_{\beta }}$

■ Scaling only r will scale r_α and r_β, but these will cancel:

$\chi =\frac{q_{\beta }\begin{array}{|c|}\hline r_a\\ \hline\end{array}}{q_a\begin{array}{|c|}\hline r_{\beta }\\ \hline\end{array}}$

■ Scaling only p will require inversely scaling q_aand r_a to keep q and r the same in Equation (12.1). These again cancel:

$\chi =\frac{q_{\beta }\begin{array}{|c|}\hline r_a\\ \hline\end{array}}{\begin{array}{|c|}\hline q_a\\ \hline\end{array}{r}_{\beta }}$

■ Scaling only s will require inversely scaling q_β and r_β. Same song, new verse:

$\chi =\frac{\begin{array}{|c|}\hline q_{\beta }\\ \hline\end{array}{r}_a}{q_a\begin{array}{|c|}\hline r_{\beta }\\ \hline\end{array}}$

From this, we can see the necessity of this arrangement of ratio of ratios in constructing a quantity that remains homogeneously meaningful.

So, even though perspective transformations do not preserve distances, or ratios of distances, they do preserve these ratios of ratios of distances.