On a two point perspective drawing of three equidistant vertical objects, what is the distance on paper between the third and second object?

Question

See this construction.

Consider a two point perspective drawing of two street lights. These are drawn as two vertical line segments, which on paper are distanced $CD$ apart. I want to draw a third street light further away, which in the real world is placed equal distance from the second as the second is from the first.

I suspected that this distance on paper $BC$ is a function of $CD$ and the distance between the first street light $AD$.

While messing around with it in a computer program, I confirmed that $BC$ depends solely on $AD$ and $CD$, but I haven't been able to derive this.

What is $BC$ as function of $AD$ and $CD$?

Answer 1

You are correct. Assuming the link I found describing perspective drawing is right, the length of $BC$ only depends on $AD$ and $CD$.

According to the link, to find $B$ you should draw a line through $G$ and $J$, where $J$ is the midpoint of $CH$. The intersection of this line with $AD$ is the point $B$. So let's find $B$, or rather $BC$.

We see that $\triangle BDG$ is similar to $\triangle BCJ$. This means we can write $$\frac{DG}{BC + CD}=\frac{CJ}{BC}$$ which gives $$BC = \frac{CJ \cdot CD}{DG-CJ} \tag {1}$$ But we can find $CJ$. We know $CJ = \frac{CH}{2}$ and we can see that $CH = DG - KG$. Since $\triangle ADG$ is similar to $\triangle HKG$ we can write $$\frac{AD}{DG} = \frac{HK}{KG} = \frac{CD}{KG}$$ which means $$KG = \frac{DG \cdot CD}{AD}$$ Inserting this in the formula for $CJ$ we find that $$CJ = \frac{DG \cdot (AD - CD)}{2 \cdot AD}$$ Finally, inserting this into equation $1$ gives the result: $$BC = \frac{(AD - CD)}{(AD + CD)} \cdot CD$$

Answer 2

Assuming that the points $A$, $B$, $C$ and $D$ are colinear, their cross-ratio $$[A,B;C,D] = {AC\over AD}\Big/{BC\over BD}$$ is projectively invariant. Computing these distances in the perspective drawing is straightforward, but $A$ is a vanishing point—its counterpart $A'$ in the scene is “at infinity”—so we’ll need to do something a bit more complicated to compute the cross-ratio of the points in the scene.

A convenient choice of coordinate system for the original line in the scene is to place the origin at $D'$ and take $C'D'$ as the unit distance along this line. The homogeneous line coordinates of the points are then $$A' = (1,0), B'=(2,1), C'=(1,1), D'=(0,1)$$ and their cross-ratio can be expressed as ratios of determinants: $$[A',B';C',D'] = {\begin{vmatrix}1&0\\1&1\end{vmatrix} \over \begin{vmatrix}1&0\\0&1\end{vmatrix}}\Big/{\begin{vmatrix}2&1\\1&1\end{vmatrix}\over\begin{vmatrix}2&1\\0&1\end{vmatrix}} = 2.$$ Equating the two cross-ratios and solving for $BC$ gives $$BC = {AC\cdot BD\over 2AD} = {AD-CD\over AD+CD} CD.$$ More generally, if $B'C':C'D'=\lambda$, then $BC = {AC\cdot BD\over AD}{\lambda\over\lambda+1}$, which I’ll leave for you to rewrite in terms of $AD$ and $CD$ only.