In his Complete Guide to Perspective Drawing, Craig Attebery defines a central concept in two-point perspective: Measuring Points, which can be used to determine depth in lines going away from the viewer.
Attebery states, without proof, that the distance from Vanishing Points to Measuring Points should be the same as the distance from Vanishing Points to the Station Point (artist’s eye). That is, Measuring Points can be obtained using a compass:
A different theory, still, comes from Joseph D’Amelio’s Perspective Drawing Handbook. From the latter’s definition, Measuring Points would be Vanishing Points themselves. As each set of parallel lines has its Vanishing Point, this would imply each Vanishing Point to have not a single, but infinite Measuring Points:
Both theories, as interpreted above, cannot be right: Vanishing Points have either a single or infinite measuring points. The question is, thus:
How many Measuring Points does each Vanishing Point has? If only one, why should Attebery’s statement hold: that the distance from Vanishing Points to Measuring Points be the same as the distance from Vanishing Points to the Station Point?
Best Answer
Each direction has a specific vanishing point.
Guide lines that have the same vanishing point, correspond to the same direction.
You can use any direction for guide lines, because the angle between the guide lines and the object edge only affects the scale.
Here is visual proof:
The black line represents a line divided into four equal length sections.
Red guide line corresponds to one measurement point, and blue guide line to another measurement point.
Both red lines and blue lines intersect the horizontal measuring line at regular intervals; only the exact scale varies.
Note that both Craig Attebery and Joseph D'Amelio both provide a way to choose a good measuring point. However, any measuring point not too close to the vanishing point of the line that is to be measured, will work.
The method shown by D'Amelio at makes the measurements along the measuring line easy, so that one can use a rule/scale without having to do any math.
An infinite number. The choice of the vanishing point only affects the scale between the measuring line and the object (line to be measured).
Since this is math.stackexchange.com, I think it is pertinent to explore the actual projective math. It is surprisingly simple, you see, if we choose a particular coordinate system.
Let Station Point (eye) be at origin $(0, 0, 0)$, and the picture plane perpendicular to the $z$ axis, at distance $d$. (You are then gazing directly along the positive $z$ axis.)
Any point $(x, y, z)$ in 3D space corresponds to point $(x^\prime, y^\prime, z^\prime)$ on the picture plane via $$\left\lbrace \; \begin{aligned} x^\prime &= x \frac{d}{z} \\ y^\prime &= y \frac{d}{z} \\ z^\prime &= z \frac{d}{z} = d \\ \end{aligned} \right.$$ Note that $x=0$, $y=0$ corresponds to the sight line; the line that passes through the station point (eye) and is perpendicular to the picture plane.
Vanishing points are points at infinite distance. You pick any line, say one that passes through $(x_0, y_0, z_0)$ in direction $(x_\Delta, y_\Delta, z_\Delta)$, $z_\Delta \gt 0$, so that the line parametrized using $t \in \mathbb{R}$ is $$\left\lbrace ~ \begin{aligned} x &= x_0 + t x_\Delta \\ y &= y_0 + t y_\Delta \\ z &= z_0 + t z_\Delta \\ \end{aligned} \right.$$ and the vanishing point corresponding to that line is $(x^\prime_v ,\; y^\prime_v ,\; z^\prime_v = d)$: $$\left\lbrace \; \begin{aligned} x^\prime_v &= \lim_{t \to \infty} \left( \bigr( x_0 + t x_\Delta \bigr) \left( \frac{d}{z_0 + t z_\Delta} \right) \right) = d \frac{ x_\Delta }{ z_\Delta } \\ y^\prime_v &= \lim_{t \to \infty} \left( \bigr( y_0 + t y_\Delta \bigr) \left( \frac{d}{z_0 + t z_\Delta} \right) \right) = d \frac{ y_\Delta }{ z_\Delta } \\ z^\prime_v &= \lim_{t \to \infty} \left( \bigr( z_0 + t z_\Delta \bigr) \left( \frac{d}{z_0 + t z_\Delta} \right) \right) = d \\ \end{aligned} \right.$$ This means that every direction $(x_\Delta, y_\Delta, z_\Delta)$ in to the picture, $z_\Delta \gt 0$, has its own explicit vanishing point at $(d x_\Delta / z_\Delta, \; d y_\Delta / z_\Delta, \; d)$.
In other words, all parallel lines do indeed have the same vanishing point.
Note that the above is an exact mathematical description of what vanishing points are. This is not any kind of approximation. This is what happens in reality, if we think of the image plane as e.g. a window, or the wall in a camera obscura with origin at the pinhole and the wall perpendicular to the $z$ axis at $(0, 0, -d)$. (The origin is technically the focal point of the lens used.)
For clarity, I use $\;^\prime$ to denote picture coordinates, although they are in the same coordinate system as everything else. $z^\prime = d$ for all points on the picture plane, so it is okay to just drop the $z$ coordinate.
We can furthermore define the rotation of the image plane around the $z$ (depth) axis, by defining horizontal parallel to the $x$ axis ($x z$ plane), and vertical parallel to the $y$ axis ($y z$ plane). Horizon plane is then $y = 0$. If we choose a left-handed coordinate system, $x$ increases right and $y$ up. If our eye is at height $h$, then the ground plane is $y = -h$ (because our eye is at origin, $(0, 0, 0)$).
Both guides OP referenced involve lines at a specific height, the same plane $y$. Because all lines on such planes have $y_\Delta = 0$, their vanishing points are on the horizon line at $x^\prime = d x_\Delta / z_\Delta$, $y^\prime = 0$. The exact height $y = h$ where the plane is, does not matter.
If we use angle $\theta$ to describe the direction of a line on such a plane (say, on the ground, or any other horizontal plane), with $\theta = 0$ being directly along the sight line, $\theta \gt 0$ right, $\theta \lt 0$ left, we can use $x_\Delta = \sin \theta$, $z_\Delta = \cos \theta$, and the vanishing point $x^\prime$ coordinate on the horizon line is $$x^\prime = d \frac{\sin\theta}{\cos\theta} = d \tan\theta$$ and inversely, if we know the $x$ coordinate $x^\prime$ (and the distance from station point to picture plane $d$), $$\theta = \arctan\left(\frac{x^\prime}{d}\right)$$
The vanishing point corresponding to a perpendicular line on the same plane ("ground level"), $\theta \pm 90°$, is $$x^\prime_\perp = d \tan(\theta \pm 90°) = -d \cot\theta = \frac{-d}{\tan\theta}$$ and therefore the distance $L$ between vanishing points corresponding to perpendicular directions is $$L = \lvert x^\prime - x^\prime_\perp \rvert = d \lvert \tan\theta + \cot\theta \rvert$$ For symmetric two-point perspective, $\theta = 45°$ and $L = 2 d$. If $\theta = 0$, the other vanishing point is at ±infinity, and vice versa. Here is a small table of angles, so you can have an idea of how $L$ varies as a function of $\theta$: $$\begin{array}{c|c} \theta & L \\ \hline 45° & 2.000 \; d \\ 40°, 50° & 2.031 \; d \\ 35°, 55° & 2.128 \; d \\ 30°, 60° & 2.309 \; d \\ 25°, 65° & 2.611 \; d \\ 20°, 70° & 3.111 \; d \\ 15°, 75° & 4.000 \; d \\ \end{array}$$
If we want to choose the vanishing point so that the distances from the point where the measuring line and the object edge intersect match along both (the measuring line and the object edge), we do need to choose a specific vanishing point. Looking from above,
we want to choose a direction $\varphi$ (and corresponding vanishing point) for the blue guide lines so that measurements along the black measuring line match the ones along the red object line. (Specifically, that distances from the point where the measuring line intersects the object line, match.)
If the object line direction is $\theta$, then the guide line direction $\varphi$ is
$$\varphi = \begin{cases} - \frac{90° - \theta}{2} = \frac{1}{2}\theta - 45°, & \theta \ge 0 \\
\frac{90° + \theta}{2} = \frac{1}{2}\theta + 45°, & \theta \lt 0 \\
\end{cases}$$
If $x^\prime$ is the $x$ coordinate for the object line vanishing point (or true vanishing point), we need to calculate $\theta$ from
$$\theta = \arctan\left(\frac{x^\prime}{d}\right)$$
and then we can calculate the measuring point $x$ coordinate $x^\prime_m$,
$$x^\prime_m = d \tan\varphi$$
Here is a small table of various object line directions, and the distance $L$ from the object line vanishing point to the optimal measurement point:
$$\begin{array}{c|c|c|c|c}
\theta & x^\prime & \varphi & x^\prime_m & L \\
\hline
0° & 0.000 \; d & -45° & -1.000 \; d & 1.000 \; d \\
5° & 0.087 \; d & -42.5° & -0.916 \; d & 1.004 \; d \\
10° & 0.176 \; d & -40° & -0.839 \; d & 1.015 \; d \\
15° & 0.268 \; d & -37.5° & -0.767 \; d & 1.035 \; d \\
20° & 0.364 \; d & -35° & -0.700 \; d & 1.064 \; d \\
25° & 0.466 \; d & -32.5° & -0.637 \; d & 1.103 \; d \\
30° & 0.577 \; d & -30° & -0.577 \; d & 1.155 \; d \\
35° & 0.700 \; d & -27.5° & -0.521 \; d & 1.221 \; d \\
40° & 0.839 \; d & -25° & -0.466 \; d & 1.305 \; d \\
45° & 1.000 \; d & -22.5° & -0.414 \; d & 1.414 \; d \\
50° & 1.192 \; d & -20° & -0.364 \; d & 1.556 \; d \\
55° & 1.428 \; d & -17.5° & -0.315 \; d & 1.743 \; d \\
60° & 1.732 \; d & -15° & -0.268 \; d & 2.000 \; d \\
65° & 2.145 \; d & -12.5° & -0.222 \; d & 2.366 \; d \\
70° & 2.747 \; d & -10.0° & -0.176 \; d & 2.924 \; d \\
75° & 3.732 \; d & -7.5° & -0.132 \; d & 3.864 \; d \\
80° & 5.671 \; d & -5° & -0.087 \; d & 5.759 \; d \\
85° & 11.430 \; d & -2.5° & -0.044 \; d & 11.474 \; d \\
\end{array}$$