Any three lines you draw are a cube

Four rungs up the axonometric ladder this site made a claim with some force in it. An orthographic projection’s three foreshortening ratios always satisfy $s_x^2 + s_y^2 + s_z^2 = 2$; cavalier projection draws all three axes at full length, so its three ratios are 1, 1 and 1 and its sum is 3; and therefore — the essay said — cavalier is not the projection of anything. The arithmetic is right. The sentence is wrong, and the way it is wrong is worth a rung of its own.

What the identity establishes is that cavalier is not an orthographic projection of anything. Orthographic projection is the case where the rays run perpendicular to the picture plane. Parallel projection is the case where the rays run parallel to each other, which is a weaker condition, and the difference between the two is the whole of this essay.

The theorem the drawing offices never heard #

In 1853 Karl Pohlke stated something that reads, the first time, as though it must be missing a hypothesis:

Three segments drawn from a common point in a plane, of any lengths whatever and in any directions whatever, provided they do not all lie on one line, are the parallel projection of three equal and mutually perpendicular segments.

Three segments of any lengths in any directions. Not three plausible-looking axes; three lines drawn at random by somebody who has never heard of a projection. That drawing is a picture of a cube — a real one, with twelve equal edges and eight right-angled corners — seen along some direction from some orientation, both of which the theorem promises exist and neither of which it says how to find.

The usual proof of Pohlke’s theorem runs through an auxiliary ellipse and a continuity argument, and it is an existence proof: it establishes that the cube is there and hands over nothing that could be drawn. That is presumably why the theorem is a footnote in descriptive geometry and absent from every practical account of axonometric drawing. It does not need to be. The recovery is four lines of arithmetic and it is exact.

Where the cube is hiding #

Write the three drawn axes as the columns of a $2 \times 3$ matrix $\mathbf{A}$ — six numbers, which is the whole input. A parallel projection onto the picture plane along a direction $\mathbf{d}$ takes a point to its picture coordinates minus however far the ray drifts on the way. For a cube of edge $L$ in an orientation $\mathbf{R}$ that is

$$\mathbf{A} = L \cdot \mathbf{P} \cdot \mathbf{R}, \qquad \mathbf{P} = \begin{bmatrix} 1 & 0 & -w_1 \\ 0 & 1 & -w_2 \end{bmatrix}$$

where $\mathbf{w} = (d_x, d_y)/d_z$ is the projection direction written as a sideways drift per unit of depth. When $\mathbf{w} = 0$ the rays are perpendicular to the picture plane and the projection is orthographic; anything else is oblique.

Three unknowns are in there — the cube’s edge, the direction, and the orientation — and the orientation is the awkward one, being a rotation matrix with its own constraints. So get rid of it. Multiply $\mathbf{A}$ by its own transpose and $\mathbf{R}$ disappears, because $\mathbf{R}\mathbf{R}^\mathsf{T}$ is the identity:

$$\mathbf{A}\mathbf{A}^\mathsf{T} = L^2\,(\mathbf{I} + \mathbf{w}\mathbf{w}^\mathsf{T})$$

The left side is a $2\times2$ matrix computable from the drawing alone: three numbers, obtained by adding up products of things the hand put on the paper. The right side is a scaled identity plus a rank-one term, and a rank-one term added to a scaled identity has an eigenvalue that can be read off. Its two eigenvalues are $L^2(1 + |\mathbf{w}|^2)$ and $L^2$.

So the cube’s edge is the square root of the smaller eigenvalue of $\mathbf{A}\mathbf{A}^\mathsf{T}$, and once $L$ is known the rest follows: $|\mathbf{w}|^2$ from the trace, the direction of $\mathbf{w}$ from the off-diagonal entry, and the orientation from the observation that two $2\times3$ matrices with the same row Gram matrix must differ by an orthogonal $3\times3$ — which is where Pohlke’s theorem is actually proved, in one line of linear algebra rather than in an ellipse.

What it says about cavalier #

Cavalier’s axes are drawn at unit length: one across, one up, one at 45° into the page. Run those six numbers through the recovery.

The cube is a unit cube and the direction is 45° off the normal. Both numbers are exact, and the second is the interesting one, because the departure the earlier essay measured is not merely correlated with it — it is it. Rearranged, the identity says

$$s_x^2 + s_y^2 + s_z^2 = L^2\,(2 + |\mathbf{w}|^2)$$

so for a system whose cube comes out at edge 1, the departure from two is $|\mathbf{w}|^2$, the squared tangent of the angle at which the rays leave the picture plane. Cavalier’s departure of 1 is $\tan^2 45°$. Cabinet’s 0.25 is $\tan^2 26.565°$.

A number that had been read as a verdict turns out to be a measurement. It was never saying this is not a projection; it was saying this projection is oblique, by this much. And the two conventional obliques are conventional in exactly the way one would hope: 45° and $\arctan\frac12$ are not arbitrary, they are what a drawing office reaches for when it wants a picture that looks reasonable and is easy to construct with a set square.

The control, which is the part that makes it a measurement #

A recovery that always finds an oblique cube would find one for isometric too, and the finding would be a property of the solver rather than of the drawing. So run the axonometric systems through the identical code.

Isometric comes back orthographic, dimetric comes back orthographic, trimetric comes back orthographic. The quantity the assertion is made on is $|\mathbf{w}|^2$ rather than the angle, and that distinction is not fussiness. $|\mathbf{w}|^2$ is a difference of two well-conditioned quantities and lands at $10^{-16}$; its square root doubles the relative error near zero, so the angle for a genuinely orthographic system comes out at about $10^{-6}$ degrees rather than $10^{-15}$. Nothing is wrong when that happens — it is what taking the square root of a quantity that vanishes at the answer costs — but a control stated on the angle would read as very slightly oblique, which is a different claim from orthographic.

Military, and a system this site did not have #

There is a fourth conventional system, and this essay’s phase added it because leaving it out made a false generalisation easy to reach. The military or planometric projection draws the ground plan true — true lengths, true angles, at whatever azimuth suits the page — and stands the verticals up from it at full length. Fortification drawings used it for three centuries.

Cavalier and military have identical axis scales, so they are the same point of the identity’s arithmetic and a different picture. That is a useful thing to be able to say, and the axis-scale triple cannot say it: what separates them is the orientation $\mathbf{R}$, which the sum of squares was never carrying.

What the drawing does not determine #

The recovery has a loose end that turns out to be the subject of its own rung. Completing the orthonormal frame in that last step involves a free sign: the third basis vector may point either way, and both choices give a legitimate cube reproducing the identical drawn axes.

So a parallel drawing is a picture of exactly two cubes, not one. They are mirror images and they project identically — not nearly, but to the last bit, over all eight vertices. Nothing in the drawing chooses. That is the Necker reversal arrived at as a count of solutions to a linear-algebra problem rather than as a fact about perception, and the drawing does not say which corner is nearer is the rung that takes it apart.

The front face, and what it costs #

One more reading of $\mathbf{w}$ is worth having, because it is the reason obliques exist at all. Under an orthographic projection, the fidelity of the frontal plane and the length of the depth axis are locked together: a circle in the front face is drawn as a circle only when the projection looks straight at that face, and then there is no depth axis left to draw.

Cavalier’s whole point is that it has both: a true front face and a full-length depth axis. No orthographic projection can do that, which is what the identity was saying. Pohlke says the price is not impossibility, it is obliquity — and the amount of obliquity is exactly the amount by which the identity is violated. Oblique is a shear, and the shear is the whole system is the rung that measures the trade.

The refusal #

Pohlke’s theorem has exactly one hypothesis — the three segments must not all lie on one line — and it earns it. If they do, $\mathbf{A}$ has rank one, its Gram matrix is singular, the smaller eigenvalue is zero, and the cube’s edge is zero with it. There is no cube, because a cube whose three axes all image onto one line is a cube seen from a direction in the plane of two of its faces, and the recovery cannot tell which of the infinitely many such cubes it was.

The implementation refuses that case rather than returning a very small number. A very small edge is a real answer to a nearby question and it would be reported with a small residual, which is the trap the applied phase left as its sharpest gotcha: a fit with fewer equations than unknowns produces a family, picks one, and fits it perfectly by construction. Here the family is visible in the arithmetic, so the refusal is cheap. It is not always.

Why the recovery is worth having beyond the correction #

The theorem’s classical statement is an existence result and this site’s version is a computation, and the difference has three practical consequences worth naming.

It gives an unnamed drawing a viewing direction. Cavalier and cabinet are conventions with names; a drawing whose depth axis happens to be at 33° and 0.62 scale is a perfectly good parallel projection that no convention covers. The recovery returns its cube and its direction anyway, so any axonometric or oblique drawing whatsoever can be told what it is a picture of.

It separates the two families by measurement rather than by name. The obliquity is zero for an orthographic system and positive for an oblique one, computed from the drawn axes with no reference to which family the system was declared to belong to. A drawing whose provenance is unknown is classified by the arithmetic.

And it counts the readings. Two cubes, always, related by a reflection — which is a fact about every parallel drawing and is invisible in every conventional description of one.

None of the three needed a new idea. What they needed was for the theorem to be arithmetic instead of an assurance, and the step that made it arithmetic is the observation that multiplying the drawn axes by their own transpose deletes the unknown orientation. That is a small step and it had been available since 1853.

What the rung is actually for #

There is a temptation, once the recovery works, to say that Pohlke rehabilitates the oblique systems — that cavalier turns out to be a respectable projection after all and the drawing offices were right. That is not what the measurement says either.

What it says is narrower and more useful. A cavalier drawing is a picture of a cube; it is a picture of a cube taken from 45° off the normal, and a picture taken from 45° off the normal is one whose rays pass obliquely through the picture plane, so the drawing carries an oblique distortion that no ordinary viewing recovers. A reader looking at a cavalier drawing head-on is not standing where its rays came from, and there is nowhere they could stand that would put them there, because parallel projection has no station point at all — which is the limit rung’s finding and is unaffected by any of this.

The correction is to one sentence and it matters because of what the sentence was doing. Not the projection of anything is a verdict; it closes the question. An oblique projection, at 45° is a measurement; it opens two more, one of which is the trade this field’s next rung is about and the other of which is the ambiguity the rung after that is about. A number that ends an inquiry is worth being suspicious of, particularly when the site made it itself.