The ball a drawing does not draw round

Every argument the parallel field has made is about a box. Isometric, the axis scales, Pohlke, the reversal, the ruler: each takes a flat-faced solid, draws it without a centre, and asks what came through.

The drawing office draws other things. A shell, a fairing, a bent plate, a tank — and, constantly, a ball. So it is worth asking what a parallel projection does to a curved surface, and the ball is where to start because the essay that put five equal balls across a wide frame has already measured what a camera does to one, and the comparison is available for nothing.

One formula, three projections #

The outline of a quadric $Q$ under a projection $P$ is

$$C \;=\; \operatorname{adj}\!\left(P \operatorname{adj}(Q) P^{\mathsf T}\right)$$

which is three matrix products and no case distinction. The essay that derived it established it for a camera; nothing in it knows whether the last row of $P$ is a camera’s or $[0,0,0,1]$.

That last row is the whole of what makes a projection parallel. With it, the homogeneous divisor does not depend on the point, so nothing recedes — and every result below that differs from the camera’s differs because of it.

Three projections, one routine:

Orthographic, along a direction perpendicular to the picture plane. Oblique, along a direction that is not — cavalier, cabinet, military, the systems a draughtsman actually uses. And a camera, for contrast.

Orthographic draws a circle, everywhere #

Project a sphere along the picture plane’s own normal and the tangent rays form a cylinder of radius $R$ whose axis is that normal. Cut a right circular cylinder by a plane perpendicular to its axis and the section is a circle.

So the outline is a circle wherever the sphere is put — at the middle of the drawing, at the corner, three metres out of the frame. The aspect comes back as one to twelve decimal places, which is an arithmetic one rather than a close one.

A camera manages a circle in exactly one place: on its own axis. Twenty-three degrees off it the same ball is an ellipse.

That contrast is the cleanest statement this site has of what a centre of projection costs. It is not that perspective is less accurate; it is that a projection through a point treats the direction to the object as special, and a parallel projection has no such direction to treat.

And it keeps the centre, exactly #

The second half is the one worth having, and it is the half a reader would assume rather than check.

The tangent cylinder’s axis passes through the sphere’s centre. The outline is where that cylinder meets the picture plane, and the section of a cylinder by a plane is centred where the axis crosses it — which is the image of the sphere’s centre.

So the outline’s centre and the image of the ball’s centre are the same point, at the arithmetic floor, in every parallel drawing including the oblique ones.

A camera does not. The centre of a drawn ellipse is not the image of the ball’s centre — the two separate as the ball moves off axis, by a real number of pixels — and the reason is that a centre is defined by midpoints of chords and a projection through a point does not keep midpoints.

Which is the same statement as a parallel projection keeping midpoints, met on an object where nobody looks for it. The field established that a parallel drawing preserves the middle of a segment; here that fact turns into the position of a conic’s centre, and the two are the same fact wearing different clothes.

An oblique drawing gives a ball an ellipse, and the aspect is exact #

Now the finding, and it is a taught construction measured.

An oblique projection projects along a direction that is not the picture plane’s normal. Cavalier draws the receding axis at full length at forty-five degrees; cabinet halves it; military uses thirty degrees.

The tangent cylinder is still a cylinder of radius $R$, but now its axis is oblique to the picture plane. Cut a right circular cylinder by a plane at an angle and the section is an ellipse with semi-minor $R$ and semi-major $R/|\hat n \cdot \hat d|$. So

$$\text{aspect} \;=\; \frac1{|\hat n \cdot \hat d|} \;=\; \sqrt{l^2 + m^2 + 1}$$

where $l$ and $m$ are how far the drawing displaces per unit of depth. For cavalier that is $\sqrt 2$. For cabinet, $\sqrt{1.25}$.

A cavalier drawing of a sphere is forty-one per cent longer along one direction than the other, and every drawing manual draws it with a circle template.

That is the same species of finding as the four-centre ellipse, which the office uses for a circle on an isometric face and which is not the ellipse. There the substitute is close and here it is not close at all — a factor of $\sqrt 2$ is not a drafting approximation, it is a different shape.

Which direction the long axis runs, and why nobody notices #

The ellipse’s long axis runs along the projection of the projection direction — which in a cavalier drawing is the receding axis, drawn at forty-five degrees.

So a sphere in a cavalier drawing should be an ellipse leaning at forty-five degrees, elongated back into the page’s depth direction. Draw a circle instead and the error is not a subtle change of proportion; it is a shape that is wrong by a factor along a diagonal.

There are two reasons this is rarely caught and both are about what oblique drawings get used for.

First, oblique projection is chosen precisely when one face matters and the depth is decoration. A cabinet drawing of a cupboard is about the front, and the receding sides are there to say it is a cupboard rather than a rectangle. Nobody measures the sphere on the shelf.

Second, cabinet’s $\sqrt{1.25}$ is twelve per cent, which is inside what an eye forgives on a small object. Cavalier’s forty-one per cent is not, and cavalier is the less-used of the two — for the unrelated reason that it makes solids look too deep.

The two routes, and why the second one had to be written #

The aspect is computed twice and the two computations share nothing.

The first is the matrix route: build $P$, build $Q$, take three products and an adjugate, read the conic’s axes. It is general — it works for any quadric under any projection — and it is opaque, in the sense that no step of it corresponds to anything a person can picture.

The second is the tangent-cylinder route: the projection direction and the picture plane’s normal make an angle, and the section of a right circular cylinder by a plane at that angle has semi-axes $R$ and $R/|\hat n \cdot \hat d|$. It is one line, it is only about spheres, and every step is a picture.

They agree to the arithmetic floor over the whole family of oblique systems. That is worth the second route’s existence for a reason this site keeps meeting: a general routine that happens to be wrong produces plausible numbers everywhere, and the only thing that catches it is a special case computed a different way. The camera version of this file caught a rank collapse the same way, and this one would have caught a sign or a transpose.

Why the oblique case is not a distortion of the orthographic one #

There is a reading available that would be wrong, and it is worth closing off because the arithmetic invites it.

Oblique is a shear — the field’s own result, that the oblique systems are shears of space along the viewing direction and the shear is the whole system. A shear of space maps a sphere to an ellipsoid, so it is tempting to say the oblique drawing draws a different solid, and the ellipse is that solid’s honest outline.

That is true and it is not the same statement. The shear reading puts the distortion in the world: a cavalier drawing is an orthographic drawing of a sheared room, and the ball in that room really is an ellipsoid.

The reading here puts it in the projection: the ball is a ball and the projection direction is oblique to the picture plane, so the tangent cylinder is cut at an angle.

Both give the same ellipse, which they must. The second is the useful one for a draughtsman because it says the aspect from two numbers on the drawing board — $l$ and $m$, which are the receding axis’s direction and scale — without needing a model of a sheared room.

The refusal, and it is the same one #

outlineUnderParallel refuses a singular quadric, and it refuses it for exactly the reason the camera’s version does.

A cone’s dual is rank one, so the dual outline collapses to a point — the image of the apex — and taking the adjugate back gives a matrix that is numerically nothing. A routine that normalised those zeros would return a perfectly plausible ellipse for a cone, and no one-sided check would notice.

That refusal was found on a camera, and it transfers here without modification, which is the thing worth reporting: the formula’s domain is a property of the quadric rather than of the projection, so removing the centre of projection removes nothing from the caveat.

What the result says about a cylinder, and it is not the same #

A ball is the easy case because it has no orientation. Every other curved surface does, and the answer changes in a way worth stating.

A cylinder’s outline under a parallel projection is a pair of straight lines — its own generators, the two that are tangent to the eye’s direction — and where they fall depends on the projection direction. Move the direction and the two generators slide round the cylinder, continuously, which is the behaviour the next rung but one is about.

What survives from this essay is only the part that came from the tangent cylinder being a cylinder. The ball’s outline was a conic because its tangent rays formed a quadric surface; a general curved surface’s tangent rays do not, so there is no one-line formula and no exact aspect. What does survive, because it comes from the last row of the matrix rather than from the sphere, is the centre: a parallel drawing of any centrally symmetric solid puts the outline’s centre on the image of the solid’s centre.

What a draughtsman should take from it #

Three things, and the third is the one this site is for.

A ball in an orthographic or isometric drawing is a circle, and a circle template is exactly right. So is measuring its diameter with a scale, since the outline’s diameter is the ball’s diameter times the drawing’s uniform scale.

A ball in an oblique drawing is an ellipse of aspect $\sqrt{l^2+m^2+1}$, leaning along the receding axis. The template is wrong, and by twelve per cent in cabinet and forty-one in cavalier.

And in either, the centre of the drawn outline is where the ball’s centre is, so a draughtsman locating a ball from its outline is doing something exact. That is a stronger guarantee than a perspective drawing offers and it is the one nobody states, because it is only interesting once the perspective case has been measured and found to fail.

What this does not settle #

It does not say what an ellipsoid does. A general quadric under an oblique projection has an outline whose axes are not the projection’s, and the formula gives it in one line without saying anything memorable about it.

It does not treat a ball that is drawn by eye rather than projected, which is what most oblique drawings of spheres actually are.

It does not treat the drawn ellipse’s construction. A draughtsman needs to put the ellipse on paper, and the four-centre approximation and its errors belong to the essay about it.

And it does not say the office is wrong to use a template. It says by how much, on which systems, and in which direction — which is what turns a habit into a decision.

A formula that covers two cases is worth more than two formulas, because the difference between the cases becomes a term rather than a paragraph. Here the term is the last row of the matrix, and everything the centre of projection costs is in it.