Jan Dibbets was an artist of the 1960s and 1970s who—in keeping with his time—was very interested in ideas, concepts, ephemeral art, and all of that good stuff which keeps the public asking “this is art?!” even after they’ve managed to swallow Mondrian. I’m familiar with Dibbets thanks to his grin-inducing *perspective correction* series. I’d like to talk a little bit about the ideas behind this work and how they relate to an old geometry called projective or sometimes perspective geometry.

So, I’ll begin with a rather innocent seeming question: Is this a picture of a square?

Certainly, there’s a square sitting on the surface of the 2-dimensional photograph. But if you were there with Dibbets and walked toward this rope constellation strung across the ground would you still see a square? Suppose you stood on the other side of the rope figure, waving at the camera.

What would you see? Maybe a trapezoid?

How can one shape be both a square and not a square—at the same time? This is the magic of projective geometry.

But just what is projective geometry? Perhaps we should start with a more familiar concept, Euclidean geometry. One of my favorite definitions of Euclidean geometry is the following:

*Euclidean geometry is the study of figures (e.g. a square or triangle) and their properties (e.g. side lengths, angles) that are invariant (i.e. do not change) when we transform them by rigid motion (rotation/translation) and uniform rescaling.*

To make this a little more concrete, here’s one example of an invariant property. The angles of triangle remain constant no matter how we rotate, translate or rescale the triangle in question. And here’s an example of an in-invariant property (maybe just a variant property). The side length of a square changes if we uniformly scale the square into a larger or smaller square. However the lengths of the sides of a square do remain equal, which gives us another invariant property. (In fact, this is the defining property of rhombi, of which the square is a special instance.)

Perhaps we can fashion a definition for projective geometry, similar to the one just given for Euclidean geometry.

*Projective geometry is the study of figures and their properties that are invariant when we transform them by projecting between surfaces.*

For instance, when we take a picture with a camera.

So the reason Jan Dibbets’ figure is both a square and not a square is because “square” doesn’t make any sense in projective geometry. “Squareness” is not invariant when we project a figure! However, Dibbets knows that we just can’t resist seeing the square. In this way, he is subtly reminding us that the photo is a two, not three dimensional space. In more artsy language, one might say that Dibbets is emphasizing the flat “ground” of the picture plane, rather than the “ground” on which the rope is strung.

What good can we put this projective geometry up to besides analyzing art? After looking at this photograph for a while I began to wonder…

Suppose I told you that I had it on Dibbets’ word of honor that the bottom edge of the quadrilateral in the photograph was exactly 3 feet long. Can you tell me how far away the camera must be from the quadrilateral (not counting vertical distance)? How? If you can’t, why not, and what other measurements would you ask for?

Feel free to post comments with thoughts, ideas or even stabs at solutions.

Hey Gilbert,

I might do the computations later, but I’m feeling ungenerous now. 🙂 I do think that it’s possible to determine everything, given the height of the observer:

Pretend that the focal plane (film plane) is between the focus and the object. Extend a pyramid with apex at the focus such that the focal plane is a cross section through the pyramid. Let cross sections of the pyramid that are parallel to the focal plane be called focal plane dilations. Horizontal lines in the picture correspond to intersections of focal plane dilations with the plane of the earth.

The distance from the focus to the front line of the square is the focal length scaled by the same factor as scales the length of the line on the image to 3 ft. This gives you the angle at which you’re looking down, which completely determines how the pyramid intersects the earth. You’re done.

* Having written this, I no longer resent Euclid and Newton for being such wordy bastards. There’s no more concise other way to do this in plain English. 🙂

Aren’t you assuming both that you know the height of the camera AND the focal length? I think you can solve the problem with one extra measurement. Also I’m not sure if or why (proof) you need any extra information whatsoever.

The distance from the focus to the front side of the square is inseparable from the focal length. If you know one, you know the other:

size on film = (focal length/distance to square) * 3 ft.

If you want to know more about the other dimensions of the square, you also need the angle at which you’re looking. For you can roll around the axis of the front line, keeping its size on the film fixed; but this would require that the other sides transform to maintain the appearance of a square.

I think that this is iff for both require measurements. How do you see it?

Hmm, I thought about this and perhaps I’m thinking of the camera as a pinhole camera (obviously not true) whereas you’re thinking of a different sort of lens?

The equation you gave:

size on file = (focal length/distance to square) * 3 ft.

only holds for a pinhole camera when the perpendicular to the film plane, passing through the pinhole intersects the front/lowest side of the square. Even then, this only gives the diagonal distance from the pinhole to the square exactly.

[…] Dilemma” (part II) 13 11 2009 Ok, getting back on the ball here, last time I closed with a problem. If I tell you that the bottom edge of the square in the following […]