Saddle Swirl

Surface · z = f(x, y, t)

z = (x · y)/6 · cos(t) + sin(x − t) · cos(y + t)

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What it draws

Two terms are added here. (x·y)/6 · cos t is a saddle: it rises along one diagonal and falls along the other. It grows with distance from the middle, so at this width it is most of what the eye sees. cos t breathes it in and out over 2π ≈ 6.28 seconds at Speed 1, flat at t = π/2 and inverted at t = π.

sin(x − t)·cos(y + t) is the second term. It is never more than 1 tall, so it reads as a fine corrugation over the middle of the saddle, where the saddle itself is shallow.

Why it swirls

That second term looks like one thing moving, and it is two. The product-to-sum identity turns it into ½sin(x + y) + ½sin(x − y − 2t). The first half does not contain t at all: it is a fixed set of ridges along the lines x + y constant, running from one corner of the plot to the other. The second half is a plane wave whose crests lie along x − y constant, the other diagonal, and it slides steadily. Its crests are 2π/√2 ≈ 4.44 apart and it travels at 2/√2 ≈ 1.41 units per second.

So the shimmer is a travelling train crossing a standing one at a right angle, and the crossings light up and go out as the moving crests pass the fixed ones. Nothing turns. The drift of the running wave across the still one is what reads as a swirl.

Try

  • Top in the deck: the two diagonal grains, one still and one sliding, are separate from above.
  • Write the ripple out as its two halves, (x·y)/6 · cos t + 0.5·sin(x + y) + 0.5·sin(x − y − 2t): the picture does not change.
  • Then drop the standing half and keep (x·y)/6 · cos t + 0.5·sin(x − y − 2t): one clean wave train sliding down the diagonal.
  • Or drop the ripple entirely, (x·y)/6 · cos t, for the bare saddle turning itself inside out.
  • Lower Span of x, y to 4: the saddle reaches only ±2.7 at the corners and the ripples take over the picture.

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Surface · z = f(x, y, t)