Trefoil Bloom
r = 1 + 0.25 · sin(3θ + t) + 0.15 · cos(6θ − t)
Open in the app The dials and keys named below are the app's.
What it draws
The leading 1 is a circle of radius one, and the two waves after it are ripples laid on that circle. 0.25·sin(3θ + t) pushes the rim out and in three times a turn. 0.15·cos(6θ − t) does the same six times a turn, at a smaller depth. Together r stays between 0.6 and 1.4, so it never reaches zero: the curve is a closed ring that bulges, and it never doubles back through the middle.
Why it turns
Each term is in a frame of its own. 3θ + t holds its value when θ = −t/3, so the three-lobed ripple slides round at 1/3 of a radian a second at Speed 1. 6θ − t holds when θ = t/6, so the six-lobed one slides the other way at 1/6. The bloom is the two drifting past each other. It returns to its opening shape after 2π ≈ 6.28 seconds, when the first has turned by 2π/3 and the second by π/3, a whole lobe each.
The threefold look is exact and never breaks. Turning the plot by 2π/3 leaves 3θ unchanged and moves 6θ by 4π, which is also nothing. So the picture has the same three arms at every moment. That is what the second harmonic buys: it sharpens and blunts the lobes without adding a count of its own.
Try
- Deepen the ripple to 1 + 0.6·sin(3θ + t) + 0.15·cos(6θ − t): the dips fall to 0.25 and the ring turns into three fat petals.
- Change the 3 to a 4. Now 6 is no harmonic of it, and the threefold symmetry drops to a twofold one.
- Raise the fine count to nine, 1 + 0.25·sin(3θ + t) + 0.15·cos(9θ − t): nine is three threes, so the three arms survive.
- Send both terms the same way, 1 + 0.25·sin(3θ + t) + 0.15·cos(6θ + t). Nothing drifts now, and the whole ring turns rigidly.