Polar Mandala
r = sin(6θ + t) + 0.4 · sin(17θ − 2t)
Open in the app The dials and keys named below are the app's.
What it draws
r is a sum of two angular harmonics and nothing else: a strong sixfold one, and a fine seventeenfold one at 0.4 of its size. The value swings between −1.4 and 1.4. Half of the angles carry a negative r, and a negative radius is plotted in the opposite direction, through the origin.
That is what fills the disc: the pen runs out along one bearing, falls back through the centre, and comes out on the far side. On its own sin(6θ) is a twelve-petal rose. The seventeenfold ripple splits and unbalances those petals, and one turn of θ now has 24 outward tips.
Why no two arms match
Turn the picture by one sixth, 2π/6. The first term does not notice, but the second shifts by 17 · 2π/6, which is 5π/3 and not a whole turn. So the fine ripple sits at a different phase on every arm. Six is not a factor of seventeen, so only a whole turn brings everything back, and the figure has no rotational symmetry at all.
In time the two terms drift against each other. 6θ + t stands still in a frame turning at 1/6 ≈ 0.17 radians a second at Speed 1. 17θ − 2t stands still in one turning the other way at 2/17 ≈ 0.12. Both terms return to a symmetry of their own after 2π ≈ 6.28 seconds, so the whole picture repeats then. In θ the sum has period 2π, so a second turn lays the same closed curve down again.
Try
- Drop Turns of θ to 1. The picture is unchanged, and the pen under trace draws it in half the time.
- Remove the ripple, sin(6θ + t): the plain twelve-petal rose it was built on.
- Make the fine term a harmonic, sin(6θ + t) + 0.4·sin(18θ − 2t): eighteen is three sixes, so the sixfold symmetry snaps back.
- Give the ripple equal weight, sin(6θ + t) + sin(17θ − 2t): the six arms stop leading and the tips scatter to every length.