Breathing Rose
r = 2 · cos(5θ + t) + 0.3 · sin(11θ − 3t)
Open in the app The dials and keys named below are the app's.
What it draws
2·cos(5θ) is the rose underneath. Five is odd, and an odd count gives that many petals, not twice as many. The five lobes where the cosine is positive and the five where it is negative land on top of each other, because a negative r is plotted in the opposite direction. The petals reach 2, and the small term 0.3·sin(11θ − 3t) is the breath: an eleven-fold ripple that rides on the petals and takes r out to ±2.29.
Why it breathes
The two terms turn at different rates and in opposite senses. 5θ + t is unchanged along θ = −t/5, so the rose creeps round at 1/5 = 0.2 radians a second at Speed 1. 11θ − 3t is unchanged along θ = 3t/11, so the ripple runs the other way at 3/11 ≈ 0.27. Their relative phase is what swells one petal while another thins. After 2π ≈ 6.28 seconds both have moved by a whole lobe of their own and the figure repeats exactly.
Five and eleven share no factor, so a fifth of a turn does not map the picture onto itself. The ripple lands differently on each petal, and no two petals are the same at any instant. There is a smaller symmetry hidden in the algebra. Both counts are odd, so r(θ + π) = −r(θ), and a negative radius at θ + π is the same point as a positive one at θ. Half a turn already draws the whole curve.
Try
- Drop Turns of θ to 1. The picture is the same, and the trace pen gets round it in half the time.
- Take the ripple away, 2·cos(5θ + t): a rigid five-petal rose, turning and not breathing.
- Make the counts even, 2·cos(6θ + t) + 0.3·sin(12θ − 3t): twelve petals instead of six, since an even count doubles.
- Deepen the breath to 2·cos(5θ + t) + 1.4·sin(11θ − 3t): the ripple is now strong enough to break the five petals up into a ragged ring.