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The cartesian equation for a rose is:

$x = rcos(k\theta)cos(\theta)$

$y = rcos(k\theta)sin(\theta)$

When $k=4$ the figure formed has 8 petals. If I wish draw those 8 petals as polygons in a graphics system such as Processing, I can iterate over $\theta$ with steps small enough to smoothly draw the vertices of the polygons. The first polygon's vertices will be drawn for $0<=\theta<\pi/4$, for instance. The next for $\pi/4<=\theta<\pi/2$.

When $k$ is not an integer we have more complex figures where the petals appear to "overlap". For instance when $k=3/2$ we have the picture below, with 6 petals which overlap their neighbour. When $k=5/2$ the situation is more complex.

For any arbitrary rational value of $k$, how do we calculate the ranges of $\theta$ which will break the rose down into distinct shapes? Is there a better way to tackle this problem?

To make my question clearer, I've added a further figure for $k=3/2$. The perimeter I can find is in red. The perimeters of polygons I'd like to find are in blue and green for this value of k.

Requested perimeters.

A rose with k=1.5

A rose with k=2.5

thenapking
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1 Answers1

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For any arbitrary $k \neq 0$ (rational, or even any real number), the values of $\theta$ for which the curve crosses the origin are the values of $\theta$ such that $x(\theta) = y(\theta) = 0$. This is equivalent to asking that $\cos (k\theta) = 0$. The cosine function vanishes at each $\frac{\pi}{2} + j \pi$. So the crossings are for $$ \theta = \theta_j = \frac{\pi}{k} \left(\frac{1}{2} + j \right). $$ In other words, you should be able to draw the $j^{th}$ petal by iterating over $\theta \in [\theta_j,\theta_{j+1}]$.

cs89
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  • Thanks for your answer! Unfort that's not the solution I am looking for, but is the solution I already have. I have added a diagram to make this clearer. The shapes are bounded by the places the curve intersects itself. By estimation I think the values for $\theta$ which make the green shape are $\pi/6<=\theta<\pi/3$ and $-\pi<=\theta<-0.83*\pi$ – thenapking Apr 18 '23 at 12:55
  • Oh! Indeed, it is clearer now. I don't know about the general case: your rose with $k=5/2$ looks hard. For the green part of the rose with $k=3/2$, the endpoint is at an angle of $\phi = - \pi/6$ (by symmetry of the figure). This corresponds to $\theta$ such that $x(\theta) > 0$ and $\tan \theta = \tan \phi$. So $\theta = -\pi/6$ and $\theta = 5\pi/6$. Since $5/6 = 0.8333$, this seems to correspond. – cs89 Apr 18 '23 at 13:23
  • OK. I've experimented a bit more when $k=3/2$. If $\phi=\pi/6$, the green leaves are drawn for two ranges for an integer $m<6$,

    $\phi+2m\pi/3<\theta<2\phi+2m\pi/3$ and $-\pi+2m\pi/3<\theta<-5\phi+2m\pi/3$

     – thenapking Apr 18 '23 at 15:26
  • The blue leaves are drawn with: $\pi/2+2m\pi/3<\theta<5\phi+2m\pi/3$, and $3\pi/2+2m\pi/3<\theta<10\phi+2m\pi/3$, and $-2\phi+2m\pi/3<\theta<-\phi+2m\pi/3$ Sorry this is a little hard to read in the comments. – thenapking Apr 18 '23 at 15:31