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How can an equation for the following curve be derived?

enter image description here

$$r=(1+0.9 \cos(8 \theta)) (1+0.1 \cos(24 \theta)) (0.9+0.1 \cos(200 \theta)) (1+\sin(\theta))$$

(From WolframAlpha)

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    $\begingroup$ I don't get your question? Are you looking for its equation? If so you can get it on the web. Do u waana its origin? $\endgroup$
    – Mikasa
    Dec 26, 2013 at 14:21
  • 2
    $\begingroup$ How do you know that it isn't hemp? $\endgroup$ Dec 26, 2013 at 16:04
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    $\begingroup$ I think someone's having a little bit too much fun for the holidays! :-) $\endgroup$
    – Lucian
    Dec 26, 2013 at 16:17
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    $\begingroup$ Too much fun? No such thing! $\endgroup$ Dec 26, 2013 at 17:40
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    $\begingroup$ This reminds me of the question about the Batman equation: Could someone make a plot of Batman smoking weed? I'll have tons of bitcoin for you. $\endgroup$
    – Red Banana
    Dec 26, 2013 at 18:38

2 Answers 2

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It can be made intuitively out of the following observations:

  • The function $\theta \mapsto 1 + \sin(\theta)$ has the property that it is closer to $0$ for $\theta\in [\pi,2\pi]$, and it is closest when $\theta = 3\pi / 2$. This is good for the shape you want, since you want it to be smaller on the "lower" part. See here

enter image description here

  • Once we know this, we still have to add the pointy parts, you can count there are $7$ corners, right? Now, how can we make $7$ corners? Note that $\theta \mapsto 1+ 0.9 \times\cos(8\theta)$ is good in that it expands the radio when $\cos(8\theta) > 0$ and it reduces it when $\cos(8\theta) < 0$. Each corner corresponds actually to a reduce,expand,reduce, and $\theta \mapsto 1+ 0.9 \times\cos(8\theta)$ has exactly $8$ regions like this, but exactly one happens where $1+\sin(\theta)\approx 0$, and so we actually have $7$ such regions. See here

enter image description here

Finally, all of this points to this

enter image description here

The remaining factors, which have much smaller periods, and expand/contract the radio much less, are there just to make the "borders" looks less regular.

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  • $\begingroup$ Now it is OK. :-) $\endgroup$
    – Mikasa
    Dec 29, 2013 at 17:09
  • $\begingroup$ @B.S. Actually, it doesn't let me delete because it was accepted. $\endgroup$ Jan 8, 2014 at 17:26
  • $\begingroup$ I wanted it to be here accepted either. :-) +1 $\endgroup$
    – Mikasa
    Jan 8, 2014 at 17:28
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You may do the same as other answer did in details by Maple:

[> with(plots):
  animate(polarplot, [(1+.9*cos(A*t))*(1+.1*cos(A*t))*(.9+0.5e-1*cos(A*t))*(1+sin(t)), t = -Pi .. Pi, thickness = 2], A = 0 .. 15);

enter image description here

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    $\begingroup$ Excellent animation +1 $\endgroup$
    – Amzoti
    Dec 26, 2013 at 15:30

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