# Cannabis Equation

How can an equation for the following curve be derived?

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

(From WolframAlpha)

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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? –  Babak S. Dec 26 '13 at 14:21
How do you know that it isn't hemp? –  Alizter Dec 26 '13 at 16:04
I think someone's having a little bit too much fun for the holidays! :-) –  Lucian Dec 26 '13 at 16:17
Too much fun? No such thing! –  Robert Lewis Dec 26 '13 at 17:40
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. –  Voyska Dec 26 '13 at 18:38

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

• 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

Finally, all of this points to this

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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Now it is OK. :-) –  Babak S. Dec 29 '13 at 17:09
@B.S. Actually, it doesn't let me delete because it was accepted. –  Pablo Rotondo Jan 8 '14 at 17:26
I wanted it to be here accepted either. :-) +1 –  Babak S. Jan 8 '14 at 17:28

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);


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Excellent animation +1 –  Amzoti Dec 26 '13 at 15:30