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I am working on automated counting and one of my solutions is the use of the template matching algorithm (specifically using Chamfer Matching Algorithm). However, granted it is a template matching approach, I need to establish my template. The objects I am trying to count are individual coconut trees. Based on the definition of coconut tree crowns, they are star-shaped with feather-like leaves. I had the idea in my mind that I could derive that given a polar equation. However, since I am not a Math major, I looked for polar equations that could describe the above mentioned description and stumbled upon the link below An equation that generates a beautiful or unique shape for motivating students in mathematics and noticed the answer given by Luscian.

My question is, is this equation

$$r(t)=|cos(nt)|^{sin(2nt)}\text{ for }2n\text{ in between 1 and 8 and }t\in(0,2\pi)$$

an already defined one? I am not familiar with this, since my background on polar equations reached only up to cardiods and rhodonea curves.

Any help is greatly appreciated. Thanks!

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  • $\begingroup$ Welcome to Math.StackExchange! Interesting Question. Please check that the light notational edits (once accepted) match your intentions. math.stackexchange.com/help/notation $\endgroup$
    – David Diaz
    Commented Jul 6, 2018 at 6:21
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    $\begingroup$ In any manner, the polar form is amazing; for $n=5$, it looks like a starfish and for $n=8$ it remembered to me how Jean Lurçat represented the sun in some of his tapestries. $\endgroup$
    – Claude Leibovici
    Commented Jul 6, 2018 at 6:47

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No, these aren't classical curves, and IMO they are uselessly contrived and asymmetric.

You could consider using a rose instead https://en.wikipedia.org/wiki/Rose_(mathematics). Considering the probable high variability/complexity of the crowns, the exact shape of the "petals" doesn't matter.

I also believe that this approach is flawed, because even when seen from above, coconut trees do not exhibit a regular pattern at all, and a regular template will overlap both leaves and voids. You will need a deformable template, for which a "nice" equation doesn't work.

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  • $\begingroup$ Oh. Thank you. I am somewhat trying to see whether I could utilise properly template matching for counting problems such as this. I already have results, and they are quite promising. But I know that there's a lot more that needs to be done. I was asking the question because I utilised the equation in the template I used. My adviser is asking me to fit the function to my empirical data (derived from the candidate edges). However, the researches I found on curve fitting doesn't give much insight on fitting given polar equations. In any case, thank you for the insight! $\endgroup$
    – htinez
    Commented Jul 6, 2018 at 9:23

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