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03122009, 10:29 PM #1
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Parametric equations for butterfly curve.
I'm trying to construct parametric equations for butterfly curve in Javascript, but it's been a long time since my math classes and I don't really remember the stuff.
Anyway, I've got this:Code:x=Math.sin(t)*( (Math.exp(Math.cos(t)))  (2*Math.cos(4*t))  (Math.sin(t/12)^5) ) y=Math.cos(t)*( (Math.exp(Math.cos(t)))  (2*Math.cos(4*t))  (Math.sin(t/12)^5) )
I'm not sure what value should it have. I think it should be some kind of range of values, but I don't know exactly what range.
I've tried just slowly incrementing value [from 0, by .01], but the curve does not look like butterfly curve, similar maybe, but certainly it isn't butterfly.Last edited by freedom_razor; 03132009 at 02:54 AM.
03122009, 11:01 PM
#2
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Wow! Even longer since I did this!
A search tells me that t is a value within the range 0 and 24*pi
Try larger values such as 1 or 2 or 10 and see what you get.
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freedom_razor (03122009)
03122009, 11:20 PM
#3
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Nice one. Do you mind posting the link, 'cos [shame on me] searching as I was, I still haven't come upon that info.
03122009, 11:20 PM
#4
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In javascript, ^ is the bitwise exclusiveor opererator. For exponentiation, use Math.pow
Code:x=Math.sin(t)*( (Math.exp(Math.cos(t)))  (2*Math.cos(4*t))  Math.pow(Math.sin(t/12),5) ) y=Math.cos(t)*( (Math.exp(Math.cos(t)))  (2*Math.cos(4*t))  Math.pow(Math.sin(t/12),5) )
The Following 2 Users Say Thank You to Shannon Blonk For This Useful Post:
freedom_razor (03122009), Philip M (03132009)
03132009, 01:06 AM
#5
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Well, finally I left the range as it was, starting at 0, each iteration +0.01. The effect [reversed 180 degrees]:
Thanks again Shannon for pointing out my mistake. It made quite a difference.
Last edited by freedom_razor; 03132009 at 02:53 AM.