Fermat's Spiral - Sketch pattern creation

[oooode]

Hi everyone,

I have an interesting design problem that at the moment is a little over my head in both solid works skills and mathematics knowledge.

I need to layout a sketch pattern / feature pattern in SolidWorks to create approx 8000 holes. The pattern of the holes follows the sun flower spiral, the layout of the seeds in a sun flower head. AKA Fermat's Spiral, this corresponds to Fibonacci numbers.

More on this can be viewed at:

http://en.wikipedia.org/wiki/Fermat's_spiral

http://ms.unimelb.edu.au/~segerman/papers/sunflower_spiral_fibonacci_metric.pdf

Has anyone experience in this? or do you think it might be possible using equations?

Any help would be much appreciated.

Thanks

 

[IRstuff]

I don't get it; your second citation gives the equation for the spiral, so what's the hangup?

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529


Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
There is a homework forum hosted by engineering.com:

http://www.engineering.com/AskForum/aff/32.aspx

 

[GregLocock]

you need a scale factor for n in that equation otherwise it is messy (or use non integer n)

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[GregLocock]

Incidentally it is a right load of old numerological nonsense, the presence of the golden ratio in the definition is practically irrelevant, you get the same shape curve whatever the relative scaling of magnitude and phasor. Here's a comparison between

x[n]=numpy.sqrt(n*nscale)*numpy.cos(2.*pi*phi*n*nscale)
y[n]=numpy.sqrt(n*nscale)*numpy.sin(2.*pi*phi*n*nscale)

and

x1[n]=numpy.sqrt(n*nscale)*numpy.cos(2.*pi*n*nscale)
y1[n]=numpy.sqrt(n*nscale)*numpy.sin(2.*pi*n*nscale)

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?