Arc Length of Sinewave calc to find amplitude? 1

[Mattm86]

Hi all,

Im trying to design a sinewave torroid (Im not sure of its actual name but look up Shark Wheels for reference). To do this I am needing to get a specific arc length across a period, by adjusting the 'amplitude' of the wave.

I know the Arc Length that I need to achieve (for 1 period of X to what would be 2 PI) which is 5.3". The length of a period (0 to 2 PI) is 2.88".

How do I work out what the amplitude y would be? i.e. to fit the Arc Length into that length of period?

Thanks

Matt

 

[GregLocock]

http://math.stackexchange.com/questions/45089/what-is-the-length-of-a-sine-wave-from-0-to-2-pi

Cheers

Greg Locock


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[JStephen]

On something like that, Simpson's rule and an Excel spreadsheet may get you the answer faster than researching the exact answer.

 

[IRstuff]

I think this is correct, but I'm not absolutely positive:

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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faq731-376 forum1529

 

[GregLocock]

Approximate the sinewave to anellipse. Circumference = π×[2.88/4+m] where m is the major radius.

Cheers

Greg Locock


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[Mattm86]

Thanks Guys!

Gregs answer gives me ~ 0.97" for the amplitude and for my purpose this is accurate enough for now. Ill investigate the more detailed answers when I get to that stage.

Thanks for all your help.

 

[rb1957]

quite different to IR's 2.3 ??

another day in paradise, or is paradise one day closer ?

 

[GregLocock]

Yes. OK, now approximate each quadrant of the ellipse to a triangle. The length of each triangle is 2.88/4, the height is .97 so the hypotenuse is sqrt(.72^2+.97^2)=1.2, so for 4 triangles we get 4.8 inches. Passes my sanity test!

Cheers

Greg Locock


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