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Calculate number of turns in Archimedes spiral

  1. Apr 29, 2012 #1
    Hi,

    I'm an engineer designing a spring system for a garage roller door. I need to know the number of turns of the door for all the size combinations.

    I've found this page which gives a good equation for finding the length if you know the number of turns, starting radius and gap between spirals:

    [​IMG]

    The equation of the spiral is r=x+yθ, so x=starting radius, y=gap/2∏, and to find L we're taking the integral from a=0 to b=2∏n (where n=turns).

    When you know n, this is straightfoward, and even I could work that out. But it's been a decade since I've done anything like this, so I was wondering if anyone could help me find an expression for n in this:

    L=[itex]^{2∏n}_{0}[/itex] [itex]\sqrt{(a+bθ)^2+b^2}dθ[/itex]

    Lord help me, my way of solving this is to find L for n=1,2,3,4,5 etc, graph it in excel and use "find trendline" to get an equation. Any help appreciated, thanks.
     
    Last edited: Apr 29, 2012
  2. jcsd
  3. Apr 30, 2012 #2
    Hey!

    If you're only looking for the answer, you can use this.

    For methodology, you can do two substitutions: first [itex]v=a+b\theta[/itex]

    Then you need to make a second substitution v=b*sinh(u). Don´t forget the derivatives. For more information you can visit this topic:

    https://www.physicsforums.com/showthread.php?t=157980

    hope that helps a bit!
     
  4. Apr 30, 2012 #3
    Ok, I'll see how I go. Thanks a lot.
     
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