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Math Help - circumference staying the same on shape change in a software???

  1. #1
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    circumference staying the same on shape change in a software???

    Hi

    I am not that good at maths, but wounderd if some one can help me.

    OK I need to know if there is a softwere that I could say place a 10mm circle image and then change its shape to any shape I want but still keep the overall circumference the same, so I can work out the sizes in sections of that shape after..

    I can work out ovals (I think, I have added an image on what I guess is right)

    I need to work out a tear drop shape?

    Thank you.

    P.S. sorry if this is the wrong place of the forum.

    Attached Thumbnails Attached Thumbnails circumference staying the same on shape change in a software???-oval.gif  
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  2. #2
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    Quote Originally Posted by madkad View Post
    OK I need to know if there is a softwere that I could say place a 10mm circle image and then change its shape to any shape I want but still keep the overall circumference the same, so I can work out the sizes in sections of that shape after..
    Hi,
    I don't know if such a software exists but what you would need is
    - a drawing tool giving you the (approximate) length of the curve you draw (made of ellipses, lines, splines,...) ; it may exist
    - a scaling tool, so that you can rescale your modified curve to make its length equal to what you want ; this surely exists.

    As for the case of the circle, your sketch is not correct. In fact, there is no simple formula to find the length of an ellipse (what you call an "oval"), but you can compute it approximately by using the formulas given in the wikipedia.
    A simpler task (in this situation) is to make the area constant. Indeed, the area of an ellipse with semi-major and -minor axes a and b is \pi a b: to keep the area constant, you must keep the product ab constant. In your case, for instance R^2=25=2.5\times 10 (in mm), so that an ellipse with semiminor axis 2.5mm and semimajor axis 10mm would do fine. But not for the circumference.

    I hope this helps you.
    Laurent.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    Hi,
    I don't know if such a software exists but what you would need is
    - a drawing tool giving you the (approximate) length of the curve you draw (made of ellipses, lines, splines,...) ; it may exist
    - a scaling tool, so that you can rescale your modified curve to make its length equal to what you want ; this surely exists.

    As for the case of the circle, your sketch is not correct. In fact, there is no simple formula to find the length of an ellipse (what you call an "oval"), but you can compute it approximately by using the formulas given in the wikipedia.
    A simpler task (in this situation) is to make the area constant. Indeed, the area of an ellipse with semi-major and -minor axes a and b is \pi a b: to keep the area constant, you must keep the product ab constant. In your case, for instance R^2=25=2.5\times 10 (in mm), so that an ellipse with semiminor axis 2.5mm and semimajor axis 10mm would do fine. But not for the circumference.

    I hope this helps you.
    Laurent.
    Hi, thanks for your reply, lol I thought I had cracked the oval shape

    OK erm I will try and find the software that could help, thanks sorry to say I didnt understand the number bits, I was never any good at school on that sort of thing.
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  4. #4
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    Just for fun I thought I'd try and see what I could do with Mathematica. The following is a Manipulate window set up to change the shape of the tear drop given by the parametric equations:

    x=\cos(t)

    y=\sin(t)\sin^{n-1}(t/2)

    as n ranges from 1.0001 to 10 by moving the slider bar. the plot shows n=6. Don't know how the circumference is changing though. Here's the code:

    Code:
    Clear[n]; 
    Manipulate[ParametricPlot[
       {Cos[t], Sin[t]*Sin[t/2]^(n - 1)}, 
       {t, 0, 2*Pi}, PlotRange -> 
        {{-1, 1}, {-1, 1}}], {n, 1.0001, 10, 1}]
    Attached Thumbnails Attached Thumbnails circumference staying the same on shape change in a software???-manipulateteardrop.jpg  
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  5. #5
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    Dang it. This is a math forum for heavens sake: I do know how the arc length is changing:

    A=\int_0^{2\pi}\sqrt{(x'(t)^2)+(y'(t))^2}dt

    Plot is below.
    Attached Thumbnails Attached Thumbnails circumference staying the same on shape change in a software???-arclengthteardrop.jpg  
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