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How to calculate the height of a parabola, given base width and arc length?

  1. Aug 23, 2012 #1
    I have a parabola centered at x=0, equation: y = a*x^2 + c, where a is always negative and c always positive.

    I need to find a way to calculate a and c, if i know: the arc length above the x axis, and the base width, knowing the base width i also know the x axis intersections x1,2 = -+base_width/2;

    The problem is the formulas for the arc length are too complicated for me to 'play around' with..

    For example the base_width/height formula from http://www.had2know.com/academics/parabola-segment-arc-length-area.html:

    arc_len = 0.5√16h²+w² + [w²/(8h)][Ln(4h + √16h²+w²) - Ln(w)]

    I tried to solve for h with wolfram alfa, and mathematica, neither helped..
     
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  3. Aug 23, 2012 #2

    Simon Bridge

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    Welcome to PF;
    It helps to be careful about specifying your information.

    if A and b are positive real constants for a parabola of form: y(x) = A(b2 - x2).

    You want to determine A and b from knowing the separation of the roots (the start and end position), which I'll call d, and the total length of the trajectory, which I'll call s.

    (notice this is deliberatley different from your own notation ... you should be able to relate the two.)

    Now you'll find it easier to talk about your problem ;)

    Presumably you can do the first bit:
    if the roots are separated by distance d,
    then b=d/2 ... see? No messing about.

    The next bit is where you are hung up ... the total length along the arc of the parabola between the roots involves some calculus (just looking up relations like this tends to lead to the kinds of confusion you are experiencing - did you do the math yourself?)

    I got: [tex]s=\int_x ds=\int_{-b}^{b} \sqrt{1+\left ( \frac{dy}{dx}\right )^2} dx[/tex]

    In your case, [itex]\frac{dy}{dx}=-2Ax[/itex] ... you can solve that with a trig substitution.
    So I am guessing you'll be stuck on extracting A from the logarithm (or sinh-1(Ad)) on the end?
    You have s(A) and you want A(s) ... inverting s(A) is going to be a pain.

    It is probably not going to be possible to produce an analytic equation for all cases from this one. I'd use an approximation from here.

    Out of interest: what is the situation? It is pretty unusual that you'll know the trajectory length without knowing at least three points on the trajectory.
     
    Last edited: Aug 24, 2012
  4. Aug 24, 2012 #3
    Thanks for your reply, your way of representing the parabola is much more intuitive.

    I did not do the integration myself, my knowledge of calculus is limited to derivatives and basic integration, i know that definite integration gives the surface under the curve, and i know that arc length is calculated using the definite integral you have shown. But i have no idea how to integrate sqrt( 1 + (-2Ax)^2 ) so i use a online integrator and i get this:

    .........2 a x Sqrt[1 + 4 a x ] + ArcSinh[2 a x]
    AL = ----------------------------------------
    ..............................4 a

    Now i have to extract a.. which i have no idea how to do.

    The reason i need this is, I am making a soft body physics simulation, and i have springs that compress, and when they compress their length is obviously smaller but it needs to stay the same, so i want to bend the spring outwards, in a parabolic shape, so that it is always of the same length. So I always know the spring base length (the distance between the particles it connects), in your equation that would be 2*b, and i know the rest length which is the length i want my spring to always be, when 2*b is smaller that this length i want to draw a parabola that has that arc length.
     
    Last edited: Aug 24, 2012
  5. Aug 24, 2012 #4

    Simon Bridge

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    You mean:[tex]s=\frac{1}{4A}\left . \left [ 2Ax\sqrt{1+(2Ax)^2} + \sinh^{-1}(2Ax) \right ] \right |_{-\frac{d}{2}}^{\frac{d}{2}}[/tex]But you are not there yet: you still have to evaluate the limits.

    Unfortunately, arcsinh(x) is an odd function so it does not subtract out :(

    Now I see... you want to know how A changed with d for a given s.
    This is actually a spring problem - that has probably been solved already: I'll have a look.

    I'm a little puzzled you are using this sort of spring though - surely longitudinal compression springs would be simpler to model?

    Of course you may have better luck expressing the integral in polar coordinates... I figured I'd better find out what this was for before going to that effort though.

    Since this is a computer simulation, you can use an approximation - what kind and how big depends on the details of your simulation ... do you expect the spring to have to bend a great deal or just flex a little for eg?

    Look at power and Taylor series expansions for answers. Or you can just solve each situation numerically - depends how fast you want the simulation to be.
     
    Last edited: Aug 24, 2012
  6. Aug 24, 2012 #5
    I'm am modeling springs using the vector form of Hooke's law, with damping. this is only for drawing, cosmetic reasons :).. It would look great if all the springs kept their length with a slight bend if necessary, my springs don't compress much maybe in the 'worst' cases they lose 50% of their rest length.

    I will try to approximate it using the series expansions you mentioned. Thanks again.
     
  7. Aug 24, 2012 #6

    Simon Bridge

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    IRL - two particles compressing a spring so it bends would also have a displacement sideways because of the changes mass distribution of the spring.

    You could just have the springs as a series of zig-zags - the corners get closer together as the spring compresses. Not only is that easier to draw, but it is truer for your model ... the leaf spring (which this is closest to) usually has a spring "constant" that varies with compression.
     
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