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Mathematical Modelling An Object for a System

  1. Jun 20, 2015 #1
    In order for me to be able to run a simulation I need to write an equation which calculates the distance from a point inside a circle of radius 'r' to any point along the circle's circumference using a point a distance 0<=d<=r from its center. The ultimate system to be simulated is a radially pre-loaded circular membrane with fixed, rigid edges that experiences a transverse deformation because of a point-force.

    Imagine drawing a circle, mark a single point inside that circle, then draw several lines going from the point to different points along the circle's circumference. (For the sake of variety, I'd ask that none of the drawn lines are parallel.) The equation would be designed to tell you the length of each of those lines. It may very well be possible to write the equation for calculating the length of a chord, which uses the circle's radius and the perpendicular distance of the chord from the circle's center:

    c=2*sqrt(r^2-d^2)
     
  2. jcsd
  3. Jun 20, 2015 #2

    mfb

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    They cannot be, they all have to cross each other in one point.

    How do you define your lines? You need some parameter that describes where the line is in order to calculate its length. I don't understand the "using a point a distance 0<=d<=r from its center" part. Is that a second point the line crosses?
    Probably, which means you can find that distance and use it.
     
  4. Jun 20, 2015 #3
    The drawing is merely for the sake of helping to better understand the visual concept of the problem.
     
  5. Jun 20, 2015 #4
    'd' is the distance from the circle's center, which is greater than or equal to 0 and less than or equal to r, the radius of the circle.
     
  6. Jun 20, 2015 #5
    ImageUploadedByPhysics Forums1434830777.077751.jpg As I said, just for the sake of variety.
     
  7. Jun 20, 2015 #6

    mfb

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    Mark the point closest to the center (extend the line if necessary). The distance from the rim to this point is half the value of your formula in post 1, the distance from your old point to the new one can be found with a right triangle. Add both (or subtract for the shorter lines) to get the total length.
     
  8. Jun 20, 2015 #7
    Dream with me, here. Speculate. An equation is what I'm seeking help to compose; an equation which doesn't use πr^2 to find the area of a circle or, similarly, πab to find the area of an oval, but chord segment length and calculus for modeling purposes. Just as the Pythagorean Theorem can be used to calculate the sides of any right triangle, so does the Law of Cosines cover both right and non-right triangles. How about this:

    The equation for chord length

    c=2*sqrt(r^2-d^2)

    can be rewritten as the sum of two individual chord segments

    c=sqrt(r^2-d^2)+sqrt(r^2-d^2)

    After a lot of experimentation I found that at least if I took one of these chord segments and either multiply or divide it by

    sqrt(r^2-d^2)/(r-d)

    I get either 'r+d' or 'r-d'.

    The problem here is that in order to move from multiplying to dividing by the above I need my exponent to be a trigonometric function, but if I do that I cannot integrate. Through much experimentation, though, I've managed to deduce an approximation of it. The graph of the original looks like a special kind of amplitude modulation.

    ex. y = (3/2)^sin(x)
    @ x = 0, y = 1
    @ x = π/2, y = 3/2
    @ x = -π/2, y = 2/3

    Going back to the original expanded equation for chord length, if we were to take

    c=sqrt(r^2-d^2)+sqrt(r^2-d^2)

    and combine it with

    sqrt(r^2-d^2)/(r-d)

    taking into consideration the necessity of an exponent that affects periodicity, we would get

    c=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(x)+sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(x-π)

    When x=π/2, you get your original equation,

    c=sqrt(r^2-d^2)+sqrt(r^2-d^2)
    =2*sqrt(r^2-d^2)

    However, a perhaps more intuitive way to write this would look something like

    c=(r-d)*(sqrt(r^2-d^2)/(r-d))^(sin(x)+1)+(r+d)*(sqrt(r^2-d^2)/(r+d))^(sin(x)+1)

    When x=π/2 you would still get the original equation.

    - - - - -

    On a different note, it also seems as if

    sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(x)

    shows promise as an equation that can cleanly translate a graph across the plane. With the addition of another component, it is capable of morphing from, for example, r=a to r=2a*sinθ, and even -2*asinθ, where in my example 'a' is any real number (though maybe it could find useful application in the complex 2D and 3D planes, for all I know). Though to date I've only tested it with the shape of a circle, I feel confident that it could be rewritten rather quickly, to be able to translate much more complicated things (I'm writing all of this from memory, being all of my own work, but I think I remember one of the variables can be substituted as any other equation). If it can morph between those three, who is to say that it can't also be used to translate the graph even farther- beyond its own diameter?
     
    Last edited: Jun 20, 2015
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