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Math Help - Ellipse Centre point calculation

  1. #1
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    Ellipse Centre point calculation

    Hi,

    I am doing a measurement of a cylindral object with might be standing veticle or tilted. A device shoots a laser and takes measurement of (HA,VA,and length from device to the point) by this process three points are measure along top cylindrical object and so on for bottom. The radius of pile is known, now i want to find center for ellipse with measurements of three points as (x1,y1),(x2,y2),(x3,y3) and radius. how can i achieve that?
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  2. #2
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    Re: Ellipse Centre point calculation

    Hey Randeep.

    For an ellipse you need to have a minor and a major axis length (so two kinds of radii).

    You can use ([x-cx]/a)^2 + ([y-cy]/b)^2 = 1 and if you three points you will get three equations with four unknowns unless you have the semi-axes a and b in which you only need two equations in two unknowns.
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  3. #3
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    Re: Ellipse Centre point calculation

    Major and minor axes will be radom as the points i shoot with generate random ellipse it wont be a measured value. solving for cx and bx will give me center but i still dont know a and b
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  4. #4
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    Re: Ellipse Centre point calculation

    Well you will need another point to get four simultaneous equations for four unknowns.

    If you do this then you can find a, b, cx, and cy.
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  5. #5
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    Re: Ellipse Centre point calculation

    i seem to know the semi-minor axis which is equal to radius of pile so i think three points are enough but not really helpful when coming to solve equation
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  6. #6
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    Re: Ellipse Centre point calculation

    Well if that is the case then simply set up three simultaneous equations get them in terms of one variable.

    If you need some guidance, please tell me which variable you know (i.e. we treat it as a constant) and we'll go through it.
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  7. #7
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    Re: Ellipse Centre point calculation

    (x1,y1)=>(1.5,1.43)
    (x2,y2)=>(1.99,1.43);
    (x3,y3)=>(2.30,1.43);
    radius of pile=> 0.3 m
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  8. #8
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    Re: Ellipse Centre point calculation

    Now I mean in terms of a and b: which one of those variables do you know (and can treat as a constant)?
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  9. #9
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    Re: Ellipse Centre point calculation

    i know b (semi-minor axis) =>0.3m
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  10. #10
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    Re: Ellipse Centre point calculation

    So we have three points and three equations which are:

    ([x1-cx]/a)^2 + ([y1-cy]/b)^2 = 1

    ([x2-cx]/a)^2 + ([y2-cy]/b)^2 = 1

    ([x3-cx]/a)^2 + ([y3-cy]/b)^2 = 1

    where x1, y1, x2, y2, x3, y3 and b are constants. Re-arranging we get:

    cy = y1 +- SQRT(b^2[1 - [(x1-cx)/a]^2])

    cy = y2 +- SQRT(b^2[1 - [(x2-cx)/a]^2])

    cy = y3 +- SQRT(b^2[1 - [(x3-cx)/a]^2])

    So now we have potential solutions that link all the cy's together. Moving on we want to eliminate one more variable:

    y1 +- SQRT(b^2[1 - [(x1-cx)/a]^2]) = y2 +- SQRT(b^2[1 - [(x2-cx)/a]^2]) = y3 +- SQRT(b^2[1 - [(x3-cx)/a]^2])

    I will just pause to ask you if you get what I'm doing and whether you want to attempt removing another variable so you get an equation in terms of one variable which you will solve via Newtons method.
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  11. #11
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    Re: Ellipse Centre point calculation

    yes i m getting it...
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  12. #12
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    Re: Ellipse Centre point calculation

    what is Newtons method?
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  13. #13
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    Re: Ellipse Centre point calculation

    I am getting nowhere can u please solve the equation
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  14. #14
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    Re: Ellipse Centre point calculation

    Newtons method is a way to numerically solve the roots of an equation by solving f(x) = 0 (there are multi-dimensional variants as well):

    Newton's method - Wikipedia, the free encyclopedia
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  15. #15
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    Re: Ellipse Centre point calculation

    hi chiro,

    I am really confused with solving the equation, i have to implementt this equation in C++ and was bit worried,so if you could help me further more i would really appreciate..: )
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