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About probability of crossing a circle circumference by a needle

  1. Aug 2, 2015 #1
    Buffon`s needle problem described the probability of crossing a set of parallel lines with a known distance between them by a needle of a known length. What is the probability of crossing a circumference of a circle with a known radius by a needle of a known length?

    I feel that a simple answer to this problem can not be found unless the problem is well defined in a geometrical sense. Unlike the parallel lines in classical Buffon`s needle problem, the geometry here is different because a 2D space can not be tiled by non-overlapped circles.
     
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  3. Aug 2, 2015 #2

    Orodruin

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    Your problem is not well defined. With a single circle you will need to make assumptions on how large the region in which you drop the needle is.
     
  4. Aug 2, 2015 #3
    Let the circle is inscribed in a square with its side equals the diameter of the circle and let the needle length is small compared with that diameter. To make the matter easier, lets ignore the position of the needle outside the circle or to consider it as a cross when the center of the needle is outside the circle.
     
    Last edited: Aug 2, 2015
  5. Aug 2, 2015 #4

    mfb

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    If the needle length is very small compared to the radius of the circle, you can approximate the circle as line with a length given by the circle circumference, as the curvature has a negligible effect. Then you are back to the classical problem.
     
  6. Aug 3, 2015 #5
    But I am actually interested in a case where l/2r is not negligible.
     
  7. Aug 3, 2015 #6
    I followed the concept of one of calculation in wikipedia. Here is the attached figure described the case. The probability of crossing the circle circumference is equal to the integration of the projection of the half needle length on the radius (l/2 cosΘ) plus the length shown in red divided by the integration of the radius of the circle where the interval of the integration is {0, π/2}.

    As expected this probability is larger than the classical probability of a needle crossing parallel lines that is because of the additional part shown in red. Now I need to calculate that length.
     

    Attached Files:

    Last edited: Aug 3, 2015
  8. Aug 3, 2015 #7
    I see that the problem can be reduced to an isosceles triangle. The problem of solving the red line in the figure of the previous post is equivalent to soling the blue side in this post. In this problem, the red line represents half the needle length and theta corresponds to the angle of inclination on the radius in the original solution. So I need to solve for the blue line as a function of theta, L/2 and r of the circle.
     

    Attached Files:

  9. Aug 3, 2015 #8
    So again the probability equals to the projection of the half needle length on the direction of the radius (l/2 cosΘ) plus the length shown in red divided by the integration of the radius of the circle where the interval of the integration is {0, π/2}. The nominator is also equal to the radius minus the length s in the attached file. S can be calculated from the law of triangle, r2=d2+s2+2ds cosθ
    this is a quadratic equation in s, s= -dcosθ±√(d2cos2θ+4(r2-d2)
    p=r-s = r-[-dcosθ±√(d2cos2θ+4(r2-d2)]
    = r+dcosθ±√(d2cos2θ+4(r2-d2)

    Then the probability of cross the circle circumference = (2/πr) ∫r+dcosθ±√(d2cos2θ+4(r2-d2) dθ
     
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