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Math Help - sin a = t.a

  1. #1
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    sin a = t.a

    I'm trying to solve the equation..

    sin a = t a ...or rewritten as... sin a - t a = 0

    solve to find 'a' for a given value of t
    where a is in radians, 0 < a < pi/2
    where t < 1, typically 0.7-0.9.

    ...Problem background...

    Taking a slice of pizza, measure around the arc of the crust, and then measure the straight chord line across the crust. Given these 2 measurements, it is possible to find the slice angle of the pizza, the angle, etc.

    c = arc length
    l = chord length

    So taking a as the unknown angle of the slice, and let t = l/c, which is a known constant, the solution reduces down to

    sin (ang/2) = t . (ang/2)

    or by making a = ang/2, the equation can be simplified to

    sin a = t.a ...or rewritten as... sin a - t.a = 0

    (a is measured in radians)
    where (2/pi < t < 1) and (0 < a < pi/2)

    However, I can only solve this by trial and error, using iterative techniques, such as Newton iteration. It would be nice if there was a precise solution for the above equation to solve for a.

    Anyone got a solution, or is this equation not possible to solve?
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  2. #2
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    e^(i*pi)'s Avatar
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    West Midlands, England
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    Quote Originally Posted by surfdabbler View Post
    I'm trying to solve the equation..

    sin a = t a ...or rewritten as... sin a - t a = 0

    solve to find 'a' for a given value of t
    where a is in radians, 0 < a < pi/2
    where t < 1, typically 0.7-0.9.

    ...Problem background...

    Taking a slice of pizza, measure around the arc of the crust, and then measure the straight chord line across the crust. Given these 2 measurements, it is possible to find the slice angle of the pizza, the angle, etc.

    c = arc length
    l = chord length

    So taking a as the unknown angle of the slice, and let t = l/c, which is a known constant, the solution reduces down to

    sin (ang/2) = t . (ang/2)

    or by making a = ang/2, the equation can be simplified to

    sin a = t.a ...or rewritten as... sin a - t.a = 0

    (a is measured in radians)
    where (2/pi < t < 1) and (0 < a < pi/2)

    However, I can only solve this by trial and error, using iterative techniques, such as Newton iteration. It would be nice if there was a precise solution for the above equation to solve for a.

    Anyone got a solution, or is this equation not possible to solve?
    Newton iteration is the best way to solve this. In theory you could use the Taylor series for sin(x) and truncate it at a suitable point but remember it'll only work for 0 < |x| < 1
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  3. #3
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    Apr 2009
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    Ok, I suspected this might be the case. Thanks!
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