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huge radius's

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  • #16
    Paul, a shortcut:-

    Given the two large triangles are isosceles (so angles 'd' and 'e' are mirrored) and that the three angles of a triangle add up to 180 :-
    B + 2d = 180
    C + 2e = 180

    Adding both sides of the above:-
    (B+C)+ 2(d+e) = 360
    A + 2F = 360
    F = 180 - A/2

    To calculate the required dimensions:-

    Where 'S' is the distance between supports, and 'D' is the circle diameter.

    A vertical dropped from the centre of the circle divides both angle A and distance S in half, giving:-
    sin(A/2) = (S/2)/(D/2) = S/D

    Now, from earlier, A/2 = 180 - F, so:-
    sin(180-F) = S/D

    (- and remembering sin(180-X) = sin(X))

    1: Given support distance and required diameter:-

    'Sled angle', F = 180 - arcsin(S/D)

    2: Given 'sled angle' and required diameter:-

    Support distance, S = D * sin(F)

    3: Given support distance and 'sled angle'

    Resulting diameter, D = S / sin(F)




    • #17
      Originally posted by Barrington View Post
      Paul, a shortcut:-

      Always more than one way to prove a theorem in math. I once saw a whole book full of different proofs of the Pythagorean theorem (a^2 + b^2 = c^2) for a right triangle. Thanks for the help.
      Paul A.
      SE Texas

      Make it fit.
      You can't win and there IS a penalty for trying!