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Thread: [RESOLVED] Create an Octagon of a Specific Size

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    Former Admin/Moderator MartinLiss's Avatar
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    Resolved [RESOLVED] Create an Octagon of a Specific Size

    I would like to fashion an octagon out of 2x4 lumber and have it have a 2.5 foot diameter (measured flat side to flat side). I know that to make an octagon I need to cut the wood at 67.5 degrees, but I need help in determining the length (outside diemension) of each piece.

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    Fanatic Member VBAhack's Avatar
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    Re: Create an Octagon of a Specific Size

    If the distance from flat to flat is 2.5, then I think each piece is 2.5/(1+2/sqrt(2)) = 1.0355 ft. (12.43 inches) in length.

    Consider the following picture:

    http://mathforum.org/library/drmath/view/54666.html

    If the length of each side is a and the side of the triangle is b, and the distance from flat to flat is d, then (if I did everything right):

    1. b^2 + b^2 = a^2 ==> b = a/sqrt(2)
    2. d = a + 2b = a + 2a/sqrt(2) = a(1 + 2/sqrt(2))
    3. a = d/(1 + 2/sqrt(2))

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    Fanatic Member damasterjo's Avatar
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    Re: [RESOLVED] Create an Octagon of a Specific Size

    opps, I though it was 2.5 inchs well you got the answer anyway ha!
    check this out for real though!
    EASY, its like cheating
    http://newmanservices.com/trig_calc/index.htm
    Last edited by damasterjo; Apr 20th, 2006 at 11:09 PM.
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