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

1. ## [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.

2. ## Re: Create an Octagon of a Specific Size

I found the answer. A very complicated formula reduces to L = 0.4142w where L is the outside length and w is the desired width , so my case L is about 12.5 inches.

3. ## 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))

4. ## 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

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