Building an octagon
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From: Lenexa, KS
Building an octagon
I have to construct a framework out of 2 x 10's in the shape of an octagon. I have cut octagons before, but I'm always guessing the lengths of the sides to come up with the interior space I need. I have to surround a tree that measures about 30" across. Does anyone know the formula for determining the lengths of the sides of an octagon to accommodate a certain radius?
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From: SW VA
Now that's funny, because I just built one today...the same size as what you need. 
You will need a miter saw to keep the angle of the 22 1/2 degree cuts even. Measure 18 inches for the long side and make yer cuts. Using an 18 inch length will give you an inner circumferance of 36 inches. You should end up with eight peices that look something like this:
___________________
\_________________/
Put em together and voloa, yer octogon sir.
You will need a miter saw to keep the angle of the 22 1/2 degree cuts even. Measure 18 inches for the long side and make yer cuts. Using an 18 inch length will give you an inner circumferance of 36 inches. You should end up with eight peices that look something like this:___________________
\_________________/
Put em together and voloa, yer octogon sir.
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From: Lenexa, KS
Did you build yours with 2 x 10's? When you say 18" for the long side do you mean the outside dimension? I figure this will change as the width of the lumber increases or decreases. Anyway, thanks for your reply.
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If you want to actually think about it, take your radius, that becomes the base of the hypothetical triangle for future use. Then the angle between the height and the hypotenuse should be 90-22.5=67.5º... use Sin(x)=opposite/hypotenuse
= Sin(67.5)=r/hyp
= r/Sin(67.5)=hyp
assuming Diameter of 30
= 15/Sin(67.5) = hyp = the inside length of the board
According to calc***** that comes to 16.2359, but of course you'd want to round up for tree growth...
I think that's right... feel free to ignore me, I'm really bored...
= Sin(67.5)=r/hyp
= r/Sin(67.5)=hyp
assuming Diameter of 30
= 15/Sin(67.5) = hyp = the inside length of the board
According to calc***** that comes to 16.2359, but of course you'd want to round up for tree growth...
I think that's right... feel free to ignore me, I'm really bored...
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From: SW VA
I built mine out of landscape timbers but according to AutoCad the measurement for a 2x10 (9 1/2 actual measurement) will be 23 13/16 on the outside and with a 22.5 degree andle cut, the inside measurement should end up at 14 15/16. Or you can just round it off to to 24 inches on the outside and have the inside circumferance just a hair over 30 inches.
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From: Lenexa, KS
Originally Posted by Wingnutt
I built mine out of landscape timbers but according to AutoCad the measurement for a 2x10 (9 1/2 actual measurement) will be 23 13/16 on the outside and with a 22.5 degree andle cut, the inside measurement should end up at 14 15/16. Or you can just round it off to to 24 inches on the outside and have the inside circumferance just a hair over 30 inches.
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