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Old 05-31-2009, 12:44 PM   #1
javajoe61
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Using landscape timbers for an octogon tree ring. Not sure how long each of the 8 pcs should be at the longest point of the angle (22.5), to have a diameter of 4'. Is there a specific formula for this ?
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Old 05-31-2009, 03:57 PM   #2
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I'm not sure if this is right or not but since no one else has chimed in I thought I would try. I found the circumference of the circle(3.141x2squared) which is 12.564'. Then I divided it by 8 which is 1.57', This isn't going to be a perfect 4 ft in diameter, it will be just a little bigger since 1.57 feet is on the rounded part of the circle and not a straight line like the parts of an octagon but I would think it will be pretty close. If it were me I would cut 2 pieces of 2x4s at 18 1/2" long with 22.5 angles on them and lay them out. When you measure to wards the "center" of the circle(these 2 pieces being 1/4 of a circle) it should be 2' where they intersect. Then you can trim them as needed to get your octagon the exact size you want. Here is a picture to hopefully help out.question-octagone.jpg
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Thanks for your help, Scott
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Old 05-31-2009, 06:58 PM   #3
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http://mathcentral.uregina.ca/QQ/dat....04/mary1.html
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Old 06-01-2009, 10:00 AM   #4
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Thanks for the info, this will give me a rough idea where to start. Will tinker with lengths till I get roughly a 4' diameter.

Pete
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Old 06-01-2009, 11:02 AM   #5
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something like this?, did this quick sketch in AutoCAD.

Jesper
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