if the octagon is 38' 3 1/2" wide what do the sides have to be to work out Thanks

the only way I know how to figure it out is to get my chalk line and tape out on a big slab

Too bad you don't have AutoCAD. It does the thinking or you. Unfortunately don't have it either kwg

15ft 10-21/64 in......I think....

I think if you use A squared plus B squared equals C squared. I am not convinced, though.

15' 10 5/16"

http://www.pagetutor.com/octagon_layout/index.html

15' 10-5/16"

HP48gx and savvy.

15' 10 5/16"

Olde School: architect's scale and graph paper

Simple math, AutoCAD not needed.

Multiply half the distance by the tangent of 22.5�, double the results.

pi R round

Most pies are round!

yes sir.

I thought it was Pi r good

Thanks !! I'm framing a pitched roof on a flat roof house that is round on both ends with a rectangle connecting them the walls are 2x6 laid flat aginst slump block so i'm doing post and beam as there isn't a top plate to tie a roof to prolly why half of it blew off in the first place

I remember those. And the 32.

I laid out a power plant with an HP32, good machine when programed correctly.

That's interesting. I was thinking about a framing square on your original problem.

I haven't thought about my HP calculator in a lot of years. Thanks for the reminder�My HP21 was not a programable calculator, but it was a dream to use when compared to my slide rule. I have just forgotten most of the formulas I used to use over the last 38 years. I always wanted a programable calculator, but by the time I could afford one, I didn't need it anymore. Heck I couldn't afford the HP21 either. I had to eat a bunch of butter sandwiches for lunch and dinner to get that thing. Still have the HP21.

That made me laugh.

I use my HP 48 every day. Had it since the early to mid 90's.

Makes me a smart sumbitch!