Help Figuring Sides Of An Octagon

house bldr

I am building a room on the side of a new home that projects 8" straight out from the wall then has three sides angling toward the center where there is a wall that runs paralell to the house then same thing back down the other side. In other words 9 sides, but not counting the 2 that are 8" that project straight out I want the other 7 sides to be equal in length. The room will be 14' wide and project 7'8" from the house or7' from where they start to angle.I have never done anything with this many sides,I'm having a brain lock here,Can anyone tell me the formula to figure this. I know in the title I called it an octagon but wasn't sure what to call something with this many sides.Although I have thought of a couple names here in the last hour I've been scratching my headThanks for any help!

Joe Wood

Here's a polygon calculator that might help you out.
http://www.1728.com/polygon.htm

here's another ..
http://www.rockler.com/articles/disp...8&cookietest=1

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curapa

From what I have read, I believe you are building half of a 16-gon. I am not sure as to the mathematical formula as I have not done this type of framing in years but you can lay this out simply. Begin with laying out your perpendicular 8" walls, layout your wall parallel to the house, bisect all of those angles the same way you would do an octagon (find the center of your overall dimension then divide by three), bisect the new angles and you will have your half of a 16-gon.

It may help to picture your polygon as a square to begin with. Start out with 45 degree angles to make your octagon then 22.5 to make your 16-gon.

Hope this helps.

Tom R

Not sure, - - I'm thinkin' he's talkin' about 7 sides of a 12-gon (dodecagon).

Other than that, - - I have a headache . . .

Maybe this will help??

http://mathworld.wolfram.com/Dodecagon.html

house bldr

Thanks for the replies, what makes this more difficult is that I am laying out footings to build this on and have nothing but dirt,strings and stakes to lay out with at this point.The polygon calculators are close but not extremely accurate because of the 8'' projection it is a little less than 1/2 of a true polygone.I drew it to scale on paper and think I have it pretty close but know there has to be an easier way.

Tom R

If I'm understanding your situation correct, - - your 8" doesn't come into play in figuring your sides, - - it's just an 'add-on'.

Anyway, - - I know I'm tired and maybe I'm crazy, - - but I come up with 48 1/2" per side, - - but I can't really say I'm all that sure . . .

Tom R

I believe you are incorrect, - - you want 7 equal sides of a 12-sided (7' radiused) polygon, - - so it is actually 'more' than half . . .

house bldr

You are correct on the 8'' not figured in but I have almost 1/2 of a 16 gon I come up with 37'' per side, you must have figured a 12 gon, or maybe I'm way off

Tom R

We need you to draw a pic, - - can't say I'm really sure in which direction the 8" parts are going, - - parallel??, - - perpendicular??, - - what?? . . .

Tom R

Oh, OK, - - I got ya' now, - - it's a 16-gon with two short (8" sides), - - gimme a few, - - and I'll try to rack my tired brain again.

neolitic

It's a footer, right? Digit like you're going to build it square, clip the corners a little with a couple of 2X's--then lay it out on the concrete when it's set.

curapa

Even though it seems to be "less than 1/2 of a true polygon" try to imagine it as a whole polygon with only 8" of two parallel sides protruding from the house..... Although considering the overall width is 14' and it protrudes 7' 8" from the house with the angles starting at 7' the numbers don't seem to add up (in my head) for this to be an equal sided polygon based on a square.

Either way find your known sides and fill in the rest.

I would like to see pictures of the plans if possible, it would help alot.

curapa

After some thought an a couple of re-reads I think you may be correct. The dimensions given just do not add up for a 16-gon.

Tom R

As I say, - - I need a pic to be sure, - - but if it is a 16-gon, - - I come up with about 32 13/16", - - if the 8" parts are just 'short sides' of an overall polygon, - - I should be correct.

I 'think' . . .

Tom R

Is the 8" part just a (straight-line) extension of the first of the three sides, - - or does it form it's 'own' (slightly angled) side??

Tom R

I believe the 8" part is added on to make the radius an actual 7', - - (beings it's an 'odd' number of sides, - - none of the actual radii of the chord intersections falls at 12 o'clock, - - so the 8" makes up the difference) . . .

neolitic

It's got 14 sides and fits in a 14' circle--the 8" bump out has nothing to do with it. But at the risk of repeating--it's a footer. Lessen you're gonna dig it by hand, the backhoe is gonna tend to make a big rectangle out of it anyway. Pour it big then lay it out when it's easier. $.02
Think bldr went to bed anyway.

house bldr

The 8'' sides project straight out square with the wall and are a small part of 2 sides of the 16 gon.I think I have it figured out! If the width is 14' the calculator tables should work I just have to back up past the point where the 8'' walls tie into the other wall before drawing the arc so I will be at the true middle of the circle. Sometimes I tend to overthink things and make them harder than they are!Thanks for all the advice

Tom R

Next time we just need one of the three Joe's around here.

Joe Bartok

Joe Fusco

Joe Carolla

Tom R

Well, - - sort of, - - the center-point of your radius should be at the center of the length and 8" (perpendicularly) away from your 14' wall (towards your new work), - - (maybe that's what you're already saying) . . .