Octagonal frame question

FramerDave

PFG, Picture Framing God
Joined
Jan 1, 2001
Posts
5,414
Loc
Houston, Texas
Ok, this one's bugging me, especially since I aced geometry in school, and I know the answer must be pretty straight forward.

Let's say you have an octagonal frame and it is to be 12 3/4 inches wide and tall. How do you determine the length of each side? I know the answer probably involves the square root of 2 since you end up with 45 degree right triangles. Beyond that I'm having a brain freeze.

I suppose I could brute force it and cut a piece of board 12 3/4 inches square and work from there. But that's not terribly elegant.

Any ideas?

UPDATE: My algebra teacher would be ashamed. I have gotten this far:

If S is the side, then then S times the square root of 2 plus 2S equals 12 3/4. Then it gets really messy.
 
According to my reference material:

With A being the Area
With W being the Width (between the parallel sides)
With S being the Side length
With P being the Perimeter

With any one parameter known:
A is .8284 times the square of the width
W is the square root of (A divided by .8284)
S is .4142 times the width
W is S divided by .4142
P is 3.3137 times the width (or 8 times S)
W is P divided by 3.3137

Hope this helps!
 
Dave,

You might want to check out the Barton's Multi-angle Framing Guide published by Shem River Publishers. I have a 1986 edition and it has all the angles, formulas, etc. for almost any type of multi-angle frame you will ever need to cut.

I think that it should be in the PPFA library or you should be able to buy it through Decor or PFM.

Framerguy

P.S. If you get stuck on the dimensions for your frame, let me know and I will figure out the dimensions for each of the sides for you.
 
We deal with these all the time. First question is do you want all of the pieces to be the same size? You can make small corners with long tops and bottoms. We usually ask customers to send us a patern. Especially when they say they are framing a mirror. To many times we have made octagons with all equal lengths and the mirror ended up having small corners instead. Either way the overall dimensions are the same.
 
Thanks for the help everyone. It will indeed be a regular octagon, which is to say all sides equal. Euclidian geometry helped me almost find the answer, but I broke down when it got to the algebra.

Let's see, I guess it's been about 20 years since high school geometry. I guess I'm doing pretty well to remember it as well as I do.
 
Originally posted by FramerDave:
Let's see, I guess it's been about 20 years since high school geometry. I guess I'm doing pretty well to remember it as well as I do.
:D Dave, it's been about 30 years since HS Geometry for me. I'm lucky I can still remember the word geometry. And yet I know people who still remember their High School locker combinations. Go figure. (No pun intended) :D
 
A simple formula for calculating the length of each side is as follows:
Formula for side of an octagon
X = D(Sq.Root of 2 - 1)
Where X = Face width of side & D = Inside diameter of octagon.

Rufus
 
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