I want to build a circular wood frame that will hold a glass table that is 44" in diameter. I will use pieces of lumber formed into a shape, probably an octagon, and use a router or jigsaw to cut out the inner and outer circles.
For the sake of a rough estimate, the outside diameter of the frame will be 45", the frame will be 2" wide, and the frame will be rabbeted at 44". I calculate the outside circumference of the 45" diameter to be a little over 141". If I start with 8 equal sized pieces of lumber using 45 degree joins, the longest part of each piece of lumber is under 18".
What is the minimum width of lumber I can use to make this octagon and have at least 2" of width in a cut-out circle all the way around? I tried to work it out on some graph paper, and it looks like the longest distance is the length of the diagonal cuts, which appears to need to be about 4". But that’s just from using graph paper. Is there a mathematical way to figure that? That number would end up being the minimum length of the 45 degree cuts. Once I have that, I can calculate the minimum width of the wood required. I’d rather not buy buy wider wood than I need, just to see it cut off into a waste pile.
If you say the join should be 2", then the circles going away from it will both move further out on the board, so you only need to make sure the outer circle doesn’t go outside the board at the midpoint between joints. The radius is perpendicular to the board at that point, so you get a right triangle between the inside of the joint, the midpoint between joints and the center, with a 22.5 angle at the center of the circle.
If the join is 2" to the outer circle and the radius of that circle is 45" the hypotenuse for that triangle is 43" and the distance between center and midpoint of the board is
43*cos 22.5 = 39.73
Since the outer circle needs to be 45" the board has to be 45-39.73 = 5.27" wide.
Not quite clear what you are asking yet, but here’s a helpful octagon calculator.
You want a circle inscribed in the octagon where the circle radius is 2" less than the octagon radius (center to midpoint of a side), is that correct? I think I’m getting what you are asking about the sides. Before I screw up the math trying to figure this out (@naita sounds right to me) I’ll suggest you not stick to the specific 2" and pick your lumber for a close match instead of a precise number.
And I’m sure you know you need to leave some extra space around the glass circle to allow for some expansion and contraction of both materials.
And finally, I’d suggest waiting a while before buying any lumber, we’re at the top of a price bubble right now.
It’s not the join that has to be 2" it’s the width of the remaining circle when it is cut out. The outside of the 2"-wide circle will touch the outside centers of the 8 boards; the inside of the circle will touch the inside of the 8 joins. So the length of the join gives me the hypotenuse of an isosceles triangle, with which I can calculate the width of the board required.
I actually figured it out this way, again using a right triangle for the calculations. The long leg of the triangle is from the center of the circle to the center of the outside edge of one of the 8 sections, which is 45". The hypotenuse is from the center of the circle to the outside point of one of the 45 degree joins for that same board. The short leg is from the the center point of the outside of that board to the outside point of the same 45 degree join; I know the length of the board, so this distance would be half of the board length. Since I know two sides I can calculate the hypotenuse. 45" on the long leg, 8.9" on the short leg, and the hypotenuse comes out to under 46". That makes the length of the join under 3" and the minimum width of the board to be the square root of 9/2 or 2.12". That does not, intuitively, seem right, so I’ll be checking my math again.
My intention is to buy the next larger size than I need.
The difference is not worth worrying about for the amount of lumber involved. This project could be done with two 2x4s (8 pieces less than 18" in length). (edited to correct stupid math error)
I’ve tried this, and it’s very useful. I need the incircle radius “r” to be 22.5", so I can fiddle around until I get that, and it will tell me the circumcircle radius (which turns out to be 24.354"). This is a different result from what I got with my math approach. It means the length of the 45 degree join needs to be 4.854" and that makes the minimum board width 3.43". That sounds better intuitively. I wonder if I’ll be able to figure out where my math went wrong. For one thing, the length of the board sections was short.
2x4s should still work, barely.
naita, that’s a handy tool too, but I’m having a hard time figuring out what you’re doing with it.
I think I have what I need, thanks folks for your help.
I understand it won’t cost much but it’s annoying me that I have to pay over $5 for the lowest quality 2x4s. Glad you’re making progress, sounds like a fun project. What’s your plan for legs?
Yes, I understand the annoyance. I will probably use redwood for outdoor durability, so it will cost even more.
I have the original metal tri-pod leg arrangement from the table in pretty good shape (the original frame for the top corroded away beyond recovery) and I might be able to re-use it, or maybe I will try to re-create it using wood.
I used a 45" radius instead of diameter. D’oh! The link is updated with that now. That gives a board with of 3.6".
Since the join in an octagon is aligned with the radius, it will be 2" after you cut off the outer circle, and shouldn’t be touched by cutting out the inner circle.
Unless I’m still misunderstanding something you’re still getting something wrong. I’ve added a trapeze shape showing the segment of the octagon this calculates, and it shows that each piece needs to be 18.6" long corner to corner to get an outside diameter of 45" and inside diameter of 43". (It might also make it easier to see what the circles and lines represent.)
Yes, another source of math error on my part. I was thinking the ends of the boards were at 45 degrees, but they are of course only 22.5 degrees, to make a 45 degree angle in the octagon…
Also, I just remembered that I have some weathered but still usable redwood boards that I can use for most or all of this, so it won’t cost that much.