Math Quandry-

[Rick Hubbard]

I’m in a bit of a math quandary.

I know that if I cut 8 pieces of identical size stock with an 11.25 degree bevel on each side I can assemble them into a perfect half-circle. What I can’t figure out though is how to determine the width of the pieces in order to come up with a pre-determined radius.

I’ll bet someone can embarrass me by pointing out something I should have learned in the sixth grade!

Thanks,


Rick

[Matt Armstrong]

You can always use trig (which, no, you probably did not learn in 6th grade) to figure this out -

sin 22.5 * radius of circle = outside length of each piece...

assuming this was the desired outcome:
http://images.rockler.com/rockler/images/24784-01-200.jpg (http://images.rockler.com/rockler/images/24784-01-200.jpg)

[Matt Armstrong]

I also have to point out the irony of our avatar juxtaposition

[Grant Morris]

Diameter is the measurement from one side of a circle to the other directly through the widest part (middle). The Radius is half of the length of the Diameter.

So if you wanted to create a 20" Diameter circle, the length of each piece would be half of that at 10" which is the Radius.

Similarly, if you wanted to end up with a 30" Diameter (15" radius), the length of each piece would be 15".

[Brian Willan]

Diameter is the measurement from one side of a circle to the other directly through the widest part (middle). The Radius is half of the length of the Diameter.

So if you wanted to create a 20" Diameter circle, the length of each piece would be half of that at 10" which is the Radius.

Similarly, if you wanted to end up with a 30" Diameter (15" radius), the length of each piece would be 15".

Actually Grant, that is not what Rick is after. He wants to know how wide his 8 boards need to be to hit a given half of a circumference.

So since the circumference of a circle is 2 * Pi * radius, half of that is clearly Pi * radius.

So what Rick is after for a given radius is

8 * width = Pi * radius

Since radius is known ahead of time.

width = (Pi * radius) / 8

Cheers

Brian

[Matt Armstrong]

brian's way is even easier... d'oh!

[Brian Walter]

Rick,
For starters 8 pieces with an 11.25 degree bevel will only give you 1/4 of a circle, not 1/2. I am assuming that when you talk about length of the pieces, you mean the distance from point to point after you assemble the 8 pieces. If this is the case, then the length or chord length = 2 x Radius x Sin(A/2) where A is the 11.25 degree angle. For example, if you wanted the outside radius of your circle to have a 40" radius, the length of the long side of your beveled pieces would be 2 x 40 x Sin(11.25/2) = 7.84".

Hope that helps.

Brian Walter

[David Christopher]

http://www.sawmillcreek.org/showthread.php?t=111665

this table will make you job easy without all the math........

[Robert McGowen]

Rick,
For starters 8 pieces with an 11.25 degree bevel will only give you 1/4 of a circle, not 1/2.


11.25 degrees is actually correct. You have to add the two cut pieces together, which gives you the 22.5 degree total.

[Doug Arndt]

A minor issue with using circumference is the boards are flat, not rounded.
Don't suppose it would make much difference however.

What you have are 8 triangles with the sides being the radius and the base being the desired "width" of the board.

A quick google of "triangle calculator" gave me this:
http://ostermiller.org/calc/triangle.html

Drop 78.75 into the 2 base angles (90 - 11.25), and the desired radius into the 2 sides, the top angle will calc out to 22.5 (correct) and give the desired base "length"

the radius dimension should be to the outside of the circle, to allow for cutting the 11.25 angle.

(this is assuming the desired outside radius is to the "points" not the flats)

[Greg Mann]

Unless I miss my guess, both Brians are wrong. First Brian is dividing half the circumference by 8. That would be the length along the arc, not along a straight line width. This is a chordal distance which is in relation to circumference/2/8 but not the same.

Second Brian is missing the fact that each bevel is 11.25 degrees, so when 2 are put together they equal 22.5 degrees. 8*22.5+180.

Matt and Brian were close but the right formula is really
length= 2(sin11.25)*rad. Matt was wrong when he said Brian's answer was easier because wrong is not easier. Other Brian was right when he said Grant was wrong.

Brian Walter's last formula was right up until he divided the sin11.25 by 2.The correct answer is 15.607"

After all this I hope I am right!:rolleyes:

[Brian Walter]

11.25 degrees is actually correct. You have to add the two cut pieces together, which gives you the 22.5 degree total.

I'm guessing we aren't talking about the same thing when we talk bevel. I'm referring to measuring the angle between opposite sides or ends of a board. If we are talking a board with an 11.25 degree bevel on each end/side which I would call a 22.5 degree bevel then you are correct, 8 boards will give you the 1/2 circle. If that is the case, then you simply wouldn't divide the angle in half in the formula that I gave.

Brian Walter

[Jeffrey Makiel]

Here's the formula. Personally, I use CAD so I don't have to think. :)

-Jeff :)

http://i3.photobucket.com/albums/y84/Beff2/HalfOctogon.jpg

[Brian Walter]

Jeff, your picture certainly makes it easier for everyone to be talking about the same thing. Greg is correct if he is measuring the bevel as Jeff shows it. I was measuring the central angle from each end of "W", not 1/2 of "W". Now wasn't that fun?:)

Brian Walter

[John Schreiber]

I just drew it in SketchUp and measured it there.

This is based on a 10" radius:

121078

[Kyle Iwamoto]

I THINK all of you missed the point. The question is, I THINK, is in order to get a diameter of X, say 30" diameter, how wide are the staves? so taking my college level skills to the test, sin 11.25 * (15) = .19 * 15 = 2.9" wide staves.

I hope no 6th graders read this forum and will prove me wrong.

[Jeffrey Makiel]

Brian Walter: "Now wasn't that fun?"

No Brian, I just hurt myself and have to go lay down now. :)


John Schreiber: "I just drew it in Sketchup..."

Sorry John, that's cheating. :)


Kyle Iwamoto: "I hope no 6th graders read this forum and will prove me wrong."

That's a risk with everything I type these days. :)

-Jeff :)

[Doug Arndt]

Sorry Kyle, but your answer doesn't fit a simple "check"

Circumference is Pi * D, 30 in dia * 3.14 = 94.2 total circle,
original post was half circle, 8 pieces, so 94.2 / 2 = 47.1 / 8 = 5.88
So the answer needs to be about 5.88

Using formula given earlier, W = .39 * R, .39 * 15 = 5.85

[Kyle Iwamoto]

Darn 6th grade graduates..... LOL

[Brian Kerley]

Ok, years of engineering, calculus, and everything else are on the line. Here's my shot at it.

First, I'm going to assume you want to assemble these pieces so you then can cut a semi-circle out of it. Therefore, the length of the side of the triangle is NOT your radius. You have a bunch of isosceles triangles, and the HEIGHT of the triangle is the radius of your circle. Therefore, the relation between the radius of the circle and the width of the pieces is via TANGENT not SINE/COSINE.

Here's the formula:
2*r*tan(11.25) = W

With that, you then will have the width of each piece required in order to be able to inscribe a circle of the radius you want. In other words, the faceted surface that if you were to put a trammel arm onto, would yield a semicircle of whatever radius you want.

[Greg Mann]

Ok, years of engineering, calculus, and everything else are on the line. Here's my shot at it.

First, I'm going to assume you want to assemble these pieces so you then can cut a semi-circle out of it. Therefore, the length of the side of the triangle is NOT your radius. You have a bunch of isosceles triangles, and the HEIGHT of the triangle is the radius of your circle. Therefore, the relation between the radius of the circle and the width of the pieces is via TANGENT not SINE/COSINE.

Here's the formula:
2*r*tan(11.25) = W

With that, you then will have the width of each piece required in order to be able to inscribe a circle of the radius you want. In other words, the faceted surface that if you were to put a trammel arm onto, would yield a semicircle of whatever radius you want.


Brian, you are exactly right if the OP wants to be able to radius off the flats to a perfect arc (which I believe he did refer to but never made clear). Sine gets the radius to the joints, Tangent gets to the flats.

[Rick Hubbard]

I just KNEW this had to be something very elementary (and something I already knew how to do), and sure enough!



width = (Pi * radius) / 8



The idea about using Sketchup to do layouts is a good one, and one I use when I have time to get to the computer and fumble around with the program. Doing it long-hand usually turns out to be a lot quicker for me, though.

Thanks everyone!!

Rick

[Jon Amsden]

I think Einstein said not to bother memorizing anything you can look up

This thing should help you out:

http://www.rockler.com/articles/display_article.cfm?story_id=98

[Paul Atkins]

I just draw it on a piece of cardboard and measure it.

[Kyle Iwamoto]

I just draw it on a piece of cardboard and measure it.


This has GOT to work! Old school.

Some of my friends buils houses ONLY with a framing square and NO math..... I asked him how he does the calculations. He says, QUOTE: "It's all right there on the square". Yeah, my college education is out the window. Took me 10 times longer to figure out a wedge, and it was STILL WRONG. As I was wrong before, in this post. :o

[Mark Norman]

I just draw it on a piece of cardboard and measure it.


Yep!! thats how I do it...not building a space shuttle.

[Prashun Patel]

Am I missing something here?

I think that answer is CLOSE, but it assumes that 8W = 1/2 circumference, which it doesn't.

I think the correct answer is:

W=2RSin(theta), where theta = 90/(# pieces in yr 1/2 circle).

In yr situation:

Brian's answer would be pi*10/8 = 3.927
The above eqn gives you: 2*10*pi*sin(11.25) = 3.902

It wouldn't make much of a diff on a circle w a LOT of pieces, bkz then the sum of the widths approaches the circumference, but on fewer # pieces, it'd make a diff, I think.

Am I wrong?

[Bill Houghton]

I've never done this - that particular sort of project remains on the "some time" list - but the literature all suggests that, since it's all but impossible to cut THAT precisely, the thing to do is glue up half of each circle and cheat the final angle by planing the meeting surfaces straight, before gluing the two halves together.