Cylinder sizing?

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• posted on September 21, 2006, 10:38 pm
I need some large quarter-round molding. The finished size must have have a "radius" of 3" - that is, the distance from the point to the cut edge must be 3". I thought of turning a 6 1/8" cylinder, and losing an eight to the kerf on the TS. Then I remembered it'd be quartered, so I'd lose another eight. Then I realized that as I take the kerfs I'm actually shortening the legs by somewhat more than an eight due to the curvature of the piece. I haven't done a google search and I haven't made any effort whatsoever to do the math. Anyone up to the task? tia JP ******************************************************* I'll be on the rowing machine if you need me.
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• posted on September 21, 2006, 10:58 pm
Jay Pique wrote:

A cylinder of 6 1/8" *will* give you sections with radius 3" when quartered, assuming a 1/8" kerf.
Molding? That's a log!
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• posted on September 21, 2006, 11:17 pm
Jay Pique wrote:

draw a cross section on a piece of paper to see your errorw.
61/8" diameter , one cut at 90degrees to diameter leaves you a length of 3" for a 1/8"kerf after the first cut, you have two half round pieces, 3"high and 6 1/8" across make a second cut and you have two quarter round pieces 3" radius. more or less.
I believe but have not drawn it , that the curve will not be a 3" radius , but will be the remainder of the 6 1/8" curve. And not a proper 90 degree circle segment.
Probably close enough for a moulding though.
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• posted on September 22, 2006, 12:38 am
snipped-for-privacy@yahoo.com wrote:

But not quite 6 1/8" across, because I've lost the 1/8" section from the middle.
snip

It's to sit vertically below another, smaller quarter-round piece. The whole assembly is for a corner detail on some wall panels.
I know I can sneak up on it, and I could rip in on the band saw for less loss, but I'm sort of keen on knowing the math behind it. Time to start digging. JP
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• posted on September 22, 2006, 1:14 am
Jay Pique wrote:

There's hardly any math involved. Quarter a 6-1/8" cylinder with a 1/8" kerf, and you come out with 4 quarter-rounds, which, when you gang them together, make a 6" cylinder.
You *do* come out with a proper 3" radius and a 90-degree arc.
Every point on that arc is 3" from the center of the new circle or the vertex of that 90-degree angle. So we say you have a 3" radius there.
Nothing to see here.
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• posted on September 22, 2006, 1:43 am

No, you don't. It's pretty close, but not exactly. The radius of curvature is 3-1/16", not 3", and the arc does not meet the flat quite at 90 degrees.

Not correct. But close enough it doesn't matter.
--
Regards,
Doug Miller (alphageek at milmac dot com)
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• posted on September 22, 2006, 5:22 pm
Doug Miller wrote:

Yeah, I wasn't thinking straight. Of course the *arc* still has a 3-1/16" radius, i.e., bisect the 90-degree angle, and the distance to the point on the arc is 3-1/16". But the distance from either leg of that right angle is only 3". I swear I got honors in plane geometry, but that was a long time ago. I was thinking more in terms of getting the piece he wanted.
So the answer is, there's really no way to section a cylinder and get a quarter-round of radius X that's exactly X on each side. You'd have to live with an arc whose radius is X + .5*kerf, or start with an X by X section and cut radius X onto it.
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• posted on September 22, 2006, 10:39 pm
The commercial technique for turners of the early part of the 20th century was to take four pieces of 3 1/8" square wood and glue them together with kraft paper (heavy paper bag materail) between. Cup centers, a dead one in the head stock and a live one in the tail, were used to turn the piece as they would hold the center tight without splitting the glue line. The cylinder was turned, split along the paper joint, sanded to tremove the paper and it was done. Today I would use hot glue to hald the wood together and separate it with paint thinner when done but the same procedure. ______ God bless and safe turning Darrell Feltmate Truro, NS, Canada www.aroundthewoods.com
wrote:

curvature is

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• posted on September 22, 2006, 1:55 am

Not exactly. Yes, the cylinder will be 6" in diameter when measured across the cut lines, but it won't be a perfect cylinder. The radius on the outside curves will still be half of 6 1/8" - JP is right. Take it a little further - say you start with the 6 1/8" cylinder, and keep cutting 1/8" sections off until the straight parts of your "quarter round" are only 1" long. Then you put the 4 quarters together, and you do NOT have a round 2" diameter circle - you have pieces with a 1" flat face, but the radius of the outside curve doesn't change. So, Jay, the question is whether it's more important that you have a radius of exactly 3", or flat sides of exactly 3". If you start with any diameter cylinder, and remove any material at all for a kerf, you can not end up with 4 pieces with 3" faces AND a 3" outside radius. Unless you're up for a lot of sanding... I hope I explained that well enough - draw it out, and keep taking away kerfs until it makes sense. If you're looking for opinions, I'd say that nobody could tell the difference between a 3" radius curve and a 3 1/16" radius curve, so I'd say the 6 1/8" cylinder would probably work. It would be even closer to use a bandsaw for the thin kerf and reduce the cylinder diameter accordingly. Good luck and let us know what works, Andy
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• posted on September 22, 2006, 2:35 am
RE: Subject
Ever consider 1/4 circle from plywood.
If interested, check out Anderson International here in SoCal.
Lew
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• posted on September 22, 2006, 3:20 am
A 6 1/8 cylinder has a radius of 3 1/16. Quarter it and each piece still has a radius of 3 1/16. You could use a 1/2" blade to cut it. The radius still won't change.

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• posted on September 23, 2006, 7:21 pm
Jay Pique wrote:

Jay,
If I do the math, I get 2.99936..." for the radius. Kind of hard to explain without using figures. If you're seriously interested, I can make an attempt, as it takes time to write a text-only explanation.
Regards,
Mark
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• posted on September 23, 2006, 8:07 pm
Forget trying to explain your math, what is it your trying to figure out?

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• posted on September 24, 2006, 4:14 am
CW wrote:

I was figuring out Jay's question, given in the original post: take a 6-1/8" diameter cylinder (or circle) and saw it into quarters, where the saw kerf is 1/8". What is the size of each quarter, measured across either of the flat faces?
At least, that is my understanding of what was asked in the original post. As a math question, the answer is 2.99936...". As a woodworking/carpentry question, 3" will suffice. Who here can position a piece to better than 0.001" on their table saw?
Mark

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• posted on September 24, 2006, 4:57 am

out?
You gave that number as a radius rather than a side length. That didn't make sense.

There was a puzzlemaker on here some time ago that claimed .0002 accuracy. After considerable time expaining why he was full of it, he finaly gave up and left. He still claims this in other places and on his website though.
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• posted on September 24, 2006, 5:28 am
CW wrote:

It's "almost" a radius, and would appear to anybody to be a radius for all intents and purposes. But it's **not** a radius because it doesn't originate from the center-of-curvature of the rounded face (the center of the original, uncut cylinder).
The original post even used the word "radius", but in quotes since it's not **really** a radius. I just figured people who had read that, and have been following this thread, would know what I meant.
Mark

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• posted on September 24, 2006, 2:17 pm
redbelly wrote:

Yeah, that's what I was trying to convey. Sorry for the confusion. I'm now going to turn, or have turned a segmented blank that will yield true quarters with 3" radii, thus eliminating the need for mental gymnastics. Thanks all for the input.
JP
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• posted on September 24, 2006, 4:58 pm
By product of the job I guess (machinist). When someone says radius, "almost" never enters my mind. It either is or isn't.

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• posted on September 24, 2006, 12:50 pm

Steve Strickland. <g>
--
Regards,
Doug Miller (alphageek at milmac dot com)
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• posted on September 22, 2006, 10:58 am
wrote:

I'd do it with a shaper.