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?
I'll be on the rowing machine if you need me.
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"
make a second cut and you have two quarter round pieces 3" radius. more
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.
But not quite 6 1/8" across, because I've lost the 1/8" section from
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
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.
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.
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
Truro, NS, Canada
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
Good luck and let us know what works,
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.
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?
You gave that number as a radius rather than a side length. That didn't make
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.
Sorry about the confusion.
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.
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.
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