Actually, he said he wanted the radius so he could draw the arc. Arc, not
chord. If he wanted to wind up with a 5' chord all he would need is a 5'
straight edge.

Actually LOL, I think Bill was filling me in with an accurate answer to
my question that I posed to Richard. I was not really asking for an
answer so to speak. I mentioned a template printed from Sketchup and
Richard questioned an 18' radius template. I believe he was thinking
about printing an 18' foot long template, maybe not. LOL
The desk I just completed I useed the printing template technique for an
8' wide arc with a 36.83 foot radius and only used 8 sheets of paper.

A few strips of hardboard screwed together, and that 18 foot template
should come together in seconds! The OP may wish to include a
micro-adjuster at one end. : )
Bill

I've seen the technique described elsewhere, but this was the first
google result on searching for "how to draw an arc of large radius"
Look at step 1 on this site:
http://www.instructables.com/id/How-to-Draw-Large-Curves/

--
There are no stupid questions, but there are lots of stupid answers.

Larry W. - Baltimore Maryland - lwasserm(a)sdf. lonestar. org

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A trip to the library or astraweb.com solves that problem.
I'm not familiar with any of the things you suggest.
Lew

I agree, the material used would have to be uniform in stiffness from
one end to the other and you would have to have pivot points beyond the
start and stop points at the same intervals AND pivot points would have
to be precisely placed to obtain a circular arc.

I'm not sure either way. If your goal was to draw a full semicircle, two
pins and a framing square (or anything that forms a right angle) would
indeed (theoretically) draw a semicircle. This is because any triangle
formed by two opposite ends of a diameter and a third point anywhere on
the circumference is a right triangle.
The question then is as follows: Can we generalize that to say that the
angle at the apex of a triangle formed by two points on opposite ends of
a (non-diameter) chord and a third point on the circumference between
them also remains fixed at some angle. If so, the method should work. I
may have to look this up. Geometry was a long time ago and circles may
have changed since then.

And yes, it turns out that the method (in theory) should work. See
http://m.everythingmaths.co.za/grade-12/09-geometry/09-geometry-02.cnxmlplus
Theorems 5 and 6.
"The angles subtended by a chord at the circumference of a circle are
equal, if the angles are on the same side of the chord."
and the converse which is more directly related to the question...
"If a line segment subtends equal angles at two other points on the same
side of the line, then these four points lie on a circle."
That may seem a little obtuse in text. The pictures on the page make it
clearer. Basically, you are making a jig whose angle remains fixed. By
sliding that jig over two fixed points, you satisfy the conditions of
Theorem 6.

The method mentioned doesn't require any bending. I do wonder how well
you could practically keep a pencil in the crook of so obtuse an angle,
but I was mostly curious about the math.

No bending involved in this method, as I understand it.
The two boards (look like hardboard in the picture) are rigidly joined
to kee a constant angle, and slid over the push-pin reference points.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

He really needs to make about 20 of them. To save time, he will array
the stock in a regular twenty-sided polygon, presumably on the
basketball court of the local high school. This will allow him to mark
all of the pieces in one step with a string and pencil, or if the
ceiling height is sufficient, a compass. :)

RE: Subject
You math wizards are making a mountain out of a mole hill.
Give me 10 minutes and some 1/4" hard board and I'll give
you a finished template.
I left my calculus in the class room the day I graduated
more years ago than I want to admit.
This is a case where a graphical solution wins hands down.
Lew

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