# For you math wizards

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• posted on April 16, 2013, 10:12 am
Bill wrote:

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.
--

dadiOH
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• posted on April 16, 2013, 11:07 am

Bill's point is that the 5' dimension is the measurement of the chord, not the arc.
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• posted on April 16, 2013, 2:05 pm
On 4/16/2013 6:07 AM, Doug Miller wrote:

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.
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• posted on April 16, 2013, 2:16 pm
On 4/16/2013 10:05 AM, Leon wrote:

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
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• posted on April 16, 2013, 2:49 pm
On 4/16/2013 9:16 AM, Bill wrote:

...

Note picture UL 2nd page... :)
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• posted on April 16, 2013, 6:16 pm
Doug Miller wrote:

Ah, OK, got it. Mea culpa.
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• posted on April 17, 2013, 2:43 am
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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• posted on April 17, 2013, 2:50 am
"Larry W" wrote:

--------------------------------------------------- That will work, but it's hard work.
Check out a PDF File "Fig 5-42" over on apbw.
Lew
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• posted on April 17, 2013, 12:25 pm

Lew, not everybody's server carries abpw. Can you post in dropbox, photobucket, flickr, or someplace similar?
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• posted on April 17, 2013, 8:21 pm

"Lew Hodgett" wrote:

----------------------------------------------- "Doug Miller" wrote:

------------------------------------------------------------------- A trip to the library or astraweb.com solves that problem.
I'm not familiar with any of the things you suggest.
Lew
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• posted on April 17, 2013, 12:18 pm
snipped-for-privacy@sdf.lNoOnSePsAtMar.org (Larry W) wrote in

I don't immediately see how to prove or disprove it... but my first impression is that that method draws an elliptical arc, not a circular arc.
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• posted on April 17, 2013, 1:24 pm
On 4/17/2013 7:18 AM, Doug Miller wrote:

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.
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• posted on April 17, 2013, 2:12 pm
On 4/17/2013 9:24 AM, Leon wrote:

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.
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• posted on April 17, 2013, 2:27 pm
On 4/17/2013 10:12 AM, Greg Guarino wrote:

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.
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• posted on April 17, 2013, 3:53 pm
On 4/17/2013 9:27 AM, Greg Guarino wrote:

Mathematically it is correct. But finding a material to bend that will bend exactly as you want it to is another matter.
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• posted on April 17, 2013, 4:23 pm
On 4/17/2013 11:53 AM, Leon wrote:

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.
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• posted on April 17, 2013, 10:45 pm
On 4/17/2013 12:23 PM, Greg Guarino wrote:

Don't you have a hand plane or a jointer? ; )
Bill
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• posted on April 17, 2013, 6:21 pm

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.

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• posted on April 16, 2013, 1:46 pm
On 4/15/2013 8:26 PM, Leon wrote:

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. :)
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• posted on April 15, 2013, 7:58 pm
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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