For you math wizards

[Leon]

I gotta ask Bill why pipe instead of string? and for that matter I have drilled a hole in a 25' tape measure and anchored a screw through that hole and held a pencil at the distance desired.

[Lew Hodgett]

-----------------------------------------------------------------

----------------------------------------------------------------- Lew Hodgett wrote:

From a previous post.

-------------------------------------------------------------- Find a copy of Fred Bingham's book, "Practical Yacht Joinery" at the library.

A very easy graphical solution is shown.

I laid out all the deck cambers for my boat using it.

Lew

[CW]

R=226" according to Sketchup. =========================================================================== My calculator says that your program is right.

[Bill]

I thought of drilling a hole in a tape measure too! String or even wire is probably too elastic. I don't thing 3/4" steel pipe is too elastic. I've got ten 4' sections of pipe, along with 10 connectors--for emergencies! So far, I've only used them pair-wise (in my pipe clamps). Evidently you were successful at 25' with your tape measure.

Bill

[Doug Miller]

Leon wrote in news: [e-mail]:

Pipe don't stretch. String do.

I found out just how much string stretches last summer when laying out a curved sidewalk. Next time I do a project like that, I'm using something rigid to mark my arcs: board, angle iron, steel rod, something like that -- but *not* string.

[Larry W]

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:

[Lew Hodgett]

--------------------------------------------------- That will work, but it's hard work.

Check out a PDF File "Fig 5-42" over on apbw.

Lew

[-MIKE-]

Many tape measures already have a hole for a finish nail exactly on the

1' mark.

[Leon]

You do have a point there.

[Lew Hodgett]

--------------------------------------------- Which is why full size arcs by swinging a radius SUCKS.

There is an easier way to run the railroad.

See Fig 5-42 over on APBW.

Lew

[Doug Miller]

[e-mail] (Larry W) wrote in news:kkl297$pbb$1 @speranza.aioe.org:

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.

[Doug Miller]

"Lew Hodgett" wrote in news:516e0ddd$0$26833$c3e8da3 $ [e-mail]:

Lew, not everybody's server carries abpw. Can you post in dropbox, photobucket, flickr, or someplace similar?

[Leon]

I'm glad you were able to verify the accuracy of your calculator. ;~)

[Leon]

is that that method

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.

[Greg Guarino]

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.

[Greg Guarino]

And yes, it turns out that the method (in theory) should work. See

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.

[Leon]

Mathematically it is correct. But finding a material to bend that will bend exactly as you want it to is another matter.

[Greg Guarino]

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.

[alexy]

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.

[Lew Hodgett]

"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