Lew Hodgett wrote:
Gramp's shop wrote:
Hah! Now all I need to do is figure out how to draw an arc with an
18 foot radius :-) I have a couple of ideas and will post pix of
the process.
==========================================
"Just Wondering" wrote:

Two lengths of PVC pipe. Drill a hole in one end, attach a pencil
to the other at 18'. Draw the arc on the ground, spiking the pivot
end in the ground and putting the wood at the pencil end.
=========================================
You've obviously never done this.

Lew

I couple of "trammel points" on a piece of steel pipe, along with two
humans, may work. What is your solution?

Bill (who just happens to have 40 feet of Schedule 40 black pipe)

On 4/16/2013 5:19 PM, Bill wrote:
Lew Hodgett wrote:
Gramp's shop wrote:
Hah! Now all I need to do is figure out how to draw an arc with an
18 foot radius :-) I have a couple of ideas and will post pix of
the process.
==========================================
"Just Wondering" wrote:

Two lengths of PVC pipe. Drill a hole in one end, attach a pencil
to the other at 18'. Draw the arc on the ground, spiking the pivot
end in the ground and putting the wood at the pencil end.
=========================================
You've obviously never done this.

Lew

I couple of "trammel points" on a piece of steel pipe, along with two
humans, may work. What is your solution?

Bill (who just happens to have 40 feet of Schedule 40 black pipe)




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.

Gramp's shop wrote:
Hah! Now all I need to do is figure out how to draw an arc with
an
18 foot radius :-) I have a couple of ideas and will post pix of
the process.
==========================================
"Just Wondering" wrote:

Two lengths of PVC pipe. Drill a hole in one end, attach a pencil
to the other at 18'. Draw the arc on the ground, spiking the
pivot
end in the ground and putting the wood at the pencil end.
=========================================
You've obviously never done this.

Lew
-----------------------------------------------------------------
I couple of "trammel points" on a piece of steel pipe, along with
two humans, may work. What is your solution?

Bill (who just happens to have 40 feet of Schedule 40 black pipe)

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

"Leon" wrote in message
news
On 4/15/2013 10:14 AM, Gramp's shop wrote:
I need to draw an arc for a piece of trim. The end points of the arc are
5 feet apart and the depth of the arc at the center point is 2 inches.
What is the radius of the circle?

There ought to be an equation for this that would be far superior to trial
and error, oui?

Larry




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

Leon wrote:
On 4/16/2013 5:19 PM, Bill wrote:
Lew Hodgett wrote:
Gramp's shop wrote:
Hah! Now all I need to do is figure out how to draw an arc with an
18 foot radius :-) I have a couple of ideas and will post pix of
the process.
==========================================
"Just Wondering" wrote:

Two lengths of PVC pipe. Drill a hole in one end, attach a pencil
to the other at 18'. Draw the arc on the ground, spiking the pivot
end in the ground and putting the wood at the pencil end.
=========================================
You've obviously never done this.

Lew

I couple of "trammel points" on a piece of steel pipe, along with two
humans, may work. What is your solution?

Bill (who just happens to have 40 feet of Schedule 40 black pipe)




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.

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

Leon lcb11211@swbelldotnet wrote in
:

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.

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.

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-...-Large-Curves/


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

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

"Larry W" wrote:

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-...-Large-Curves/
---------------------------------------------------
That will work, but it's hard work.

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

Lew

On 4/16/13 8:17 PM, Bill wrote:
Leon wrote:
On 4/16/2013 5:19 PM, Bill wrote:
Lew Hodgett wrote:

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.

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


Many tape measures already have a hole for a finish nail exactly on the
1' mark.


--

-MIKE-

"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com

---remove "DOT" ^^^^ to reply

Doug Miller wrote:
Leon lcb11211@swbelldotnet wrote in
:

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.

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.

You do have a point there.

Doug Miller wrote:

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.
---------------------------------------------
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

(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.

"Lew Hodgett" wrote in news:516e0ddd$0$26833$c3e8da3
:


"Larry W" wrote:

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-...-Large-Curves/
---------------------------------------------------
That will work, but it's hard work.

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

Lew

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

On 4/16/2013 7:57 PM, CW wrote:


"Leon" wrote in message
news
On 4/15/2013 10:14 AM, Gramp's shop wrote:
I need to draw an arc for a piece of trim. The end points of the arc
are 5 feet apart and the depth of the arc at the center point is 2
inches. What is the radius of the circle?

There ought to be an equation for this that would be far superior to
trial and error, oui?

Larry




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


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

On 4/17/2013 7:18 AM, Doug Miller wrote:
(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.



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.

On 4/17/2013 9:24 AM, Leon wrote:
On 4/17/2013 7:18 AM, Doug Miller wrote:
(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.



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.

On 4/17/2013 10:12 AM, Greg Guarino wrote:
On 4/17/2013 9:24 AM, Leon wrote:
On 4/17/2013 7:18 AM, Doug Miller wrote:
(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.



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...y-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.

On 4/17/2013 9:27 AM, Greg Guarino wrote:
On 4/17/2013 10:12 AM, Greg Guarino wrote:
On 4/17/2013 9:24 AM, Leon wrote:
On 4/17/2013 7:18 AM, Doug Miller wrote:
(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.



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...y-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.


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

On 4/17/2013 11:53 AM, Leon wrote:
On 4/17/2013 9:27 AM, Greg Guarino wrote:
On 4/17/2013 10:12 AM, Greg Guarino wrote:
On 4/17/2013 9:24 AM, Leon wrote:
On 4/17/2013 7:18 AM, Doug Miller wrote:
(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.



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...y-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.


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

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.

Leon lcb11211@swbelldotnet wrote:

On 4/17/2013 9:27 AM, Greg Guarino wrote:
On 4/17/2013 10:12 AM, Greg Guarino wrote:
On 4/17/2013 9:24 AM, Leon wrote:
On 4/17/2013 7:18 AM, Doug Miller wrote:
(Larry W) wrote in news:kkl297$pbb$1
@speranza.aioe.org:

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-...-Large-Curves/

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.



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...y-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.


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

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.

"Larry W" wrote:

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-...-Large-Curves/
---------------------------------------------------

"Lew Hodgett" wrote:

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

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

Lew, not everybody's server carries abpw. Can you post in dropbox,
photobucket, flickr, or
someplace similar?
-------------------------------------------------------------------
A trip to the library or astraweb.com solves that problem.

I'm not familiar with any of the things you suggest.

Lew

On 4/17/2013 12:23 PM, 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.

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.

Don't you have a hand plane or a jointer? ; )

Bill

On 4/18/2013 1:03 PM, Edward A. Falk wrote:
In article ,
Greg Guarino wrote:

My sense of the physics involved tells me that that method will not
produce an arc of a circle.

I agree, and my own experiments have borne this out.

The bending stress will be highest at the center, and tend
toward zero at the ends. I think you'll get a hyperbola.

That's OK. There's a simple outpatient procedure for hyperbola these days.