My plan exactly, Larry. I'm going to start with drilling screws into the waste side of the end points and the top of the arc, attach a thin strip of one-by with spring clamps and then add a couple of screws/clamps along the arc.
The other Larry
On Monday, April 15, 2013 11:43:27 AM UTC-5, Larry Blanchard wrote:
> On Mon, 15 Apr 2013 09:10:54 -0700, 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.
>
>
>
> Bend a piece of thin cutoff around the points?
>
>
>
> --
>
> When fascism comes to America, it will be wrapped in the flag and
>
> carrying a cross.
Doug Miller <[email protected]> wrote:
> Leon <lcb11211@swbelldotnet> wrote in
> news:[email protected]:
>
>> 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.
On 4/15/2013 9:14 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
>
http://www.mathopenref.com/arcradius.html
On 4/15/2013 9:14 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
>
http://www.handymath.com/cgi-bin/rad2.cgi?submit=Entry
http://www.ehow.com/how_7846775_radius-arc.html
http://mathcentral.uregina.ca/QQ/database/QQ.09.98/riddleberger1.html
http://www.woodweb.com/knowledge_base/Radius_of_an_arc.html
"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?
--------------------------------------------------------------
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
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
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?
------------------------------------------------
"dadiOH" wrote:
> Trial and error isn't needed, neither is knowing the radius of the
> arc's circle.
>
> Put two nails 5' apart in a piece of ply. Put another nail 2" above
> the line formed by the first two. Take a batten, bend it between
> the nails and draw the arc.
---------------------------------------------------------------
Nice try but no cigar.
The end result needed is a cambered beam shape which requires
more than the three points you suggest.
Bingham outlines the method that works in his book.
Have used the method to define the deck cambers which varied from
10' to 16' in length for the boat I built.
BTW, a batten is needed, I used a 3/4" x 3/4" x 1/16" x 96" aluminum
angle which provides a knife edge for fairing out the profile with a
fairing
board.
Lew
On 4/15/2013 10:10 AM, 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.
>
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.
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
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
"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-to-Draw-Large-Curves/
---------------------------------------------------
That will work, but it's hard work.
Check out a PDF File "Fig 5-42" over on apbw.
Lew
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:
>>
>>> 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/
>> ---------------------------------------------------
"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
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.
On Monday, April 15, 2013 10:14:29 AM UTC-5, 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
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:
>>>> [email protected] (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-to-Draw-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-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.
>
>
Mathematically it is correct. But finding a material to bend that will
bend exactly as you want it to is another matter.
On 4/16/2013 7:57 PM, CW wrote:
>
>
> "Leon" wrote in message
> news:[email protected]...
>
> 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 4/15/2013 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.
On 4/17/2013 7:18 AM, Doug Miller wrote:
> [email protected] (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-to-Draw-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 4/15/2013 12:20 PM, Leon wrote:
> On 4/15/2013 11:10 AM, 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.
>>
>> On Monday, April 15, 2013 10:14:29 AM UTC-5, 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
>>
>
> !8' string with a pencil died around one end. Nail in the ground on the
> other end.
>
> Or use sketchup to print out a template.
An 18 foot radius template?
On 4/16/2013 6:07 AM, Doug Miller wrote:
> "dadiOH" <[email protected]> wrote in news:[email protected]:
>
>> Bill wrote:
>>> Leon wrote:
>>
>>>> Did he say wanted the full circle or the 5' arc?
>>>
>>> He said he wanted the 5' Chord!
>>
>>
>> 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.
>>
> Bill's point is that the 5' dimension is the measurement of the chord, not the arc.
>
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.
On 4/15/2013 11:10 AM, 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.
>
> On Monday, April 15, 2013 10:14:29 AM UTC-5, 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
>
!8' string with a pencil died around one end. Nail in the ground on the
other end.
Or use sketchup to print out a template.
On 4/15/2013 12:47 AM, Richard wrote:
> On 4/15/2013 12:20 PM, Leon wrote:
>> On 4/15/2013 11:10 AM, 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.
>>>
>>> On Monday, April 15, 2013 10:14:29 AM UTC-5, 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
>>>
>>
>> !8' string with a pencil died around one end. Nail in the ground on the
>> other end.
>>
>> Or use sketchup to print out a template.
>
>
>
> An 18 foot radius template?
Did he say wanted the full circle or the 5' arc?
"Greg Guarino" wrote in message news:[email protected]...
>On 4/15/2013 3:26 PM, Bill wrote:
>> Greg, We both drew the same picture. How much math background do you
>> have (if you don't mind me asking)?
>
>> Bill
>
>Nothing too advanced. Algebra, geometry and trig in high school, a
>little calculus in college. That would all have been in the '70s.
Here's where my kids would ask "1870s or 1970s?" ;~)
John
...of about the same vintage
"Gramp's shop" <[email protected]> wrote in news:384921f5-292d-40cc-9e3e-
[email protected]:
> 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?
224 inches
>
> There ought to be an equation for this that would be far superior to trial and error, oui?
There is.
radius squared = (radius minus height) squared + (half the distance between endpoints)
squared
In this case:
r^2 = (r - 2)^2 + 30^2
r^2 = r^2 -4r +4 + 900
4r = 904
r = 226
"dadiOH" <[email protected]> wrote in news:[email protected]:
> Bill wrote:
>> Leon wrote:
>
>>> Did he say wanted the full circle or the 5' arc?
>>
>> He said he wanted the 5' Chord!
>
>
> 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.
>
Bill's point is that the 5' dimension is the measurement of the chord, not the arc.
Leon <lcb11211@swbelldotnet> wrote in
news:[email protected]:
> 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.
[email protected] (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-to-Draw-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" <[email protected]> wrote in news:516e0ddd$0$26833$c3e8da3
[email protected]:
>
> "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-to-Draw-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 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.
On 4/15/13 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
>
According to those cool calculators, 18'10".
--
-MIKE-
"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com
[email protected]
---remove "DOT" ^^^^ to reply
On Mon, 15 Apr 2013 09:10:54 -0700, 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.
Bend a piece of thin cutoff around the points?
--
When fascism comes to America, it will be wrapped in the flag and
carrying a cross.
On 4/15/2013 11:38 AM, Leon wrote:
> 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.
Trigonometry concurs. :)
http://www.flickr.com/photos/gdguarino/8651887015/in/photostream/lightbox/
Thanks to the O.P. for a pleasant lunchtime puzzle.
On 4/15/2013 12:43 PM, Larry Blanchard wrote:
> On Mon, 15 Apr 2013 09:10:54 -0700, 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.
>
> Bend a piece of thin cutoff around the points?
>
My sense of the physics involved tells me that that method will not
produce an arc of a circle. Will it matter? Might the resultant curve be
subtly nicer than a circular arc? That's up to the designer.
On 4/15/2013 1:41 PM, Greg Guarino wrote:
> On 4/15/2013 11:38 AM, Leon wrote:
>> 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.
>
> Trigonometry concurs. :)
>
> http://www.flickr.com/photos/gdguarino/8651887015/in/photostream/lightbox/
>
> Thanks to the O.P. for a pleasant lunchtime puzzle.
It was more fun to try to work it out for myself, but I just looked it
up and there is (of course) a more direct way to find it.
Radius = H/2 + W^2/8H
Where H = the height of the arc (2")
and W= the width of the base (60")
In our example that's:
2/2 + 60^2/8(2)= 1 + 3600/16 = 1 + 225 = 226
On 04/15/2013 01:47 PM, Doug Miller wrote:
> "Gramp's shop" <[email protected]> wrote in news:384921f5-292d-40cc-9e3e-
> [email protected]:
>
>> 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?
>
> 224 inches
>>
>> There ought to be an equation for this that would be far superior to trial and error, oui?
>
> There is.
>
> radius squared = (radius minus height) squared + (half the distance between endpoints)
> squared
>
> In this case:
>
> r^2 = (r - 2)^2 + 30^2
> r^2 = r^2 -4r +4 + 900
> 4r = 904
> r = 226
>
Now we know how long a piece of string is!
On 4/15/2013 2:47 PM, Doug Miller wrote:
> "Gramp's shop" <[email protected]> wrote in news:384921f5-292d-40cc-9e3e-
> [email protected]:
>
>> 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?
>
> 224 inches
>>
>> There ought to be an equation for this that would be far superior to trial and error, oui?
>
> There is.
>
> radius squared = (radius minus height) squared + (half the distance between endpoints)
> squared
>
> In this case:
>
> r^2 = (r - 2)^2 + 30^2
> r^2 = r^2 -4r +4 + 900
> 4r = 904
> r = 226
>
I just drew that out. Clever. It's much more elegant solution than the
one I posted. Good work.
Greg Guarino wrote:
> On 4/15/2013 11:38 AM, Leon wrote:
>> 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.
>
> Trigonometry concurs. :)
>
> http://www.flickr.com/photos/gdguarino/8651887015/in/photostream/lightbox/
>
>
> Thanks to the O.P. for a pleasant lunchtime puzzle.
Greg, We both drew the same picture. How much math background do you
have (if you don't mind me asking)?
Bill
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
Pythagorean Theorem: 30^2 + (r-2)^2 = r^2.
Solution is 226" exactly, which is 18' 10", as hasalready been
disclosed,I believe.
Didn't even need trig. (which surprised me).
On 4/15/2013 3:26 PM, Bill wrote:
> Greg Guarino wrote:
>> On 4/15/2013 11:38 AM, Leon wrote:
>>> 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.
>>
>> Trigonometry concurs. :)
>>
>> http://www.flickr.com/photos/gdguarino/8651887015/in/photostream/lightbox/
>>
>>
>> Thanks to the O.P. for a pleasant lunchtime puzzle.
>
>
> Greg, We both drew the same picture. How much math background do you
> have (if you don't mind me asking)?
>
> Bill
>
Nothing too advanced. Algebra, geometry and trig in high school, a
little calculus in college. That would all have been in the '70s.
In article <[email protected]>,
Leon <lcb11211@swbelldotnet> wrote:
>
>!8' string with a pencil died around one end. Nail in the ground on the
>other end.
String is too stretchy. Wire might work.
>Or use sketchup to print out a template.
That's what I typically do. There are plugins that will do this for you
with registration marks, but I've done it successfully by adding my own
marks and then using the default print dialog.
On one occasion, I simply calculated what the height of the arch
would be at 1' intervals from the endpoints and made marks on
a sheet of plywood.
http://www.efalk.org/Vardo/bigs/vardo082.jpg.html
You can't see the marks themselves, but there's a block of
wood screwed to the plywood at each mark.
--
-Ed Falk, [email protected]
http://thespamdiaries.blogspot.com/
In article <[email protected]>,
Richard <[email protected]> wrote:
>
>An 18 foot radius template?
Sure. You don't have to print the whole 5'x18' template
on paper, just the 5'x2" part you need to lay out your
arc.
This rib sections have a 104" radius of curvature, but I
only needed to print enough paper to cover their actual
span. It was like four sheets of paper.
http://www.efalk.org/Vardo/bigs/vardo087.jpg.html
--
-Ed Falk, [email protected]
http://thespamdiaries.blogspot.com/
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?
Trial and error isn't needed, neither is knowing the radius of the arc's
circle.
Put two nails 5' apart in a piece of ply. Put another nail 2" above the
line formed by the first two. Take a batten, bend it between the nails and
draw the arc.
--
dadiOH
____________________________
Winters getting colder? Tired of the rat race?
Taxes out of hand? Maybe just ready for a change?
Check it out... http://www.floridaloghouse.net
On 4/15/13 4:32 PM, dadiOH wrote:
> 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?
>
> Trial and error isn't needed, neither is knowing the radius of the arc's
> circle.
>
> Put two nails 5' apart in a piece of ply. Put another nail 2" above the
> line formed by the first two. Take a batten, bend it between the nails and
> draw the arc.
>
The potential flaws I see in that method are...
...you might get a peak/angle in the curve at the center nail
...you can't always count on getting equal bending at every point along
the length of a piece of wood.
I know it's a standard method to use a long, flexible piece and
something to mark out a curve so I'm certain it works. I would just want
to double check and practice a few times to make sure the bendable thing
was bending equally.
--
-MIKE-
"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com
[email protected]
---remove "DOT" ^^^^ to reply
"-MIKE-" <[email protected]> wrote in message
news:[email protected]...
> On 4/15/13 4:32 PM, dadiOH wrote:
>> 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?
>>
>> Trial and error isn't needed, neither is
>> knowing the radius of the arc's
>> circle.
>>
>> Put two nails 5' apart in a piece of ply. Put
>> another nail 2" above the
>> line formed by the first two. Take a batten,
>> bend it between the nails and
>> draw the arc.
>>
>
> The potential flaws I see in that method are...
> ...you might get a peak/angle in the curve at
> the center nail
> ...you can't always count on getting equal
> bending at every point along the length of a
> piece of wood.
>
> I know it's a standard method to use a long,
> flexible piece and
> something to mark out a curve so I'm certain it
> works. I would just want
> to double check and practice a few times to make
> sure the bendable thing
> was bending equally.
> -MIKE-
>
Perhaps the curve could be checked by flipping the
layout
stick end for end?
Leon wrote:
> On 4/15/2013 12:47 AM, Richard wrote:
>> On 4/15/2013 12:20 PM, Leon wrote:
>>> On 4/15/2013 11:10 AM, 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.
>>>>
>>>> On Monday, April 15, 2013 10:14:29 AM UTC-5, 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
>>>>
>>>
>>> !8' string with a pencil died around one end. Nail in the ground on the
>>> other end.
>>>
>>> Or use sketchup to print out a template.
>>
>>
>>
>> An 18 foot radius template?
>
>
> Did he say wanted the full circle or the 5' arc?
He said he wanted the 5' Chord!
On 4/15/13 7:01 PM, Phil Kangas wrote:
>>> Put two nails 5' apart in a piece of ply. Put
>>> another nail 2" above the
>>> line formed by the first two. Take a batten,
>>> bend it between the nails and
>>> draw the arc.
>>>
>>
>> The potential flaws I see in that method are...
>> ...you might get a peak/angle in the curve at
>> the center nail
>> ...you can't always count on getting equal
>> bending at every point along the length of a
>> piece of wood.
>>
>> I know it's a standard method to use a long,
>> flexible piece and
>> something to mark out a curve so I'm certain it
>> works. I would just want
>> to double check and practice a few times to make
>> sure the bendable thing
>> was bending equally.
>> -MIKE-
>>
>
> Perhaps the curve could be checked by flipping the
> layout
> stick end for end?
>
Good advice.
--
-MIKE-
"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com
[email protected]
---remove "DOT" ^^^^ to reply
"Gramp's shop" <[email protected]> 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
No trig or anything real advanced needed here. Merely that the square
of the hypotenuse of a right triangle = the sum of the squares of the
other two sides
Picture your arc, and a line between the ends of the arc. Now draw
another line from the center of your circle to the midpoint of the
arc, and a third line from the circle center to one of the endpoints.
Now if you have a circle of radius R, you have just drawn a right
triangle with hypotenuse R, one side of R-2, and the other side of 30
(converting the 5-ft width to inches and dividing by 2.
30^2 + (R-2)^2 = R^2
900 + R^2 -4R + 4 = R^2
904 = 4R
R=226"
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
Bill wrote:
> Leon wrote:
>> Did he say wanted the full circle or the 5' arc?
>
> He said he wanted the 5' Chord!
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
____________________________
Winters getting colder? Tired of the rat race?
Taxes out of hand? Maybe just ready for a change?
Check it out... http://www.floridaloghouse.net
On 4/15/2013 8:26 PM, Leon wrote:
> On 4/15/2013 12:47 AM, Richard wrote:
>> On 4/15/2013 12:20 PM, Leon wrote:
>>> On 4/15/2013 11:10 AM, 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.
>>>>
>>>> On Monday, April 15, 2013 10:14:29 AM UTC-5, 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
>>>>
>>>
>>> !8' string with a pencil died around one end. Nail in the ground on the
>>> other end.
>>>
>>> Or use sketchup to print out a template.
>>
>>
>>
>> An 18 foot radius template?
>
>
> Did he say wanted the full circle or the 5' arc?
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. :)
On 4/16/2013 10:05 AM, Leon wrote:
> On 4/16/2013 6:07 AM, Doug Miller wrote:
>> "dadiOH" <[email protected]> wrote in news:[email protected]:
>>
>>> Bill wrote:
>>>> Leon wrote:
>>>
>>>>> Did he say wanted the full circle or the 5' arc?
>>>>
>>>> He said he wanted the 5' Chord!
>>>
>>>
>>> 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.
>>>
>> Bill's point is that the 5' dimension is the measurement of the chord,
>> not the arc.
>>
>
>
> 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
On 4/16/2013 9:16 AM, Bill wrote:
> On 4/16/2013 10:05 AM, Leon wrote:
...
>> ... 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. : )
<http://www.finewoodworking.com/how-to/articles/easier-joinery-for-curved-drawer-fronts.aspx>
Note picture UL 2nd page... :)
--
Doug Miller wrote:
> "dadiOH" <[email protected]> wrote in
> news:[email protected]:
>
>> Bill wrote:
>>> Leon wrote:
>>
>>>> Did he say wanted the full circle or the 5' arc?
>>>
>>> He said he wanted the 5' Chord!
>>
>>
>> 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.
>>
> Bill's point is that the 5' dimension is the measurement of the
> chord, not the arc.
Ah, OK, got it. Mea culpa.
--
dadiOH
____________________________
Winters getting colder? Tired of the rat race?
Taxes out of hand? Maybe just ready for a change?
Check it out... http://www.floridaloghouse.net
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)
>
>
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
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
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
[email protected]
---remove "DOT" ^^^^ to reply
On 4/17/2013 9:24 AM, Leon wrote:
> On 4/17/2013 7:18 AM, Doug Miller wrote:
>> [email protected] (Larry W) wrote in news:kkl297$pbb$1
>> @speranza.aioe.org:
>>
>>> I've seen the technique described elsewhere, but this was the first
>>> google result on searching for "how to draw an arc of large radius"
>>>
>>> Look at step 1 on this site:
>>>
>>> http://www.instructables.com/id/How-to-Draw-Large-Curves/
>>
>> I don't immediately see how to prove or disprove it... but my first
>> impression is that that method
>> draws an elliptical arc, not a circular arc.
>>
>
>
> I agree, the material used would have to be uniform in stiffness from
> one end to the other and you would have to have pivot points beyond the
> start and stop points at the same intervals AND pivot points would have
> to be precisely placed to obtain a circular arc.
I'm not sure either way. If your goal was to draw a full semicircle, two
pins and a framing square (or anything that forms a right angle) would
indeed (theoretically) draw a semicircle. This is because any triangle
formed by two opposite ends of a diameter and a third point anywhere on
the circumference is a right triangle.
The question then is as follows: Can we generalize that to say that the
angle at the apex of a triangle formed by two points on opposite ends of
a (non-diameter) chord and a third point on the circumference between
them also remains fixed at some angle. If so, the method should work. I
may have to look this up. Geometry was a long time ago and circles may
have changed since then.
On 4/17/2013 10:12 AM, Greg Guarino wrote:
> On 4/17/2013 9:24 AM, Leon wrote:
>> On 4/17/2013 7:18 AM, Doug Miller wrote:
>>> [email protected] (Larry W) wrote in news:kkl297$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-to-Draw-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-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.
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:
>>>>> [email protected] (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-to-Draw-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-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.
>>
>>
> 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:
>>>>> [email protected] (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-to-Draw-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-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.
>>
>>
>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.
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
In article <[email protected]>,
Greg Guarino <[email protected]> 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.
--
-Ed Falk, [email protected]
http://thespamdiaries.blogspot.com/
On 4/18/2013 1:03 PM, Edward A. Falk wrote:
> In article <[email protected]>,
> Greg Guarino <[email protected]> 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.
:)
"Larry Blanchard" wrote in message news:[email protected]...
>On Mon, 15 Apr 2013 09:10:54 -0700, 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.
>Bend a piece of thin cutoff around the points?
+1
"Leon" wrote in message
news:[email protected]...
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.
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My calculator says that your program is right.