I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long way
removed from Mr. Cole's high school trig class.

tia

jc

Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long way
removed from Mr. Cole's high school trig class.

Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection
above the chord subtending a length of 1-5/16"?

--

"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long
way removed from Mr. Cole's high school trig class.

Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection above
the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe

Joe wrote:
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long
way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection above
the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

That's easier than a couple of alternatives...

No real way for ASCII art for the circle, so I'll try just the algebra
from a verbal description. Consider the triangle of the radius
perpendicular to the chord and another radius to one end of the chord.
That's a right triangle whose hypotenuse is R, one leg is your desired
L/2 and the other side is the radius R-1/8.

So, letting x = L/2, r = R and h the projection desired, Pythagorus says

x^2 + (r-h)^2 = r^2

x^2 + r^2 - 2rh + h^2 = r^2

x^2 - 2rh + h^2 = 0

r = (x^2 - h^2)/2h

Substituting in your values I get r = 6.828 -- 6-53/64", approx.

--

"Joe" wrote in message
. net...

"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm a
long way removed from Mr. Cole's high school trig class.

Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection above
the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe

The diameter would be 3.570". Thank God for AutoCAD.

todd

On Oct 8, 11:48 am, "Joe" wrote:
"dpb" wrote in ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long
way removed from Mr. Cole's high school trig class.

Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection above
the chord subtending a length of 1-5/16"?

--

There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe

Hi Joe,

Radius: 1.7852"
Diameter 3.5704"

Ed Bennett


http://www.ts-aligner.com
Home of the TS-Aligner

todd wrote:
"Joe" wrote in message
. net...
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm a
long way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection above
the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe

The diameter would be 3.570". Thank God for AutoCAD.

Dang! I'll have to check the calculator, then...

Oh, I forgot to divide the chord length...but I get 3.32", not 3.57"

Wonder where the difference is coming from?

--

"dpb" wrote in message ...
todd wrote:
"Joe" wrote in message
. net...
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm
a long way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection
above the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe

The diameter would be 3.570". Thank God for AutoCAD.

Dang! I'll have to check the calculator, then...

Oh, I forgot to divide the chord length...but I get 3.32", not 3.57"

Wonder where the difference is coming from?

Well, the last bit should be r=(x^2 + h^2)/2h. Plug those numbers in and it
works out.

todd

"dpb" wrote in message ...
Joe wrote:
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm a
long way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection
above the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

That's easier than a couple of alternatives...

No real way for ASCII art for the circle, so I'll try just the algebra
from a verbal description. Consider the triangle of the radius
perpendicular to the chord and another radius to one end of the chord.
That's a right triangle whose hypotenuse is R, one leg is your desired L/2
and the other side is the radius R-1/8.

So, letting x = L/2, r = R and h the projection desired, Pythagorus says

x^2 + (r-h)^2 = r^2

x^2 + r^2 - 2rh + h^2 = r^2

x^2 - 2rh + h^2 = 0

r = (x^2 - h^2)/2h

Substituting in your values I get r = 6.828 -- 6-53/64", approx.

--

Don't have AutoCad; so did about the same calculation as you did.
In your last formula, it should be "+h^2". Then the result matches
AutoCad's.
Kerry

todd wrote:
"dpb" wrote in message ...
todd wrote:
"Joe" wrote in message
. net...
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm
a long way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8" projection
above the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe
The diameter would be 3.570". Thank God for AutoCAD.
Dang! I'll have to check the calculator, then...

Oh, I forgot to divide the chord length...but I get 3.32", not 3.57"

Wonder where the difference is coming from?

Well, the last bit should be r=(x^2 + h^2)/2h. Plug those numbers in and it
works out.

Yeah, I saw I didn't transpose the one sign when I did the copy and
paste from one line to the next after I too briefly looked at it
earlier...dang bifocals! (Got to blame something, can't be I just made a
misteak! )

--

Thank you everyone for the answer. I have a bending form to make.

be well, work wood

jc

dpb wrote:
todd wrote:
"dpb" wrote in message ...
todd wrote:
"Joe" wrote in message
. net...
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I
need the diameter (or radius, it doesn't matter) based on the
following

The diameter of a circle formed by an arc with a 1/8" apex over a
1 5/16" base length. Yeah, I know some of the terms are wrong,
but I'm a long way removed from Mr. Cole's high school trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8"
projection above the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe
The diameter would be 3.570". Thank God for AutoCAD.
Dang! I'll have to check the calculator, then...

Oh, I forgot to divide the chord length...but I get 3.32", not 3.57"

Wonder where the difference is coming from?

Well, the last bit should be r=(x^2 + h^2)/2h. Plug those numbers in
and it works out.

Yeah, I saw I didn't transpose the one sign when I did the copy and
paste from one line to the next after I too briefly looked at it
earlier...dang bifocals! (Got to blame something, can't be I just made a
misteak! )

--
Following this thread has reminded my why I was a liberal arts major.

"Joe" wrote in
. net:

I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. Yeah, I know some of the terms are wrong, but I'm
a long way removed from Mr. Cole's high school trig class.

tia

jc



It's been a while for me too, but I think a diameter of 3 9/16 is close
to what you want. It seems small to me, but I did it twice and checked,
so like I said it should be close (I got 3.57 inches for the diameter,
which is just a hair over 3 9/16.)

The setup is to draw the chord and radii from each endpoint and one from
the center of the chord - that gives four right triangles, two big ones
with a radius as hypotenuse and two little ones. Solve the small
triangle (you know the two sides, 1/8 and 21/32), then figure out how
the little triangle is related to the larger one with which it shares a
side. I figured out that the central angle of the larger right triangle
is just twice the smaller angle in the little triangle. I got
r sin 2(arctan {(1/8)/(21/32)}) = 21/32
and solved for r. You can check me on this, but I think it's right.

Jim Willemin wrote:
"Joe" wrote in
. net:
....
The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16" base length. ...
....
The setup is to draw the chord and radii from each endpoint and one from
the center of the chord - that gives four right triangles, two big ones
with a radius as hypotenuse and two little ones. Solve the small
triangle (you know the two sides, 1/8 and 21/32), then figure out how
the little triangle is related to the larger one with which it shares a
side. I figured out that the central angle of the larger right triangle
is just twice the smaller angle in the little triangle. I got
r sin 2(arctan {(1/8)/(21/32)}) = 21/32
and solved for r. You can check me on this, but I think it's right.

You got the answer (and I didn't until somebody pointed out the
algebraic error in sign I made, so take this for what that's worth ),
but in this particular case you can solve for the radius directly from
one of the large (right) triangles as two sides are known in terms of
one unknown (the radius) without even having to solve the quadratic as
the quadratic terms end up canceling...

--

"Joe" wrote in message
. net...
I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long way
removed from Mr. Cole's high school trig class.

tia

jc



Drawing that on AutoCAD that comes to a diameter of 3-9/16 rounded to the
nearest 1/16th.

"Joe" wrote:

I need some trig help. I need to create a bending form but I need
the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1
5/16"
base length.

Life is short, the problem is very straight forward using a graphical
camber layout using the info given.

I'm reminded of the time I took the PE exam oh those many years ago.

I'm a young, pimple faced, smart assed engineering guy with the
fastest slide rule on the planet.

Seated at a desk a few rows in front of me was a gray haired man,
perhaps 50, with a small drafting board complete with triangles,
pencils and scales.

This old man was going to try to compete with the fastest slide rule
on the planet using graphical solutions.

Never heard if he passed the exam or not, but he completed the exam
before I did.

The older I get, the more respect I gain for that man and his
approach. He has has probably long since departed.

Lew

"Joe" wrote

I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long way
removed from Mr. Cole's high school trig class.

I think that Mr Cole and his fellow geometricians would have called Joe's
'apex' the 'sagitta'.

Jeff

--
Jeff Gorman, West Yorkshire, UK
email : Username is amgron
ISP is clara.co.uk
www.amgron.clara.net

Joe wrote:
| I need some trig help. I need to create a bending form but I need
| the diameter (or radius, it doesn't matter) based on the following
|
| The diameter of a circle formed by an arc with a 1/8" apex over a 1
| 5/16" base length. Yeah, I know some of the terms are wrong, but
| I'm a long way removed from Mr. Cole's high school trig class.

Sorry to have arrived so late. This gets asked enough that I put up a
web page with diagram and step-by-step algebra at the link below.

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/CNC/cove_geom.html

Lew Hodgett wrote:
I'm a young, pimple faced, smart assed engineering guy with the
fastest slide rule on the planet.

Seated at a desk a few rows in front of me was a gray haired man,
perhaps 50, with a small drafting board complete with triangles,
pencils and scales.

This old man was going to try to compete with the fastest slide rule
on the planet using graphical solutions.

Never heard if he passed the exam or not, but he completed the exam
before I did.

The older I get, the more respect I gain for that man and his
approach. He has has probably long since departed.

Decartes proved, oh so many years ago, that geometry and algebra were
mathematically equivalent - anything doable in one was doable in the other.

Charlie M. 1958 wrote:
dpb wrote:
todd wrote:
"dpb" wrote in message ...
todd wrote:
"Joe" wrote in message
. net...
"dpb" wrote in message ...
Joe wrote:
I need some trig help. I need to create a bending form but I
need the diameter (or radius, it doesn't matter) based on the
following

The diameter of a circle formed by an arc with a 1/8" apex over
a 1 5/16" base length. Yeah, I know some of the terms are
wrong, but I'm a long way removed from Mr. Cole's high school
trig class.
Before I go too far, to be sure...

You mean find a circle whose diameter will yield an 1/8"
projection above the chord subtending a length of 1-5/16"?

--
There's those terms I can't remember!

Yes, that's exactly what I was trying to say.

Joe
The diameter would be 3.570". Thank God for AutoCAD.
Dang! I'll have to check the calculator, then...

Oh, I forgot to divide the chord length...but I get 3.32", not 3.57"

Wonder where the difference is coming from?

Well, the last bit should be r=(x^2 + h^2)/2h. Plug those numbers in
and it works out.

Yeah, I saw I didn't transpose the one sign when I did the copy and
paste from one line to the next after I too briefly looked at it
earlier...dang bifocals! (Got to blame something, can't be I just made
a misteak! )

--
Following this thread has reminded my why I was a liberal arts major.

Well, we were allowed to count any math theory course as a liberal arts
elective...

Of course, that was in the era when a programming language qualified as
the foreign language requirement in a PhD program, too...

--

"HeyBub" wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

But not necessarily with the same degree of ease.

Lew

HeyBub wrote:
Lew Hodgett wrote:
I'm a young, pimple faced, smart assed engineering guy with the
fastest slide rule on the planet.

Seated at a desk a few rows in front of me was a gray haired man,
perhaps 50, with a small drafting board complete with triangles,
pencils and scales.

This old man was going to try to compete with the fastest slide
rule
on the planet using graphical solutions.

Never heard if he passed the exam or not, but he completed the exam
before I did.

The older I get, the more respect I gain for that man and his
approach. He has has probably long since departed.

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.

--
--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.

You are right about "squaring the circle." Can't be done since pi is a
trancendental number. Can't be done with algebra either, unless you simply
express the area as an equation. In other words, it can't be "solved" for a
value.

Possibly. I was drunk the day all that was covered.


--

"HeyBub" wrote in message
...
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.

You are right about "squaring the circle." Can't be done since pi is a
trancendental number. Can't be done with algebra either, unless you simply
express the area as an equation. In other words, it can't be "solved" for
a value.

Possibly. I was drunk the day all that was covered.


--


HeyBub,
Yes, Archimedes trisected an angle, but his method requires putting marks on
a straight edge, which advocates of pure geometry don't allow.
Kerry

In article , "HeyBub" wrote:
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle.

Not *validly*.

You really should keep up.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

"Kerry Montgomery" wrote in message
...

"HeyBub" wrote in message
...
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.

You are right about "squaring the circle." Can't be done since pi is a
trancendental number. Can't be done with algebra either, unless you
simply express the area as an equation. In other words, it can't be
"solved" for a value.

Possibly. I was drunk the day all that was covered.


--


HeyBub,
Yes, Archimedes trisected an angle, but his method requires putting marks
on a straight edge, which advocates of pure geometry don't allow.
Kerry

While you guys were trisecting and trancendentaling (sounds painful), I went
ahead and built my bending form. Came out nicely. Thanks again for the
help.

jc

In article ,
HeyBub wrote:
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.
Ah, but I bet he used more than just a compass and straightedge.
And yes, I know of the marked straightedge method. It works, but purists
don't like it.

You are right about "squaring the circle." Can't be done since pi is a
trancendental number. Can't be done with algebra either, unless you simply
express the area as an equation. In other words, it can't be "solved" for a
value.
Actually, you can square the circle provided you're not limited to
a straight edge and compass.

Method:
1. Make a wheel the size of the circle to be squared.
2. Mark a point on the wheel.
3. Use wheel to measure distance on line equal to circumference.
Finally, use the standard method of computing the square root of
the length of the line segment you got in step 3.

See? It's easy provided you're not restricted to just a compass and straight
edge.

In article , dpb wrote:
todd wrote:

Yeah, I saw I didn't transpose the one sign when I did the copy and
paste from one line to the next after I too briefly looked at it
earlier...dang bifocals! (Got to blame something, can't be I just made a
misteak! )

Todd
It *IS* a well-known fact that -you- don't make mistakes.

When a 'oopsie!' is committed by someone of your gender, it's a mister-stake.

HeyBub wrote:
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in
the
other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.

Yes, he did, but he wasn't using geometry. He was doing something
which might to a layman look like geometry but he was not following
the rules that define geometry.

You might find http://www.jimloy.com/geometry/trisect.htm to be of
interest.

You are right about "squaring the circle." Can't be done since pi is
a
trancendental number. Can't be done with algebra either, unless you
simply express the area as an equation. In other words, it can't be
"solved" for a value.

The equation is the general solution. Particular values are of little
interest in mathematics. Squaring the circle was the first great
triumph of analytic geometry.

Possibly. I was drunk the day all that was covered.

Very likely.


--

--
--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

John Cochran wrote:
In article ,
HeyBub wrote:
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in
the other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.
Ah, but I bet he used more than just a compass and straightedge.
And yes, I know of the marked straightedge method. It works, but
purists don't like it.

You are right about "squaring the circle." Can't be done since pi
is
a trancendental number. Can't be done with algebra either, unless
you simply express the area as an equation. In other words, it
can't
be "solved" for a value.
Actually, you can square the circle provided you're not limited to
a straight edge and compass.

Method:
1. Make a wheel the size of the circle to be squared.
2. Mark a point on the wheel.
3. Use wheel to measure distance on line equal to circumference.
Finally, use the standard method of computing the square root of
the length of the line segment you got in step 3.

See? It's easy provided you're not restricted to just a compass and
straight edge.

And what exactly does that have to do with geometry? Your method may
be perfectly valid mathematically but that doesn't make it "geometry".
And that assumes that it in fact squares the circle, which it does
not--the challenge in squaring the circle is to find a square with the
same area and I don't see how knowing the square root of the
circumference helps you in that endeavor. It gives you s=SQRT(2*pi*r)
and what you need is s=r*SQRT(pi).


--
--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

Kerry Montgomery wrote:
"HeyBub" wrote in message
...
J. Clarke wrote:

Decartes proved, oh so many years ago, that geometry and algebra
were
mathematically equivalent - anything doable in one was doable in
the other.

So figure out the dimensions of a square with the same area as a
circle or trisect an angle geometrically.


Hmm. Archimedes trisected an angle. You really should keep up.

You are right about "squaring the circle." Can't be done since pi
is
a trancendental number. Can't be done with algebra either, unless
you simply express the area as an equation. In other words, it
can't
be "solved" for a value.

Possibly. I was drunk the day all that was covered.


--


HeyBub,
Yes, Archimedes trisected an angle, but his method requires putting
marks on a straight edge, which advocates of pure geometry don't
allow.

"Advocates of pure geometry"? The challenge issued in ancient times
was to square the circle using Euclidean geometry. Euclidean geometry
is a game with certain rules, one of which is no marks on the
straightedge. If as a practical matter you need to square the circule
then there are many ways to do it, however none of them answer the
original challenge.

This isn't a matter of "advocacy". It's a matter of obeying the rules
of the game. If you don't want to obey the rules of the game that's
fine, but then you're playing som other game, not "Euclidean
Geometry".

--
--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

"Robert Bonomi" wrote in message
...
In article , dpb wrote:
todd wrote:

Yeah, I saw I didn't transpose the one sign when I did the copy and
paste from one line to the next after I too briefly looked at it
earlier...dang bifocals! (Got to blame something, can't be I just made a
misteak! )

Todd
It *IS* a well-known fact that -you- don't make mistakes.

When a 'oopsie!' is committed by someone of your gender, it's a
mister-stake.

I'm not sure I'm inferring the intended tone of this post. What I am sure
of, however, is that I did not write the part above that you attributed to
me. That was written by dpb. Clip with care!

todd

On Mon, 08 Oct 2007 17:37:53 GMT, "Joe" wrote:

I need some trig help. I need to create a bending form but I need the
diameter (or radius, it doesn't matter) based on the following

The diameter of a circle formed by an arc with a 1/8" apex over a 1 5/16"
base length. Yeah, I know some of the terms are wrong, but I'm a long way
removed from Mr. Cole's high school trig class.

tia

jc



Where A is the apex (or sagitta) and B is the chord (A = .125 and B =
1.3125 in the question) then the Radius R is given by:

R =(A^2 + (B/2)^2)/(2A)

derived from:

R(1-cosT) = A
RsinT = B/2

where T is the half angle of the arc.


Tom Veatch
Wichita, KS
USA