Burt wrote:

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

perimeter = pi * diameter = 3.1415926 * 3 = 9.42478 ft

Or, just measure around and divide by 10...

Burt wrote:

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

perimeter = pi * diameter = 3.1415926 * 3 = 9.42478 ft

Or, just measure around and divide by 10...

Burt (in ) said:

| I can't remember the formula for the life of me.
| If a dish is almost 3 ft across and I want to segment it like an
| orange into 10 segments how do I calculate how wide each will be at
| the rim?
| So I end up with a dish that has 10 sides.
|
| I'm math clueless.

Burt...

Each of the sides will be 36" * sin(360 degrees / 10) or approximately
21.16"

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

Try these sites, hope something here helps.

http://www.verifiedsoftware.com/goodturns/plans.htm
http://www.delorie.com/wood/segturn.html

"Burt" wrote in message
...

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange
into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

Each segment will be 11.125" wide.

"Burt" wrote in message
...
On Sun, 26 Jun 2005 20:20:38 -0500, Duane Bozarth
wrote:

Burt wrote:

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange
into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

perimeter = pi * diameter = 3.1415926 * 3 = 9.42478 ft

Or, just measure around and divide by 10...

Thanks. I can't measure around because it ain't made yet.g
I need to cut ten pieces of steel to form a ten sided form that will fit
exactly
inside a 3 foot circle. I need the distance between the points as a
straight
line. so if it section is shaped like a bow I need the length of the
string.
Does this make any sense?

"Burt" wrote in message
...

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange
into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

The arc length = (diameter*pi)/10
The chord length = (sin 18 degrees)*diameter
which is .309016994*diameter

For other circle splitting, the arc length is (diameter*pi)/ # of equal
segments.
The chord length is (sin 180/# of equal segments)*diameter.

For the 3' example the arc length is .942477796... feet or 11.30973355...
inches. The chord lengths are .92705098... feet and 11.1246118... inches.
The units and accuracy you require are up to you.

I hope I got all the parentheses right.

TW

"Nate Weber" wrote in message
...
Correction:

4. The triangle has a angle of 18 degrees and hypotenuse of 18"
5. Sin (18) * 18 = 5.56" which is the opposite side of the triangle and
1/2 the point to point distance.
6. 5.56 * 2 = 11.12"

LOL... well at least you kept at until you got it right. I used AutoCAD
for the answer. :~)

Burt wrote:
I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.


If you need the straight line distance between the points here is how I
would solve it.

1. Find the Radius, 36" /2 = 18"
2. Find the angle of the wedge, 360 degrees / 10 segments = 36 degrees
3. Divide that wedge in half, using the resulting triangle you can find
1/2 the point to point distance.

4. The triangle has a angle of 18 degrees and adjacent side of 18"
5. Tan (18) * 18 = 5.85" which is the opposite side of the triangle and
1/2 the point to point distance.
6. 5.85 * 2 = 11.70"

Nate

--
Http://www.Weber-Automation.net:8000

Nate Weber wrote:



If you need the straight line distance between the points here is how I
would solve it.

1. Find the Radius, 36" /2 = 18"
2. Find the angle of the wedge, 360 degrees / 10 segments = 36 degrees
3. Divide that wedge in half, using the resulting triangle you can find
1/2 the point to point distance.

4. The triangle has a angle of 18 degrees and adjacent side of 18"
5. Tan (18) * 18 = 5.85" which is the opposite side of the triangle and
1/2 the point to point distance.
6. 5.85 * 2 = 11.70"

Nate


bah to all that, I'm wrong.

Nate

--
Http://www.Weber-Automation.net:8000

Nate Weber wrote:
Burt wrote:

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.



If you need the straight line distance between the points here is how I
would solve it.

1. Find the Radius, 36" /2 = 18"
2. Find the angle of the wedge, 360 degrees / 10 segments = 36 degrees
3. Divide that wedge in half, using the resulting triangle you can find
1/2 the point to point distance.


Correction:

4. The triangle has a angle of 18 degrees and hypotenuse of 18"
5. Sin (18) * 18 = 5.56" which is the opposite side of the triangle and
1/2 the point to point distance.
6. 5.56 * 2 = 11.12"

Nate


--
Http://www.Weber-Automation.net:8000

"CW" wrote in message
ink.net...
Each segment will be 11.125" wide.

I get 11.3 inches.

Burt (in ) said:

| I can't remember the formula for the life of me.

Me too. I've stashed a number of "cheat sheets" on the web so I can
look up what I can't remember.

Now all I have to do is remember that they're at
http://www.iedu.com/DeSoto/CNC/.

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

In article ,
Burt wrote:

I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.


*sigh*

The length of a side of an "n-ogon" inscribed in a circle is:
2*sin(180/n)

If you consider the angle out from the center of the circle, to the ends of
the section (which is called the 'chord') it's easily remembered as:
"twice the sine of half the angle".

How to confuse people -- note that you scale the above by the radius of
the circle. *BUT* there is that little '2x' factor sitting in front of
things. 2x the radius is the diameter. so you can use
diameter*sin(angle/2)
and seriously confuse the spectators.

*GRIN*

On Sun, 26 Jun 2005 18:17:56 -0700, Burt
wrote:


I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

Pi (3.14) x diameter will give you the circumference. Divide that by
ten, and it's the number you need.

In your example, 3.14 x 36" = 113.04" This is the total length of the
outside edge. If you divide that into ten equal segments, each
outside edge will be 11.3" inches long. That is the length of the
curve.

On Mon, 27 Jun 2005 06:52:16 -0500, "Morris Dovey"
wrote:

Burt (in ) said:

| I can't remember the formula for the life of me.

Me too. I've stashed a number of "cheat sheets" on the web so I can
look up what I can't remember.

Now all I have to do is remember that they're at
http://www.iedu.com/DeSoto/CNC/.

You saved me a little work there- I'm trying to teach one of the guys
at work trig, and that will come in very handy. Any chance anyone has
a link to a printable version of the old-style sine/cosine/tangent
tables? We don't have those umm... "fancy" calculators in the shop

"bw" wrote in message
...

"CW" wrote in message
ink.net...
Each segment will be 11.125" wide.

I get 11.3 inches.

11.3 is 1/10th of the circumference of the circle, not the width of that
segment.

On Sun, 26 Jun 2005 18:17:56 -0700, Burt
wrote:


I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

You've received a lot of replies to sort through. Personally, if i had
your apparent background in math, I'd go for using a CAD program
[recommend DeltaCad as good and very intuitive]. However, it never
hurts to know how and why: Here's one mo

The general formula for the chord length, or side of the polygon, is C
= D*sin(180/N) [angle in degrees] where D is the diameter, and N is
the number of sides.

For N = 10, D = 72", you have C = 72*sin(18)

On the calculator use the following order of keypress:

18
sin
x
72
=

Ans: 22.249...

Now subtract 22

-
22
=

Now change the decimal to 16ths

x
16
=
Ans: 3.98...

That is, about 4-16ths, or 1/4"

So .... 22 1/4", near as dammit is to swearing.

"Prometheus" wrote in message
...
On Sun, 26 Jun 2005 18:17:56 -0700, Burt
wrote:


I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange
into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

Pi (3.14) x diameter will give you the circumference. Divide that by
ten, and it's the number you need.

He does not want the circumference of the circle or the length of arc of
that segment. He wants the total width of that segment which is less than
the length of the arc.

Prometheus (in ) said:

| Any chance
| anyone has a link to a printable version of the old-style
| sine/cosine/tangent tables?

That could represent a /lot/ of bandwidth and either server processing
time or file space if you want full tables at degree, minute, and
seconds. Would you settle for an application that creates the file on
your machine?
If so, how much precision do you want?

You might consider hunting down a copy of Richard S. Burington's
_Handbook_of_Mathematical_Tables_and_Formulas_ (McGraw-Hill). I think
you can still find copies for less than US$10 on-line. It's one of my
most-used shop tools.

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

"Guess who" wrote in message
...

The general formula for the chord length, or side of the polygon, is C
= D*sin(180/N) [angle in degrees] where D is the diameter, and N is
the number of sides.

For N = 10, D = 72", you have C = 72*sin(18)

On the calculator use the following order of keypress:

18
sin
x
72
=

Ans: 22.249...

Now subtract 22

-
22
=

Now change the decimal to 16ths

x
16
=
Ans: 3.98...

That is, about 4-16ths, or 1/4"

So .... 22 1/4", near as dammit is to swearing.


You might want to reread the question. The diameter is 36" not 72".

Robert Bonomi (in ) said:

| *sigh*
|
| The length of a side of an "n-ogon" inscribed in a circle is:
| 2*sin(180/n)

I /almost/ hate to do this to you, but the length of a side of an
"n-gon" inscribed in a circle of radius r is:
2*r*sin(180/n)

| *GRIN*

:-)

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

I can confirm with AutoCAD your findings on the different setups if you like
or can send you a PDF file with drawings using your sizes.


"Burt" wrote in message
news
On Mon, 27 Jun 2005 13:28:41 GMT, "Leon"
wrote:


"bw" wrote in message
...

"CW" wrote in message
ink.net...
Each segment will be 11.125" wide.

I get 11.3 inches.

11.3 is 1/10th of the circumference of the circle, not the width of that
segment.

Thanks for the responses guys. The circles are all less than my example of
36",
the biggest is just over 35 3/4 on the inside.
Some are as small as 12"

I will try some of the formulas and see if I can figure out which one is
easy to
use.
I also need to calculate on 6, 8 and 12 segments.

So .... 22 1/4", near as dammit is to swearing.


You might want to reread the question. The diameter is 36" not 72".
Scales linearly. Divide by 2, get the same result as many others. Hasn't
this been beaten to death yet?

Steve

On Mon, 27 Jun 2005 13:55:42 GMT, "Leon"
wrote:

ut 4-16ths, or 1/4"

So .... 22 1/4", near as dammit is to swearing.


You might want to reread the question. The diameter is 36" not 72".

Sigh. Thanks. I was half awake, I guess, so half right. I had
started to use 36" then decided that was the radius. To the OP: Just
put 36 in place of 72 and use the same method.

On Mon, 27 Jun 2005 08:22:48 -0500, Prometheus
wrote:

You saved me a little work there- I'm trying to teach one of the guys
at work trig, and that will come in very handy. Any chance anyone has
a link to a printable version of the old-style sine/cosine/tangent
tables?

This might be of some help:
http://www.uni.edu/darrow/new/geodes...ings/trig.html

"Burt" wrote in message

I'm sure somewhere there was a term posted for the straight line distance
between two points on the edge of a circle but I'm still unsure what that
term
is.

A "chord", as opposed to the curved part, which is generally called the
"arc".

An old Artilleryman will tell you that one mil of angle will subtend an
"arc" of 1 meter at 1000 meters .. but, as in your case, it is really the
"chord" that is the distance on the ground you're after when adjusting
artillery fire. With the roughly 50 meter effective zone of a HE 105mm
round, the difference between the "chord" and the "arc: is moot ... but you
need a bit more precision than that.

.... I mean, ya gotta put this stuff in perspective with those things of
which you are intimately familiar.

--
www.e-woodshop.net
Last update: 5/14/05

On Sun, 26 Jun 2005 18:17:56 -0700, the opaque Burt
spake:


I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

Buy Charlie Self's new book and figure it out yourself.
Woodworker's Pocket Reference : Everything a Woodworker Needs to Know
at a Glance ($10 + s/h at Amazon.com)


From a chart in Lee Valley's copy of Handyman In-Your-Pocket,
the chord is 0.3090170 times the diameter of the circle, or
11.124612".

On a lighter note, read this:

Teaching Math
-------------

Teaching Math in 1950:
A logger sells a truckload of lumber for $100. His cost of
production is 4/5 of the price. What is his profit?

Teaching Math in 1960:
A logger sells a truckload of lumber for $100. His cost of
production is 4/5 of the price, or $80. What is his profit?

Teaching Math in 1970:
A logger exchanges a set "L" of lumber for a set "M" of money.
the cardinality of set "M" is 100. Each element is worth one dollar.
Make 100 dots representing the elements of the set "M". The set "C",
the cost of production contains 20 fewer points than set "M."
Represent
the set "C" as a subset of set "M" and answer the following question:
What is the cardinality of the set "P" for profits?

Teaching Math in 1980:
A logger sells a truckload of lumber for $100. Her cost of
production is $80 and her profit is $20.
Your assignment: Underline the number 20.

Teaching Math in 1990:
By cutting down beautiful forest trees, the logger makes $20.
that do you think of this way of making a living?
Topic for class participation after answering the question: How
did the forest birds and squirrels feel as the logger cut down the
trees? There are no wrong answers.

Teaching Math in 1996:
By laying off 40% of its loggers, a company improves its stock
price from $80 to $100. How much capital gain per share does the CEO
make by exercising his stock options at $80? Assume capital gains are
no
longer taxed, because this encourages investment.

Teaching Math in 1997:
A company outsources all of its loggers. The firm saves on
benefits, and when demand for its product is down, the logging work
force can easily be cut back. The average logger employed by the
company
earned $50,000, had three weeks vacation, a nice retirement plan and
medical insurance. The contracted logger charges $50 an hour. Was
outsourcing a good move?

Teaching Math in 1998:
A laid-off logger with four kids at home and a ridiculous
alimony from his first failed marriage comes into the logging-company
corporate offices and goes postal, mowing down 16 executives and a
couple of secretaries, and gets lucky when he nails a politician on
the
premises collecting his kickback. Was outsourcing the loggers a good
move for the company?

Teaching Math in 1999:
A laid-off logger serving time in Folsom for blowing away
several people is being trained as a COBOL programmer in order to work
on Y2K projects. What is the probability that the automatic cell doors
will open on their own as of 00:01, 01/01/2000?

-
DANCING: The vertical frustration of a horizontal desire.
---------------------------------------------------------
http://diversify.com Full Service Web Programming

LOL. Too much truth

Steve

"Larry Jaques" wrote in message
...
On Sun, 26 Jun 2005 18:17:56 -0700, the opaque Burt
spake:


I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange
into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

Buy Charlie Self's new book and figure it out yourself.
Woodworker's Pocket Reference : Everything a Woodworker Needs to Know
at a Glance ($10 + s/h at Amazon.com)


From a chart in Lee Valley's copy of Handyman In-Your-Pocket,
the chord is 0.3090170 times the diameter of the circle, or
11.124612".

On a lighter note, read this:

Teaching Math
-------------

Teaching Math in 1950:
A logger sells a truckload of lumber for $100. His cost of
production is 4/5 of the price. What is his profit?

Teaching Math in 1960:
A logger sells a truckload of lumber for $100. His cost of
production is 4/5 of the price, or $80. What is his profit?

Teaching Math in 1970:
A logger exchanges a set "L" of lumber for a set "M" of money.
the cardinality of set "M" is 100. Each element is worth one dollar.
Make 100 dots representing the elements of the set "M". The set "C",
the cost of production contains 20 fewer points than set "M."
Represent
the set "C" as a subset of set "M" and answer the following question:
What is the cardinality of the set "P" for profits?

Teaching Math in 1980:
A logger sells a truckload of lumber for $100. Her cost of
production is $80 and her profit is $20.
Your assignment: Underline the number 20.

Teaching Math in 1990:
By cutting down beautiful forest trees, the logger makes $20.
that do you think of this way of making a living?
Topic for class participation after answering the question: How
did the forest birds and squirrels feel as the logger cut down the
trees? There are no wrong answers.

Teaching Math in 1996:
By laying off 40% of its loggers, a company improves its stock
price from $80 to $100. How much capital gain per share does the CEO
make by exercising his stock options at $80? Assume capital gains are
no
longer taxed, because this encourages investment.

Teaching Math in 1997:
A company outsources all of its loggers. The firm saves on
benefits, and when demand for its product is down, the logging work
force can easily be cut back. The average logger employed by the
company
earned $50,000, had three weeks vacation, a nice retirement plan and
medical insurance. The contracted logger charges $50 an hour. Was
outsourcing a good move?

Teaching Math in 1998:
A laid-off logger with four kids at home and a ridiculous
alimony from his first failed marriage comes into the logging-company
corporate offices and goes postal, mowing down 16 executives and a
couple of secretaries, and gets lucky when he nails a politician on
the
premises collecting his kickback. Was outsourcing the loggers a good
move for the company?

Teaching Math in 1999:
A laid-off logger serving time in Folsom for blowing away
several people is being trained as a COBOL programmer in order to work
on Y2K projects. What is the probability that the automatic cell doors
will open on their own as of 00:01, 01/01/2000?

-
DANCING: The vertical frustration of a horizontal desire.
---------------------------------------------------------
http://diversify.com Full Service Web Programming

"Guess who" wrote in message
...
Sigh. Thanks. I was half awake, I guess, so half right. I had
started to use 36" then decided that was the radius. To the OP: Just
put 36 in place of 72 and use the same method.



LOL. I thought you may have done what I have been known to do and read the
3' diameter as the radius.

"Swingman" wrote in message
...
A "chord", as opposed to the curved part, which is generally called the
"arc".

An old Artilleryman will tell you that one mil of angle will subtend an
"arc" of 1 meter at 1000 meters .. but, as in your case, it is really the
"chord" that is the distance on the ground you're after when adjusting
artillery fire. With the roughly 50 meter effective zone of a HE 105mm
round, the difference between the "chord" and the "arc: is moot ... but
you
need a bit more precision than that.

... I mean, ya gotta put this stuff in perspective with those things of
which you are intimately familiar.


And if the gun was on a ship....... LOL. I don't think I will ever forget
the length of the equation we used in Physics when determining when to pull
the trigger and when will it hit if the seas were rough and the ship was
traveling.

On Mon, 27 Jun 2005 09:49:46 -0700, "Burt"
wrote:

thanks.
I'll try to digest some of this.
The problem for me is the terminology of math.
Financial formulas I can use but I have no clue what the trig terms mean.

Don't worry about it. Aside from a stupid error reading the original
post, I usually do math like you can breathe, but put a dollar sign in
front, and my eyes glaze over.

If you want to *understand* trig, you should first study ratio and
proportion and then similarity, triangles being the simplest example.
Then move into trig ratios, involving sides and angles of a special
triangle, the right triangle.

Keep an eye on alt.binaries.pictures.woodworking your quick formula will =
show up there as soon as I have it written.

--=20
PDQ

"Burt" wrote in message =
...
| On Sun, 26 Jun 2005 20:20:38 -0500, Duane Bozarth =

| wrote:
|=20
| Burt wrote:
| =20
| I can't remember the formula for the life of me.
| If a dish is almost 3 ft across and I want to segment it like an =
orange into 10
| segments how do I calculate how wide each will be at the rim?
| So I end up with a dish that has 10 sides.
| =20
| I'm math clueless.
|
| perimeter =3D pi * diameter =3D 3.1415926 * 3 =3D 9.42478 ft
|
| Or, just measure around and divide by 10...
|=20
| Thanks. I can't measure around because it ain't made yet.g
| I need to cut ten pieces of steel to form a ten sided form that will =
fit exactly
| inside a 3 foot circle. I need the distance between the points as a =
straight
| line. so if it section is shaped like a bow I need the length of the =
string.
| Does this make any sense?

In article ,
Morris Dovey wrote:
Robert Bonomi (in ) said:

| *sigh*
|
| The length of a side of an "n-ogon" inscribed in a circle is:
| 2*sin(180/n)

I /almost/ hate to do this to you, but the length of a side of an
"n-gon" inscribed in a circle of radius r is:
2*r*sin(180/n)

The dimension of a circle is *always* "1", when the unit of measure is a
"radius", and thus it drops out of the formula. *GRIN*

On Mon, 27 Jun 2005 21:17:05 -0000,
(Robert Bonomi) wrote:

|
| The length of a side of an "n-ogon" inscribed in a circle is:
| 2*sin(180/n)

I /almost/ hate to do this to you, but the length of a side of an
"n-gon" inscribed in a circle of radius r is:
2*r*sin(180/n)

The dimension of a circle is *always* "1", when the unit of measure is a
"radius", and thus it drops out of the formula. *GRIN*

What do you mean by the "dimension of a circle"? A circle of radius
1[arbitrary] unit is called the "unit circle". Dropping it here would
give the distance of a hord in the unit circle. Then you could use
similarity for the one in question ...a fancier way of saying multiply
by 36. I think I got it right this time. :-)

Can't help that. 11.1246 is the correct number.

"bw" wrote in message
...
I get 11.3 inches.

A good calculator will run about $14.00 these days. Much handier (and
faster) than those tables.
"Prometheus" wrote in message
...

You saved me a little work there- I'm trying to teach one of the guys
at work trig, and that will come in very handy. Any chance anyone has
a link to a printable version of the old-style sine/cosine/tangent
tables? We don't have those umm... "fancy" calculators in the shop

On Mon, 27 Jun 2005 13:32:38 GMT, "Leon"
wrote:


"Prometheus" wrote in message
.. .
On Sun, 26 Jun 2005 18:17:56 -0700, Burt
wrote:


I can't remember the formula for the life of me.
If a dish is almost 3 ft across and I want to segment it like an orange
into 10
segments how do I calculate how wide each will be at the rim?
So I end up with a dish that has 10 sides.

I'm math clueless.

Pi (3.14) x diameter will give you the circumference. Divide that by
ten, and it's the number you need.

He does not want the circumference of the circle or the length of arc of
that segment. He wants the total width of that segment which is less than
the length of the arc.

That became apparent a little later in the thread. I was going to
throw the chord length in there, but I don't have a scientific
calculator handy, and I was too lazy to dig around on the net for the
value of sin(18)!

Of course, others nailed it, so there is no point in my posting it as
well.

On Mon, 27 Jun 2005 08:50:35 -0500, "Morris Dovey"
wrote:

Prometheus (in ) said:

| Any chance
| anyone has a link to a printable version of the old-style
| sine/cosine/tangent tables?

That could represent a /lot/ of bandwidth and either server processing
time or file space if you want full tables at degree, minute, and
seconds. Would you settle for an application that creates the file on
your machine?
If so, how much precision do you want?

Whew, just the whole degrees! I'm not doing global surveys or
anything... Actually, after posting this, I found one online and
posted it into an excel spreadsheet. It's just a quick reference to
hang on the bandsaw at work, while I help a guy learn trig. (He's
pretty sharp, but never took it in school) We just double check the
prints to make sure the first piece in a run doesn't end up as scrap.
Sometimes the engineers mess up the short length on a print with
angled sides- kinda hard to do with AutoCAD, I'd think, but some of
them manage to do it anyhow.

You might consider hunting down a copy of Richard S. Burington's
_Handbook_of_Mathematical_Tables_and_Formulas_ (McGraw-Hill). I think
you can still find copies for less than US$10 on-line. It's one of my
most-used shop tools.

I'll keep an eye out for it! I had a lot of math in HS and college,
but it's been a while now, and it's all getting a little fuzzy. A
reference tool probably wouldn't hurt...

On Mon, 27 Jun 2005 12:12:14 -0400, Guess who
wrote:

On Mon, 27 Jun 2005 08:22:48 -0500, Prometheus
wrote:

You saved me a little work there- I'm trying to teach one of the guys
at work trig, and that will come in very handy. Any chance anyone has
a link to a printable version of the old-style sine/cosine/tangent
tables?

This might be of some help:
http://www.uni.edu/darrow/new/geodes...ings/trig.html

I think it may, thanks! Saves me a whole lot of typing, at the very
least.

On Tue, 28 Jun 2005 01:24:35 GMT, "CW" wrote:

A good calculator will run about $14.00 these days. Much handier (and
faster) than those tables.

Hmm... yeah. But then what happens when you don't have a calculator
handy? People like to walk off with the $2 ones as it is! Nobody
wants to steal a trig table, though if they do, you can print another
off for almost nothing.

That, and dependancy on calculators seems to interfere with people
fully understanding the math. If it's important to do a thousand
calculations very quickly, then it's important to have automated help.
But I'm talking about one or two problems a day- so the time savings
is trivial.