On Sun, 26 Jun 2005 21:28:11 -0500, "Morris Dovey" <[email protected]> wrote:
>Phil (in [email protected]) said:
>
>| "Morris Dovey" <[email protected]> wrote in message
>| news:[email protected]...
>|| Burt (in [email protected]) 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 / 20) or
>|| approximately 11-1/8"
>||
>|| --
>|| Morris Dovey
>|| DeSoto Solar
>|| DeSoto, Iowa USA
>|| http://www.iedu.com/DeSoto/solar.html
>||
>||
>|
>| If he were building a 10 segment, flat circle you would be right on
>| but he said "dish". I suspect he needs the other sides's dimension
>| as well. More information is necessary to figure it out. How deep
>| is the dish and does it have an elliptical section or is it part of
>| a sphere? What is the dish for exactly? There are myriad
>| possibilities
>| when you say dish so there is no way to give a (complete) correct
>| answer...
>
>Phil...
>
>Burt also wrote:
><< 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. >>
>
>I interpreted this to mean he wanted the straight line distance
>between adjacent points on a circle. Since circles are planar objects,
>I don't think deformations of the 10 segments enter into /this/
>calculation, but I could be wrong ( and frequently am :-)
Morris that is exactly what I need.
Beyond my figuring skills and it has to be exact. :)
Each of the points has to touch the inside of the 3 foot circle.
"Nate Weber" <[email protected]> wrote in message
news:[email protected]...
> 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. :~)
On Wed, 29 Jun 2005 23:19:51 GMT, "Ace" <[email protected]>
wrote:
>Wish I had one of those back when..... However, it occurs to me that the
>real challenge is
>stating the "problem" in the correct mathematical form.
I do see your point, in the context of the OP, you guys have got me
dead in the water- I was referring (out of context) to all the people
I've seen using a calculator for things they should have known right
off the top of their head if they got past the 3rd grade. After
seeing people run around looking for a calculator to work a problem
like 9 x 8 or 10+25+3, I just got to thinking that maybe the little
buggers aren't so good for us after all. Most of the things an
average calculator gets used for are the equivilent of killing a
sparrow with a cannon.
>Just a thought,
>Ace
In article <[email protected]>,
Burt <[email protected]> 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*
In article <[email protected]>,
Morris Dovey <[email protected]> wrote:
>Robert Bonomi (in [email protected]) 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*>
In article <[email protected]>,
Morris Dovey <[email protected]> wrote:
>Prometheus (in [email protected]) said:
>
>| But that's the extreme case- I know a lot of people who can't do
>| long division, and don't care to know how because they have a
>| calculator. But then when they don't have a calculator handy,
>| they're lost. That would indicate to me that they don't understand
>| the math, they just know how to operate a calculator.
>
>Hmm. I studied math through an advanced course in partial differential
>equations - and, when it comes to solving even simple trig problems,
>I'm one of those who're lost without a calculator.
>
>I suppose I could do a little (lot of?) review and calculate my own
>trig function values:
>
> sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
>
>[ Anyone who did this in their head for sin(18 degrees = pi/10
>radians) to solve Burt's problem can pat themselves on the back and
>disregard the remainder of this post. All of those who evaluated /pi/
>with a series approximation have a surplus of brain cells and a
>serious need to "get a life".]
*snort* who needs a series approximation?? I've had the numerical value of
pi, out to _20_ decimal places, memorized for more than 35 years.
I have, however, *rarely* needed more than 6-place accuracy for same.
Now, the numerical value for 'cornbread', THAT's a different matter. <grin>
Note; I also used to have a handful of common log values memorized.
And a dozen or so trig values. With that, and a 'half-angle' formula, you
can do faily impressive pencil-and-paper 'approximtions'.
In article <[email protected]>,
Morris Dovey <[email protected]> wrote:
>Robert Bonomi (in [email protected]) said:
>
>| *snort* who needs a series approximation?? I've had the numerical
>| value of pi, out to _20_ decimal places, memorized for more than 35
>| years.
>|
>| I have, however, *rarely* needed more than 6-place accuracy for
>| same.
>|
>| Now, the numerical value for 'cornbread', THAT's a different
>| matter. <grin>
>|
>| Note; I also used to have a handful of common log values memorized.
>| And a dozen or so trig values. With that, and a 'half-angle'
>| formula, you can do faily impressive pencil-and-paper
>| 'approximtions'.
>
>Lucky you! I've never been able to memorize stuff like that; but
>somehow managed to remember basic formulas like the series
>approximations. I can remember thinking early on that the actual
>numbers weren't as important as the relationships that produced them.
>Now I see 'em as two ends of the same stick that different people feel
>comfortable grasping in different places.
>
>Heh heh. Just remembered the physics prof who got repeated cases of
>the heebeegeebies because I started nearly all problem solutions with
>"F = ma" and derived whatever I needed from that. It was my first real
>clue that we're not all wired alike.
It is said that you only need to know two things to be an engineer;
1) F = ma
2) you can't push on a rope.
"Guess who" <[email protected]> wrote in message
news:[email protected]...
> 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.
"Morris Dovey" <[email protected]> wrote in message
news:[email protected]...
> Phil (in [email protected]) said:
>
> | "Morris Dovey" <[email protected]> wrote in message
> | news:[email protected]...
> || Burt (in [email protected]) 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 / 20) or
> || approximately 11-1/8"
> ||
> || --
> || Morris Dovey
> || DeSoto Solar
> || DeSoto, Iowa USA
> || http://www.iedu.com/DeSoto/solar.html
> ||
> ||
> |
> | If he were building a 10 segment, flat circle you would be right on
> | but he said "dish". I suspect he needs the other sides's dimension
> | as well. More information is necessary to figure it out. How deep
> | is the dish and does it have an elliptical section or is it part of
> | a sphere? What is the dish for exactly? There are myriad
> | possibilities
> | when you say dish so there is no way to give a (complete) correct
> | answer...
>
> Phil...
>
> Burt also wrote:
> << 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. >>
>
> I interpreted this to mean he wanted the straight line distance
> between adjacent points on a circle. Since circles are planar objects,
> I don't think deformations of the 10 segments enter into /this/
> calculation, but I could be wrong ( and frequently am :-)
>
> --
> Morris Dovey
> DeSoto Solar
> DeSoto, Iowa USA
> http://www.iedu.com/DeSoto/solar.html
>
>
Morris,
You are correct and it seems you are not to often incorrect. I
often read your posts and you have some good solutions to the
various problems posted.
While a few others here were using their math skills, I cheated
and used my CAD software to draw the problem and arrived
at the length of (rounded) 11.125".
Phil
In article <[email protected]>, Morris Dovey
<[email protected]> wrote:
> Burt has a formula and, presumably, a calculator - and can grind out
> whatever chord lengths he wants. He doesn't /need/ to understand the
> mathematics. It wouldn't hurt if he did, but all the understanding in
> the world won't produce any better results.
Nobody needs to understand math. Well, that's just...
Hm... Hum... What to say?
I fart in your general direction. Your mother was a hamster and your
father smelled of elderberries.
I despise your comment as quoted above, and the attitude that allowed
it to be verbalized/typalized.
It's, simply, despicable. IMO.
No more to say on this topic, except "SOHCAHTOA".
--
~ Stay Calm... Be Brave... Wait for the Signs ~
------------------------------------------------------
One site: <http://www.balderstone.ca>
The other site, with ww links<http://www.woodenwabbits.com>
Each segment will be 11.125" wide.
"Burt" <[email protected]> wrote in message
news:[email protected]...
> On Sun, 26 Jun 2005 20:20:38 -0500, Duane Bozarth <[email protected]>
> 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?
Can't help that. 11.1246 is the correct number.
"bw" <[email protected]> wrote in message
news:[email protected]...
> I get 11.3 inches.
>
>
On 2005-06-27, Morris Dovey <[email protected]> wrote:
> Burt (in [email protected]) 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 / 20) or approximately
> 11-1/8"
Too much work and I don't get the same answer anyway. (PI * 36 / 10) =
PI * 3.6 = 11.3097312 etc. etc. etc. Significantly more than 1/8 inch
difference, it's over 11 1/4 inches.
--
I can find no modern furniture that is as well designed and emotionally
satisfying as that made by the Arts and Crafts movement in the early years
of the last century.
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...
On Mon, 27 Jun 2005 06:52:16 -0500, "Morris Dovey" <[email protected]>
wrote:
>Burt (in [email protected]) 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 :)
On Thu, 30 Jun 2005 00:08:01 -0500, "Morris Dovey" <[email protected]>
wrote:
>| No more to say on this topic, except "SOHCAHTOA".
>
>I had to go to Google for that one. Neat. I wish you'd told me about
>that a half century ago, when I was struggling to keep all of those
>straight in my head.
You didn't have that? Ah, that's just criminal. When one of the
other guys at work found out that I was helping one of our co-workers
learn trig, he screwed up his face a bit, thought about it for a
minute, and then came up with "Do you mean SOHCAHTOA?" It was
hilarious. Useful acronym, though.
On Mon, 27 Jun 2005 19:08:49 -0400, Guess who
<[email protected]> wrote:
>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.
...or even a chord.
Prometheus (in [email protected]) 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
Not true at all. You can have any device you want, that will give you any
answer you want, but it will just sit there and act stupid if you don't know
what to ask it. Even then, you have to now if it is giving you the right
answer. Take the segmented circle problem under discussion. There is no way
that any calculator is going to figure that out for you. You have to know
what you are wanting to do, lay it out and devise a formula (whether it is a
standard one or one you devised yourself). Only then, fed the correct
formula, is that calculator of any use and all it is then doing is replacing
the trig tables that yo feel are better. When ariving at and asnswer, before
commiting time and wood to that number, you will want to turn that formula
around, work it backwards and see if it is still correct. Many mistakes can
be found this way. The calculator is a dumb slave to the human brain
commanding it. If you don't know what you are doing, that device will do
nothing.
"Prometheus" <[email protected]> wrote in message
news:[email protected]...
> On Tue, 28 Jun 2005 13:56:34 GMT, "CW" <[email protected]> wrote:
>
> >Not.
>
> You don't think so? I've got a calculator from 1996 laying around
> somewhere that can solve just about any calculus problem by typing in
> solve( and then the problem. Same for algebra, trig, or any other
> branch of math you care to name. I don't imagine that they've gotten
> less powerful over time.
>
> But that's the extreme case- I know a lot of people who can't do long
> division, and don't care to know how because they have a calculator.
> But then when they don't have a calculator handy, they're lost. That
> would indicate to me that they don't understand the math, they just
> know how to operate a calculator.
>
> >"Prometheus" <[email protected]> wrote in message
> >news:[email protected]...
> >
> >> On Tue, 28 Jun 2005 01:24:35 GMT, "CW" <[email protected]> wrote:
> >>> That, and dependancy on calculators seems to interfere with people
> >> fully understanding the math.
> >
>
On Thu, 30 Jun 2005 06:16:50 -0500, Prometheus
<[email protected]> wrote:
>Same logic applies to calipers- Sure, it's easier to use a dial or
>digital caliper when measuring, but I still use a vernier. It's not
>because it's inherantly better, it's just better for the environment
>I'm using it in, and lasts a heck of a lot longer!
That argument is always good. If you use a vernier constantly, you'll
be just as good with that as someone with an electronic instrument.
That doesn't mean everyone should dash out and buy an abacus.
OK, if people promise not to throw things ...I used to teach math, and
at years end would show others how to use their calculator
effectively, or how to use a spreadsheet or suitable program, or make
up one for them if they felt the need but didn't know how. I used a
slide-rule myself. Why? I could use all of the above and more, but
had used the slide-rule so often that it was second nature; much
easier for me in the long run, if not for the others.
Moral: Use whatever works! Just build the bloody thing. It's
woodworking, not brain surgery.
LOL. Too much truth
Steve
"Larry Jaques" <novalidaddress@di\/ersify.com> wrote in message
news:[email protected]...
> On Sun, 26 Jun 2005 18:17:56 -0700, the opaque Burt
> <[email protected]> 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
"Prometheus" <[email protected]> wrote in message
news:[email protected]...
>
> 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.
>
The question required a bit of thought. I was going with the 11.3" until I
saw the other answers being lower and ended up solving the problem with a
drawing on AutoCAD.
"Ed Clarke" <[email protected]> wrote in message
news:[email protected]...
>
> Too much work and I don't get the same answer anyway. (PI * 36 / 10) =
> PI * 3.6 = 11.3097312 etc. etc. etc. Significantly more than 1/8 inch
> difference, it's over 11 1/4 inches.
>
Your answer is to a question that was not asked. Had the OP asked what the
length of the exterior of the segment was you would have been correct.
However the OP asked how wide the segment would be. The widest part would
be the distance between the two closest points of that triangle shaped
segment. That distance between those two points is approximately 11.125".
On Mon, 27 Jun 2005 08:50:35 -0500, "Morris Dovey" <[email protected]>
wrote:
>Prometheus (in [email protected]) 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...
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" <[email protected]> wrote in message
news:[email protected]...
> On Mon, 27 Jun 2005 13:28:41 GMT, "Leon" <[email protected]>
> wrote:
>
>>
>>"bw" <[email protected]> wrote in message
>>news:[email protected]...
>>>
>>> "CW" <[email protected]> wrote in message
>>> news:[email protected]...
>>>> 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.
>
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
Try these sites, hope something here helps.
http://www.verifiedsoftware.com/goodturns/plans.htm
http://www.delorie.com/wood/segturn.html
"Burt" <[email protected]> wrote in message
news:[email protected]...
>
> 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.
>
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
"Swingman" <[email protected]> wrote in message
news:[email protected]...
> 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.
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" <[email protected]> wrote in message =
news:[email protected]...
| On Sun, 26 Jun 2005 20:20:38 -0500, Duane Bozarth =
<[email protected]>
| 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?
Robert Bonomi (in [email protected]) 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
Robert Bonomi (in [email protected]) said:
| *snort* who needs a series approximation?? I've had the numerical
| value of pi, out to _20_ decimal places, memorized for more than 35
| years.
|
| I have, however, *rarely* needed more than 6-place accuracy for
| same.
|
| Now, the numerical value for 'cornbread', THAT's a different
| matter. <grin>
|
| Note; I also used to have a handful of common log values memorized.
| And a dozen or so trig values. With that, and a 'half-angle'
| formula, you can do faily impressive pencil-and-paper
| 'approximtions'.
Lucky you! I've never been able to memorize stuff like that; but
somehow managed to remember basic formulas like the series
approximations. I can remember thinking early on that the actual
numbers weren't as important as the relationships that produced them.
Now I see 'em as two ends of the same stick that different people feel
comfortable grasping in different places.
Heh heh. Just remembered the physics prof who got repeated cases of
the heebeegeebies because I started nearly all problem solutions with
"F = ma" and derived whatever I needed from that. It was my first real
clue that we're not all wired alike.
Pencil and paper works - but (IMO) there are better tools like slide
rule, calculator, and computer to make calculations faster and easier.
A side note: a couple of years ago I wrote a tiny/fast sine/cosine
subroutine that divided a quadrant (quarter circle) into 256 parts and
used a table of 256 16-bit numerators and a common denominator of
65535. The subroutine folded all angles into the first quadrant and
interpolated to produce sine and cosine values accurate to +/-
0.0000005; it made me wish I could memorize the table.
Cornbread is good. I have the formula around here somewhere...
:-)
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html
"Prometheus" <[email protected]> wrote in message
news:[email protected]...
> On Sun, 26 Jun 2005 18:17:56 -0700, Burt <[email protected]>
> 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.
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
Prometheus (in [email protected]) said:
| But that's the extreme case- I know a lot of people who can't do
| long division, and don't care to know how because they have a
| calculator. But then when they don't have a calculator handy,
| they're lost. That would indicate to me that they don't understand
| the math, they just know how to operate a calculator.
Hmm. I studied math through an advanced course in partial differential
equations - and, when it comes to solving even simple trig problems,
I'm one of those who're lost without a calculator.
I suppose I could do a little (lot of?) review and calculate my own
trig function values:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
[ Anyone who did this in their head for sin(18 degrees = pi/10
radians) to solve Burt's problem can pat themselves on the back and
disregard the remainder of this post. All of those who evaluated /pi/
with a series approximation have a surplus of brain cells and a
serious need to "get a life".]
Ah. Glad I'm not alone :-)
It's a Good Thing, IMO, to understand the math - but I don't think
it's bad to not understand the math. Mathematics *and* calculators are
both tools.
Burt has a formula and, presumably, a calculator - and can grind out
whatever chord lengths he wants. He doesn't /need/ to understand the
mathematics. It wouldn't hurt if he did, but all the understanding in
the world won't produce any better results.
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html
Burt (in [email protected]) 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
On Thu, 30 Jun 2005 01:38:15 -0500, "Morris Dovey" <[email protected]>
wrote:
>A side note: a couple of years ago I wrote a tiny/fast sine/cosine
>subroutine that divided a quadrant (quarter circle) into 256 parts and
>used a table of 256 16-bit numerators and a common denominator of
>65535. The subroutine folded all angles into the first quadrant and
>interpolated to produce sine and cosine values accurate to +/-
>0.0000005; it made me wish I could memorize the table.
See, somehow I knew you saw it my way- even if you don't agree with
the calculator bit, you just proved you're math nerd. Reminds me of
my TI-OS program for simulating synthetic division. Worked great, but
everyone who saw it thought I was nuts.
>Cornbread is good. I have the formula around here somewhere...
>
>:-)
On Mon, 27 Jun 2005 12:12:14 -0400, Guess who
<[email protected]> wrote:
>On Mon, 27 Jun 2005 08:22:48 -0500, Prometheus
><[email protected]> 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/geodesic/drawings/trig.html
I think it may, thanks! Saves me a whole lot of typing, at the very
least. :)
On Sun, 26 Jun 2005 20:20:38 -0500, Duane Bozarth <[email protected]>
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?
On Sun, 26 Jun 2005 18:17:56 -0700, the opaque Burt
<[email protected]> 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
On Mon, 27 Jun 2005 09:49:46 -0700, "Burt"
<[email protected]> 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.
"CW" <[email protected]> wrote in message
news:[email protected]...
> Each segment will be 11.125" wide.
I get 11.3 inches.
"bw" <[email protected]> wrote in message
news:[email protected]...
>
> "CW" <[email protected]> wrote in message
> news:[email protected]...
>> 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.
In article <[email protected]>,
Prometheus <[email protected]> wrote:
> That, and dependancy on calculators seems to interfere with people
> fully understanding the math.
Can't use a calculator unless you understand the math.
Dave Balderstone (in
290620052217077499%dave@N_O_T_T_H_I_S.balderstone.ca) said:
| In article <[email protected]>, Morris Dovey
| <[email protected]> wrote:
|
|| Burt has a formula and, presumably, a calculator - and can grind
|| out whatever chord lengths he wants. He doesn't /need/ to
|| understand the mathematics. It wouldn't hurt if he did, but all
|| the understanding in the world won't produce any better results.
|
| Nobody needs to understand math. Well, that's just...
...not what I said at all. :-)
Not all people can be all things; and everyone has different aptitudes
and talents to develop. It took me a long time to internalize that not
everyone needed to become a programmer in order to use a computer to
good advantage. The same applies to mathematics.
| Hm... Hum... What to say?
|
| I fart in your general direction. Your mother was a hamster and your
| father smelled of elderberries.
Well, ok - but the smell hasn't reached Iowa yet. In case you hadn't
noticed, the prevailing wind is from the southwest this time of year -
I would caution you against over-exerting yourself. :-D
My mother would have smiled and assured you that she did her best to
be a proper /Scottish/ hamster. From what I've been told, I think my
father would have grinned and said that he'd been told worse. (I wish
he'd had a chance to become elder-than-26.)
| I despise your comment as quoted above, and the attitude that
| allowed it to be verbalized/typalized.
|
| It's, simply, despicable. IMO.
|
| No more to say on this topic, except "SOHCAHTOA".
I had to go to Google for that one. Neat. I wish you'd told me about
that a half century ago, when I was struggling to keep all of those
straight in my head.
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html
"Burt" <[email protected]> wrote in message
news:[email protected]...
>
> 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
On Mon, 27 Jun 2005 14:32:00 GMT, "Leon" <[email protected]> wrote:
>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.
>
Thanks I'll keep that in mind if I can't figure this out.
Like pumpkins the sizes will always change.
I would like to be able to do segments that are not equal as well.
If I can't figure out the formulas I will fire up a cad program and see if it
helps.
It's the math terminology that drives me crazy.
Beyond the terms radius and circumference I'm clueless.
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.
I haven't had time to digest all the posts yet.
On Sun, 26 Jun 2005 18:17:56 -0700, Burt <[email protected]>
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 more:
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.
"Morris Dovey" <[email protected]> wrote in message
news:[email protected]...
> Burt (in [email protected]) 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 / 20) or approximately
> 11-1/8"
>
> --
> Morris Dovey
> DeSoto Solar
> DeSoto, Iowa USA
> http://www.iedu.com/DeSoto/solar.html
>
>
If he were building a 10 segment, flat circle you would be right on
but he said "dish". I suspect he needs the other sides's dimension
as well. More information is necessary to figure it out. How deep
is the dish and does it have an elliptical section or is it part of a
sphere? What is the dish for exactly? There are myriad possibilities
when you say dish so there is no way to give a (complete) correct answer...
Phil Davis
247PalmBeachRE.com
On Tue, 28 Jun 2005 13:56:34 GMT, "CW" <[email protected]> wrote:
>Not.
You don't think so? I've got a calculator from 1996 laying around
somewhere that can solve just about any calculus problem by typing in
solve( and then the problem. Same for algebra, trig, or any other
branch of math you care to name. I don't imagine that they've gotten
less powerful over time.
But that's the extreme case- I know a lot of people who can't do long
division, and don't care to know how because they have a calculator.
But then when they don't have a calculator handy, they're lost. That
would indicate to me that they don't understand the math, they just
know how to operate a calculator.
>"Prometheus" <[email protected]> wrote in message
>news:[email protected]...
>
>> On Tue, 28 Jun 2005 01:24:35 GMT, "CW" <[email protected]> wrote:
>>> That, and dependancy on calculators seems to interfere with people
>> fully understanding the math.
>
On Mon, 27 Jun 2005 21:17:05 -0000, [email protected]
(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. :-)
On Mon, 27 Jun 2005 13:32:38 GMT, "Leon"
<[email protected]> wrote:
>
>"Prometheus" <[email protected]> wrote in message
>news:[email protected]...
>> On Sun, 26 Jun 2005 18:17:56 -0700, Burt <[email protected]>
>> 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 Wed, 29 Jun 2005 23:04:47 -0500, "Morris Dovey" <[email protected]>
wrote:
>I suppose I could do a little (lot of?) review and calculate my own
>trig function values:
>egrees = pi/10
>radians) to solve Burt's problem can pat themselves on the back and
>disregard the remainder of this post. All of those who evaluated /pi/
>with a series approximation have a surplus of brain cells and a
>serious need to "get a life".]
Okay, I fold. I still like the trig table in the context I needed it
for though. Let me fill it in a little, so I'm not completely nuts to
you all. We're working these problems on the motor cover of a huge
vertical CNC bandsaw amid tons of coolant and steel swarf. The
employer provides regular four-function calculators- not scientific
versions. A calculator only lasts a couple of weeks, on average. A
trig table lasts for 6 months or better once it's laminated (if the
fraction-decimal charts are anything to go by), and it allows us to
solve the problems in question without having to buy a much more
expensive calculator every couple of weeks because some dummy knocked
it into the coolant or the keys got jammed up by tiny steel chips.
The guy I'm teaching appreciated it as well- even though he is aware
that calculators that will evaluate sin/cos/tan values exist, he's got
4 kids at home that (evidently) like to break things, so he doesn't
have one of them.
Same logic applies to calipers- Sure, it's easier to use a dial or
digital caliper when measuring, but I still use a vernier. It's not
because it's inherantly better, it's just better for the environment
I'm using it in, and lasts a heck of a lot longer!
>Ah. Glad I'm not alone :-)
>
>It's a Good Thing, IMO, to understand the math - but I don't think
>it's bad to not understand the math. Mathematics *and* calculators are
>both tools.
>
>Burt has a formula and, presumably, a calculator - and can grind out
>whatever chord lengths he wants. He doesn't /need/ to understand the
>mathematics. It wouldn't hurt if he did, but all the understanding in
>the world won't produce any better results.
I suppose there's an argument for just dropping in here again and
asking if the application changes, but to me it's one of those Give a
man a fish V. Teach a man to fish senarios. And, if the equation is
not being used properly because of some difference in the problem, it
can lead to some pretty large errors pretty quickly because he didn't
know how to double-check it. As another poster noted, you can have
any machine you want, but it won't do anything if you don't know what
to put in it.
I see your point, though.
On Sun, 26 Jun 2005 18:17:56 -0700, Burt <[email protected]>
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.
Not.
"Prometheus" <[email protected]> wrote in message
news:[email protected]...
> On Tue, 28 Jun 2005 01:24:35 GMT, "CW" <[email protected]> wrote:
>> That, and dependancy on calculators seems to interfere with people
> fully understanding the math.
On Mon, 27 Jun 2005 12:12:14 -0400, Guess who <[email protected]> wrote:
>On Mon, 27 Jun 2005 08:22:48 -0500, Prometheus
><[email protected]> 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/geodesic/drawings/trig.html
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.:)
On Mon, 27 Jun 2005 08:22:48 -0500, Prometheus
<[email protected]> 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/geodesic/drawings/trig.html
A good calculator will run about $14.00 these days. Much handier (and
faster) than those tables.
"Prometheus" <[email protected]> wrote in message
news:[email protected]...
>>
> 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 :)
>
>
Burt (in [email protected]) 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 / 20) or approximately
11-1/8"
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html
Phil (in [email protected]) said:
| "Morris Dovey" <[email protected]> wrote in message
| news:[email protected]...
|| Burt (in [email protected]) 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 / 20) or
|| approximately 11-1/8"
||
|| --
|| Morris Dovey
|| DeSoto Solar
|| DeSoto, Iowa USA
|| http://www.iedu.com/DeSoto/solar.html
||
||
|
| If he were building a 10 segment, flat circle you would be right on
| but he said "dish". I suspect he needs the other sides's dimension
| as well. More information is necessary to figure it out. How deep
| is the dish and does it have an elliptical section or is it part of
| a sphere? What is the dish for exactly? There are myriad
| possibilities
| when you say dish so there is no way to give a (complete) correct
| answer...
Phil...
Burt also wrote:
<< 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. >>
I interpreted this to mean he wanted the straight line distance
between adjacent points on a circle. Since circles are planar objects,
I don't think deformations of the 10 segments enter into /this/
calculation, but I could be wrong ( and frequently am :-)
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html
Burt (in [email protected]) 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
On Sun, 26 Jun 2005 22:10:09 -0400, "Phil" <[email protected]> wrote:
>
>"Morris Dovey" <[email protected]> wrote in message
>news:[email protected]...
>> Burt (in [email protected]) 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.:)
>If he were building a 10 segment, flat circle you would be right on
>but he said "dish". I suspect he needs the other sides's dimension
>as well. More information is necessary to figure it out. How deep
>is the dish and does it have an elliptical section or is it part of a
>sphere? What is the dish for exactly? There are myriad possibilities
>when you say dish so there is no way to give a (complete) correct answer...
Nah, you're making it too complex. The difference in the two answers
given is that one is the length from point to point on the circle
(Morris's answer, approx 11.125"), useful if the OP is making ten
segments with straight edges, and the other is the length of the curve
(approx 11.3"), useful if the segments need tobe curved to fit the rim
exactly. We don't need to know how deep the dish is to figure it out.
On Tue, 28 Jun 2005 01:24:35 GMT, "CW" <[email protected]> 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.
On Thu, 30 Jun 2005 00:08:01 -0500, "Morris Dovey" <[email protected]>
wrote:
>| No more to say on this topic, except "SOHCAHTOA".
>
>I had to go to Google for that one. Neat. I wish you'd told me about
>that a half century ago, when I was struggling to keep all of those
>straight in my head.
The struggle was worth it. Those who memorise the "trick" never
really understand the main ideas. You are a lot better off having
learned to look for the sides and angles and ratios in each working
problem. You might not realise it, but you gained a lot more insight,
never mind the struggle.
On Mon, 27 Jun 2005 13:28:41 GMT, "Leon" <[email protected]> wrote:
>
>"bw" <[email protected]> wrote in message
>news:[email protected]...
>>
>> "CW" <[email protected]> wrote in message
>> news:[email protected]...
>>> 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.
Wish I had one of those back when..... However, it occurs to me that the
real challenge is
stating the "problem" in the correct mathematical form.
Just a thought,
Ace
"Prometheus" <[email protected]> wrote in message
news:[email protected]...
> On Tue, 28 Jun 2005 13:56:34 GMT, "CW" <[email protected]> wrote:
>
>>Not.
>
> You don't think so? I've got a calculator from 1996 laying around
> somewhere that can solve just about any calculus problem by typing in
> solve( and then the problem. Same for algebra, trig, or any other
> branch of math you care to name. I don't imagine that they've gotten
> less powerful over time.
>
> But that's the extreme case- I know a lot of people who can't do long
> division, and don't care to know how because they have a calculator.
> But then when they don't have a calculator handy, they're lost. That
> would indicate to me that they don't understand the math, they just
> know how to operate a calculator.
>
>>"Prometheus" <[email protected]> wrote in message
>>news:[email protected]...
>>
>>> On Tue, 28 Jun 2005 01:24:35 GMT, "CW" <[email protected]> wrote:
>>>> That, and dependancy on calculators seems to interfere with people
>>> fully understanding the math.
>>
>
On Mon, 27 Jun 2005 13:55:42 GMT, "Leon"
<[email protected]> 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.
"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
Ed Clarke (in [email protected]) said:
| On 2005-06-27, Morris Dovey <[email protected]> wrote:
|| Burt (in [email protected]) 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 / 20) or
|| approximately 11-1/8"
|
| Too much work and I don't get the same answer anyway. (PI * 36 /
| 10) = PI * 3.6 = 11.3097312 etc. etc. etc. Significantly more than
| 1/8 inch difference, it's over 11 1/4 inches.
Not too much work if your calculator has trig functions. My Windows
calculator came up with 11.124611797498107267682563018581", which
misses 11-1/8 by only 0.0004".
Pi * 36 / 10 would be the arc length of the segment, while 36*sin(18)
is the chord length. The difference is
0.18512175542514839078295316122472", somewhere near 3/16" - so the
extra effort may be worthwhile :-)
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html
Send me an email and I will send you a CAD drawing. Exact to 8 decimal
places (though the answers you got here are correct). Thought you might want
something on "paper". Will send as PDF.
"Burt" <[email protected]> wrote in message
news:[email protected]...
> Beyond my figuring skills and it has to be exact. :)
> Each of the points has to touch the inside of the 3 foot circle.
On Wed, 29 Jun 2005 05:02:39 -0500, Prometheus
<[email protected]> wrote:
> I've got a calculator from 1996 laying around
>somewhere that can solve just about any calculus problem by typing in
>solve( and then the problem. Same for algebra, trig, or any other
>branch of math you care to name. I don't imagine that they've gotten
>less powerful over time.
Perhaps they aid in some, but calculators don't "solve the calculus
problems" I've seen. That's just it; they are an aid, not "the
answer." Solution of problems in calculus involves a pretty thorough
knowledge of calculus and all that precedes it, and most of that is
done with the calculator we're born with.
>That
>would indicate to me that they don't understand the math, they just
>know how to operate a calculator.
That's what I mean.
All said and done, the calculator is a tremendous asset *after* an
understanding of basic principles of the subject it is supposed to
assist. Same with woodworking [thought I'd throw that in.] The tools
don't make the master woodworker, but good tools sure help. Lots of
people have much better tools than some superb craftsmen stuck with
less, but all they do is build stepstools. The calculator [ or CAD
program or whatever] is good in the right hands and next to useless in
the wrong ones, and really not all that necessary, as you pointed out.
"Guess who" <[email protected]> wrote in message
news:[email protected]...
>
> 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".