# I need a formula for segmenting a circle

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• posted on June 27, 2005, 2:28 am

Phil (in f4Jve.12876\$ snipped-for-privacy@bignews6.bellsouth.net) said:
| 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
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• posted on June 27, 2005, 3:24 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.
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• posted on June 27, 2005, 3:42 am
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.

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• posted on June 27, 2005, 3:45 am
said:

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
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• posted on June 27, 2005, 1:27 pm

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.
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• posted on June 27, 2005, 3:44 am

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
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• posted on June 27, 2005, 4:19 am
Burt 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
--
Http://www.Weber-Automation.net:8000

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• posted on June 27, 2005, 4:23 am
Nate Weber wrote:

bah to all that, I'm wrong.
Nate
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Http://www.Weber-Automation.net:8000

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• posted on June 27, 2005, 4:26 am
Nate Weber wrote:

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

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• posted on June 27, 2005, 3:47 am

LOL... well at least you kept at until you got it right. I used AutoCAD for the answer. :~)
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• posted on June 27, 2005, 11:52 am
Burt (in snipped-for-privacy@4ax.com) 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
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• posted on June 27, 2005, 1:22 pm
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? We don't have those umm... "fancy" calculators in the shop :)
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• posted on June 27, 2005, 1:50 pm
Prometheus (in snipped-for-privacy@4ax.com) 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
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• posted on June 28, 2005, 10:55 am
wrote:

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.

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...
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• posted on June 27, 2005, 4:12 pm
On Mon, 27 Jun 2005 08:22:48 -0500, Prometheus

This might be of some help: http://www.uni.edu/darrow/new/geodesic/drawings/trig.html
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• posted on June 27, 2005, 4:49 pm

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.:)
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• posted on June 27, 2005, 8:47 pm
On Mon, 27 Jun 2005 09:49:46 -0700, "Burt"

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.
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• posted on June 28, 2005, 10:58 am
On Mon, 27 Jun 2005 12:12:14 -0400, Guess who

I think it may, thanks! Saves me a whole lot of typing, at the very least. :)
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• posted on June 28, 2005, 1:24 am
A good calculator will run about \$14.00 these days. Much handier (and faster) than those tables.

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• posted on June 28, 2005, 11:11 am

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.