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Segmented Ring Size

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I have a problem with my segmented rings coming out larger than I want them to be.

I use caliper to measure cut size, and I cut my segments with and Incra mitre gauge, but my rings always come out bigger than they should.
Example. 12 sided 6 inch ring: Diameter (6 inches) times .26795 = 1.6077:
I cut 12 segments to size @ 15 Degrees. Ring turns out to b 6.5 inches instead of 6.0 inches.
Where am I going wrong. This oversize happens on all size of rings. Not only on 6 inch rings.

Thanks for the help.

Allen
 
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segment length problem

Here is a table that I ran across to determine segment lengths. I have not used it but it makes sense. Hopefully the pdf is attached. If not, I need help in how to attach it.

Rod
 

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  • SegmentEdgeEstimationTable.pdf
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Thanks Rod,
I appreciate the help. I have that table and that is the one I am using but my rings still come out anywhere from quarter inch to a half inch or more larger than the formula.

Will keep trying.

Allen
 
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Segment Length

It appears to me that your segments are too long. I don't recognize the formula you are using, but here's one that works.

Desired Ring Outside Diameter = 6"
Calculated Ring Circumference = pi x ring outside diameter = 3.142 x 6" = 18.850"
Segment Length = circumference/number of segments = 18.850/12 = 1.571"

Above segment length is shorter than your calculation and should yield a 6" outside diameter ring. - John

PS - This calculation matches the segment lengths in Segmented Edge Length Estimation Table provided by Rod Smith. - J
 
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First: The table probably has rounded-up values from the exact calculation, which is not so bad, because it's easier to make large pieces smaller than to make small pieces larger.

Second: You might be using the "wrong" diameter. For the outside of the ring, the target diameter should be across the flats, not across the corners. The corners get cut away when turning, and the center of the flats is the critical dimension. For the inside, the target diameter should be across the corners, because the middle of the segment gets cut during turning. For both of these, you'll benefit by adding material away from the final diameters, to allow for minor eccentricity in assembly, as well as sanding allowance. In other words, make the outside diameter larger than necessary, and make the inside diameter smaller than necessary.

Third: If gang cutting the segments, the kerf allowance should match the tool. Circular saws usually leave a kerf of about 1/8"; back saws and band saws usually closer to 1/16".

The exact calculation, from one of my engineering manuals:
Regular Polygon, n = number of sides.
Phi = 1/2 angle subtended by the segment = 180/n
s = length of chord/flat (= length of segment) = 2 * sqrt (R*R - R1*R1)
R = radius to the corner = s / (2 sin Phi)
R1 = radius to the flat = s / (2 tan Phi)
from these, s = 2 R sin Phi, or s = 2 R1 tan Phi.

or, if you prefer, D = 2R and D1 = 2R1.

Computations of circumference are immaterial for a low number of segments. These provide arc lengths, not chord lengths.

If all else fails, you can draw the segment layout using pencil and paper, a protractor, compass and straightedge; and test the results before cutting wood. You don't need to be a professional draftsman, and you don't need CAD for this.
 

john lucas

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Being Math inhibited I simply draw my piece. Then lay out the rings and draw them. Then I draw in the angle for the number of sides. All I have to do after that is measure where the angles cross the circles. It also gives me drawing to set the segments on to check for size and placement.
OK I know it's slower but if I'm doing segmented work they usually take more planning and time so I don't mind. I just look at it as part of the process.
 
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Segment Calculator

I have an executable file that I have used for a number of years which works pretty well. You might compare it to your existing method of calculation and see if the segments copme out slightly shorter.

Since I can't [post an *.exe file in this forum, I will try to send you an email.

Jerry
 
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I'll try to find a postable link.

:confused:
 
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Thanks for finding it, Jerry. Since the length units are all the same, you can use any measurements, as long as they're consistent - inches, feet, furlongs, millimeters, etc. Even set the fudge factor to zero, if you like.
 
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Smoots

Joe - I like to work a a unit of measure called Smoots. Probably 60 years ago, a guy named Smoot was laid off end to end across the Mass. Ave. bridge in Cambridge as a fraternity prank. Every year, the Smoot marks are repainted. Now, if I could just find a lathe that works in Smoots, I'd be all set.:)
 

john lucas

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Jerry just buy Joe Hermann's Conover lathe for sale in the classifieds. Then you can make the ways any length you want. I think 2 smoots would be about right.
 
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John:
Great idea!! There is a pro-turner in the Adirondacks, north of here who has an antique lathe apparently used for turning columns, pillars, etc. Its bed is made of two quite long 10" square hand hewn beams. It is a precursor to Ernie's approach with lathes. The guy also has a home built bowl lathe which will handle about 40" diameter, if not more. It is driven by a 5hp electric motor through a 4 or 5 speed transmission and a truck rear end differential. He says he sells 1 or 2, 30 - 36" bowls per year. Whoever buys those must set a Thanksgiving table that I would like to sit at!!

Jerry
 
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Thanks very much to all who replied to my request for help with ring sizes.

I have had to put my turning on hold until August 1. I have grandchildren with me until then, and they are not yet ready to participate in power tool activities.

Thanks again. I will put all the advice I received to good use.

Allen Pousson
 
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