On 12/11/2011 10:26 AM, Dan Coby wrote:
On 12/11/2011 10:13 AM, Dan Coby wrote:
On 12/10/2011 8:11 PM, Kevin Miller wrote:
The usual way to do segmented turnings is to cut trapezoids and glue
them up in a ring then stack the rings up to make the vessel.
Typically, one cuts the same angle at both ends
of the segment.

When doing reverse segments, one cuts one end to the appropriate
angle, and leaves the other end at 90 degrees. The angled edge of the
segment is oriented outwards rather than
inwards. See my facebook post for an example of a ring being glued up:
https://www.facebook.com/media/set/?...l=f 7a342b1e7


When cutting normal segments, one sizes them for a given outside and
inside diameter. There's lots of segment calculators on the web that
will give you the width of the board to
use, and the length of each segment. However I can't find any
calculators that will determine the dimensions for a reverse segment
except I have a spreadsheet that calculates the
width and length for an eight sided ring using the following formula:

OR = outside radius
IR = inside radius

Width = OR - (.924* IR)
Length = (.541*IR+Width)/.707

The segment length seems to come out a little long but that gives
some fudge factor so that's fine.

What I'm looking for is the formula to enter the number of segments,
the inside and outside diameter and for it to calculate the length of
the long edge of the segment. Any math
whiz out there that can clue me in?

As you said the equations that you show give a result that is a little
long.
This is because they calculate the width (and from it the length) a
little
long. They calculate the width measuring the OR perpendicular to an edge
of a segment. A more accurate (and less wasteful) value uses a diagonal.
The error in their method will get worse as you increase the number of
segments.


Making the same error that they did:

theta (the angle) = 360 degrees / number of segments

width = OR - cos(theta/2) * IR

length = 2*sin(theta/2)*IR + width/sin(theta)

or length = (2*sin(theta/2)*sin(theta)*IR + width) / sin(theta)

Check for number of segments = 8:
theta = 45 degrees
theta/2 = 22.5 degrees
cos(theta/2) = .9238 (check)
sin(theta/2) = .38268
sin(theta) = .7071 (check)
2*sin(theta/2)*sin(theta) = 2*.38268*.7071 = .5411 (check)

... snip

I need to add a correction to my last posting: IGNORE THE SECOND SET OF
EQUATIONS

I did not consider the case when the outside radius OR is only a little
larger than the inside radius IR. I suspect that this is commonly the case
for most segmented turnings. In this case you need to use the first set of
equations (which I called an error and wasteful). What can I say, it was
early in the morning here on the west coast when I wrote that. I am not a
morning person. A shower helped to clear my mind.

In that situation I'd probably use a traditional segment. The selling
point of the reverse segment method is the joints are angled rather than
pointing towards the center of the piece, giving a pin-wheel effect.
That is most dramatic on a wide segment and best viewed from the top
(see my posts today for an example). The effect would be pretty much
lost on a half inch wide piece. On a lid or a closed form however which
utilizes a wider 'face', your more accurate method will be a definite plus.

With or without a shower, it's all a muddle to me. Sure glad there's
further. That should givfolks like you that grasp this stuff!


Your example picture has a large difference between the two radii. As a
result, the segments ware very wide and there is a large joint length.
My second set of equations works fine for this case.

The piece I showed will be the lid to a bowl I made, thus needed to be
wide. It will have a solid cherry center and probably a finial of
whatever dark wood I can find laying around the shop.


With narrower segments (i.e. a small difference between IR and OR) we have
to make sure that when we get away from the joints that we still have
enough width to get the desired outside radius. The first set of equations
guarantees that.

A more accurate analysis would say the there are situations in which each
set of equations is better to be used. However, unless you indicate a
burning need, I am not going to bother to determine the cross over
criteria.

What you've provided is great. The amount of wood wasted using the
least optimal formula is pretty negligible as it is so no need to
optimize further. That should give me enough to play with quite
handily. I assume that any modern spreadsheet ought to have sin, and
cosine type functions so I should be able to plug in the values w/o much
trouble.

Thanks much Dan! Appreciate the input...

....Kevin
--
Kevin Miller - http://www.alaska.net/~atftb
Juneau, Alaska
In a recent survey, 7 out of 10 hard drives preferred Linux
Registered Linux User No: 307357, http://linuxcounter.net