I have 12 bottles with an o.d. of 3.625.I need to make a 12 pocket
star wheel for these jars,and I can have no more than a .250 space
between pockets.My question is ,what is the formula for figuring the
o.d. and i.d.

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Request for Question Clarification by kyrie26-ga on 14 Dec 2002 20:39 PST

Hello ccoral-ga,

What are OD and ID respectively? Are they "outer diameter" and "inner
diameter"? What is a 12 pocket star wheel?

Thank you,

kyrie26-ga

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Clarification of Question by ccoral-ga on 15 Dec 2002 08:16 PST

Yes it does mean outer diameter and inner diameter.A 12 pocket
starwheel is usually 1/2" thick aluminum circle with 12 pockets cut
into itthat would be able to fit the 3.625 diameter of the jar.Every
30 degrees a pocket will be cut.If you put the jars in a perfect
circle I would know that the o.d.would be right in the middle of the
jar and the i.d. would be the inside of the pocket.The pocket for the
jar only needs to accept half of the jar,so essentially you have an
aluminum circle with 12 half round pockets every 30 degrees around the
outer edge.I thought that 12x3.625 divided by 3.1416 would give me the
o.d.,then subtract half the diameter of the 3.635 jar would give me
the i.d but it doesn't seem to work out,see what you can come
with.thanks

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Request for Question Clarification by kyrie26-ga on 15 Dec 2002 15:34 PST

Hi ccoral-ga,

OK, thanks for the extra information. Now, it may be just me, but I
think actually seeing an image of what you have in mind, would be
immensely helpful, and will definitely speed up an answer to your
question. If you could point us to any images, that would be great.
Just a recommendation in case you don't get any further responses from
the researchers (don't think we don't want to help) -being a visual
person, it would at least help me a lot. Thanks!


Regards,

kyrie26-ga

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Request for Question Clarification by raisingmyhand-ga on 15 Dec 2002 16:24 PST

ccoral, 

I believe there are two ways that one could describe the distance
between the pockets--either on a straight line from the edge of one
pocket to the edge of the next, or along the curve of the wheel from
the edge of one pocket to the next. If one measures along the curve,
the distance will be a little bit longer than if one imagines the
straight line between the two pockets.

Can you tell us whether the .250 space is calculated along the outer
curve or along a straight line?

thanks, 
RMH

Hi, ccoral-ga:

I think there is a basic misconception about the geometry of the "star
wheel" and its pockets.  If the teeth of this "gear" are to end at the
points of closest approach between the circular jars, then the pockets
will not be semi-circular.  Each will hold less than "half" the
circumference of the jars.

GEOMETRY

Consider a regular dodecagon (12-sided convex polygon), whose sides
are each equal in length to the diameter of a jar plus the gap space
between:

X = 3.625 + 0.25 = 3.875 (or smaller gap space if desired)

Let X be the length of an edge of a regular dodecagon.  Each edge is
the base of an isoceles triangle with vertex at the center of the
dodecagon (and at the center of the star wheel).  Each such isoceles
triangle can be bisected into two right triangles, each with a 15
degree angle at the center (and 75 degree angle at the corner of the
dodecagon).  The hypotenuse of each such right triangle is a radius of
the dodecagon's circumscribed circle, while the shorter leg (opposite
the 15 degree angle) is half an edge (so length X/2) and the longer
leg is a radius of the dodecagon's inscribed circle.

If the jars are positioned at the corners of the dodecagon, and the
respective circular arcs of their outlines removed, what remains is
what (I think) you want for the "star wheel".  A little plane geometry
shows that each pocket subtends 150 degrees of arc around each jar,
not 180 degrees, so the pockets go only 5/12ths of the way around each
jar instead of halfway around.

The outer diameter of the "star wheel" is then the diameter of the
dodecagon's inscribed circle (which passes through the tip of each
tooth in the gear).  The inner diameter of the "star wheel" is
actually the diameter of the dodecagon's circumscribed circle (which
passes through the corners on which the jars are centered) minus the
diameter of a jar.

CALCULATIONS

The outer diameter of the star wheel is the dodecagon's inscribed
diameter

 = X * cotangent(15 deg)

 = 14.462 [rounded]

The inner diameter of the star wheel is the dodecagon's circumscribed
diameter minus one jar diameter

 = X * cosecant(15 deg) - 3.625

 = 11.347 [rounded]

regards, mathtalk-ga

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Clarification of Answer by mathtalk-ga on 15 Dec 2002 22:16 PST

Perhaps this "thought experiment" will help to clarify the design.

Suppose the "gap" between jars were reduced to zero, so that the jars
actually touch (their circles are tangent).  Assuming the jars'
centers are equidistant from the "star wheel" center, they would form
12 corners of a regular dodecagon.

Increasing the gap spaces between the jars equally amounts to moving
these jars outward from the center of the star wheel, retaining the
dodecagon formation.

-- mathtalk-ga

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Request for Answer Clarification by ccoral-ga on 16 Dec 2002 16:37 PST

You obviously know your math,and I believe the answer is
correct,however what I really need to know is a written formula for
finding the o.d. and i.d.as I will be using this formula for different
sized bottles and star wheels that have less than 12 pockets.You may
have given the formula in your answer and I didn't understand it.It
was kind of over my head,If you could simplify it for someone in a
machine shop I would greatly appreciate it,thank you very much

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Clarification of Answer by mathtalk-ga on 16 Dec 2002 17:38 PST

Hi, ccoral-ga:

Glad to clarify.  I think you've identified these "inputs":

J = diameter of jar
G = width of gap
P = number of pockets (at least 3)

and these "outputs":

O = outer diameter of star wheel
I = inner diameter of star wheel

The formulas are:

O = (J+G) * cotangent( 180/P )
 = (J+G) / tangent( 180/P )

I = [(J+G) * cosecant( 180/P )] - J
 = [(J+G) / sine( 180/P )] - J

If I were trying to fabricate such pieces, I think I'd find it useful
to have also the circumscribed diameter of the P-sided polygon (in
which the star wheel is inscribed), since the centers for the pockets
sit on that "outermost" circle (the corners of the P-sided polygon):

D = circumscribed diameter of polygon

D = (J+G) * cosecant( 180/P )
 = (J+G) / sine( 180/P )

Note that D = I + J.

regards, mathtalk-ga

very precise