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Posted by Robert Swinney on May 3, 2008, 11:39 pm
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Paul sez: Now I'll have to get a copy of Law's book and see if my formulas
match his
numbers. Good go, Paul. While you're at it, see if you can dupe the math that
correlates the radii
of standard end mills with that of gear cutters. I haven't tried to reconcile
the math yet but I
recently made some 45 DP gear cutters (form) with the radius of a 3/32" end
mill. This was
according to a Jerry Keiffer article in Home Shop Machinist. I spoke briefly
with Jerry at the
Sherline booth at N.A.M.E.S. He put me onto the idea that most gear DPs can be
matched with the
radii of standard end mills. Thus far, I've only tried 56DP with a 3/32 end
mill. It runs
perfectly wtih the gears of my Sherline lathe.
Bob Swinney
>I have ivan Law's book, "Gears and Gear Cutting" and a copy of John
>Stevenson's paper, "Cutting
> involute Gears with Form Tools". In each are detailed instructions for
> making "button type" form
> tools for the fabrication of gear cutters. Choose any specific gear and
> pressure angle and the 2
> sets of instructions show different sizes of buttons, pin centers, and
> etc. Can anyone elaborate on
> why this is so? I would think the specifications for making precision
> gear cutters would be
> consistent among "authorities".
>
> Bob Swinney
Years back, as a mental exercise I worked out the math to duplicate
Stevenson's chart for 14 1/2 deg pa.
When I cross checked my math, I found some discrepancies with his 20 deg pa
chart. It wasn't much, but I lost interest and never followed it up. It
might have been a clearance issue or simply which tooth count is really the
average of the cutter range.
Paul K. Dickman
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Posted by Paul K. Dickman on May 4, 2008, 12:04 pm
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> Paul sez: Now I'll have to get a copy of Law's book and see if my
> formulas match his
> numbers. Good go, Paul. While you're at it, see if you can dupe the math
> that correlates the radii
> of standard end mills with that of gear cutters. I haven't tried to
> reconcile the math yet but I
> recently made some 45 DP gear cutters (form) with the radius of a 3/32"
> end mill. This was
> according to a Jerry Keiffer article in Home Shop Machinist. I spoke
> briefly with Jerry at the
> Sherline booth at N.A.M.E.S. He put me onto the idea that most gear DPs
> can be matched with the
> radii of standard end mills. Thus far, I've only tried 56DP with a 3/32
> end mill. It runs
> perfectly wtih the gears of my Sherline lathe.
>
>
> Bob Swinney
Am I right in assuming that you are using the side of the endmill to cut the
concavity on the flank of the form tool?
While some may work perfectly, the main reason it would be useful for other
ranges, is that for most of what you or I would use gears for, close enough
is good enough.
Heck, the gears in windmills were made out of logs and sticks. If backlash,
perfectly smooth power transmission, and high speed operation are taken out
of the requirements, all gears have to do is mesh.
The formulas I came up with are these:
First, you needed to figure the pitch radius of the gear from the standard
formulas.
The base circle radius (from which the arc of the involutes is defined)
BCR = PR*cos(PA)
Then the tooth radius (radius of the form cutter)
TR = tan(PA)*BCR
This is based on 1DP so you have to divide by the DP you are using.
Also, remember those numbers are radii not diameters.
I have a formula for the center to center spacing of the circular pins, but
it contains some scrawled additions, that I don't remember the logic behind,
and this is where the discrepancy showed up
You could take the numbers on Stevenson's chart and divide them by standard
cutter diameters, the ones with quotients close to whole numbers, would be
the DPs they would work for.
Paul K. Dickman
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Posted by Robert Swinney on May 4, 2008, 10:59 pm
Please log in for more thread options Paul sez:
"Am I right in assuming that you are using the side of the endmill to cut the
concavity on the flank of the form tool?"
Precisely. Unlike Ivan Law's and John Stevenson's method, the form tool is not
used to make a gear
cutter. The form tool unto itself is the gear cutter, used in fly-cutter
fashion.
Paul, please elaborate a bit on your equations: BCR = PR*cos(PA) and TR =
tan(PA)*BCR In each,
please explain what the term (PA) represents.
Thank you,
Bob Swinney
> Paul sez: Now I'll have to get a copy of Law's book and see if my
> formulas match his
> numbers. Good go, Paul. While you're at it, see if you can dupe the math
> that correlates the radii
> of standard end mills with that of gear cutters. I haven't tried to
> reconcile the math yet but I
> recently made some 45 DP gear cutters (form) with the radius of a 3/32"
> end mill. This was
> according to a Jerry Keiffer article in Home Shop Machinist. I spoke
> briefly with Jerry at the
> Sherline booth at N.A.M.E.S. He put me onto the idea that most gear DPs
> can be matched with the
> radii of standard end mills. Thus far, I've only tried 56DP with a 3/32
> end mill. It runs
> perfectly wtih the gears of my Sherline lathe.
>
>
> Bob Swinney
While some may work perfectly, the main reason it would be useful for other
ranges, is that for most of what you or I would use gears for, close enough
is good enough.
Heck, the gears in windmills were made out of logs and sticks. If backlash,
perfectly smooth power transmission, and high speed operation are taken out
of the requirements, all gears have to do is mesh.
The formulas I came up with are these:
First, you needed to figure the pitch radius of the gear from the standard
formulas.
The base circle radius (from which the arc of the involutes is defined)
Then the tooth radius (radius of the form cutter)
This is based on 1DP so you have to divide by the DP you are using.
Also, remember those numbers are radii not diameters.
I have a formula for the center to center spacing of the circular pins, but
it contains some scrawled additions, that I don't remember the logic behind,
and this is where the discrepancy showed up
You could take the numbers on Stevenson's chart and divide them by standard
cutter diameters, the ones with quotients close to whole numbers, would be
the DPs they would work for.
Paul K. Dickman
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Posted by Paul K. Dickman on May 5, 2008, 11:25 am
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> Paul sez:
>
> "Am I right in assuming that you are using the side of the endmill to cut
> the
> concavity on the flank of the form tool?"
>
> Precisely. Unlike Ivan Law's and John Stevenson's method, the form tool
> is not used to make a gear
> cutter. The form tool unto itself is the gear cutter, used in fly-cutter
> fashion.
>
> Paul, please elaborate a bit on your equations: BCR = PR*cos(PA) and TR
> = tan(PA)*BCR In each,
> please explain what the term (PA) represents.
>
>
> Thank you,
>
>
> Bob Swinney
>
PA stands for pressure angle.
Let us say that you have decided to make a 37-47 metric transposition set of
gears in 56DP 20deg PA.
For that you would need a #3 cutter which will cut 35 to 54 tooth gears.
Cutters are sized for the lowest in the range, so we design it for 35 tooth
gears.
First work your calculations like you were making a 1DP gear to simplify the
calculations.
35 tooth gear 1dp has a pitch diameter of 35 inches or a pitch radius of
17.5.
Base circle radius or BCR = PR*cos(PA) = 17.5 * cos(20deg) = 17.5 * .9396 =
16.443
Tooth radius (the radius of the pins in Stevenson's method) or TR =
tan(PA)*BCR = tan(20deg) * 16.443 = 5.9848
Diameter if the pins is 2* TR or 11.9696. Stevenson's chart shows 11.97.
Divide that by your desired DP 11.9696/56 and you need pins of .2137
diameter
For 14.5 deg PA gears, substitute the cosine and tangent of 14.5 des for
those of 20 deg.
The formula I had for the pin spacing was 2(cos(PA+(1/4 the angular spread
of the teeth))*TR)+ the chordal thickness from tables.
This is where it fell apart and I can't find the drawings I used to work out
the formula, so I can't really explain the logic I used to derive it.
Paul K. Dickman
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Posted by Don Foreman on May 5, 2008, 12:07 pm
Please log in for more thread options On Sun, 4 May 2008 21:59:43 -0500, "Robert Swinney"
>
>Paul, please elaborate a bit on your equations: BCR = PR*cos(PA) and TR =
tan(PA)*BCR In each,
>please explain what the term (PA) represents.
>
>
>Thank you,
>
>
>Bob Swinney
I'll bet PA represents pressure angle.
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>Stevenson's paper, "Cutting
> involute Gears with Form Tools". In each are detailed instructions for
> making "button type" form
> tools for the fabrication of gear cutters. Choose any specific gear and
> pressure angle and the 2
> sets of instructions show different sizes of buttons, pin centers, and
> etc. Can anyone elaborate on
> why this is so? I would think the specifications for making precision
> gear cutters would be
> consistent among "authorities".
>
> Bob Swinney