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gear docs seq
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2022-08-26
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Louis Ricker
PO BOX 917
ITALY TX 76651
USA
SELECTING GEAR TRAINS FROM A SET
OF LATHE CHANGE GEARS
This is an attempt at making gear train calculating more
productive. Both a non-computer method and a computer method will be
considered. For the benefit of both methods I have taken a look at
fundamentals and have jettisoned some lumber that has encumbered
method and mathematics.
For this undertaking one should have at hand a calculator which
has the logarithm, reciprocal, and change sign functions. Some
calculators give only natural (base e) logarithms. Others give both
natural and common (base 10) logs. Either can be used, but only one
kind of log should be used throughout a process. The reciprocal of a
number is one divided by that number. The 1/x key gives the reciprocal
of the number in the calculator display. The change sign key, +/- ,
changes the sign of the number in the display (+ to - or - to +). If a
common log is in the display the 10^x (10 raised to the x power) key
gives the number whose log is in the display. The e^x key gives the
anti log of a natural log in the display.
In addition to a calculator access to a computer is desirable. We
can manage without a computer, but it is not possible to duplicate the
work of the computer using only a calculator, because the amount of
work involved is beyond human capacity.
Although the methods discussed here can be applied to gear trains
in general, we shall be thinking only of the change gear lathe and
three applications of gear trains - screw cutting, spindle indexing,
and linear indexing. It should be noted that our methods produce many
approximate gear trains. The user must decide whether this is
tolerable.
Gear train solvers have always been plagued by multiple unknowns.
In finding a three-pair compound train one must solve a problem having
one known quantity (the ratio of the train) and six unknowns (the
Louis Ricker, Gear Trains 2
gears). There are two ways to do this. One way is to examine all the
possible combinations of a set of gears. My set consists of sixteen
gears having fourteen sizes. For one, two, or three-pair trains there
are hundreds of thousands of combinations. This must be done by
computer. The other way is to doodle. Doodling works, but it is not
altogether satisfactory. It simply leaves too many solutions
undiscovered. On the other hand, I have already received a complaint
about "the vast amount of output" of correct solutions by the
computer. I have responded by tightening the tolerances in the
programs. This has reduced the number of trains printed. In some cases
a tight tolerance will exclude all trains, and there will be no trains
printed. Then the tolerance must be widened.
In order to control the doodling some methods apply many rules.
Others add known factors to the formulas. This tampering only serves
to restrict the number of possible solutions. Often failure to
properly label values has led to logical leaps. It is one thing to do
arithmetic correctly, but quite another to give mathematics meaning.
To do this we must be careful with language.
Consider the following problem: "How many millimeters are
contained in 0.125 inch ?" The answer to this problem is "how many",
not "how many millimeters"; we already know it is millimeters. Now let
us state the problem mathematically:
Let x represent "how many".
x mm = 0.125 in.
If a quantity is divided by its equal the quotient is one:
x mm
--------- = 1 = the number of times 0.125 in. is contained in x mm .
0.125 in.
By law: 1 in. = 25.4 mm .
This is called a "conversion factor". It is not a factor.
One divided by 25.4 does not equal one, but 1 in. divided by 25.4
mm does.
1 in. 25.4 mm
------- = 1 = -------
25.4 mm 1 in.
x mm 1 in. 25.4 mm
--------- = 1 = ------- = -------
0.125 in. 25.4 mm 1 in.
We want x or "how many"-1alone-0.
0.125 in. x mm 25.4 mm 0.125 in.
--------- X --------- = ------- X ---------
1 0.125 in. 1 in. 1
Louis Ricker, Gear Trains 3
(If equals are multiplied by equals the results are equal.)
1 x mm 25.4 mm 0.125 1
---- X ---- = ------- X ----- X ----
1 mm 1 1 1 1 mm
All labels cancel.
x = 3.175 = "how many".
Another "conversion factor" is Pi or 3.1416 .
Properly labeled it becomes:
1 circumference = 3.1416 diameters.
3.1416 diameters
or ---------------- = 1
1 circumference
Pi and 25.4 are often injected into gear train formulas in the
form of gear pairs. The arithmetic is correct, but many other pairs
are excluded by this practice. For our purposes conversions should be
done up front once and for all, not hundreds of thousands of times.
In this article the gear on the spindle is a driveR ; the gear on
the lead screw is driveN.
Let "DRIVERS" equal the product of the numbers of teeth of the
driver gears. Let "DRIVEN" equal the product of the numbers of teeth
of the driven gears.
Formula 1 is for gear trains between spindle and lead screw:
Lead of thread to be cut DRIVERS
RATIO = ------------------------ = -------
Lead of leadscrew DRIVEN
Formula 2 is for spindle indexing gear trains.
Let DIV = Number of divisions of one spindle rotation.
Let INDEX = Number of teeth indexed.
Formula 2 :
DRIVERS DRIVERS
RATIO = DIV = ------- = --------------
DRIVEN INDEX X DRIVEN
DRIVERS
INDEX = ------------
DIV X DRIVEN
Louis Ricker, Gear Trains 4
Notice that the number of teeth indexed is equivalent to a gear
of that number of teeth making one revolution.
Formula 3 is for linear indexing of the leadscrew.
Required advance of carriage by leadscrew
RATIO = ----------------------------------------
Lead of leadscrew
DRIVERS INDEX X DRIVERS
RATIO = ------- = ---------------
DRIVEN DRIVEN
Formula 3 :
RATIO X DRIVEN
INDEX = --------------
DRIVERS
The ratio is the number of turns of the leadscrew per index. The
number of turns can be fractional. Notice that INDEX is a driver in
Formula 3 while INDEX was driven in Formula 2 .
THE NON-COMPUTER METHOD
The chart on my Myford 7 lathe calls for the following gear train
for a 2.25 mm thread lead:
45 X 60 X 21 DRIVERS
------------ = ------- = ACTUAL RATIO = 0.70875
40 X 40 X 50 DRIVEN
My lead screw has a lead of 0.125 in.
LEAD OF THREAD TO BE CUT 2.25 mm
DESIRED RATIO = ------------------------ = --------
LEAD OF LEAD SCREW 3.175 mm