In train A—­F: 

n            n’                         n’
--- = 1 = --------; m’ = 0, [therefore] ---- = 1, or n’ = -a;
m         m’ — a                        -a

  In train A’—­F’: 

n’           n’                         n’
--- = 1 = --------; m’ = 0, [therefore] ---- = 1, or n’ = -t;
m         m’ — t                       -t

In combining, we have in the latter train m’ = 0, t = -a, whence

n           n’           n’
--- = 1 = -------- gives ---- = 1, or n’ = a, as before.
m         m’ — t         +a

Now it happens that the only examples given by Prof.  Willis of incomplete trains in which the axis of a planet-wheel whose motion is to be determined is not parallel to the central axis of the system, are similar to the one just discussed; the wheel in question being carried by a secondary train-arm which derives its motion from a wheel of the primary train.

The application of his general equation in these cases gives results which agree with observed facts; and it would seem that this circumstance, in connection doubtless with the complexity of these compound trains, led him to the too hasty conclusion that the formula would hold true in all cases; although we are still left to wonder at his overlooking the fact that in these very cases the “absolute” and the “relative” rotations of the last wheel are identical.

[Illustration:  PLANETARY WHEEL TRAINS.  Fig. 21]

In Fig. 21 is shown a combination consisting also of two distinct trains, in which, however, there is but one train-arm T turning freely upon the horizontal shaft OO, to which shaft the wheels A’, F, are secured; the train-arm has two studs, upon which turn the idlers B B’, and also carries the bearings of the last wheel F’; the first wheel A is annular, and fixed to the frame of the machine.  Let it be required to determine the results of one revolution of the crank H, the numbers of teeth being assigned as follows: 

A = 60, F = 30, A’ = 60, F’ = 10.

We shall then have, for the train ABF (Eq.  I.),

n       60          n’ — a
--- = - ---- = -2 = --------, in which n’ = 1, m’ = 0,
m       30          m’ — a’

1 — a                            1
whence -2 = -------, 2a = 1 - a, 3a = 1, a = ---.
-a                             3

And for the train A’B’F’ (Eq.  II.),

n     60           n’                   1
--- = ---- = 6 = --------, in which a = ---, m’ = 1,
m     10         m’ — a’                3

n’
whence            6 = -----------, or n’ = 4.
1 — (1/3)

That is, the last wheel F’ turns four times about the axis LL during one revolution of the crank H. But according to Profs.  Willis and Goodeve, we should have for the second train: 

n     60         n’ — a                 1
--- = ---- = 6 = --------, in which a = ---, m’ = 1,
m     10         m’ — a’                3

n’ — (1/3)
which gives 6 = -----------, n’ - (1/3) = 4, n’ = 4-1/3,
1 — (1/3)

or four and one-third revolutions of F’ for one of H.