unts to 5 1/2deg. + the angle of the
loss; the angle W A U embraces the above angles besides 1/4deg. for run.
If the locks are equal on each pallet, it proves that the lifts are also
equal. This gives us a practical method of proving the correctness of
the drawing; to do so, place the dividers on the locking circle M N at
the intersection of T A and V A with it, as this is the extent of
motion; transfer this measurement to N, if the _actual_ lift is the same
on each pallet, the dividers will locate the point which the locking
corner N will occupy _when locked_; this, in the present case, will be
at an angle of 1 3/4deg. below the tangent D A. By this simple method,
the correctness of our proposition that the loss of lift should be
measured from the outside circle of the wheel, can be proven. We often
see the loss measured for the engaging pallet on the primitive
circumference G H, and on the real circumference for the disengaging; if
one is right then the other must be wrong, as there is a noticeable
deviation of the tangent C A from the primitive circle G H at the
intersection of the locking circle M N; had we added this amount to the
lifting angle V' A V of the engaging pallet, the result would have been
that the discharging edge O would be over 1deg. below its present
location, thus showing that by the time the lift on the engaging pallet
had been completed, the locking corner N of the disengaging pallet would
be locked at an angle of 2 3/4deg. instead of only 1 3/4deg. Many
watches contain precisely this fault. If we wish to make a draft showing
the pallets at any desired position, at the center of motion for
instance, with the fork standing on the line of centers, we would
proceed in the following manner: 10 1/4deg. being the total motion,
one-half would equal 5 1/8deg.; as the total lock equals 1 3/4deg., we
deduct this amount from it which leaves 5 1/8 - 1 3/4 = 3 3/8deg., which
is the angle at which the locking corner M should be shown above the
tangent C A. Now let us see where the locking corner N should stand; M
having moved up 5 1/8deg., therefore N moved down by that amount, the
lift on the pallet being 5 1/2deg. and on the tooth 3deg. (which is
added to the tangent D A), it follows that N should stand
5 1/2 + 3 - 5 1/8 = 3 3/8deg. above D A. We can prove it by the lock,
namely: 3 3/8deg. + 1 3/4 = 5 1/8deg., half the remaining motion. This
shows how simple it is to draft pallets in various positions,
rememb