*Project Gutenberg*. Public domain.

The club-tooth lever escapement is really the most complicated escapement made. We mean by this that there are more factors involved in the problem of designing it correctly than in any other known escapement. Most—we had better say all, for there are no exceptions which occur to us—writers on the lever escapement lay down certain empirical rules for delineating the several parts, without giving reasons for this or that course. For illustration, it is an established practice among escapement makers to employ tangential lockings, as we explained and illustrated in Fig. 16.

Now, when we adopt circular pallets and carry the
locking face of the entrance pallet around to the
left two and a half degrees, the true center for the
pallet staff, if we employ tangent lockings, would
be located on a line drawn tangent to the circle *a
a* from its intersection with the radial line *A
k*, Fig. 21. Such a tangent is depicted at
the line *s l’*. If we reason on the
situation, we will see that the line *A k* is
not at right angles to the line *s l*; and, consequently,
the locking face of the entrance pallet *E* has
not really the twelve-degree lock we are taught to
believe it has.

[Illustration: Fig. 21]

We will not discuss these minor points further at
present, but leave them for subsequent consideration.
We will say, however, that we could locate the center
of the pallet action at the small circle *B’*
above the center *B*, which we have selected
as our fork-and-pallet action, and secure a perfectly
sound escapement, with several claimed advantages.

Let us now take up the delineation of the exit pallet.
It is very easy to locate the outer angle of this
pallet, as this must be situated at the intersection
of the addendum circle *i* and the arc *g*,
and located at *o*. It is also self-evident
that the inner or locking angle must be situated at
some point on the arc *h*. To determine this
location we draw the line *B c* from *B*
(the pallet center) through the intersection of the
arc *h* with the pitch circle *a*.

Again, it follows as a self-evident fact, if the pallet
we are dealing with was locked, that is, engaged with
the tooth *D’’*, the inner angle
*n* of the exit pallet would be one and a half
degrees inside the pitch circle *a*. With
the dividers set at 5”, we sweep the short arc
*b b*, and from the intersection of this arc
with the line *B c* we lay off ten degrees, and
through the point so established, from *B*, we
draw the line *B d*. Below the point of
intersection of the line *B d* with the short
arc *b b* we lay off one and a half degrees, and
through the point thus established we draw the line
*B e*.