here 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_.

LOCATING THE INNER ANGLE OF THE EXIT PALLET.