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.

The intersection of the line _B e_ with the arc _h_, which we will term
the point _n_, represents the location of the inner angle of the exit
pallet. We have already explained how we located the position of the
outer angle at _o_. We draw the line _n o_ and define the impulse face
of the exit pallet. If we mentally analyze the problem in hand, we will
see that as the exit pallet vibrates through its ten degrees of arc the