from the escape-wheel center _A_, as shown, by laying off thirty degrees
on each side of the intersection of the vertical line _i_ (passing
through the centers _A B_) with the arc _a_, and then laying off two and
a half degrees on _a_ and establishing the point _f_, and through _f_
from the center _A_ draw the radial line _A f'_. Through the point _f_
we draw the tangent line _b' b b''_, and at the intersection of the line
_b_ with _i_ we establish the center of our pallet staff at _B_. At two
and a half degrees from the point _c_ we lay off two and a half degrees
to the right of said point and establish the point _n_, and draw the
radial line _A n n'_, which establishes the extent of the arc of angular
motion of the escape wheel utilized by the pallet arm.

[Illustration: Fig. 90]

We have now come to the point where we must exercise our reasoning
powers a little. We know the locking angle of the escape-wheel tooth
passes on the arc _a_, and if we utilize the impulse face of the tooth
for five degrees of pallet or lever motion we must shape it to this end.
We draw the short arc _k_ through the point _n_, knowing that the inner
angle of the pallet stone must rest on this arc wherever it is situated.
As, for instance, when the locking face of the pallet is engaged, the
inner angle of the pallet stone must rest somewhere on this arc (_k_)
inside of _a_, and the extreme outer angle of the impulse face of the
tooth must part with the pallet on this arc _k_.

HOW TO LOCATE THE PALLET ACTION.

With the parts related to each other as shown in the cut, to establish
where the inner angle of the pallet stone is located in the drawing, we
measure down on the arc _k_ five degrees from its intersection with _a_,
and establish the point _s_. The line _B b_, Fig. 90, as the reader will
see, does not coincide with the intersection of the arcs _a_ and _k_,
and to conveniently get at the proper location for the inner angle of
our pallet stone, we draw the line _B b'_, which passes through the
point _n_ located at the intersection of the arc _a_ with the arc _k_.
From _B_ as a center we sweep the short arc _j_ with any convenient
radius of which we have a sixty-degree scale, and from the intersection
of _B b'_ with _j_ we lay off five degrees and draw the line _B s'_,
which establishes the point _s_ on the arc _k_. As stated above, we
allow one degree for lock, which we establish on the arc _o_ by laying
off one degree on the arc _j_