Before we proceed to delineate the exit pallet of our escapement, let us reason on the relations of the several parts.
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.
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_.