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.

LOCATING THE INNER ANGLE OF THE EXIT PALLET.