With a fifteen-tooth escape wheel each tooth occupies twenty-four degrees, and from the point f to e would be two and one-half tooth-spaces. We show the dotted points of four teeth at D D' D'' D'''. To establish the center of the pallet staff we draw a line at right angles to the line p e' from the point e so it intersects the line f h at k. For drawing a line at right angles to another line, as we have just done, a hard-rubber triangle, shaped as shown at C, Fig. 7, can be employed. To use such a triangle, we place it so the right, or ninety-degrees angle, rests at e, as shown at the dotted triangle C, Fig. 6, and the long side coincides with the radial line p e'. If the short side of the hard-rubber triangle is too short, as indicated, we place a short ruler so it rests against the edge, as shown at the dotted line g e, Fig. 7, and while holding it securely down on the drawing we remove the triangle, and with a fine-pointed pencil draw the line e g, Fig. 6, by the short rule. Let us imagine a flat surface placed at e so its face was at right angles to the line g e, which would arrest the tooth D'' after the tooth D resting on f had been released and passed through an arc of twelve degrees. A tooth resting on a flat surface, as imagined above, would also rest dead. As stated previously, the pallets we are considering have equidistant locking faces and correspond to the arc l l, Fig. 6.

In order to realize any power from our escape-wheel tooth, we must provide an impulse face to the pallets faced at f e; and the problem before us is to delineate these pallets so that the lever will be propelled through an arc of eight and one-half degrees, while the escape wheel is moving through an arc of ten and one-half degrees. We make the arc of fork action eight and one-half degrees for two reasons—(1) because most text-books have selected ten degrees of fork-and-pallet action; (2) because most of the finer lever escapements of recent construction have a lever action of less than ten degrees.