. . .s, 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 line B d and B c change places, and the tooth D'' locks one and a half degrees. To delineate the locking face of the exit pallet, we erect a perpendicular to the line B e from the point n, as shown by the line n p.

From n as a center we sweep the short arc t t, and from its intersection with the line n p we lay off twelve degrees, and through the point so established we draw the line n u, which defines the locking face of the exit pa. . . Read More