I thought we had an article in the Technical Issues area of our Home Page site that explains this but apparently we don't so here goes nothin':
Think "fulcrum" and "lever arm." The force applied to any object in rotation is equal to the force applied to the end of the lever times the length of the lever from the axios of rotation. That's why torque wrenches are calibrated in (typically, for automotive work) foot-pounds. If you have a force of 100 pounds applied to the end of a lever that's 12 inches (one foot) long, you have a torque of 100 foot pounds. Cut that lever to 6 inches in length and apply the same 100 pounds of force to the end, and you now only have a torque of 50 foot-pounds.
Now consider a 1911 hammer. In cocking, you have two levers acting on it. The hammer rotates around the hammer pin. (:duh. That's the axis. The mainspring acts on the hammer though the hammer strut, and the hole for the hammer strut pin is 0.23255 inches from the center of the hammer pin. That dimension is fixed.
In order to cock the hammer, a force has to be applied somewhere on the hammer to counteract the force being applied by the mainspring. If you cock the hammer by hand, your thumb rests somewhere on the hammer spur, which is considerably farther away from the hammer pin axis than the hammer strut pin hole, so cocking is comparatively easy. But what happens when the hammer is cocked by racking the slide -- either manually, or by firing the pistol? The slide now cocks the hammer, but what's the lever arm distance? It's the distance from the hammer pin axis to wherever the point of contact is between the slide and the face of the hammer. It's actually not the slide itself that makes the contact, it's the firing pin retainer. And the larger the radius on the bottom of the firing pin retainer, the higher up on the face of the hammer is the point of contact. Longer lever arm ==> reduced cocking effort.
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