>The issue is really one of the "moment of inertia" which is a measure of >the resistance of an object to changes in its rotational motion, or torque. >It is not quite the same as Newton's second law f=ma. > True. You need to use the rotational version: torque = inertia * angular_acceleration. They are analagous though. >If I recall my basic physics (and there's no guarantee of that), you are >calculating the torque as a function of the angular acceleration, the >distance from the axis of rotation and the mass. Torque is analogous to force. Measured as force * radius from center of rotation (balance rail and center pins). >In the key, the axis of >rotation is clearly at the balance rail. In the hammer shank assembly, I >don't recall whether you calculate the axis of rotation from the flange >center pin, or from the knuckle mounting, Hammer center would be the center of rotation, torque would be knuckle radius (e.g. 17mm) X force on knuckle. >but I believe it is from the >knuckle mounting since that is where the force is being applied in the >second class lever. >Similarly the wippen lever must be taken into account. >The formula goes something like t = m*r^2*a. >Torque equals mass x distance from the axis of rotation >x angular acceleration. But I'm not >sure about this. Just force X distance. >The torque in a system of compound levers, I believe, is >a simple sum. Take input torque, multiply by product of all the levers. I think this is effectively input_torque/SWR. >The question would be, then, how coordinated changes in the >set of levers such that the overall action ratio remains equal effect the >moment of inertia in each system. In more practical language: what happens >to the torque when the knuckle is at 16mm and the key ratio is at .48 >versus when the knuckle is at 17mm and the key ratio is at .52; all other >things being equal. The issue become complex because each lever has mass, >distance from the axis of rotation and acceleration which are distintly >different from the other (the key's angular acceleration is much different >from that of the hammer shank). What will be changing in this experiment >is only the r (or does the angular acceleration of each component change, >though the relationship between them remains the same??? Not sure.) >Anyway, since the input "r" is squared, it's hard to imagine that changes >in three unequal levers that keep the overall action ratio the same will >not yield differences in the moment of inertia when all components are >added together. > >I'll take my answer on the air--and good luck. Still working on it... but let's start with this example: 1st put 50g on key to overcome BW and friction. Any additional force will move key. Take hammer ratio = 130 / 17 wippen ratio = 92 / 60 key ratio = 116 / 223 If we put 100 g at key front, then torque will be 100*223=22300 g mm. Force at capstan will be 22300/116 = 192 g. (same torque, smaller radius, higher force). Now put 192 g force on wippen to get 60*192=11520 g mm torque on wippen. Output force at jack will be 11520/92=125 g. Torque on shank is 125 * 17 = 2125g mm. Effective force on shank at radius of hammer is 2125/130=16 g. -Mark
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