Found a web page that describes how to calculate moment of inertia of two gears: http://www.tech.plym.ac.uk/sme/desnotes/gears/gearaccel.htm Basically, if inertias are Ia and Ib, and relative speed of second gear is n, then total inertia is: Ia + n^2*Ib Can use this to find total key/wippen/hammer inertia as follows: Let key, wippen, and hammer moments be Ik, Iw, Ih Further let Ko = key "out" lever arm = balance rail to capstan Wi = wippen "in" lever arm = capstan to wippen center Wo = wippen center to jack/knuckle contact Hi = jack/knuckle contact to hammer center. Relative angular speed of wippen to key would be Ko/Wi. Relative angular speed of hammer to wippen would Wo/Hi. Then total inertia would be Ik + (Ko/Wi)^2*(Iw + (Wo/Hi)^2*Ih) or: Ik + (Ko/Wi)^2*Iw + ((Ko/Wi)*(Wo/Hi))^2*Ih -Mark
This PTG archive page provided courtesy of Moy Piano Service, LLC