This is a multi-part message in MIME format. ---------------------- multipart/mixed attachment Don A. Gilmore wrote: > Hi John: > > That formula you are using won't even get you anywhere close to the total > moment of inertia. The second part is probably all right. You're assuming > that the leads are small enough that they will perform like point masses, > which will probably get you pretty close. But the first part (1/2ML^2) is > wayyyy off. > > The formula you should use for a rectangular shaft that pivots at its center > is > > I = 1/12 * m * (h^2 + L^2) > > where h is the height of the key, bottom to top. Thanks, I meant it to read 1/12ML^2. I forgot the 1 in the denominator. I changed the drawing to reflect your suggestion to include the height. It's good to have someone on the list to help with this stuff. I sent two other drawings with simple equations. Did you have a chance to view them? If there is something wrong with them it would be good to clear it up now before moving on. Thanks for taking the time to help. John Hartman RPT John Hartman Pianos [link redacted at request of site owner - Jul 25, 2015] Rebuilding Steinway and Mason & Hamlin Grand Pianos Since 1979 Piano Technicians Journal Journal Illustrator/Contributing Editor [link redacted at request of site owner - Jul 25, 2015] ---------------------- multipart/mixed attachment A non-text attachment was scrubbed... Name: MOI-of-key.jpg Type: image/jpeg Size: 28178 bytes Desc: not available Url : https://www.moypiano.com/ptg/pianotech.php/attachments/dc/01/d8/29/MOI-of-key.jpg ---------------------- multipart/mixed attachment--
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