Page discussing the unbalanced moment of inertia and linear out-of-balance using mathematical formulas.
| Identifier | WestWitteringFiles\M\2October1924-December1924\ Scan5 | |
| Date | 1st October 1924 | |
| EFCl/T6.10.24. -13- Contd. This couple can be looked upon as that due to the unbalanced moment of inertia about G of masses Mₓ - Mx {John H Maddocks - Chief Proving Officer} = (k² - ab)/(a(a + b)) M = - b/(a + b) Mg at X and Mₚ - Mp = (k² - ab)/(a(a + b)) M = - a {Mr Adams} /(a + b) Mg at P. which moment of inertia is (Mₓ - Mx {John H Maddocks - Chief Proving Officer} ) a² + (Mₚ - Mp) b² = M (k² - ab) by substitution. If linear balance be now carried a step farther, so as to balance to the extent of Mx {John H Maddocks - Chief Proving Officer} and Mp (these being greater than Mₓ and Mₚ in the present case) this con. rod couple would be provided for and the outstanding out-of-balance would then be linear, viz. that of the negative massive particle at G of magnitude (ab - k²) / ab Note. Though we can speak correctly of a pure couple out-of-balance, because such can only be balanced by another couple, we cannot speak correctly of a pure linear out-of-balance, because the equivalent of a single linear force may be another linear force plus a couple, and therefore the linear force may be balanced by a linear force and a couple, as well as by a single linear force. To this extent the^any out-of-balance referred to^as linear is equivalent to simultaneous linear and couple out-of-balances. Contd. | ||
From the Rolls-Royce experimental archive: a quarter of a million communications from Rolls-Royce, 1906 to 1960's. Documents from the Sir Henry Royce Memorial Foundation (SHRMF).