Diagram and calculations for determining the center of gravity of a vehicle's sprung and unsprung mass.
| Identifier | ExFiles\Box 107\4\ scan0084 | |
| Date | 28th November 1927 guessed | |
| contd :- -2- A. {Mr Adams} is front axle of wt. ma B. is rear " " mb G. {Mr Griffiths - Chief Accountant / Mr Gnapp} is c.g. of whole car including axles. E. {Mr Elliott - Chief Engineer} is c.g. of axles (together). F. {Mr Friese} is c.g. of sprung part of car. F. {Mr Friese} will lie on the line joining E. {Mr Elliott - Chief Engineer} and G. {Mr Griffiths - Chief Accountant / Mr Gnapp} c/d = mb/ma = 1.59 ∴ d = 4.17' c = 6.63' E G = √ DE² + DG² = 1.66' Then, sprung wt. X F G = unsprung wt. X E G. {Mr Griffiths - Chief Accountant / Mr Gnapp} ∴ F G = .408' Also k²F = k² about F. {Mr Friese} kG = " " G. {Mr Griffiths - Chief Accountant / Mr Gnapp} And since the k² about its c.g. is the minimum k² that any body can have in any particular plane, ∴ k²G = k²F + FG² 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).