Page of calculations and a diagram concerning vehicle center of gravity and mass distribution.
| Identifier | ExFiles\Box 107\4\ scan0089 | |
| Date | 16th January 1928 guessed | |
| contd :- -2- A. {Mr Adams} is front axle of wt. m_a B. is rear " " " m_b 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/a = m_b/m_a = 1.59 ∴ d = 4.17' c = 6.63' E G = sqrt(DE^2 + DG^2) = 1.66' Then, sprung wt. X F G = unsprung wt. X E G. {Mr Griffiths - Chief Accountant / Mr Gnapp} ∴ F G = .408' Also k^2_F = k^2 about F. {Mr Friese} k^2_G = " " G. {Mr Griffiths - Chief Accountant / Mr Gnapp} And since the k^2 about its c.g. is the minimum k^2 that any body can have in any particular plane, ∴ k^2_G = k^2_F + FG^2 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).