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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).
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 :-
  
  


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