Analysis of transverse axis net couple and frame twist, including calculations and a diagram of a rear suspension layout.
| Identifier | ExFiles\Box 26\1\ Scan041 | |
| Date | 19th March 1965 | |
| - 4 - Tln/CJC. 3.FL. 19.3.65 About transverse axis, net couple = We₂ - We₁ ~ twisting frame in opposite direction to longitudinal axis couple. If a = b, the twist of the frame will be zero when (e₂ - e₁) = b. Putting figures to this, if we assume that b = centroidal distance between sills = 60", the crossmembers would be conveniently situated at the dash and rear seat heelboard. The minimum value of e₁ possible = 15" so for zero twist e₂ = 75". This results in a wheelbase of 150" which is rather long. Any departure from the above figures will result in some twist but much less than with a conventional arrangement. The ideal set-up is a front engine front wheel drive car, the load crossing obviously taking place at the rear end. One possible suspension layout which achieves the desired frame loading is shown below:- FIGURE 5 STABILISING RODS. REAR WHEELARCH DAMPER FRAME FORWARD LONGITUDINAL TORSION BAR BEARING TO REDUCE BM ON TORSION BAR DUE TO CORNERING FORCE. FORGED H.T.S. LEVERS. -NATURAL FREQUENCY OF UNDAMPED LENGTH >1,000 C.P.M. Continued......... | ||
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).