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 for spring load, section moduli, and stiffness.
| Identifier | ExFiles\Box 43\2\ Scan198 | |
| Date | 28th September 1937 guessed | |
| - 3 - W = W₁ + W₂ = M/L₁ + M/L₂ = (L₁+L₂)/(L₁ L₂) M If L₁ = 21" W = .092 M L₂ = 22¼" Or, for each plate W = .092 x f (bt{Capt. J. S. Burt - Engineer}²/6) Say b = 2". bt{Capt. J. S. Burt - Engineer}²/6 = for .285 plate - .027 } - Main Plate. .250 - .0208} .218 - .0158} Section Moduli. .187 - .0117} If f.{Mr Friese} is taken as 100,000 lbs/in² on maximum bump, one can then find the max.load produced by each plate thickness thus:- .285 plate W = .027 x 9200 = 248 lbs. .250 " W = .0208 x 9200 = 191 lbs. .218 " W = .0158 x 9200 = 145 lbs. .187 " W = .0117 x 9200 = 108 lbs. STIFFNESS OF SPRING. The stiffness of the spring is found by figuring the load at normal riding position (say ½" neg.) and deducting this from the load at max. bump, and dividing the increase in load by the deflection from normal to maximum positions. The stress on each plate at normal position is first found by the same method as the stress at max. bump. The radius at normal position R₀ = (L₁ L₂)/(2d) = (L₁ L₂)/1 (for ½" neg.) Rₓ for each plate is known (see above) | ||