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).
Calculations and specifications for vehicle road spring stiffness, deflection, and stress.
| Identifier | ExFiles\Box 43\2\ Scan265 | |
| Date | 12th December 1930 guessed | |
| - 2 - We started by making these springs 3¾" deflection (½ negative camber). Stiffness = 1150 / 3.75 = 305 and this has since been specified as 315 ± 20 lbs./in. 2. Rear Springs (a) Length at front 26.5 Length at rear 29.5 Ratio center load to rear end load = 4 x (29.5 / 26.5) = 4.45 Weight on rear tires = W Axle weight 680 lbs. 4 passengers and 200 lbs. luggage 730 lbs. Weight on springs at end = W + 50 Originally working with 10.5" deflection at rear this gave stiffness at center = ((W + 50) / (2 x 10.5)) x 4.45 = .212 (W + 50). Later the stiffness was reduced and is now calculated to the formula Stiffness = .199 (W + 50) which it will be found agrees with the chart. Note that working loads are maintained at (4.45 / 2) (W + 50). Our method of calculating a road spring is comparatively simple thus: 1. From consideration of the extreme curvatures - i.e. free and maximum bump - calculate total change of curvature from free to maximum stressed condition of each plate. For example a 45" long spring, 4" free camber, curvature is (8 h / L²) = (8 x 4 / 45²) = .0158 when flat. From flat to 3" bump, curvature is = .0119 Total .0277 Allowing a maximum stress 100,000 lbs./sq. in. then the maximum thickness of plate we can use through this range is (2 x stress) / (Elasticity x curvature) = 200,000 / (30,000,000 x .0277) = .241 so under these conditions the thickness of plate must not exceed .240. (Continued) | ||