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).
The formula for calculating spring deflection with worked examples.
| Identifier | ExFiles\Box 158\4\ scan0040 | |
| Date | 12th June 1936 guessed | |
| -3- Deflection: After obtaining the correct wire size, the next thing to calculate is the rate of the spring. The usual formula for deflection is: F = (8 P D³ N) / (G d⁴) Wherein: F = deflection in inches P = load D = mean dia. of spring, i.e., O.D. minus the wire size N = Number of active coils G = Modulus of the material in torsion = 11,500,000 for steel d = Wire diameter in inches Let us now continue with the problem and assume the deflection desired to be 1/2" for the 5-lb. load. Substituting: 1/2 = (8 x 5 x (1)³ x N) / (11,500,000 x (.0625)⁴) 87.6 = 40 N N = 2.19 coils The spring would be made with 2.25 coils and if squared and ground ends were desired two more inactive coils would be added for this, making a spring of 4¼ coils. If the solid height of a spring is a determining factor, this formula can have other substitutions. For instance, suppose 7/16" was the maximum solid height, this would allow 6.8 coils of 1/16" wire or practically 5 active coils. The rate of the spring becomes: F = (8 x 5 x (1)³ x 5) / (11,500,000 x (.0625)⁴) F = 200 / (11,500,000 x (.0625)⁴) F = 200 / 175 = 1.14" | ||