Calculation of the moment of inertia (k²) for car model 11-G-111 with a landaulet body by the swinging method.
| Identifier | ExFiles\Box 5\2\ 02-page128 | |
| Date | 1st December 1927 guessed | |
| X5830 4538 V5830 TO FIND k² FOR 11-G-111 BY SWINGING. (LANDAULET BODY, SPARE WHEEL AT REAR). Let k²_GT = k² of total car (including axles) about c.g. of total car. = (T²pg / 4π²) - p² ft² Where p = distance of c.g. from point of support when swinging = 10.167' T = 3.9 secs. Ther'f k²_GT = 22.6 ft². Weight of total car = 4004 lbs. ∴ I_GT = 4004 X 22.6 = 90,500 lbs.ft². From this must be subtracted the axle inertias about the same point. Weights are ( Front 305 lbs. k² = 1² + 6.083² ( Rear 485 lbs. k² = 1² + 4.708² Inertia for two = 305(1² + 6.083²) + 485(1² + 4.708²) = 22,840 lbs.ft². ∴ I_G = Inertia of sprung part about c.g. of sprung + unsprung parts = 67,660 lbs.ft². Weight of sprung part = 3214 lbs. ∴ k_G² = 67,660 / 3214 = 21.06 ft². To find position of c.g. of sprung part. contd :- | ||
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).