Calculations to find the radius of gyration squared for an 11-G-111 landaulet body by swinging.
| Identifier | ExFiles\Box 107\4\ scan0083 | |
| Date | 28th November 1927 guessed | |
| X4538 TO FIND K² FOR 11-G-111 BY SWINGING. (LANDAULET BODY. SPARE WHEEL AT REAR). X5830 Let K²_GT = K² of total car (including axles) about c.g. of total car. = T² p g / 4 π² - p² ft² Where p = distance of c.g. from point of support when swinging = 10.167' T = 3.9 secs. Then 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 spring + 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).