Speed in miles per hour V Resistance of air (according to the received empirical formula) 1 = --- (frontage area) × V^{2} 400 Frontage area of train in square feet 63 Frontage area of portion of train unprotected by the tender, in square feet 24 For a locomotive train therefore 24 F = R + WK + --- V^{2} + (50 + W) 2240 G. 400 For a system that dispenses with the locomotive 63 Tractive force = WK + --- V^{2} + W 2240 G. 400 Therefore W (K + 2240 G) + ·1575 V^{2} = the useful tractive force, and R + 112000 G - ·0975 V^{2} = the tractive force wasted by the use of the locomotive. Therefore F={R + 112000 G-·0975 V^{2}} + {W (K + 2240 G) +·1575 V^{2}} and the useful load (F- R - 112000 G - ·06 V^{2}) W = ----------------------------- K + 2240 G. The values which Mr. Gooch’s experiments give for the two selected speeds are as follows[74]:-- +---------------+---------+-----------------+----------+ |Miles per Hour | R (lbs.)| K (lbs. per ton)| F (lbs.) | +---------------+---------+-----------------+----------+ | 40 | 1500 | 12·5 | 5200 | | 60 | 2100 | 18·6 | 4900 | +---------------+---------+-----------------+----------+ Using these values, the results in the following table are obtained, being the conditions appropriate to the two speeds at successive ascending gradients:--