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604 TRANSACTIONS OF THE A.S.M.E. OCTOBER, 1941We should consider then the approximate formulaThe average of these three cases gives y = 0.0847.The case of empty freight cars has not been considered, as itis unlikely that empty freight cars would be transported bymodern high-speed locomotives.Equation [1] is written in absolute units of foot, pound, andsecond—see Lionel S. Marks, Mechanical Engineer’s Handbook,first edition, 1916, page 73 (symbols in lower-case letters). Forunits which are customary in railroad engineering, capital lettersare used; for miles of length, one mile equals 5280 ft; for speed3600 1F in miles per hour V = rrion X v — - - g(T X v, where v is in fps.5280 ' ' ' 1.4671For acceleration in miles per hour per second A =1.467 X a,where a is in fpsps; and for mass M in tons of 2000 lb of weight =32.17 1—---- = ------ tons of weight.2000 62.17 6For these latter units, a constant C must be introduced in the1right side of Equation [1] (10, p. 19; 25, p. 49) equal to C =62.17 X1 1 dV------ = ------ , if V is in miles per hour, t in seconds, — in miles1.467 91.18 dtper hour per second, (T — R) in lb of weight, and M in tons ofweight.dsRemembering that v = — and canceling ds, Equation [11 willdtas close to the average conditions of modern trains. However,if the consist and y of the train are known in advance, a more accuratecoefficient can be figured out and formula [la] should beused instead of [2a],If the time-speed curve, namely, speed versus time, is thedVone sought, then — is the tangent to this curve. Equation [2]shows that the tangent is represented by a simple relation, thedifference between tractive effort and train resistance in poundsper ton of train weight divided by 100 for American units, mile,hour (respectively, second), and ton. If curves of Fig. 1 representtractive effort of the locomotive and R the train resistanceversus speed V, then the ordinates of the shaded area ABCDArepresent in a certain scale the right-hand side of Equation [2],The desired acceleration curve will be found if a curve is so builtthat the tangents to it are equal to the ordinates divided by98.69 or 100, as the case may be.Suppose that a train stands in a station and the locomotivestarts to accelerate it from standstill on a level. The excess oftractive effort over the train resistance on a level at low speedsfrom zero will be represented by the shaded area, and under theinfluence of this difference of forces the train will be acceleratedfrom zero speed until the balanced speed Vo is reached (Fig. 1),at which point the two forces (tractive effort and train resistance)readcLVwhere V is speed in miles per hour, and t time in seconds; — acdtceleration in miles per hour per second.Since y for trains of different consist varies from 0.04 to 0.1233,with an average of 0.0847, the author assumed that Equation[la] can, for average conditions, be rewrittenwhere — is acceleration in miles per hour per second: t and rdtare tractive effort and train resistance in pounds per ton of theirweight, and the coefficient corresponds to 91.18(1 + y) = 91.18X 1.0847 = 98.69, or approximately 100. F i g . 1 T r a c t i v e E f f o r t a n d T k a i n R e s i s t a n c e