# Page:EB1911 - Volume 22.djvu/861

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844
[LOCOMOTIVE POWER
RAILWAYS

carriage bodies. The two following expressions are given in the Bulletin of the International Railway Congress (vol. xii. p. 1275), by Barbier, for some experiments made on the Northern railway of France with a train of 157 tons mean weight; they are valid between 37 and 77 m. per hour:—

$r_{v}=3.58+{\frac {1.65{\mbox{S}}\left(1.61{\mbox{S}}+50\right)}{1000}}{\mbox{ for 4-wheel coaches,}}$ (13)

$r_{v}=3.58+{\frac {1.64{\mbox{S}}\left(1.61{\mbox{S}}+10\right)}{1000}}{\mbox{ for bogie coaches.}}$ (14)

The Baldwin Locomotive Company give the formulae

$r_{v}=3.36+{\frac {0.56{\mbox{S}}}{3}}$ (15)

and

$r_{v}=1.68+0.224{\mbox{S for speeds from 47 to 77 m. per hour.}}$ (16)

All the above formulae refer to carriage stock. The resistance of goods wagons has not been so systematically investigated. In the paper above quoted Aspinall cites a case where the resistance of a train of empty wagons 1830 ft. long was 18·33 ℔ per ton at a speed of 26 m. per hour, and a train of full wagons 1045 ft. long gave only 9·12 ℔ per ton at a speed of 29 m. per hour. The resistance found from the above expressions includes the components 1, 2 and 4 of § 4. The resistance caused by the wind is very variable, and in extreme cases may double the resistance found from the formulae. A side wind causes excessive flange friction on the leeward side of the train, and increases the tractive resistances therefore very considerably, even though its velocity be relatively moderate. The curves corresponding to the above expressions are plotted in fig. 17, four values of L being taken for formula (12) corresponding to trains of 5, 10, 15 and 20 bogie carriages.

The resistance at starting is greater than the running resistance at moderate speeds. From Aspinall’s experiments it appears to be about 17 ℔ per ton, and this value is plotted on the diagram.

The resistance to motion round a curve has not been so systematically studied that any definite rule can be formulated applicable to all classes of rolling stock and all radii of curves. A general result could not be obtained, even from a large number of experiments, because the resistance round curves depends upon so many variable factors. In some cases the gauge is laid a little wider than the standard, and there are varying amounts of super elevation of the outer rail; but the most formidable factor in the production of resistance is the guard-rail, which is sometimes put in with the object of guiding the wheel which runs on the inner rail of the curve on the inside of the flange.

§ 6. Engine Resistance.—From experiments made on the North-Eastern railway (see a paper by W. H. Smith on “Express Locomotive Engines,” Proc. Inst. Mech. Eng., October 1898), it appeared that the engine resistance was about 35% of the total resistance, and in the train-resistance experiments on the Lancashire & Yorkshire railway quoted above the engine resistance was also about 35% of the total resistance, thus confirming the North-Eastern railway results. Barbier (loc. cit.) gives as the formula for the engine resistance

$r_{e}=8.51+3.24{\mbox{S}}\left(1.61{\mbox{S}}+30\right)/1000$ (17)

where S is the speed in miles per hour. This formula is valid between speeds of 37 and 77 m. per hour, and was obtained in connexion with the experiments previously quoted on the Northern railway of France with an engine and tender weighing about 83 tons. Barbier’s formula is plotted in Fig. 17, together with a curve expressing generally the results of some early experiments on the Great Western railway carried out by Sir D. Gooch. The extension of the Barbier curve beyond the above limits in fig. 17 gives values which must be regarded as only very approximate.

Fig. 17

§ 7. Rate at which work is done against the resistances given by the curves in fig. 17.—When the weight of the engine and tender and the weight of the vehicles are respectively given, the rate at which work must be done in the engine cylinders in order to maintain the train in motion at a stated speed can be computed by the aid of the curves plotted in fig. 17. Thus let an engine and tender weighing 80 tons haul vehicles weighing 200 tons at a uniform speed on the level of 40 m. per hour. As given by the Barbier curves in fig. 17, the engine resistance at 40 m. per hour is 20 ℔ per ton, and the vehicle resistance 8·5 ℔ per ton at the same speed. Hence

 Engine resistance, Re = 80 × 20 = 1600 ℔ Vehicle resistance, Rv = 200 × 8·5 = 1700 ” Train resistance, R = 3300 ”

The speed, 40 m. per hour, is equal to 58·6 ft. per second; therefore the rate of working in foot-pounds per second is 3300 × 58·6, from which I.H.P. = (3300 × 58·6)/550 = 354. This is the horse-power, therefore, which must be developed in the cylinders to maintain the train in motion at a uniform speed of 40 m. per hour on a level straight road with the values of the resistances assumed.

§ 8. Rate at which work is done against a gradient.—Gradients are measured either by stating the number of feet horizontally, G say, in which the vertical rise is 1 ft., or by the vertical rise in 100 measured horizontally expressed as a percentage, or by the number of feet rising vertically in a mile. Thus a gradient of 1 in 200 is the same as a half per cent. grade or a rise of 26·4 ft. per mile. The difference between the horizontal distance and the distance measured along the rail is so small that it is negligible in all practical calculations. Hence if a train is travelling up the gradient at a speed of V ft. per second, the vertical rise per second is V/G ft. If W₁ is the weight of the train in pounds, the rate of working against the gradient expressed in horse-power units is,

${\mbox{H.P.}}={\mbox{W}}_{1}{\mbox{V}}/550{\mbox{G}}.$ (18)

Assuming the data of the previous section, and in addition that the train is required to maintain a speed of 40 m. per hour up a gradient of 1 in 300, the extra horse-power required will be

${\mbox{H.P.}}={\frac {280\times 2240\times 58.6}{300\times 550}}=223.$ This must be exerted in addition to the horse-power calculated in the previous section, so that the total indicated horse-power which must be developed in the cylinders is now 354+223 = 577. If the train is running down a gradient this horse-power is the rate at which gravity is working on the train, so that with the data of the previous section, on the assumption that the train is running down a gradient of 1 in 300, the horse-power required to maintain the speed would be 354−223 = 131.

§ 9.—Rate at which work is done against acceleration.—If W₁ is the weight of the train in pounds and a the acceleration in feet per second, the force required to produce the acceleration is

$f={\mbox{W}}_{1}a/g$ (19)

And if V is the average speed during the change of velocity implied by the uniform acceleration a, the rate at which work is done by this force is

$f{\mbox{V}}={\mbox{W}}_{1}{\mbox{V}}a/g$ (20)

or in horse-power units

${\mbox{H.P.}}={\mbox{W}}_{1}{\mbox{V}}a/550g.$ (21)

Assuming the data of § 7, suppose the train to change its speed from 40 to 41 m. per hour in 13 seconds. The average acceleration in feet per second is measured by the fraction

${\frac {\mbox{Change of speed in feet per sec.}}{\mbox{Time occupied in the change}}}={\frac {60.07-58.6}{13}}=0.113$ Therefore the horse-power which must be developed in the cylinders to effect this change of speed is from (21)

${\mbox{H.P.}}={\frac {280\times 2240\times 0.113\times 59}{550\times 32}}=237.$ The rate of working is negative when the train is retarded; for instance, if the train had changed its speed from 41 to 40 m. per hour in 13 seconds, the rate at which work would have to be absorbed by the brake blocks would represent 237 H.P. This is lost in heat produced by the friction between the brake blocks and the wheels, though in some systems of electric driving some of the energy stored in the train may be returned to the central station during retardation. The principal condition operating in the design of locomotives intended for local services with frequent stops is the degree of acceleration required, the aim of the designer being to produce an engine which shall be able to bring the train to its journey speed in the shortest time possible. For example, suppose it is required to start a train weighing 200 tons from rest and bring it to a speed of 30 m. per hour in 30 seconds. The weight of the engine may be assumed in advance to be 80 tons. The acceleration, a, which may be supposed uniform, is 1·465. The average velocity is 15 m. per hour, which is equal to 22 ft. per second; therefore the tractive force required is, from (19),

(280×2240×1·465)/32 = 28,720 ℔,

and the corresponding horse-power which must be developed in the cylinders is, from (20), fV/550, and this is with f and V equal to the above values, 1149. To obtain the tractive force the weight on the coupled wheels must be about five times this amount—that is, 