80,400 ------ = 10,242 pounds, 7.85 or double what it was before. It will be seen, then, that the tractive force of a locomotive is dependent upon (1) the average steam pressure in the cylinders, (2) the area, (3) the stroke of the pistons, and (4) the diameter of the driving-wheels. QUESTION 311. _How is the tractive power of a locomotive calculated?_ _Answer._ BY MULTIPLYING TOGETHER THE AREA OF THE PISTON IN SQUARE INCHES, THE AVERAGE STEAM PRESSURE IN POUNDS PER SQUARE INCH ON THE PISTON DURING THE WHOLE STROKE, AND FOUR TIMES THE LENGTH OF THE STROKE OF THE PISTON,[78] AND DIVIDING THE PRODUCT BY THE CIRCUMFERENCE OF THE WHEEL. The result will be the tractive power exerted in pounds. The adhesion must of course always exceed the tractive force, otherwise the wheels will slip. [78] This length may be taken in feet, inches or any other measure, but in making the calculation the circumference of the wheel must be taken in the _same_ measure as the stroke of the piston. QUESTION 312. _How is the locomotive made to advance by causing the wheels to revolve?_ _Answer._ The pressure of steam in the cylinders is exerted in one direction against the piston, and in the opposite direction against the cylinder head, as shown in fig. 192, in which the steam is represented by the dotted shading in the back end of the cylinder, and the direction of the pressure by the darts _s_, _s_. The pressure against the piston is communicated by the connecting-rod to the crank-pin _E_, and that on the cylinder-head is carried to the axle by the frame _F F′_, and the direction of the two forces is indicated by the two darts, _a_ and _b_. We may now regard the spokes of the wheels as acting as levers, and assume that the fulcrum is either at the centre _G_ of the axle, or at _B_, the point of contact of the wheel with the rail.[79] We will assume that it is at the centre _G_ of the axle and for the sake of even figures that the wheel is six feet in diameter and cylinders have two feet stroke. We will also suppose that the engine is supported so that the wheels do not touch the rails, and that a chain or rope passing over a pulley _C_ is attached to the wheels at _B_ and with a weight at _D_. We now have a force, _a_, of 10,000 pounds exerted on the crank-pin, or at the end of the short arm _E G_ of the lever _E G B_. As _E G_ is one foot and _G B_ three feet long, 10,000 would be balanced by 10,000 × 1 ---------- = 3,333 pounds, 3 at _B_. In other words, it would require 3,333 pounds suspended from the chain at _D_ to resist the strain at _E_. But when this is the case, the pressure of the axle at the fulcrum, in the direction of the dart _c_, is equal to the pressure against the crank-pin _E_ added to that exerted by the weight _D_ at _B_, or 10,000 + 3,333 = 13,333 pounds.