_Answer._ The force required to make them slip will, as already stated, vary very much with the condition of the rails. If they are quite dry and clean it will require a force equal to about one-fourth the weight on the wheels. That is, supposing we have a wheel, _A B_, fig. 192, attached to a frame which is fastened so that it cannot move, and that the wheel rests on a rail and is loaded with say 10,000 pounds, if now a rope or chain could be attached at a point, _B_, exactly at the tread of the wheel, and carried over a pulley, _C_, then it would require a weight, _D_, of about 2,500 pounds attached to the end of the rope to make the wheel slip. If the rails were sanded, the adhesion would be somewhat greater, and if they were wet or muddy or greasy, considerably less. For ordinary circumstances the adhesion may safely be assumed to be _one-fifth of the weight of the driving-wheels_. Of course the total weight on all the driving-wheels must be taken in calculating the adhesion. Thus, if a locomotive has four driving-wheels and each one of them bears a load of 10,000 pounds, then the total weight on the driving-wheels, or _adhesive weight_, as it is called, will be 10,000 × 4 = 40,000 pounds, and the adhesion will be 40,000 ------ = 8,000 pounds. 5 QUESTION 310. _What is meant by the tractive power of a locomotive?_ _Answer._ It is the force with which the locomotive is urged in a horizontal direction by the pressure of the steam in the cylinders, and which therefore tends to move the locomotive and draw the load attached to it. The tractive power is of course due to the pressure of steam on the piston, and therefore its amount is dependent upon the average steam pressure in the cylinder, on the area of the piston, and also on the distance through which the pressure is exerted, or, in other words, on the stroke of the piston. Thus if we have a cylinder 16 inches in diameter and two feet stroke and an average steam pressure of 50 pounds per square inch, then as the area of such a piston would be 201 square inches, the average pressure on it would be 201 × 50 = 10,050 pounds, and as each piston moves through four feet during one revolution of the wheels, the number of foot-pounds of energy exerted by it would be 10,050 × 4 = 40,200, and for the two cylinders of a locomotive double that amount, or 80,400 foot-pounds. Now if the driving-wheels are five feet in diameter, their circumference will be 15.7 feet, and therefore the locomotive will move that distance on the rails during one revolution, if the wheels do not slip. The 80,400 foot-pounds of energy is therefore exerted through a distance of 15.7 feet, and therefore 80,400 ------ = 5,121 pounds, 15.7 which is the force exerted through each foot that the circumference of the wheel revolves and the locomotive moves. If the wheels were only half the diameter, or 2¹⁄₂ feet, then their circumference would be 7.85 feet and the tractive power would be