engine with 20 tons adhesive weight one inch, if we divide 100 by 20 we will get the cylinder capacity needed for each ton. That is, 100 ÷ 20 = 5 CUBIC IN. CYLINDER CAPACITY PER TON (of 2,000 lbs.) OF ADHESIVE WEIGHT IS NEEDED TO MOVE ANY LOCOMOTIVE ONE INCH. This quantity we have named the _modulus of propulsion_. [101] The cylinder capacity is the space swept through by the two pistons. In the above illustrations what is meant is, that the average space in the cylinder swept through by the piston is 100 cubic inches for each inch that the locomotive advances. Supposing now that it is required to calculate the cylinder capacity for a locomotive with 15 tons adhesive weight, and wheels 4¹⁄₂ feet or 54 in. in diameter. We will first multiply 15 by the modulus of propulsion, 15 × 5 = 75 = the number of cubic inches of cylinder capacity required to move such a locomotive one inch. Multiplying the length of the circumference of the wheels, which in this case is 169.6 in. by 75, will give us the total cylinder capacity for one revolution. That is 169.6 × 75 = 12,720 cubic inches of cylinder capacity, or the space which should be swept through by the two pistons. Dividing this by 4 will give us the cubical contents in inches of one of the cylinders. Thus, 12,720 ÷ 4 = 3,180 cubic inches = the capacity of one cylinder. Now as the capacity of a cylinder is calculated by multiplying the area of the piston by the length of the stroke, if we have the one we can easily determine the other. Thus, supposing it was intended to make the stroke of the pistons 22 in., dividing 3,180 by 22 will give us the area of the piston. Thus, 3,180 ÷ 22 = 144.5 square inches. Now by the well-known rule in mensuration, if we DIVIDE THE AREA OF A CIRCLE BY 0.7854, THE SQUARE ROOT OF THE QUOTIENT WILL BE THE DIAMETER OF THE CIRCLE. Thus, 144.5 ÷ 0.7854 = 183.9. The square root of 183.9 is 13¹⁄₂ nearly, which should be the diameter of the cylinder. Instead of calculating the diameter of the circle, a more convenient way is to refer the area to a table of areas, and from it find the diameter. Of course if we have the diameter of the piston and want to get the stroke, we DIVIDE THE CUBICAL CONTENTS OF THE CYLINDER BY THE AREA OF THE PISTON. Thus, in the present illustration, if it was intended to have the piston 13¹⁄₂ in. diameter, we would have divided 3,180 by the area of a piston 13¹⁄₂ in. diameter, which is 143.1, so that we would have 3,180 ÷ 143.1 = 22 nearly, = inches of stroke of piston. From the above considerations we can deduce the following RULE FOR CALCULATING THE CAPACITY OF THE CYLINDERS WHEN THE ADHESIVE WEIGHT IS KNOWN: MULTIPLY THE TOTAL WEIGHT ON THE DRIVING-WHEELS IN TONS (of 2,000 lbs.) BY 5, AND THEN BY THE CIRCUMFERENCE OF THE WHEELS IN INCHES, AND DIVIDE BY 4. THE QUOTIENT WILL BE THE CUBICAL CONTENTS IN INCHES OF EACH CYLINDER. From this, if either the diameter or stroke is given the other can easily be found, as has been explained.