If the dynamo be wound directly on the axle, it must be designed to exert the couple, L, corresponding to the maximum load, when revolving at an angular velocity, w, the difference of potential between the terminals being the available E.M.F. of the conductor, and the current the maximum the armature will safely stand. This will be the case in the Charing-cross Electrical Railway. But when the dynamo is connected by intermediate gear to the driving wheels only, the product of L and _w_ remains constant, and the two factors may be varied. In the present case L is diminished in the ratio of 7 to 1, and _w_ consequently increased in the same ratio. Hence the dynamo, with its maximum load, must revolve at 588 revolutions per minute, and exert a couple of forty-seven foot-pounds. Let E be the potential of the conductor from which the current is drawn, measured in volts, C the current in amperes, and E1 the E.M.F. of the dynamo. Then E1 is proportional to the product of the angular velocity, and a certain function of the current. For a velocity [omega], let this function be denoted by _f_(C). If the characteristic of the dynamo can be drawn, then _f_(C) is known. We have then w E1 = -------- f [Omega] (1.) If R be the resistance in circuit by Ohm's law, E - E1 C = -------- R w = E ------- f(C) [Omega] ---------------- R and therefore [Omega](E - CR) (2.) w = ----------------- f(C) Let _a_ be the efficiency with which the motor transforms electrical into mechanical energy, then-- Power required = L w = a E1 C w = a C ------- f(C) [Omega] Dividing by _w_, a C f(C) L = -------- . (3.) [Omega]