The slope E P being 4 feet, the slope E′ P′ is consequently 2 feet; the mean slope for the entire mass is therefore 3 feet. The mean amount of work performed by the descent of the mass will of course be 3 foot-pounds per pound of water. The amount of work performed by the vertical descent of P′ P ought therefore to be one foot-pound per pound. That this is the amount will be evident thus:—The transference of the one foot of water from the equatorial column to the polar disturbs the equilibrium by making the equatorial column too light by one foot of water and the polar column too heavy by the same amount of water. The polar column will therefore tend to sink, and the equatorial to rise till equilibrium is restored. The difference of weight of the two columns being equal to 2 feet of water, the polar column will begin to descend with a pressure of 2 feet of water; and the equatorial column will begin to rise with an equal amount of pressure. When the polar column has descended half a foot the equatorial column will have risen half a foot. The pressure of the descending polar column will now be reduced to one foot of water. And when the polar column has descended another foot, P′ will have reached P, and E′ will have reached E; the two columns will then be in equilibrium. It therefore follows that the mean pressure with which the polar column descended the one foot was equal to the pressure of one foot of water. Consequently the mean amount of work performed by the descent of the mass was equal to one foot-pound per pound of water; this, added to the 3 foot-pounds derived from the slope, gives a total of 4 foot-pounds. In whatever way we view the question, we are led to the conclusion that if 4 feet represent the amount of slope between the equatorial and polar columns when the two are in equilibrium, then 4 foot-pounds is the total amount of work that gravity can perform upon a pound of water in overcoming the resistance to motion in its passage from the equator to the pole down the slope, and then in its vertical descent to the bottom of the ocean. But it will be replied, not only does the one foot of water P′ P descend, but the entire column P O, 10,000 feet in length, descends also. What, then, it will be asked, becomes of the force which gravity exerts in the descent of this column? We shall shortly see that this force is entirely applied in work against gravity in other parts of the circuit; so that not a single foot-pound of this force goes to overcome cohesion, friction, and other resistances; it is all spent in counteracting the efforts which gravity exerts to stop the current in another part of the circuit.