John Sims's Problem. 1830
John Sims, a Welshman, furnished the following suggested device to "Mechanics' Magazine" in 1830:
Let us suppose an apparatus to be constructed of the description represented in the annexed engraving: a is a water cistern, whence water is to be raised by the pump b, to supply the cistern; c d is a small pipe with a stop-cock at e, which lets the water from cistern c into a strong water-tight bellows f. The bellows have no valve, but a cock g to let out the water into cistern a; h is a weight, and i a rack on the top of the bellows which works in the cogs on the axle of the large cog-wheel j; j turns the little cog-wheel k, that gives motion to the arm l, and works the pump-handle m; n is an upright rod on the end of the lever o, which rod has a turn at p and q for the top of the bellows to press against in ascending and descending. The water being let into the bellows from the pipe d, will cause the top of the bellows, with the weight and rack, to ascend till the former reaches and presses p, which will move the lever o and the arm or rod r; by which means the stop-cock e of the pipe will be shut, and the cock g opened, and the water let in from the bellows into the cistern a. The top of the bellows will now descend till it comes down and presses the turn q, which will again shut the cock g and open e, on which the water will again flow from the pipe into the bellows, and cause the top with the rack to ascend.
Now it is generally known that the power of an hydrostatic bellows is thus calculated:—
As the area of the orifice or section of the pipe,
To the area of the bellows:
The weight of water in the pipe is,
To the weight the bellows will sustain on the top-board.
We will suppose, therefore, the pipe d to be 10 feet high, with a bore equal to 1 square inch, which would give 120 cubic inches, and about 4¼ lbs. of water. Let us suppose, also, the boards of the bellows to be 20 inches square, which gives 400 square inches. When the water is let from the pipe into the bellows, there will be a pressure of 4¼ lbs. on every square inch, which on the whole will amount to 1,700 lbs. Now take half of this force and place it on the top of the bellows; there will then be a working power of 850 lbs. up and down, and allowing the bellows to raise one foot, it will contain about 20 gallons of water. Now the question is, will not the machinery, with a moving power of 2 feet and 850 lbs., raise 20 gallons of water 10 feet, which would, of course, cause the motion to be perpetual?—John Sims.
Pwllheli, North Wales, Dec. 11, 1829.
The foregoing device brought from another correspondent the following:
Had Mr. Sims gained the power exerted by the descending weight on his bellows, he would have been fortunate indeed; but it unfortunately happens that its returning power (or an equivalent) was expended in raising it.
With respect to his question, whether a circulation of water would be kept up by the arrangement, I answer, no; as the velocities will be in the inverse ratios to the forces, and the descending column of 120 inches must expend itself forty times to raise the ascending one to the height of twelve inches, as proposed:—
10 ft. or 120 in. × 40 = 4,800, lifting force or power.
400 in. × 12 = 4,800, opposing force, resistance, or weight.
Here is an equilibrium, and nothing gained to overcome friction or the weight of the atmosphere on the piston of the pump. Were it possible to annihilate both friction and atmospheric weight, even then, unless the power exceed the weight, the power would not be a moving one.