The general arrangement of the system is shown in the engraving.  Behind the return pulleys, D D, are attached cords, A A, with some parachutes strung upon them.  These present their openings to the current and preserve the tension of the connecting ropes.  At the further end of each cord is a board, B, which is kept in a vertical plane, but lying at a slight angle to the direction of the current; and this acts to keep the two moving ropes apart from each other.  The two return pulleys are, however, connected by a line, E, which can be shortened or lengthened from the pontoon, and in this way the angle of inclination between the two ropes can be varied if required.  A grooved pulley presses upon the trailing span at the moment before it reaches the circumference of the drum.  It is mounted on a screwed spindle, which is depressed by a nut, and thus makes the wet rope grip the outside of the drum in a thoroughly efficacious manner.

The author has made a theoretical investigation of the power which may be developed by the system, and has worked out tables by which, when the velocity of the current and the other elements of the problem are known, the power developed by any given number of parachutes can be at once determined.  We do not reproduce this investigation, which takes account of the resistance of the returning parachutes and other circumstances, but will content ourselves with quoting the final equation, which is as follows:  T = 0.328 S V cubed.  Here T is the work done in H.P., S is the total working area in sq. m., and V is the velocity of the current in m. per sec.  Taking V = 1, and S = 1 sq. m., which is by no means an impracticable quantity, we have T = 0.328 H.P. per sq. m.  We may check this result by the equation given, in English measures, by Rankine—­“Applied Mechanics,” p. 398—­for the pressure of a current upon a solid body immersed in it.  This equation, F = 1.8 m A v squared / 2g, where m is the weight of a unit of volume of the fluid—­say 62 lb.—­A is the area exposed, and v the relative velocity of the current.  Mr. Jagn finds that the maximum of efficiency is obtained when the rope moves at one-third the velocity of the stream.  If this velocity be 3 feet per second, we shall have v = 2. and we then get F = 7 lb. per sq. ft. very nearly.  Now 1 sq. meter = 10.76 sq. ft., and a speed of 1 ft. per second (which is that of the rope) is 60 ft. per minute.  Hence the H.P. realized in the same case as that taken above will be 7 x 10.76 x 60 / 33,000 = 0.137 H.P.  The difference between the two values is very large, but Rankine, of course, depends entirely on the value of the constant 1.8, which is quite empirical, and is for a flat band instead of a hollow parachute.  Taking, however, his smaller figure, and an area of 544 square inches, which Mr. Jagn has actually employed, we get a gross power of = 0.137 x 544 = 7.43 H.P.  Hence it will be seen that the amount of power which can be realized by the system is far from being inconsiderable.