Scientific American Supplement, No. 275, April 9, 1881

This cogged wheel is worked by clockwork inclosed between two disks, and would rotate continuously were it not for the catch, G, working in and out of the cogs. Through this catch, G, the wheel is dependent on the movement of electro-magnet. This cogged wheel is a double one, consisting of two wheels coupled together, exactly similar one with the other, and so fixed that the cogs of the one correspond with the void between the cogs of the others. As the catch, G, moves down it frees a cog in first wheel, and both wheels begin to turn, but the second wheel is immediately checked by catch, G, and the movement ceases. A catch again works the two wheels, turn half a cog, and so on. Each wheel contains as many cogs as there are contacts on transmitter disk, consequently as many as on circular disk, A, Fig. 3, and on brass disk within camera.

FIG. 4

FIG. 5

Having now described the several parts of the apparatus, let us see how it works. All the contacts correspond one with the other, both on the side of selenium current and that of the motive current. Let us suppose that the slide of transmitter is on contact No. 10 for instance; the selenium current starting from No. 10 reaches contact 10 of rectangular transmitter, half the slide bearing on this point, as also on the parallel rail, communicates the current to said rail, thence to line, from the line to axis of cogged wheel, from axis to contact 10 of circular fixed disk, and thence to contact 10 of receiver. At each selenium contact of the rectangular disk there is a corresponding contact to the battery and electro-magnet. Now, on reaching contact 10 the intermission of the current has turned the wheel 10 cogs, and so brought the small contact, O, Fig. 5, on No. 10 of the fixed circular disk.

As may be seen, the synchronism of the apparatus could not be obtained in a more simple and complete mode–the rectangular transmitter being placed vertically, and the slide being of a certain weight to its fall from the first point of contact sufficient to carry it rapidly over the whole length of this transmitter.

The picture is, therefore, reproduced almost instantaneously; indeed, by using platinum wires on the receiver connected with the negative pole, by the incandescence of these wires according to the different degrees of electricity we can obtain a picture, of a fugitive kind, it is true, but yet so vivid that the impression on the retina does not fade during the relatively very brief space of time the slide occupies in traveling over all the contacts. A Ruhmkorff coil may also be employed for obtaining sparks in proportion to the current emitted. The apparatus is regulated in precisely the same way as dial telegraphs, starting always from first contact. The slide should, therefore, never be removed from the rectangular disk, whereon it is held by the grooves in the brass rails, into which it fits with but slight friction, without communicating any current to the line wires when not placed on points of contact.

[Continued from SUPPLEMENT No. 274, page 4368.]

THE VARIOUS MODES OF TRANSMITTING POWER TO A DISTANCE.[2 - A paper lately read before the Institution of Mechanical Engineers.]

By ARTHUR ACHARD, of Geneva

But allowing that the figure of 22 H. P., assumed for this power (the result in calculating the work with compressed air being 19 H. P.) may be somewhat incorrect, it is unlikely that this error can be so large that its correction could reduce the efficiency below 80 per cent. Messrs. Sautter and Lemonnier, who construct a number of compressors, on being consulted by the author, have written to say that they always confined themselves in estimating the power stored in the compressed air, and had never measured the gross power expended. Compressed air in passing along the pipe, assumed to be horizontal, which conveys it from the place of production to the place where it is to be used, experiences by friction a diminution of pressure, which represents a reduction in the mechanical power stored up, and consequently a loss of efficiency.

The loss of pressure in question can only be calculated conveniently on the hypothesis that it is very small, and the general formula,

, is employed for the purpose, where D is the diameter of the pipe, assumed to be uniform, L the length of the pipe, p

the pressure at the entrance, p the pressure at the farther end, u the velocity at which the compressed air travels, Δ its specific weight, and f(u) the friction per unit of length. In proportion as the air loses pressure its speed increases, while its specific weight diminishes; but the variations in pressure are assumed to be so small that u and Δ may be considered constant. As regards the quantity f(u), or the friction per unit of length, the natural law which regulates it is not known, audit can only be expressed by some empirical formula, which, while according sufficiently nearly with the facts, is suited for calculation. For this purpose the binomial formula, au + bu², or the simple formula, b

u², is generally adopted; a b and b

being coefficients deduced from experiment. The values, however, which are to be given to these coefficients are not constant, for they vary with the diameter of the pipe, and in particular, contrary to formerly received ideas, they vary according to its internal surface. The uncertainty in this respect is so great that it is not worth while, with a view to accuracy, to relinquish the great convenience which the simple formula, b

u², offers. It would be better from this point of view to endeavor, as has been suggested, to render this formula more exact by the substitution of a fractional power in the place of the square, rather than to go through the long calculations necessitated by the use of the binomial au + bu². Accordingly, making use of the formula b

u², the above equation becomes,

; or, introducing the discharge per second, Q, which is the usual figure supplied, and which is connected with the velocity by the relation,

, we have

. Generally the pressure, p

, at the entrance is known, and the pressure, p, has to be found; it is then from p

that the values of Q and Δ are calculated. In experiments where p

and p are measured directly, in order to arrive at the value of the coefficient b

, Q and Δ would be calculated for the mean pressure ½(p

+ p). The values given to the coefficient b

vary considerably, because, as stated above, it varies with the diameter, and also with the nature of the material of the pipe. It is generally admitted that it is independent of the pressure, and it is probable that within certain limits of pressure this hypothesis is in accordance with the truth.

D'Aubuisson gives for this case, in his Traité d'Hydraulique, a rather complicated formula, containing a constant deduced from experiment, whose value, according to a calculation made by the author, is approximately b

= 0.0003. This constant was determined by taking the mean of experiments made with tin tubes of 0.0235 meter (15/16 in.), 0.05 meter (2 in.), and 0.10 meter (4 in.) diameter; and it was erroneously assumed that it was correct for all diameters and all substances.

M. Arson, engineer to the Paris Gas Company, published in 1867, in the Mémoires de la Société des Ingénieurs Civils de France, the results of some experiments on the loss of pressure in gas when passing through pipes. He employed cast-iron pipes of the ordinary type. He has represented the results of his experiments by the binomial formula, au + bu², and gives values for the coefficients a and b, which diminish with an increase in diameter, but would indicate greater losses of pressure than D'Aubuisson's formula. M. Deviller, in his Rapport sur les travaux de percement du tunnel sous les Alpes, states that the losses of pressure observed in the air pipe at the Mont Cenis Tunnel confirm the correctness of D'Aubuisson's formula; but his reasoning applies to too complicated a formula to be absolutely convincing.

Quite recently M. E. Stockalper, engineer-in-chief at the northern end of the St. Gothard Tunnel, has made some experiments on the air conduit of this tunnel, the results of which he has kindly furnished to the author. These lead to values for the coefficient b

appreciably less than that which is contained implicitly in D'Aubuisson's formula. As he experimented on a rising pipe, it is necessary to introduce into the formula the difference of level, h, between the two ends; it then becomes

. The following are the details of the experiments: First series of experiments: Conduit consisting of cast or wrought iron pipes, joined by means of flanges, bolts, and gutta percha rings. D = 0.20 m. (8 in.); L = 4,600 m. (15,100 ft,); h= 26.77 m. (87 ft. 10 in.). 1st experiment: Q = 0.1860 cubic meter (6.57 cubic feet), at a pressure of ½(p

+ p), and a temperature of 22° Cent. (72° Fahr.); p

= 5.60 atm., p =5.24 atm. Hence p

– p = 0.36 atm.= 0.36 x 10,334 kilogrammes per square meter (2.116 lb. per square foot), whence we obtain b

=0.0001697. D'Aubuisson's formula would have given p

– p = 0.626 atm.; and M. Arson's would have given p

– p = 0.9316 atm. 2d experiment: Q = 0.1566 cubic meter (5.53 cubic feet), at a pressure of ½(p

+ p), and a temperature of 22° Cent. (72° Fahr.); p

= 4.35 atm., p = 4.13 atm. Hence p

– p = 0.22 atm. = 0.22 X 10,334 kilogrammes per square meter (2,116 lb. per square foot); whence we obtain b

= 0.0001816. D'Aubuisson's formula would have given p

– p = 0.347 atm; and M. Arson's would have given p

– p = 0.5382 atm. 3d experiment: Q = 0.1495 cubic meter (5.28 cubic feet) at a pressure of ½(p

+ p) and a temperature 22° Cent. (72º Fahr.); p

= 3.84 atm., p = 3.65 atm. Hence p

– p = 0.19 atm. = 0.19 X 10,334 kilogrammes per square meter (2.116 lb. per square foot); whence we obtain B

= 0.0001966. D'Aubuisson's formula would have given p

– p = 0.284 atm., and M. Arson's would have given p

– p = 0.4329 atm. Second series of experiments: Conduit composed of wrought-iron pipes, with joints as in the first experiments. D = 0.15 meter (6 in.), L – 0.522 meters (1,712 ft.), h = 3.04 meters (10 ft.) 1st experiments: Q = 0.2005 cubic meter (7.08 cubic feet), at a pressure of ½(p