THE ATMOSPHERIC RAILWAY SYSTEM-IMPROVED PLAN. SIR,-I beg leave to forward for insertion in your pages (if acceptable) a plan for an atmospheric railroad, which seems to promise all the advantages of a locomotive railway, with more than its ordinary safety. I am acquainted with Mr. Pilbrow's plan, as far as your work gives it; but I must acknowledge I cannot perceive what is to prevent the piston outstripping the loaded carriages. The Dalkey Railroad, I think, possesses no arrangement for crossing on the level, nor any that allows of the trains stopping at intermediate stations, or going the whole length of the line without stopping, and in either case of travelling without attention from any party but the conductor. The "hot iron" of the Clegg and Samuda system is certainly no recommendation; and if it could be dispensed with, there would, doubtless, be a great advantage gained. I remain, yours respectfully, Description. B. C. G. It is proposed to have one centre main exhausting tube, for the whole length of the line, or such length as the power of the stationary engines will admit of, &c. The working tube to be in sections, as required, and connected with the main as shown in the sketches.-W is the main ; o, the main continued under a roadway; B, the working tube; F, sliding valve, shutting off communication with the atmosphere; while, at the same time, by means of the wheel R, the valve G opens a communication with the main exhausting tube, W, through the connecting pipe a. At a proper part on each side of the leading carriage would be fixed the wheels X, to travel the slides d d, (the ends, dd, being on their centres at fig. 1, rr,) thus raising the rack e upon the wheel R, as required, to act upon the valves F and G. Fig. 3 is an inside view of the valve F; H is a plate, ground true on one side, to slide over the opening upon the leather I, (which might be united to the casting by india rubber) and not quite flush with the inside of the tube. The top, v, of the plate H, is intended to fit close under the leather of the long valve. Fig. 4 shows the long valve raised by the piston slide L. Fig. 5 is a side view of the piston, and the slide L, which is intended to lift the valve; showing also the place of connexion, E, with the carriage. It will be observed, that should any accident prevent the opening of the valve H, the carriages would proceed notwithstanding, without the possibility of incurring any damage. Any shock that might arise from the percussion of the piston against the valve, as it is but a light weight, would be obviated by a spiral spring inside the hollow piston. To ensure the safe passage of the piston from one section to the other, it is proposed to enlarge the ends of the tubes, say to 4 in. inside diameter more than the size of the piston, as shown in fig. 2, and gradually to decrease the diameter until it assumes its proper size, and fits the piston, &c. a short distance before the longitudinal valve commences at C, fig. 1 and 2; which valve is continued some 6 or 8 feet, unconfined by any pressure of the atmosphere, till it reaches the section valve, thus permitting an easy entrance for the piston slide under the long valve, and providing against the ingress of air in front of the piston at the opening of the section valve. The long valve may be effectually protected from the weather by a kind of tarpauling cloth, as indicated by the dotted line, fig. 4. The other valves are necessarily protected by their peculiar construction. All being thus secured from injury by the weather, there does not appear anything to prevent them from wearing well-if they can be made air-tight. Some good self-lubricating plan would probably effect that object, and under that idea, it is proposed that soft grease should exude from conveniently-sized holes, n n, round the fore part of the piston, and on both sides the piston slide, as shown at one side marked n n n. The hollow piston being provided with another hollow cylinder inside, as far as y, and the space between the two, holding the lubricating matter, a piston, v, aided by a slight spiral spring at s, so that the matter may be kept up to the surface, the frictional demand would induce the required supply. By this means it is supposed that the tube and valves would be sufficiently and equally greased; for the suction and connection valves, when once done, would supply themselves by their own action from the waste from the tube. Thus the long valve, by bearing on the convex edge, CASE IN RAILWAY ENGINEERING. is at A and the radius the direct distance from A to D. This inference leads us directly to the method of resolving the problem; for it is a well-known principle in geometry, that when two straight lines are drawn from two points any how placed without the circle, to another point in the circumference, the sum of these lines is a minimum, when they make equal angles with the tangent at the point of concourse, and as a consequence, with the radius of the circle to which the tangent is drawn. With the three given distances construct the right-lined triangle ABC, making A B=25 miles; A Č=31 miles, and B C 22 miles; then with the radius A D=12 miles and centre A, describe the circle MD N. Let D be the point in the circumference sought, and through the point D, draw the tangen RD S, intersecting the distance A Cin the point H. From the given points B and C to the point of contact D, draw the straight lines B D and CD; then if the angles B DR, CD S, or AD B, A DC, R B THE CASE IN RAILWAY ENGINEERING (PAGE 111.) It appears from the conditions of the problem, that the distance of the town A from the station of supply at D, is given in magnitude, but not in position as regards the direction of the distances from A to B, and from A to C respectively; for it is stated in the enunciation, that the engineer was at liberty to lay it down in any position he pleased, limited, of course, to the angle contained between the direct distances of A from B, and of A from C. From this we infer, that the station D is situated in the circumference of a circle, of which the centre are equal between themselves, the straight lines D B, D C, when taken conjointly, shall be less than if they were drawn to any other point in the circumference; this, therefore, is the condition that the problem requires to be satisfied. From the point B let fall the perpendicular B P, meeting A C in the point P, and through D draw DQ and D G respectively parallel to B P and A C; so shall D G be perpendicular to B P and DQ to A C. Through the point H, where the distance A C is intersected by the tangent R S and parallel to the radius A D, draw the straight line EHF meeting B D produced in F; then from the figure thus constructed are the principles of solution to be deduced. Since the straight line B P is perpendicular to A C, we have by the property of the triangle, A B2-B C2-A P2~ C P2; from which equation the segments A P and C P can readily be found, we therefore consider A Pas known. Because the straight line H E is parallel to the radius A D, the triangles A CD and H CE are similar to one another; and because the straight line D F, which is the production of B D, falls upon the parallel lines A D and F E, it makes the alternate angles A DF and D F E equal between themselves; but the angles A ID and H I F, are equal, being opposite angles; consequently, the triangles À I D and HIF are similar to one another. Therefore, by the property of similar triangles we obtain the following analogies, viz., from the first pair of triangles A CD and HCE, it is AC: AD::HC: HE; and from the second pair, viz. AID and HI F, it is AI: AD::HI: HF. But HE and H F are obviously equal, for the angle FDH is equal to the angle BDR, being opposite angles, and by the nature of the problem, the angles BDR and C D S are equal; therefore, the angle EDH is equal to the angle A D2 A C+AC A Q :ACAQ. FDH and DH is common to the two triangles E DH and, F D H, so that H E and H F are equal; from this it follows, that there is an antecedent and consequent, the same in each of the foregoing analogies, and by expunging those terms, the remaining ones, by the comparison of ratios, give AC: HC:: AI: IH; from which, by composition, we obtain A C+HC: AC:: AI+IH : A I. Since D Q is perpendicular to A C, and ADH a right angle, the triangles AD Q and A H D are similar; from which we obtain AQ AD: AD: A H = A D2 A Q But B G is manifestly equal to the difference between B P and G P, or between BP and D Q; that is, B G=BP-DQ; and furthermore D G=A P-A Q; therefore, by substituting these values of BG and D G in the above expression for IQ, we obtain IQ=DQ (AP-A Q) BP-BQ Now A I is equal to A Q-IQ; that is, A I=A QDQ (A P-A Q). BP-DQ and H C is equal to A C-A H; that is, HC-AC-AD2 A Q Let these values of A I and H C be substituted instead of them in the analogy obtained by composition, and it becomes DQ (A P-AQ) BP-B Q DQ (A P-A Q) BP-DQ In order to simplify the succeeding steps of the investigation, and to avoid confusion in the result, it becomes necessary to substitute symbols for the several quantities or lines in the figure; and for this purpose, Put b=A C, the greatest distance, or base of the triangle A B C ; dA P, the distance from the centre A to the point where the perpendicular BP meets the base A C; +IH: A Q p=B P, the perpendicular from the angle B on the base A C; 7A D, the radius of the circle, or the distance from A to the station of supply at D; x=A Q, the abscissa, or distance from A, where the perpendicular from D meets the base A C; And y=D Q, the ordinate or perpendicular from D on A C. Then, by substituting each of these 66 symbols in the above analogy for the 2bx - 72 : b :: 2. px-dy. : ; p-y and this, by equating the products of the extremes and means, becomes (2bx − r2) (px−dy) =br2, p-y from which, by expansion, we obtain 2bpx2-pr2x-2bdxy + br2y + dr3y=bpr. Now this is evidently an equation belonging to a conic section; and by calling to mind the nature of these curves, we find it to be the equation of a hyperbola between the asymptotes. Consequently, if this hyperbola be constructed according to the rules laid down for that purpose by the writers on conic sections, one of its branches will intersect the circle in the point D. It would, however, render the figure by far too complicated to show the method of construction in this place; we shall therefore omit it, and proceed to determine the things required by means of the equation alone. In order to eliminate y from the above equation, we must endeavour to express it in terms of x and known quantities, and by referring to the figure, it will readily be perceived how that is to be done; for, by the property of the rightangled triangle, we have A D2=A Q2+D Q2; that is, 2x2+y2; consequently, by substituting in this equation the square of the value of y deduced from the preceding one, we get x2+4bp2 (bx4-r2x3—br2x2) + p2rs (b + x)2 ̧ 4bd (bdx2 - br2x − dr2x) + (b + d)2p4 an equation of the fourth degree, having all its terms, and in order to prepare it for solution, we must restore the numerical values of the several quantities which it contains. From the expression A B2-B C2=A P2 - P C2 we get A P=d=17·7742, and B P2-p2-309-0778; consequently, by introducing these numbers, together with the values of band the powers of r, as expressed in the equation, it will be found, after reduction, that one of the roots or values of x AD C=A D B=66° the angle of direction sought. = is equal to 11 very nearly. Taking it at 11, to avoid fractions, and simplify the calculation, we shall have D Q=4·7958; therefore, by trigonometry, we get AD: AQ:: rad. : sin. A D Q; that is, 12 11 :: 1 : 0·91667, which corresponds to the sine of 66° 27′ very nearly. Since A Q-11, and A C=31, it follows by subtraction, that C Q==20; and by trigonometry it is DQ: CQ :: rad. : tan. C Q D; that is, 4.7958: 20 :: 1: 417029, which corresponds to the tangent of 76° 31', taking it to the nearest minute. Therefore, by addition, it is 27′+ 76° 31′=142° 58′, it is sin. 76° 31′: 20 :: 1: CD=20·5669 miles; and again we have 25 sin. 142° 58′ :: 12: sin. 16° 48′-angle A B D ; Hence again it is sin. 16° 48′ 12 :: and the sum of the two distances is 34.9255 miles. Some of our readers will probably recognize in the above problem the one which was proposed by Alhazen, the celebrated Moorish astronomer, and which was solved in all its cases by the French nobleman, the Marquis de l'Hospital. We believe that the above solution is drawn out on principles something similar to those employed by the Mar sin. 20° 14′ : D B=14.3586 miles; We do not presume to say that the several results are rigorously correct, the numerical co-efficients being so complicated, and the quantity of calculation so great, that verification was wholly out of the question; the errors, however, if any exist, will, it is hoped, be found very trifling. |