АВХАС 2 BD =EC. A B+CB -CF; add these together, and AB(AC+CB)_EC+CF. Multiply off by 2 BD, and 2 BD A B(A C+C B)=EC+C FX2 BD; but A C+C B=A B, therefore A B2=EC+CF × 2B D. Extract the square root of both sides A B=√(EC+C F) × 2 B D. Having found A B, we have, as shown above, A C=? 2 ECXBD A B PROBLEM XIII. Given the line A B, which joins the fixed tangent points A and B, the angles D A B and A B G, to find the common radius E C=C F of a serpentine or reversed curve to unite the tangents H A and B L. (cos. A K E— (DAB+A BG)) cos. (DAB-A BG)' EXAMPLE. Given A B=1500, D A B = 18°, and A B G = 6°, to find EC CF. = Given the length of the common tangent D G, and the angles of intersection CDI and B G N, to find the common radius CE CF of a reversed curve to unite the tangents H A and B L. (See last figure.) CE CF= D G cos. & C D I cos. BGN sin. (C DI+BGN) The points A and B are found by measuring from D a distance A D=CE, tan. CDI, and from G a distance B G=CE, tan.BG N. EXAMPLE. Given D G 800, C D I=14° and B G N = = 10°, to find VERTICAL CURVES. Vertical curves are used to round off the angles formed at the junction of two gradients. Let A C and C B be two gradients meeting at C. The rates of inclination of the gradients are assumed to be known. Thus, starting from A, the rates of inclination of A C and C B may be denoted respectively by r and r'; that is, r denotes what is added to the height at every station on A C, and r' denotes what is added to the height at every station on C B; but as C B is a descending gradient, the quantity is therefore minus. By the principles of the parabola we are enabled readily to unite any two gradients by a vertical curve. PROBLEM XV. Given r equal the rate of inclination of the gradient A C, and the rate of inclination of C B, and the number of stations from A and B, the tangent points to C equal, to unite the tangent points by a parabolic vertical curve. Let A E B be the parabolic curve required. Through A draw the horizontal line A K; let fall B K perpendicular on A K; produce BK to meet A C produced in F. From C let fall CH perpendicular on A K. Then, since the distance from C to A and B is measured horizontally, A H is equal H K, and therefore AD=D B. The vertical line C D is therefore a diameter of the parabola and the distances T M, T′ M′, &c., to the curve in a vertical direction from the stations on the tangent A C are to each other as the squares of the number of stations from A. That is, if d represent this distance at T, the first station from A, the distance at T', the second station, would be Γ E M 0" 4d, at the third 9d, and at B, which is 2n would be 4n'd; that is, F B=4n2d, or d = stations from A, it FB As FB is still an unknown quantity, it is requisite to find it first in order to find d. Through C draw C G parallel to A K. Then the triangles C F G and A C H are equal, and F G=CH. C H is the rise in n stations from A to C; that is, C H=nr, But or F G=n r. And G B is the rise in the second gradient C B in n stations; but as is minus, G B——n r'. Therefore F B=F G+G B=n r—n r'. If we substitute this value of F B in the foregoing formula for d, we obtain d = n r―n r 4n2 r—p The value of d being thus found, all the distances d, 4d, 9d, 16d, &c., from the tangent A F to the curve, are known. If T and T be the first and second stations on the tangent, and the vertical lines T O and T′ O′ be drawn - to the horizontal line A K, the height T O of the first station above A is r, the height T′ O′ of the second station above A is 2r, and so on for successive stations we should find the heights 3r, 4r, &c. As we have found T M=d, T′ M'=4d, &c., we shall have for the heights of the curve above the level of A, MO-TOTM—r—d, T0′ = M′ O'—T′ M'—2r—4d, and in like manner for the succeeding heights 3r-9d, 4r-16d, &c. Then to find the heights for the curve at the successive stations from A, that is, the rise of each height over the preceding height, we must subtract each height from the next following, thus: (r-d) —0—r—d, (2r—4d) — (r—d) —r—3d, (3r—9d) — (2r4d)—r—5d, (4r—16d)—(3r—9d)—r—7d, &c. The heights of the successive stations for the vertical curve are therefore r-d, r-3d, r-5d, r-7d, &c. In finding these heights close attention should be given to the algebraic signs. EXAMPLE. Let the number of stations on each side of A C be 4, and A C ascend 8 per station, and C B descend 6 per station. In this 1.4 = 16 =0875, and the heights from A to B are TURNOUTS AND CROSSINGS. The ordinary mode of turning off from a main track is switching a pair of rails in the main track, and laying a turnout curve tangent to the switched rails, placing a frog where the outer rail of the turnout crosses the rail of the main track. One of the rails of the main track switched is represented by A B; BF represents the outer rail of the turnout curve, tangent to A B; and F shows the position of the frog. The switch angle is D A B, formed by the switched rail A B with A D, its former position in the main track. GF M is the frog angle made by the crossing rails, the direction of the turnout rail at F being the tangent F M at that point. The gauge of the track D C, and the range D B of the switched rail A B, are supposed to be known. The switch angle D A B, the sine of which by formula 1, p. 160, is equal D B divided by the length of the switched rail. If the switched rail should be 20 feet long, and DB5 of a foot, then the angle D A B = 1° 41′. PROBLEM XVI. Given the radius BE of the centre line of a turnout, to find the frog angle G F M and the chord B F. By this formula GFM may be found by a table of natural cosines. Given D C, the width of the gauge, equal 4 feet 8 inches, or 4.708 feet, DB equal half a foot, or 5', the angle D AB D B = -1° 41′, and the radius B E 500, to find the angle G F M = and the chord B F. = |