. Field-book for railroad engineers. Containing formulas for laying out curves, determining frog angles, levelling, calculating earth-work, etc., etc., together with tables of radii, ordinates deflections, long chords, magnetic variation, logarithms, logarithmic and natural sines, tangents, etc., etc . f the main track. A B (fig. 14) repre-sents one of the rails of the main track switched, B /represents theouter rail of the turnout curve, tangent to A B, and E shows the posi-lion of the frog The switch angle, denoted by S, is the angle DAB,^urnled by the switched rail A B with A D, its former
. Field-book for railroad engineers. Containing formulas for laying out curves, determining frog angles, levelling, calculating earth-work, etc., etc., together with tables of radii, ordinates deflections, long chords, magnetic variation, logarithms, logarithmic and natural sines, tangents, etc., etc . f the main track. A B (fig. 14) repre-sents one of the rails of the main track switched, B /represents theouter rail of the turnout curve, tangent to A B, and E shows the posi-lion of the frog The switch angle, denoted by S, is the angle DAB,^urnled by the switched rail A B with A D, its former position in themain track. The frog angle, denoted by E, is the angle G EM madeIjy the crossing rails, the direction of the turnout rail at 7^ being thetangent EM at that jjoint. In the problems of this article the gaugeot the track D C. denoted by g, and the distance D B, denoted by dare supposed to be known. The switch angle S is also supposed tobo known, since its sine (Tab. X. 1) is equal to d divided by the lengtu Ori CIRCULAR CURVES. of the switched rail. If, for example, the rail is 18 feet in lengih andi = .42, we have S == P 20. A. Turnout from Straight Lines. 50. ProfolCBll. Given the radius R of the centre line of a tur*-oui(JiQ. U), to fold the frog angle G FM = F and the chard B Solution. Through die centre E draw E K parallel to the n :track. Draw Ci/and FK perpendicular to E K, and join L , since E Fis perpendicular io F?,J and FK is perpendicular toFG, the angle E rK = G FIvl = F; and since E B and B H arerespectively perpendicular to A B and A D, the angle E B H ^ DAB = S. Now the tianglc E FEgives (Tab. X. 2) cos. E F K = f-^ But E F, the radius of the outer rail, is equal to R -\- ^ g, andFK=CH=Bn— BC=B E cos. E B H — B C = ,R-\-^ g)cos. S — (g — d). Substituting these values, we have cos. E FK ~IK + iff) COS. 5 -{g — d) B + is IS^ ,or cos. F = COS. S — 9 — ^ RTT9 From thin formula Fmay be found by the table of natural cosinesTo adapt
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectrailroadengineering
