. 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 . e have s — a = , s — h = 87,and s — c = Then bv means of the logarithms just used, wefind ^FEK= 3^ 2 45. Sul)tnicting ^ B E K = W 48, we have^BEF ^ 2° 27 57. Lastly. BF = 1368 94 sin. 2^ 27 57 = The formula ^J^ = {F+ S) (§ 5 ote) would give BF =58 906, and this value
. 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 . e have s — a = , s — h = 87,and s — c = Then bv means of the logarithms just used, wefind ^FEK= 3^ 2 45. Sul)tnicting ^ B E K = W 48, we have^BEF ^ 2° 27 57. Lastly. BF = 1368 94 sin. 2^ 27 57 = The formula ^J^ = {F+ S) (§ 5 ote) would give BF =58 906, and this value is even nearer the truth than that just found,owing, however, to no eiTor in the formulfe, but to inaccuracifs inci-dent to the calculation. 61. If the turnout is to reverse, in order to join a track parallel tothe main track, as A CB (fig. 20), it will be necessary to determinethe reversing points C and B. These points will be detennined, if wefind the angles A E C and B F C, and the chords A C and CB. 62 Problem. Given the radius D K = R {Jig 20) of the centrtline of the main truck the common radius E C = CF = R of the centreline of a turnout, and the distance B G = b between the centre lines of the^parallel tracks, to find the central angles A E C and B F C and the chordaA C and Solution. In the triangle A E K fitrd the angle AEK and the side CROSSINGS ON CURVES. i5 e K For tliis purpose we have AE = R, A K = R — d, and tlicincluded angle E A K =- S. Or, if the frog angle has been previouslycalculated by § GO, the values of A E K and E K are already known.*Find in the triangle EFK the amjles E FKand FE K For thispurpose we have E K^ as just found, E F ^ 2 A, and FK = A -^-R — h. Then AE C = AEK — FEK, and BFC ^ , (§69) ^^ AC^2Rs^AEC; C B = 2 Rsin. ^ B F C. This solution, with a few obvious modifications, will apply, whenthe turnout is from the outside of a curve. D. Crossings on Curves. 63. When a turnout enters a parallel main track by a seco
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectrailroadengineering
