. 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 . Fig. 17. Whcu the two radii are equal, the same formulae apply by makingR = R. In this case, we have COS. (E-\-S) = COS. S — ~^ ; 2 72 AC= BC= 2Rs.^E. Example. Given d = .42, g = , 5=1° 20, 6 = 11, and the an-gles of the two frogs each 7°, to find A C = B C =^A B. Thecommon radius 72, corresp
. 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 . Fig. 17. Whcu the two radii are equal, the same formulae apply by makingR = R. In this case, we have COS. (E-\-S) = COS. S — ~^ ; 2 72 AC= BC= 2Rs.^E. Example. Given d = .42, g = , 5=1° 20, 6 = 11, and the an-gles of the two frogs each 7°, to find A C = B C =^A B. Thecommon radius 72, corresponding to F = 7°, is found (^ 51) to Then 2 72= 1187, 6 — 2 (/= , and ^-1187 =. Therefore, nat. cos. {E -\-S) = .99973 — .00856 = .99117 ;whence E-^ S = 1°Z1 15. Subtracting S, we have E = 6° 17 15Next 2 72 = 1187 i i?; = 3° 8 37^ sin. ^ C= ! 813557 38 CIRCULAR CURVES. C. Turnout from Curves. 57. Problem. Given the radius R of the cadre line of the mairtrack and the frog angle F, to determine the position of the frog by meansof the chord B F {figs. 18 and 19), and to find the radius R of the centre line of the Fig. 18. Solution, I. When the turnout is from the inside of the cunrt(fig. 18). Let A G and CF be the rails of the main track, AB theswitch rail, and the arc ^i^the outer rail ot the turnout, crossing theinside rail of the main track sliF. Then, since the angle E FK has itssides perpendicular to the tangents of the two curves at F, it is equal tothe acute angle made by the crossing rails, that is, E FK = F. AlsoE B L ^ S. The first step is to find the angle B KF denoted by find this angle, we have in the triangle B FK{Tab. X. 14), BK-\-KF:BK—KF= tan ^ (B FK -{- FB K): tan. ^ (BFK— F B K).But B K = R -\- ^ g — d, and K F =^ R — ^ g. Therefore, B K-^KF = 2R — d, and BK — KF =■- g - d. Moreover, BFK =BFE + EFK= BFE + F, and FBK= EBF—EBK =BFE — S. Therefore, BFK—FBK = F-^ S. Las
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectrailroadengineering
