. 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 . rossing, to find the chords A C and B C Solution. Draw E G perpendicular to the main track, and A LCM, and B D parallel to it. Denote theangle A E C by E. Then,since the angle A E L = AUG = S, we have CE L = E -\- S,and in the right triangle CE jV (Tab. X. 2), CE cos. OEM =R cos. {E -]- S)=^ EM=
. 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 . rossing, to find the chords A C and B C Solution. Draw E G perpendicular to the main track, and A LCM, and B D parallel to it. Denote theangle A E C by E. Then,since the angle A E L = AUG = S, we have CE L = E -\- S,and in the right triangle CE jV (Tab. X. 2), CE cos. OEM =R cos. {E -]- S)=^ EM= EL — L M. But EL = AE cos. A EL= R cos. S, and L M : L M = A C : B C Now AC: B C ^E C: CF= R: R>. Therefore, L M: LM = R: R\ or L M: LM■\- LM= R: R + R; that is, L M: b — 2d = R : R -\-R, whence L M = j^ , „, -. Substituting these values of E L and L Mm the equation for R cos. {E + S), we have R cos. {E -\- S) = R cos. S —■R{b — 2d) R4- R ■> CROSSINGS ON STRAIGHT LINES. S? G^ / n 1 e\ c b — 2d COS. {L + S) = COS. o — — A + R Having thus found jE + S, we have the angle E and also its equalVFB. Then (§ 69) irr- ^C= 2i2sin. iJE;; Z5 C = 2 72sin. ^ ^. We have also A D = A C-\- B C, since .4 C and Z? C are in theEcme straight line (§ 32), or .d C = 2 (i? + 72) sin ^ ^.. 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 cad
Size: 2113px × 1183px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1800, bookdecade1870, booksubjectrailroadengineering
