. 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 . em. Given the radius DK= R {Jig- 21) of the cetitteline of the main track, the common radius E C = C F = R of the centreline of a crossing, and the distance D G = b between the centre lines of theparallel tracks, to find the central angles AE Cand B F Cand the chordsA Cand CB. Solution. In the t
. 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 . em. Given the radius DK= R {Jig- 21) of the cetitteline of the main track, the common radius E C = C F = R of the centreline of a crossing, and the distance D G = b between the centre lines of theparallel tracks, to find the central angles AE Cand B F Cand the chordsA Cand CB. Solution. In the triangle AEK find the angle AE K and the sideE K. For this purpose we have A E = R, A K = R — d, and theincluded angle E Ax K = S. Find in the triangle B FK the angle B FK and the side FK. Forthis purpose we have B F ^ R, B K= R — h + d, and the included&ng\QFBK= 180=^ — 6. Find in the trianale EFK the angles FE Kand EFK. For this * The triangle AEK does not correspond precisely with BEKm^ ^, A beingon the centre line and B on the outer rail ; but the difference is too slight to affectthe calculations. 16 CIRCULAR CURVES. purpose we have E K and FK just found, and E F —- 2 W. rhet>AEC =^ AEK— FEK, and BFC^EFK—B FK. Lastlv(§ 69,) AC=^2R< C, CB == 2 Rsin. ^ BF Fig. 21. Article IV, — Miscellaneous Problems. 65, Problem. Given A B = a [Jig. 22) and the perpendicularB C = b, to Jind the radius of a curve that shall pass through C and thetangent point A. Solution. Let 0 be the centre of the curve, and draw the radii A 0and C 0 and the line CD parallel to A B. Then in the right triangleCOD we have 0 C^ = CD + OD^ But 0 C = R, CD = a, andOD = AO — AD = R — b. Therefore, R = a-{- {R — 6)» =a^ + R^ — 2 Rb -\- b\ or 2 Rb = a^ -{- b^; 2 b Example. Given a = 204 and b = 24, to find R. Here R »- 204-2 242X-24 + 2 = «67 + 12 = 879. iillSCELLANEOUS PROBLEMS. 47 C6. Corollary 1. If R and b are given to find A B = a, thatvs, to determine the tangent
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