. 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 . ctionangle ^ K. Having found the point D, measure the perpendicular dis-tance D M = b between the parallel tangents. ■ The distance DB = 2DC = a may then be obtained from the for-muln r§ , Cor.) l^ a = 2 ^ITb . The second tangent point B and the reversing point Care now ue-tcrniined. The dire
. 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 . ctionangle ^ K. Having found the point D, measure the perpendicular dis-tance D M = b between the parallel tangents. ■ The distance DB = 2DC = a may then be obtained from the for-muln r§ , Cor.) l^ a = 2 ^ITb . The second tangent point B and the reversing point Care now ue-tcrniined. The direction o( D B or the angle B DNmnj also I)e ob-tained ; for sin BDN sin, DBM = TTiF,, or sin. BDN b a 38. Problem. Given the line A B = a {fig. 9) which Joins thefixed tangent points A and B, the angles HAB = A and ABL = B,tnd the first radius A E = R, to find the second radius B F = R of aTeversfd curve to unite the tangents H A and B K. First Solution. With the given radius run the carve to the point Z),ohere the tangent D N becomes parallel to B K. The point D is found 20 CIRCULAR CURVES. thus. Since the angle H G N, which is double HA D (§ 2, II.), isequiil to J. CO S, lay off from HA the angle HA D — ^ (At^ B), andmeasure in this direction the chord A D = 2 R sin. ^ (A&o^) (§ 69). Setting the instrument at Z), run the curve to the reversing point C in theline from D to B {^ 32), and measure D C and C B. Then the similartriangles DEC and BFC give DC:DE =^ CB : B F, or D C: Ji^ CB:R; CB .R^. DC X R- Second Solution. By this method the second radius may bt founuby calculation alone. The figure being drawn as above, we have, inthe triangle A BD, AB = a, AD = 2R sin. ^ (A — B), and theincluded angle DAB = HA li ~ HAD = A — h (A — B) ^^ {A -\- B). Find in this triangle (Tab. X. 14 and \2) BD and theangle ABD. Find also the angle DBL^B-\-ABD. Then the chord CB = 2 R sin. hBFC =2R sin. DB L, and the chord D GCB = BD —DBL, = 2R sin. ^DE C = 2R sin. DBL (§ 69). ButD C
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectrailroadengineering
