. 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 . nd a, a, and /?. Here a = ^ = 90, a = 160 — 90 = 70, and R< = 160 X 70 _^^-2X8 =700. 35. Corollary 1. If 6, a, and a are given, to find a. A, and A,we have (§ 34) ^ a = a + a ; R=—; R = 1^. 2 6 26 Example. Given 6 = 8, a = 90, and a = 70, to find a, A, and R Here a = 90 -f 70 = 160, A = -g j
. 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 . nd a, a, and /?. Here a = ^ = 90, a = 160 — 90 = 70, and R< = 160 X 70 _^^-2X8 =700. 35. Corollary 1. If 6, a, and a are given, to find a. A, and A,we have (§ 34) ^ a = a + a ; R=—; R = 1^. 2 6 26 Example. Given 6 = 8, a = 90, and a = 70, to find a, A, and R Here a = 90 -f 70 = 160, A = -g j I -ni aai + aa a (a-fa) a2wc have (§ 35), R -j- R = ^b— ~ —2b— ~ 2b- Therefore ..a = y2 6(A-t-A).Having found a, we have (§ 34) a a Example. Given A = 900, A = 700 and 6 = 8, to find a, a, aim a. Here a = ^2 X 8 (900 + 700) = ^16 X 1600 ^- 160, a =- 8 X 900 X 8 ^^ - ,, 2X700X8 _—jg^j = 90, and a = ^ = 70. REVERSED CURVES. 19 37. Problem. Given the angle A K B = K, which shows thechange of direction of two tangents HA and B K {fig. 8), to unitr. thesetangents by a reversed curve of given common radius R, starting from a giv-en tangent point 3l B K ^^ Fig. 8. Solution. With the given radius run the curve to the point Z), where thetangent D Nbecomes parallel to D K The point D is found thus. Sincethe angle N G K, which is double the angle IIA D {(j 2, II.), is to bemade equal to A KB = K, lay off from FIA the angle HA D=\EMeasure in the direction thus found the chord AD = 2R sin. ^ 75:This will be shown (§ 69) to be the length of the chord for a deflectionangle ^ 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
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