. 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 . and B K. Solution. Let ACB be the required curve. Since the radii are REVERSED CURVES. n equal, and the angle AE C = BFC, the triangles AE Cand B FCare equal, and A C = CB ^ ^a. The reversing point C is, therefore,the middle point of A B. To find R, draw E G perpendicular to A C. Then the right
. 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 . and B K. Solution. Let ACB be the required curve. Since the radii are REVERSED CURVES. n equal, and the angle AE C = BFC, the triangles AE Cand B FCare equal, and A C = CB ^ ^a. The reversing point C is, therefore,the middle point of A B. To find R, draw E G perpendicular to A C. Then the right tri-itngles AEG and BAD are similar, since (§ 2, III.) the angleBAD = hAEC^ AEG. Therefore A & -. A G ^. AB : BD,or ii : ^ a = a : 6 ; 46 Corollary. If R and h are given, to find a, the equation 72 = j^gives a = 4 Rb; a 2 JR b. Examples. Given 6 = 12, and a =^ 200, to determine R. Here2002 10000 12 ~ 833|. /i = 4X12 Given R = 675, and b = 12, to find a. Here a = 2^675 X 12 =2y8T00 == 2 X 90 = 180. 34. Protolem. Given the perpendicular distance between two par-allel tangents B D = b {fig- 7), the distance between the two tangent pointsA B = a, and the first radius E C = R of a reversed curve uniting thetangents HA and B K. to find the chords A C ~ a and C B = a, andthe second radius CF = Solution. Draw the perpendiculars E G- and FL. Then the righttriangles A B D and E A G are similar, since the angle B AD ^ i8 CIRCULAR CURVES. iAEC= AE G. Therefore AB : BD = E A : A G, or a : b 2Rb a Since a and a are (§ 32) parts of a, we have a = a — a. To find R the similar triangles A B D and F B L give A B : B D= F B : B L,ox a :b = R. ^ a; a a Example. Given 6 = 8, a = 160, and R = 900, to find 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
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