. 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 . T- Solution. By § 4 wc have DC = R tan. | /, and CG= Rtan. | /.whence R (tan. ^ / + tan. hi) = D C -{- C G = a, or R = tan. ^ / + ti^n- k ^This formula may be adapted to calculation by logarithms; for we have (Tab. X. 35) tan. ^7+ tm.^P = ^T.^^jcot fj- Substitutingthis value, we get rw R - «gos.
. 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 . T- Solution. By § 4 wc have DC = R tan. | /, and CG= Rtan. | /.whence R (tan. ^ / + tan. hi) = D C -{- C G = a, or R = tan. ^ / + ti^n- k ^This formula may be adapted to calculation by logarithms; for we have (Tab. X. 35) tan. ^7+ tm.^P = ^T.^^jcot fj- Substitutingthis value, we get rw R - «gos. ^7cos. ^7^ (^+/0 The tangent points A and B are obtained by measuring from D aiistance J. Z) = 72 tan |- 7, and from G a distance B G — R tan. \ I, Example. Given a = 600, 1 = 12°, and F = 8° to find R. Here a = 600 i7=6° cos. f 7 = 4° cos. R = 22 CIRCULAR CURVES 40. Problem. Given the line AB = a {fig- 10), which jchis thefixed tangent points A and B, the angle DAB = A, and thr angleA B G = B,io Jind the common radius E C = CF = Rof ar, versed■:urre to unite the tangents HA and B Solution. Find Jirst the auxiliai-y angle A KE = B KF, ivhich inmjbe denoted by K. For this purpose the triangle A E K gives A E: E K= sin. K: sin. E A AT. Therefore E K sin. K = A E sin. E A K ~R cos. A, since EAK = 90^ — A. In like manner, the triangleBFK gives FK sin K= BF sin. FBK = R cos. B. Addingthese equations, we have {EK-\- FK) sin. K= R (cos. J. -\- cos. B),or, since E K + FK = 2 R, 2 R sin. K = R (cos. A + cos. B)Therefore, sin. K = ^ (cos. A -\- cos. B). For calculation by loga-rithms, this becomes (Tab. X. 28) sin K = cos. i(A-\- B) cos. ^(A — B). Having found K, we have the angle AEK = E = 18(P — K —EAK= \%QP — K — (90^ — yl) = 90° + ^ — Z; and the angleBFK= F= 180°—K—FBK = 180° — TT—(90=—J5) = 90--{- B — K Moreover, the triangle A E K gives Ah A K =sin. K: sin. E,or R si
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectrailroadengineering
