. 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 . s subtend equal angles at the centre of a circle, andalso at the circumference, if the angles are inscribed in similar seg-ments. Thus, AOD = DOE, and D A E = E A F. VI. The angle of intersection of two tangents is equal to the cen-tral angle subtended by the chord which unites the tangent point
. 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 . s subtend equal angles at the centre of a circle, andalso at the circumference, if the angles are inscribed in similar seg-ments. Thus, AOD = DOE, and D A E = E A F. VI. The angle of intersection of two tangents is equal to the cen-tral angle subtended by the chord which unites the tangent , KGB = AOb 3. In order to unite two straight lines, as GA and BH, by a curve,the angle of intersection is measured, and then a radius for the curvemay be assumed, and the tangents calculated, or the tangents may beassumed of a certain length, and the radius calculated. * Some engineers prefer a chain 50 feet in length, and measuie the length cf :ienrve by chords of 50 instead of 100 feet. The chord of 100 feet has been adopteiithroughout this article; but the formulae deduced may be very readily modified t(.Buit chords of any length. See also ^ 13. SIMPLE CURVES. ti 4. Probleni. Given the angle of intersection K C B — 1 fjig \)and the radius A 0 = R, tojind the tangent A C = T. 1-iy Solution. ])niw CO. Then in the right triangle AOC we lia«, iTab. X. 3) 4-;- = tan. AO C, or, since A 0 0=^1 a 2, VI.)A O - = tan. 2 /; T = R tan. ^ /. Example. Given 7 = 22== .52, and /? = 3000, to find T. Here A= 3000 ^7=11° 26 tan. T= 606 72 »0 .5. Problem. Given the angle of intersection KCB = I {fg. I),ind the tangent A C — 1\ to find thp radius A 0 =-. R. 4 CIRCULAR CURVES. Solution. In tlie right triangle A 0 C we have (Tab. X. 61 — = cot. A O C. ov — = cot. h i;AC r ^ !^= ,. R== Tcot. i/. Example. Given 7 = 31° 16 and r= 950, to find 72. Here r=950 ^1= 15° 38 cot. R = 6. The decree of a curve is determined by
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