. In this case the radiating point will be changed from M towardsA and B, the radius shortened, and the point A moved towards K. Let the required distance between tangents, the given radius,and curvature be .as in Proposition II., then we have by logarithms:As the external secant of 40° . 9-484879 ? ;Is to radius 10-000000 So is 30 feet rr: 1-477121 ? To difference of radii = 98*23 . 1 992242 By natural external secants = - 30 •305407 = 98- — And 1146—98 = 1048 = radius of a 5 28 , as 1146 : 1048:: 800 : 782 = length of 5°
. In this case the radiating point will be changed from M towardsA and B, the radius shortened, and the point A moved towards K. Let the required distance between tangents, the given radius,and curvature be .as in Proposition II., then we have by logarithms:As the external secant of 40° . 9-484879 ? ;Is to radius 10-000000 So is 30 feet rr: 1-477121 ? To difference of radii = 98*23 . 1 992242 By natural external secants = - 30 •305407 = 98- — And 1146—98 = 1048 = radius of a 5 28 , as 1146 : 1048:: 800 : 782 = length of 5° 28 (natural tangent of 40°= 83910) = 82 tangent 82 feet from A to K, and curve from thence 732feet of a 5° 28 curve. a 2* V J 3Y8 Formula for Running Lines, PROPOSITION^ r\^. Fij. 3. Raving located a curve with a given terminating in a givenpoint, it is required to change the origin of the curve, also theradius, so as to pass through the same terminating point, with adifferent direction of tangent. .4 Let the given radius MB equal 2292 feet; the given arc BDequal 1000 feet, containing 25 of curvature; the given tangentsD F and D E make an angle of (say) 4°, DF being 400 feet, and EF= 28 feet 28 4° = angle E D F, consequently the angle 9-0982298-9717033-3602173-233991 We have , ^ 4 X 1-75 OLD = 25= +4°= 29°.By logarithms: As versed sine 29° . Is to versed sine 25° So is radius given B M = 2292 To radius required C L = 1714 tables 1714 feet = radius of 3° 20^ curve. PROPOSITION Y. Fig. 4. Having produced the two tangerUs to their intersection, it is requiredto connect them by a curve passing a given distance from the the angle LCB = 31° 44, and C E = 50 feet, to find the I Locating Side Tracks, Etc. 379 radius MA, By geometry, the angle A M E = | L C B = 15° 52. IlgA, ^y ^ m By logarithms we have : As external secant 15° 52 = ^ L C B S-aOlTSO Is to 50 1-698970 So is R. . . . 10-000000 To M A=126
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Keywords: ., bookcentury1800, bookdecade1850, booksubjectenginee, bookyear1856
