. common to both. Suppose R to be the point in the first curve, and S the point inthe second. There being obstructions in the way, we will run thezigzag line R L P S, making R L tangent to R, and P S Suppose R L Q — 20°and TPS=:15°; let R L r= 1100 feet, L P = 1300, and P S = radius N R as a meridian; that is, suppose N R to bedue north. Then will R L be due west, LP south 70° west, PSsouth 85° west, and radius S M north 5° west. These artificialcourses, then, will show the relative bearings, with which we ob
. common to both. Suppose R to be the point in the first curve, and S the point inthe second. There being obstructions in the way, we will run thezigzag line R L P S, making R L tangent to R, and P S Suppose R L Q — 20°and TPS=:15°; let R L r= 1100 feet, L P = 1300, and P S = radius N R as a meridian; that is, suppose N R to bedue north. Then will R L be due west, LP south 70° west, PSsouth 85° west, and radius S M north 5° west. These artificialcourses, then, will show the relative bearings, with which we obtainthe following traverse: Course. Distance. Northinff. Southing. Easting. Westing. North . .West . S. 70° 85° 5° W. 20001100130014002605 20000000000000002595 07 0000 0000444-6312202 0000- 00000000000000000000 000011001221-601394-66227 05 4595-07 566-65 0000 3943-31 ^84 FoRMUL.« FOR Running Lines, then Difference northing and southing (4595*07 — 566-65) = 4028*42 ;8943*31 4028*42 = -97882 — natural tangent R N G = 44 2o — course. of NM=:K 44^ 23 west, and angle SMD = 39-23, or 44°23 —6^ Locating Side Tracks, Etc. 3g5 To calculate MN make the difference of latitude 4028*42 = cosine44° 23, and the required distance N M =r radius. Then we have . 4028-42 4028-42 by natural cosines cosine 44* 23 •71468 = 5636-7 =MK Or bv logarithms: As cosine R N M = 44° 23 . . 9-854109 Is to R 10-000000 So is difference of latitude 4028*42 . 3605134 To N M = 5636-7 .... 3^1025 The triangles A N I and B M I being similar, we have by loga-rithms (Davies Legendre, book II., prop. X)—that is, by compo-sition and division: As :N^M = 5636-7 8-751025 Is to R 10-000000 So is sum of radii -4605 = (2000 + 2605| . 3*663230 Tocosine ANI = BMIr=:35° IS . 9-912205Having now determined the angle R I^ I = 44° 23, and theangle A N I = 35° 13, the angle R N A becomes = to their differ-ence = 9° 10. Therefore continue the curve from R towards A, 9° 10 of curva-ture, and we ha
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