Horizontal Curves

578r 57295781000 5 43 46 5 43 46 l 100 id 100

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Unformatted text preview: ” 56° 48’ 20” 40° 10’ 20” 16° 38’ 00” D = 5729.578/R = 5729.578/1000 = 5° 43’ 46” 5° 43’ 46” L = 100 (I/D) = 100 (16.63333/5.72944444) = 290.31’ 290.31’ T = R tan (I/2) = 1000 tan (16.63333/2) = 146.18’ 146.18’ LC = 2R sin (I/2) = 2(1000) sin (16.63333/2) = 289.29’ 289.29’ E = R [ 1/cos (I/2) -1 ] = 1000 [ 1/cos 16.63333/2) – 1 ] = 10.63’ 10.63’ M = R [ 1 – cos (I/2) ] = 1000 [ 1 – cos (16.63333/2) ] = 10.52’ 10.52’ 4 Example Problem (continued) A tangent with a bearing of N 56° 48’ 20” E meets another tangent with a bearing 56° 48’ 20...
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This note was uploaded on 04/05/2014 for the course CE 115 taught by Professor Mcginni during the Fall '14 term at Christian Brothers.

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