Question

5. For a horizontal curve, A=82'30', the station of the PC is 6+14.00, and terrain conditions require the minimum radius permitted by the specifications of, 100 ft. (are definition). Calculate (15 points): a. The stationing of PI and PT, and the external (E) and middle ordinate distances (M) and chord length (LC) for this curve. b. Compute sub-deflection angles and sub-chords to stake out this curve. Use quarter-stations 1251 (Note: In the example we solved together in class, we used full (100 ft.) stationing and for homework you used half (50 ft.) stationing)

Answer 1

Page - Given : 0 - 82030' - 82.50 Station of PC = 6+14.00 Radius of horizontal curve (R)=100ft @ Tangent length (T) = R tan (4) = loox tan( 8925) = 87070 ft → station of PI = station of PC XT = (6+14.00) +(87.70) = (614)+ (8779) = 701.70 = 7+01.70 tength of curve (L) =R xax The = 100x 82.5x The = 143.99 so station of PT = station of PCAL = 614 + 143.99 = 757.99 = 7+57.99

Page 2 External distance (E) Os ] - = loow conocen -) = 33.01 €7 Mieldle ordinate (m) = Rx (1-COS (11) = 100x (1-cos( 82.5}} = 24.82 ht length of chorl (2c) = QR sin (A) = 2x100x sinc 82,5) z 131.87&t

Page 3 ② station length from Deflection angle fromPC chool length from Pelet PC PC >6+14 . 6+25 (625-614 =1187) 11x (8205) 143.99 2 x 100x sin ( 30159 - 10.99 = 3.150 6650 650-614 = 36kt (8205) 36* 143.99 2x100x sin(0.38 = 35079 6+75 = 10.30 17.48° 24.640 61ht 86 lt 60.07 too 83038 7+25 lift 105.39 31080° 38.960 1+so 13684 125076 7+57.99 143.99 41.25° 131087