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Chainage HIP 1 'IN Start Chainage 0.0 m 1600m HIP 1 Bearing 90 Deg + TP 1 NOTE: Distance HIP1 to HIP2 is 2578.358m TP 2 Curve 2 Curve 1 Radius Radius 650m TP 3 600m TP 4 Bearing 90 Deg HIP 2 Diagram 1: Bearing 140 Deg You are working in the Northern Territory for Main Roads. You have been given the job of upgrading the road between Tanami Airport and Rabbit Flat Airport, line distance of approximately 40 km. on Route 5, a straight Your field crew have provided preliminary alignment bearing and distance information, and you have made an initial curve radii selection as shown in Diagram 1 Calculate and Report: (To 3 places of decimal, e.g. 700.123m) Chainage at TP1 Chainage at TP2 at TP3 Chainage Chainage Chainage HIP1 (out) Chainage HIP2 (in) Chainage HIP2 (out) at TP4 (100 marks) Chainage at TP1 1320.215m Chainage at TP2 =1843.814m Chainage at TP3 -3839.287m Chainage at TP4 =4406.519m Chainage HIP1 (out) 1564.029m Chainage HIP2 (in) 4142.387m Chainage HIP2 (out) 4103.419m

Answer 1

Date Chaina ge I600.00 Chamage Beari 0.0 HIP TP2 Culue Radius R,= 650m. Bearing 140 Curve radius RI600M HiP 2 TPa 69

Ch. 1600.0 m Angle of deftecHon (A) 1Seari ef 2nd tine searing af fine = 1 40- 9o' HIPL-In TP + Ch. 0.0 m A=50 R-60om TP2 A = 50 TT R, A TTX600Xち0 -523·599 r1 A Leng th circular curve 1g0 ,a81 523.599 11 261.799 mm Half length of circular Curve 2- - R,tan 600.tan(5) Back tangent lengph TRH = 279.785 m Back tangent lengh chainage of HIP-m chainage of TP 1320.215 m 1600-279.185 m

Ch. 1600.0 m Angle of deftecHon (A) 1Seari ef 2nd tine searing af fine = 1 40- 9o' HIPL-In TP + Ch. 0.0 m A=50 R-60om TP2 A = 50 TT R, A TTX600Xち0 -523·599 r1 A Leng th circular curve 1g0 ,a81 523.599 11 261.799 mm Half length of circular Curve 2- - R,tan 600.tan(5) Back tangent lengph TRH = 279.785 m Back tangent lengh chainage of HIP-m chainage of TP 1320.215 m 1600-279.185 m

chainage of TP chainage ef TR+ legth af circalar Curse 1320.215+ a 523.599 1843. 814 m. = Distance behoeen TP to HP,= Distance(1)Tangent logth = 2578.358-279.85 2298.S73 HIP-Im Chainage of Chaineuge ef TP+ 2298.573 1843.814 2298.573 11 4142.387.

1843 81 4 TP2 TP3 R2= 650m A=50 TA HIPIM Ch. 4142 384 longth of back tangent R tan 650.tan2s 303.099 m- Baek Chainage et T3 = Chaimage ef HIP2-Im 414-2. 387-303.099 1 3839. 287m 7T R2A TT XGSoX5o length ef circular curve 567.232M 1 180 ef T+ length ef circular Curne Chainage ef Thh Chainage 3839.287 t 567.232 11 4406.519m

data for hip out is not clear.