latihan desain alinmen horizontal
DESCRIPTION
aTRANSCRIPT
DESAIN ALINEMEN HORIZONTAL
Note: Koordinat koordinat'Radiant Degree x y x y
1 57.2957 488963.3 9192077 <==> 0 03.14159701 180 489237.8 9192068 <==> 274.5 -8.35
489523.7 9192625 <==> 560.45 547.94489862.2 9192840 <==> 898.92 763.33
<==><==>
Delta ∆ = …… ˚V.renc = 80 km/jam
Perhitungan DELTA koordinat'x y radiant Degree
0 0 0274.5 -8.35 274.627 -0.03041 -1.742337
560.45 547.94 625.4806898.92 763.33 401.1917
1301.299Tikungan 1 ∆ DELTA 1 Tikungan 2sudut 0_1 1.7423373935 sudut 0_1 + sudut 1_2= 64.53771 sudut 1_2'sudut 1_2 62.795368797 sudut 2_1
x y ∆x ∆y radiant degree x274.5 -8.35
-285.95 -556.291.095987 62.79537 = sudut 1_2 274.5
560.45 547.94 560.45
x560.45898.92
Tikungan 1∆ DELTA R LS ∆e Le Penentuan Bentuk tikungan
64.5377062 205 80 11.17971 42.17829 150.8343 Bisa F-C Bisa S-C-S1.12639703 0.195123 0.736151
p Penentuan Bentuk Tikungan Ye Xe P K Ts1.30081301 S-C-S 5.203252 79.69542 1.313138 39.94855 170.217602
Tikungan 2∆ DELTA R LS ∆e Le Penentuan Bentuk tikungan
30.3242216 205 80 11.17971 7.964807 28.48303 Bisa F-C Bisa S-C-S0.52925824 0.195123 0.139012
Øs
Øs
p Penentuan Bentuk Tikungan Ye Xe P K Ts1.30081301 S-C-S 5.203252 79.98625 1.313138 40.23938 96.1470248
SCATER KOORDINAT'
Tikungan 2 ∆ DELTA 227.20451 sudut 1_2' - sudut 2_1= 30.3242257.52873
y ∆x ∆y radiant degree-8.35
-285.95 -556.290.474809 27.20451 = sudut 1_2'
547.94
y ∆x ∆y radiant degree547.94
-338.47 -215.391.004067 57.52873 = sudut 2_1
763.33
Es Jarak Minimum:38.99823 D - Ts1 - Ts2 =
625.4806 - 170.2176 - 96.14702 = 359.116
0 100 200 300 400 500 600 700 800 900 1000-100
0
100
200
300
400
500
600
700
800
900
1; 0 2; -8.34999999962747
3; 547.939999999479
4; 763.330000000075
y
∆ DELTA 1
∆ DELT
A
2
sudut 0_1 sudut 0_2
sudut 2_1
sudut 1_2'
Es8.754007
L total:
DESAIN ALINEMEN HORIZONTAL
Note: Koordinat koordinat'Radiant Degree x y x y
1 57.2957 562228 9114253 <==> 0 03.14159701 180 562070 9114301 <==> -158 48
562100 9114601 <==> -128 348562273 9114609 <==> 45 356
<==><==>
Delta ∆ = …… ˚V.renc = 80 km/jam
Perhitungan DELTA koordinat'x y radiant Degree
0 0 0-158 48 165.1303 0.294937 16.89863-128 348 301.4963
45 356 173.1849639.8114
Tikungan 1 ∆ DELTA 1 Tikungan 2sudut 0_1 16.898625243 180-sudut 0_1 + sudut 0_2= 78.81208 sudut 1_2'sudut 0_2 84.289289889 sudut 2_1
x y ∆x ∆y radiant degree x-158 48
-30 -3001.471128 84.28929 = sudut 0_2 -158
-128 348 -128
x-128
45Tikungan 1
∆ DELTA R LS ∆e Le Penentuan Bentuk tikungan78.8120849 120 70 16.71133 45.38942 95.01518 Bisa F-C Bisa S-C-S1.37553228 0.291668 0.792196
p Penentuan Bentuk Tikungan Ye Xe P K Ts1.70138889 S-C-S 6.805556 69.40451 1.737418 34.89847 134.91641
Tikungan 2∆ DELTA R LS ∆e Le Penentuan Bentuk tikungan
81.6416639 110 70 18.23055 45.18057 86.6965 Bisa F-C Bisa S-C-S1.42491782 0.318183 0.788551
Øs
Øs
p Penentuan Bentuk Tikungan Ye Xe P K Ts1.85606061 S-C-S 7.424242 68.65365 1.902822 34.24105 130.904195
SCATER KOORDINAT'
Tikungan 2 ∆ DELTA 25.710585 sudut 1_2' - sudut 2_1= 81.6416687.35225
y ∆x ∆y radiant degree48
-30 -3000.099669 5.710585 = sudut 1_2'
348
y ∆x ∆y radiant degree348
-173 -81.524586 87.35225 = sudut 2_1
356
Es Jarak Minimum:37.55503 D - Ts1 - Ts2 =
301.4963 - 134.9164 - 130.9042 = 35.67566
-200 -150 -100 -50 0 50 1000
50
100
150
200
250
300
350
400
y
∆ DELT
A
2
∆ DELTA 1
sudut 0_1
sudut 0_2
sudut 1_2'
sudut 2_1
Es37.87158
L total:d01-Ts1+Ls1+Lc1+Ls1+sisa+Ls2+Lc2+Ls2+d23-Ts2698.1933 m