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MODEL GREENSHIELD:
HUBUNGAN KECEPATAN (S) – KERAPATAN (D):
Sf
DJ
Y= A.D + Sf
Y= A.X + B
Dm
Sm
Y= A.D + Sf D=0 == S= Sf = B
S=0 == D= DJ === A= -SF / DJ
SEHINGGA:
S = SF – (SF/Dj). D ……….(1)
VM = CAPASITAS= DM . SM
22.
..
XXn
YXXYnA B = Y - A. X
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HUBUNGAN KECEPATAN (S) – VOLUME (V):
Sf
Sm
Vm=CAPASITAS
S = SF – (SF/Dj). D ……….(1)
DIMANA: D = V/S MAKA:
S = SF – (SF/Dj). V/S
S2 = S.SF – (SF/Dj). V
S.SF – S2 =(SF/Dj). V
V= DJ .SF – (DJ / SF ). S2 ……….(2)
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HUBUNGAN VOLUME (V) – KERAPATAN (D):
DJDM
VMS = SF – (SF/Dj). D ……….(1)
S = V/D
V/D = SF – (SF/Dj). D
V = SF..D – (SF/Dj). D2 ………(3)
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Sf
DJDm
Sm
Sf
Sm
Vm=CAPASITAS
DJDM
VM
A
B
DBDA
TERDAPAT KONDISI DIMANA VOLUME LALU LINTAS SAMA TETAPI KERAPATAN YANG BERBEDA SEHINGGA TINGKAT PELAYANAN KONDISI “A” LEBIH BAIK DARI KONDISI “B”
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CONTOH: CARILAH MODEL HUBUNGAN S – D – V DENGAN MEMAKAI MODEL GREENSHIELD DARI DATA LALU LINTAS BERIKUT:
No.Volume Lalu lintas.(smp/jam), V = X.Y
Kecepatan Kend.
(km/jam), S = Y
D= V/S X X2
1 277 45 6.16 37.89
2 444 46 9.65 93.16
3 848 46 18.43 339.84
4 1270 45 28.22 796.49
5 1344 44 30.55 933.02
6 1296 43 30.14 908.39
7 1154 44 26.23 687.87
8 1256 41 30.63 938.45
9 1283 42 30.55 933.16
10 1244 40 31.10 967.21
11 285 41 6.95 48.32
12 218 44 4.95 24.55
TOTAL 10919 521 253.56 6708.36
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067,0.
..22
XXn
YXXYnA
B= SF = 43,42 – (-0,067) . 21,13 = 44,82 Km/jam X = 253,56/12 =21,13
Y = 521/12 =43,42
DJ = - Sf / A = - 44,82 / -0,067 = 630,87 smp /km
DENGAN DEMIKIAN DIDAPAT HUBUNGAN:
1. KECEPATAN (S) – KERAPATAN (D): S = SF – (SF/Dj). D
S = 44,82 – (44,82/630,87) . D
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2. VOLUME (V) – KECEPATAN (S):
V= DJ .SF – (DJ / SF ). S2
V= 630,87 x 44,82 – (630,87/44,82). S2
V = SF.D – (SF/Dj). D2
V = 44,82. D – (44,82/630,87) . D2
3. VOLUME (V) – KERAPATAN (D):