Geometrais.xlsx
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Tikungan 1 perencanaan untuk kelas II A Diketahui : Vr = Rc = � = C
90 Km/Jam 480 60 o 0.4 (perubahan kecepatan dari 0,1 - 0,4)
=
Fm
=
0.00125
=
0.35
=
1432.39 Rc
=
1432.39 480
=
2.98
Nilai e Ls
= =
0.081 80
Ls min
=
0.022
Vr3 Rc.C
-
x =
0.022
729000 192
-
Maka D
x
x
90
(digunakan Untuk interpolasi)
=
84 -
=
34
50
Syarat
:
Ls min
<
34
<
80
=
Ls
x x
90 Rc
=
80 3.14
x x
90 480
θs
π
=
+
Ls
5
Syarat
2θ
� 10
θc
�
< <
60
(Aman)
= =
-
2θs
60
-
=
60
-
=
50
o
> >
0 0
2
x 10
Syarat 50
θc Lc
=
θc
Syarat Lc 422
P
x
2
50 360
x
2
=
422
m
≥ ≥
20 20
=
Ls2 6 xθs Rc
=
=
K
(Aman)
π
x
360
(Aman)
-
6400 2880
-
=
2.22
-
=
0.55
m
=
Ls
=
80
=
80
x
Rc
480
1.67
Ls3
θs
-
40 x Rc 2 512000 9216000 -
40
= TS
=
(
=
440
=
θs
+ -
0.55
x
0.87
+
m ( Rc + P ) Cos 1/2 B
)
480
480.55 0.87
=
75
m
=
LC
+
=
422
+
=
490
< <
2 Ts 879
θc
-
+ Cos 1/2 x
=
Kelas jalan IIA �
+k
480
=
(
490
)
480.55
=
Syarat L
( Rc + P ) tan 1/2 B
=
Es
L
40
0.55 60
-
- 480
2 Ls 68
(Aman)
Vr Rc
= =
90 480
km/jam m
60 80
o
LS
= = =
5
o
n dari 0,1 - 0,4) 0.24 2.98 3.00
-
2.50 2.50
=
0.48 0.50 0.081
2.727
x
Vr x e C
2.727
x
7.2615 0.4
=
5
Rc
3.14
(1- cos
x
480
Sin
5
)
0.0035
Rc Sin
480
tan 1/2
60
+
40
40
RC
480
km/jam
= Lc =
50 o 422
P = K =
0.55 40
Ts =
440
e 0.081 e 0.010
-
0.071 0.071 0.071
Tikungan 1 perencanaan untuk kelas II A Diketahui : Vr = Rc = � = C
90 Km/Jam 500 60 o 0.4 (perubahan kecepatan dari 0,1 - 0,4)
=
Fm
=
0.00125
=
0.35
=
1432.39 Rc
=
1432.39 500
=
2.86
Nilai e Ls
= =
0.078 80
Ls min
=
0.022x
Vr3 Rc.C
-
2.727
x
Vr x e C
=
0.022x
729000 200
-
2.727
x
7.046604 0.4
Maka D
x
90
0.24
(digunakan Untuk interpolasi)
=
80 -
=
32
48
Syarat
:
Ls min
<
32
<
80
θs
=
Ls
x x
90 Rc
=
80 3.14
x x
90 500
=
+
Ls
π
4.586
Syarat
2θ
9.172
θc
< <
�
=
�
=
60
(Aman) -
60
-
2θs 2
=
60
-
=
50.828
o
θc
> >
0 0
Lc
=
θc
x
2
=
50.828 360
x
2
=
443
Syarat Lc 443
≥ ≥
20 20
P
=
Ls2
x
4.586
9.17197
Syarat 50.828
360
6 x Rc
K
(Aman)
π x x
Rc
3.14
x
500
Sin
4.586
m
(Aman)
-
=
6400 3000
-
=
2.13
-
=
0.53
m
=
Ls
Rc
(1- cosθs )
500
0.0032
1.60
Ls3 40 x Rc
-
2
=
80
512000 10000000
=
80
-
40
-
Rc Sin θs 500
= TS
= = = =
Es
L
Syarat L 508
40 ( Rc + P ) tan 1/2 B ( 500
-
459.7
m
=
0.53 )
+
500.53
+k
0.87
( Rc + P ) Cos 1/2 B
=
( 500 + Cos 1/2 x
=
500.53 0.87
=
78
m
=
LC
+
=
443
+
=
508
< <
2 Ts 919.4
x
tan 1/2
+
40
LS
θs
+
40
- RC 0.53 ) - 500 60
- 500
2 Ls 64
(Aman)
Kelas jalan IIA Vr Rc �
60
= =
90 500
km/jam m
= =
60 80
o
=
4.586
o
θc
= Lc =
50.828 o 443
P = K =
0.53 40
Ts =
459.7
2.86 3.00
-
2.50 2.50
0.36 0.50 0.078
=
e 0.081
=
e 0.010
-
0.071 0.071 0.071
PERENCANAAN TIPE TIKUNGAN S-S Diketahui
Vr = Rc = β=
90 520 45
km/jam m
π=
3.14
o
SILAHKAN DI ISI DATA DIATAS θs = = = Ls =
1 2 1 2 22.50 θs
x
β
x
45
o
x
π
x
Rc
x
520
90 22.50
x
408
m
Ls 6
^ x
2 Rc
-
Rc
(
1
-
cos
=
408 6
^ x
2 520
-
520
(
1
-
cos
=
14
k=
Ls
-
Ls 40
^ x
3 Rc
^
2
-
Rc
x
sin
=
408
-
408 40
^ x
3 520
^
2
-
520
x
sin
=
203
= = p=
3.14 90
θs
22.50 )
Ts =
(
Rc
+
P
)
x
tan
θs
+
=
(
520
+
14
)
x
tan
22.50
+
=
424
Rc
+
P
)
-
Rc
Es =
(
)
k 203
Es =
= =
cos (
θs
520 cos
+
14 22.50
TS 424
> >
LS 408
)
-
Rc
-
520
58
SYARAT
AMAN
RC Maka di peroleh e tabel e
= = =
sin
θs
sin
22.50
203
520 0.07 0.07
PERENCANAAN TIPE TIKUNGAN S-S Diketahui
Vr = Rc = β=
90 530 45
km/jam m
π=
3.14
o
SILAHKAN DI ISI DATA DIATAS θs = = = Ls =
1 2 1 2 22.50 θs
x
β
x
45
o
x
π
x
Rc
x
530
90 22.50
x
416
m
Ls 6
^ x
2 Rc
-
Rc
(
1
-
cos
=
416 6
^ x
2 530
-
530
(
1
-
cos
=
14
k=
Ls
-
Ls 40
^ x
3 Rc
^
2
-
Rc
x
sin
=
416
-
416 40
^ x
3 530
^
2
-
530
x
sin
=
207
= = p=
3.14 90
θs
22.50 )
Ts =
(
Rc
+
P
)
x
tan
θs
+
=
(
530
+
14
)
x
tan
22.50
+
=
432
Rc
+
P
)
-
Rc
Es =
(
)
k 207
Es =
= =
cos (
θs
530 cos
+
14 22.50
TS 432
> >
LS 416
)
-
Rc
-
530
59
SYARAT
AMAN
RC Maka di peroleh e tabel e
= = =
sin
θs
sin
22.50
207
530 0.07 0.07
PANJANG " Ls " UNTUK VARIASI KECEPATAN D (°) 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.50 3.00 3.50 4.00 4.50 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 15.00 16.00 17.00 18.00
R (m) 5730 2865 1910 1432 1146 955 819 716 573 477 409 358 318 286 239 205 179 159 143 130 119 110 102 95 90 84 80
V = 40 km/jam e Ls LN 35 LN 35 LN 35 LN 35 LN 35 LP 35 LP 35 LP 35 LP 35 LP 35 0.023 35 0.026 35 0.029 35 0.032 35 0.038 35 0.043 35 0.049 35 0.053 35 0.056 35 0.062 35 0.066 35 0.070 35 0.073 35 0.077 35 0.080 35 0.082 35 0.085 40
V = 50 km/jam e Ls LN 30 LN 30 LN 30 LP 30 LP 30 LP 30 LP 30 0.021 40 0.026 40 0.030 40 0.035 40 0.039 40 0.044 40 0.048 40 0.055 40 0.062 40 0.068 40 0.074 40 0.079 40 0.084 40 0.088 40 0.091 40 0.094 50 0.097 50 0.098 50 0.100 50 0.100 50
V = 60 km/jam e Ls LN 50 LN 50 LP 50 LP 50 LP 50 0.023 50 0.026 50 0.029 50 0.036 50 0.042 50 0.048 50 0.054 50 0.059 50 0.064 50 0.072 50 0.080 50 0.086 50 0.091 60 0.095 60 0.098 60 0.099 60 Dmaks = 12,78
V = 70 km/jam e Ls LN 60 LP 60 LP 60 0.021 60 0.025 60 0.030 60 0.035 60 0.039 60 0.047 60 0.055 60 0.062 60 0.068 60 0.074 60 0.079 60 0.088 60 0.094 60 0.098 70 0.099 70 0.100 70 Dmaks = 9,13
V = 80 km/jam e Ls LN 70 LP 70 0.020 70 0.027 70 0.033 70 0.038 70 0.044 70 0.049 70 0.059 70 0.068 70 0.076 70 0.082 70 0.088 70 0.093 70 0.099 70 0.100 80 Dmaks = 6,82
V = 90 km/jam e Ls LN 80 LP 80 0.025 80 0.033 80 0.040 80 0.047 80 0.054 80 0.060 80 0.071 80 0.081 80 0.089 80 0.095 80 0.098 80 0.100 90 Dmaks = 5,12
V = 100 km/jam e LP 0.021 0.031 0.040 0.049 0.057 0.065 0.072 0.085 0.094 0.099 0.100 Dmaks = 3,91
19.00 20.00 21.00 22.00 23.00 24.00 25.00 26.00 27.00 28.00 29.00 30.00
75 72 69 65 62 60 57 55 53 51 49 48
0.087 40 0.090 40 0.092 45 0.093 45 0.095 45 0.096 45 0.097 45 0.098 50 0.099 50 0.0996 50 0.0999 50 Dmaks = 29,90
Dmaks = 18,85
V = 100 km/jam Ls 90 90 90 90 90 90 90 90 90 90 90 90 Dmaks = 3,91
V = 120 km/jam e Ls LP 100 0.030 100 0.043 100 0.055 100 0.066 100 0.076 100 0.090 100 0.096 100 0.100 100 Dmaks = 2,40
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN A. Tikungan 1 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
90 480
=
km/ Jam
x
Vr2 127
x Rc
=
8100 60960
-
=
0.005
= =
0.24 90
+
-
Fm
0.24
0.128
e < e tabel 0.005
<
Landai Relatif
=
(
0.02
+
=
(
0.02
+
=
e maks
0.081
< <
0.081 90
127
Vr2 x
Rc
127
8100 x
480
=
Syarat e maks 0.005
e tabel vr
x
3.5 )
x
3.5
-
Fm
-
0.1275
0.0039
=
=
(Aman)
0.005
e tabel 0.081
(Aman)
)
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN B. Tikungan 2 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
90 500
=
km/ Jam
x
Vr2 127
x Rc
=
8100 63500
-
=
0.000
= =
0.24 90
+
-
Fm
0.24
0.128
e < e tabel 0.000 Landai Relatif
<
0.02
+
(
0.02
+
=
=
=
< <
e tabel vr 0.078 90
x
3.5 )
x
3.5
-
Fm
-
0.1275
0.0038
127
Vr2 x
Rc
127
8100 x
500
=
Syarat e maks 0.000
(Aman)
= ( =
e maks
0.078
0.000
e tabel 0.078
(Aman)
)
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN C. Tikungan 3 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
90 520
=
km/ Jam
x
Vr2 127
x Rc
=
8100 66040
-
=
-0.005
= =
0.24 90
+
-
Fm
0.24
0.1275
e < e tabel -0.005 Landai Relatif
)
<
0.02
+
(
0.02
+
=
=
=
< <
e tabel vr 0.071 90
x
x
0.0035
127
Vr2 x
RC
127
8100 x
520
=
Syarat e maks -0.005
(Aman)
= ( =
e maks
0.071
-0.005
e tabel 0.071
(Aman)
-
-
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN D. Tikungan 4 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
=
3.5
km/ Jam
=
8100 67310
-
=
-0.007
-
0.1275
e < e tabel -0.007 Landai Relatif
)
<
0.02
(
0.02
=
e maks
0.071
= (
=
Fm
90
Vr2 x Rc
=
0.24
x
127
= 3.5 )
90 530
vr
90 0.0035
=
Vr2 127
=
0.1275
8100 127
= Syarat e maks -0.007
< <
-0.007
e tabel 0.071
+
0.24
Fm
(Aman) +
e tabel vr
+
0.071 90
Vr2 x
RC
8100 x
530
(Aman)
x
3.5 )
x
3.5
-
Fm
-
0.1275
)
pelebaran pada tikungan Rumus : B
= n x ( B' + C ) + ( n - 1 ) x Td + z
B n B' Td C P z A L
= = = = = = = = =
Keterangan : Lebar perkerasan jalan pada tikungan jumlah jalur lalu lintas lebar lintas truk pada tikungan lebar melintang akibat tanjakan depan (m) kebebasan samping ( 0,4 - 0,8 ) jarak roda depan dengan roda belakang ( 3,4 m ) Lebar tambahan karena kelalaian dalam mengemudi (m) Jarak ujung mobil dengan ban depan (0,9 m) Lebar mobil (2,1)
P A L
Tikungan 1 Diketahui : Rc Vr
= =
480 90
m km/ jam
B'
=
L
=
2.1
+ (Rc
=
2.1
=
2.1
=
2.1
=
2.112042
Td
= √ Rc
2
- √ Rc2 - P2)
+ ( 480
-
+ ( 480
-
+
( 480
-
11.56
)
x
3.40
+
√230400
- √230388.4 )
+ 0.012042
479.988 )
+ n ( 2 x P x A ) - Rc
= √ 230400
+
= √230415.4 =
480.016 -
=
0.016041 m
2
x
480 480
( 2
0.9
) -
480
Z
B
=
0.105
x Rc
Vr
=
0.105 x √ 480
90
=
9.45 21.909
=
0.431
= n x ( B' + C ) + ( n - 1 ) x Td + z =
x ( 2.112
2
= 5.024084 + 0.016041 =
0.4 )
+ +
0.431
m <
7m
+ (
2
-
x
0.016
+
5.471 Karena B =
5.471
Maka tidak perlu di adakan pelebaran jalan
Tikungan 2 Diketahui : Rc Vr
= =
500 90
m km/ jam
B'
=
L
=
2.1
+ (Rc
=
2.1
=
2.1
=
2.1
=
2.11156
Td
1 )
= √ Rc √
2
- √ Rc2 - P2)
+ ( 500
-
+ ( 500
-
+
+
( 500
√250000
- √249988.4 )
0.01156
499.9884 )
+ n ( 2 x P x A ) - Rc (
-
11.56
)
= √ 250000
+
2
= √250015.4 -
Z
B
=
500.0154 -
=
0.0154 m
500
0.105
x Rc
Vr
=
0.105 x √ 500
90
9.45 22.361
=
0.423
( 2
x
3.40
+
2
-
0.9
) -
500
500
=
=
x
= n x ( B' + C ) + ( n - 1 ) x Td + z = = =
x ( 2.112
2
5.02312 +
0.4 )
+
0.0154
+
0.423
5.461
m <
7m
+ (
1 )
x
0.015
+
5.461 Karena B =
Maka tidak perlu di adakan pelebaran jalan
Tikungan 3 Diketahui : Rc Vr
= =
520 90
m km/ jam
B'
=
L
=
2.1
+ (Rc
=
2.1
=
2.1
=
2.1
- √ Rc2 - P2)
+ ( 520
-
+ ( 520
-
+
( 520
+ 0.011116
√270400
- √270388.4 )
519.9889 )
-
11.56
)
=
Td
2.111116
= √ Rc
2
+ n ( 2 x P x A ) - Rc
= √ 270400
+
2
= √270415.4 -
Z
B
=
520.0148 -
=
0.014807 m
520
0.105
x Rc
Vr
=
0.105 x √ 520
90
9.45 22.804
=
0.414
( 2
x
3.40
+
2
-
0.9
) -
520
520
=
=
x
= n x ( B' + C ) + ( n - 1 ) x Td + z =
x ( 2.111
2
= 5.022231 + 0.014807 =
+
0.4 )
+
0.414
m <
7m
+ (
1 )
x
0.015
+
5.451 Karena B =
5.451
Maka tidak perlu di adakan pelebaran jalan
Tikungan 4 Diketahui : Rc Vr
= =
530 90
m km/ jam
B'
=
L
+ (Rc (
- √ Rc2 - P2)
)
=
2.1
=
2.1
=
2.1
=
2.1
=
Td
+ ( 530
-
+ ( 530
-
+
+
= √ Rc
2
)
x
0.00
+
0
2
-
1 )
- √ 280900 ) 530
0
)
+ n ( 2 x P x A ) - Rc +
= √ 280900 -
B
0
2.1
= √ 280900
Z
( 530
-
√280900
=
530 -
=
0 m
2 530
0.105
x Rc
Vr
=
0.105 x √ 530
90
9.45 23.022
=
0.410
( 2
) -
530
530
=
=
x
= n x ( B' + C ) + ( n - 1 ) x Td + z =
2
=
5
=
5.410 Karena B =
x ( 2.100
+
+
0.4 )
0
+
0.410
5.410
m <
7m
+ (
x
0.000
+
Maka tidak perlu di adakan pelebaran jalan
3.4 0.9 2.1
0.431
0.423
0.414
0.410
Perhitungan Landai Relatif pada lengkung Horisontal
Tikungan 1 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
480 60 80 0.081 5.471
=
e x B Ls
=
8.1
=
0.552
8.1 %
x 80
Jarak Pandang Lengkung Horisontal R' = Rc - B = 480 = 474.529 m θ = =
S
m
1 2 1 2
=
30.0
=
�
5.5
5.471
� x
60
2 x ∏ x R'
360
x
=
60 360
x
=
496.673 m
2
= S x ( 1 - Cos θ )
x
=
496.673 x ( 1 - cos 30.0 )
=
496.673
x
=
66.542
m
0.13
3.14
x 474.529
Tikungan 2 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
500 60 80 0.078 5.461
=
e x B Ls
=
7.8
=
0.534
7.8 %
x 80
Jarak Pandang Lengkung Horisontal R' = Rc - B = 500 = 494.539 m θ = =
S
m
1 2 1 2
=
30.0
=
�
5.5
5.461
� x
60
2 x ∏ x R'
360
x
=
60 360
x
=
517.617 m
2
= S x ( 1 - Cos θ )
x
=
517.617 x ( 1 - cos 30.0 )
=
517.617
x
=
69.348
m
0.13
3.14
x 494.539
Tikungan 3 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
520 45 408.20 0.071 5.451
=
e x B Ls
=
7.1
=
0.095
7.1 %
x 408.2
Jarak Pandang Lengkung Horisontal R' = Rc - B = 520 = 514.549 m θ = =
S
m
1 2 1 2
=
22.5
=
�
5.5
5.451
� x
45
2 x ∏ x R'
360
x
=
45 360
x
=
403.921 m
2
= S x ( 1 - Cos θ )
x
=
403.921 x ( 1 - cos 22.5 )
=
403.921
x
=
30.747
m
0.08
3.14
x 514.549
Tikungan 4 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
530 45 416.05 0.071 % 5.410
=
e x B Ls
=
7.1
=
0.092
x 416.05
Jarak Pandang Lengkung Horisontal R' = Rc - B = 530 = 524.590 m θ = =
S
m
1 2 1 2
=
22.5
=
�
7.1
5.4
5.410
� x
45
2 x ∏ x R'
360
x
=
45 360
x
=
411.803 m
2
= S x ( 1 - Cos θ )
x
=
411.803 x ( 1 - cos 22.5 )
=
411.803
x
=
31.347
m
0.08
3.14
x 524.590
PERHITUNGAN JARAK PANDANG PADA LENGKUNG HORIZONTAL 1.
TIKUNGAN 1 Dik :
Rc β Ls B
= = = =
480 60 80 5.471
m Π = 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
R'
=
Rc
-
=
480
-
=
477.264
penye : -
-
θ
= = =
-
S
2.
m
*
β
*
60
=
β * 360 60 * 360 499.5366 m
= = = =
S *( 499.5366 * ( 499.5366 * 66.925
= =
-
1 2 1 2 30
B 2 5.471 2
2
*
Π
*
R'
2
*
3.14
*
477.264
1 1 0.134
cos θ ) cos
30 )
TIKUNGAN 2 Dik :
Rc β Ls B
= = = =
500 60 80 5.461
m Π 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
penye : -
-
R'
θ
=
Rc
-
=
500
-
=
497.269
= = =
-
S
3.
m
*
β
*
60
⁰
=
β * 360 60 * 360 520.4753 m
= = = =
S *( 520.4753 * ( 520.4753 * 69.73047 m
= =
-
1 2 1 2 30
B 2 5.461 2
2
*
Π
*
R'
2
*
3.14
*
497.269
1 1 0.134
cos θ ) cos
30 )
TIKUNGAN 3 Dik :
Rc β Ls B
= = = =
520 45 80 5.451
m Π 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
R'
=
Rc
-
=
520
-
=
517.274
penye : -
-
θ
=
1
*
B 2 5.451 2
β
-
θ
= = =
-
S
4
m
*
β
*
45
⁰
=
β * 360 45 * 360 406.0603 m
= = = =
S *( 406.0603 * ( 406.0603 * 30.9095 m
= =
-
2 1 2 22.5
2
*
Π
*
R'
2
*
3.14
*
517.274
1 1 0.0761
cos θ ) cos
22.5 )
TIKUNGAN 4 Dik :
Rc β Ls B
= = = =
530 45 80 5.410
m Π 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
R'
=
Rc
-
=
530
-
=
527.295
penye : -
-
θ
= = =
-
S
= =
1 2 1 2 22.5
β 360 45 360
B 2 5.410 2
*
β
*
45
*
2
*
Π
*
R'
*
2
*
3.14
*
527.295
⁰
-
m
=
413.9264 m
= = = =
S *( 413.9264 * ( 413.9264 * 31.50827 m
1 1 0.0761
cos θ ) cos
22.5 )
Jarak Pandang Henti Rumus Dh Dp
= =
Dr
=
Dh Dp Dr t Fm
= = = = =
Dh + Dr 0.278
x Vr x t
Vr2 254 x Fm
keterangan Jarak Pandang Henti Jarak dari saat melihat rintangan sampai menginjak pedal rem Jarak pengereman waktu siap ( 2,5 detik ) koefisien gesek minimum antara ban dan muka jalan dalam arah memanjang jalan
Daerah Datar Diketahui Vr t Fm
: = = =
90 2.5 0.285
km/ jam detik
Dp
= = =
0.278 0.278 62.55
x x m
Dr
=
Vr2 254 x Fm
= 254
Dh
Vr 90
8100 x 0.285
=
111.894 m
= = =
Dp + Dr 62.55 + 111.894 174.444 m Daerah Bukit
Diketahui Vr t Fm
: = = =
70 2.5 0.300
km/ jam detik
x x
t 2.5
Dp
= = =
Dr
=
0.278 0.278 48.65
254
Dh
= = =
Vr 70
x x
t 2.5
x x
t 2.5
Vr2 254 x Fm
=
=
x x m
4900 x
0.3
64.304 m Dp + Dr 48.65 + 64.304 112.954 m Daerah Gunung
Diketahui Vr t Fm
: = = =
50 2.5 0.330
km/ jam detik
Dp
= = =
0.278 0.278 34.75
x x m
Dr
=
Vr2 254 x Fm
= 254 = Dh
= = =
Vr 50
2500 x
0.33
29.826 m Dp + Dr 34.75 + 29.826 64.576 m
jarak pandang menyiap rumus D D1=+ D2 + D3 + D4 D1 = 0.278 x t1 x
(
T1 x Vr - m + a - t1 2 D2 = 0.278 x Vr + t2 D3 = 30-100 meter D4 = 2 d2 3
)
Daerah datar diketahui Vr = m =
90 15
km/ jam
data dari tabel jp 1 di peroleh t1 t2 a
= = =
4.72 11.36 2.412
D1
=
0.278
x
4.72
=
1.31216
x
73.846
=
96.90
=
0.278
x
Vr
x
t2
=
0.278
x
90
x
11.36
=
284.2272
m
D2
D3
=
60
D4
=
2 3
=
2
detik detik km/ jam
m
(
90
- 15
(di peroleh dari tabel jp 2) d2
284
+
2 - 4.72 2
)
3 =
189.485
284 m
Jadi D = D1 + D2 + D3 + d4 D
=
96.90
+
=
630.610 m
284.2
+
60
+
189.485
Daerah Bukit diketahui Vr = m =
70 15
detik
data dari tabel jp 1 di peroleh t1 t2 a
= = =
4.2 10.4 2.34
D1
=
0.278
x
=
1.1676
x
=
63.13
=
0.278
x
Vr
x
t2
=
0.278
x
70
x
10.4
=
202.384
m
D2
detik detik km/ jam 4.2
=
40
D4
=
2 3
d2
=
2 3
202
=
134.923
m
70
- 15
54.07
m
D3
(
(di peroleh dari tabel jp 2)
Jadi D = D1 + D2 + D3 + d4
+
2 - 4.2 2
)
D
=
63.13
+
=
440.439 m
202.4
+
40
+
134.923
Daerah gunung diketahui Vr = m =
50 15
km/ jam
data dari tabel jp 1 di peroleh t1 t2 a
= = =
3.68 9.44 2.268
D1
=
0.278
x
3.68
=
1.023
x
34.294
=
35.08
=
0.28
x
Vr
x
t2
=
0.28
x
50
x
9.44
=
131.22
m
D2
detik detik km/ jam
m
D3
=
60
D4
=
2 3
d2
=
2 3
131
=
87.477
m
(
50
- 15
+
2 - 3.68 2
)
(di peroleh dari tabel jp 2)
Jadi D = D1 + D2 + D3 + d4 D
=
35.08
+
131.2
+
60
+
87.477
=
313.777 m
Derajat lengkung maksimum Tikungan 1 Rumus : Dmax
=
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max = f max = Vr =
0.005 0.128 90
Maka Dmax
=
=
Tikungan 2 Rumus : Dmax
=
181913.53 (0.005374
+ 8100
0.1275 )
2.984
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max = f max = Vr =
0.000 0.128 90
Maka Dmax
=
=
Tikungan 3 Rumus : Dmax
=
181913.53 (5.91E-05 2.865
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max =
+ 8100
-0.005
0.1275 )
f max Vr
= =
0.128 90
Maka Dmax
=
=
Tikungan 4 Rumus : Dmax
=
181913.53 (-0.00485
+ 8100
0.1275 )
2.755
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max = f max = Vr =
-0.007 0.128 90
Maka Dmax
=
=
181913.53 (-0.00716 2.703
+ 8100
0.1275 )
Perhitungan Derajat lengkung cekung
a.
1. P6-p8
Dik :
200
q1 q2 PPV LV
= = = =
0.23 0.17 25.02 300
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.17
*
100
*
100
* 800 * 800
LV
p6 p7 p8
= = =
24.79 25.02 25.19
100
R
= = =
EV
= = =
Y
= = =
0.23 0.06 LV A 300 0.06 477551 A 0.06 0.02 A 200 0.06 200 0.06
* * * * * 60000
q2
300
X² LV X² 300 X²
2. P14-P16
Dik :
200
q1 q2 PPV LV
= = = =
0.11 0.11 26.36 300
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.11
*
100
*
100
P14 P15 P16
100
R
= = =
0.11 0.00 LV A 300 0.00 31500000
q2
= = =
26.25 26.36 26.47
EV
= = =
Y
= = =
3
* 800 * 800
0.00 0.00 A 200 0.00 200 0.00
* * * * * 60000
LV 300
X² LV X² 300 X²
P21-P23
Dik :
100
A
q1 q2 PPV LV
= = = =
0.00 0.04 26.80 200
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.04
*
100
*
100
* 800 * 800
LV
P21 P22 P23
100
R
= = =
EV
= = =
Y
= = =
0.00 -0.04 LV A 200 -0.04 -500175 A -0.04 -0.01 A 200 -0.04 200 -0.04
* * * * * 40000
q2
200
X² LV X² 200 X²
= = =
26.81 26.80 26.84
JARAK LANGSUNG 0 100 200 250 300
Tinggi stasiun 24.79 24.91 25.02 25.11 25.19
X 0 100 200 50 0
X² 0 10000 40000 2500 0
Y 0.00 0.01 0.04 0.00 0.00
Tinggi total patok 24.79 24.90 24.98 25.11 25.19
JARAK LANGSUNG 1550 1650 1750 1800 1850
Tinggi stasiun 26.25 26.31 26.36 26.41 26.47
X 0 100 200 50 0
X² 0 10000 40000 2500 0
Y 0 0.00 0.00 0.00 0
Tinggi total patok 26.25 26.31 26.36 26.41 26.47
JARAK LANGSUNG 0 50 100 150 200
Tinggi stasiun 26.81 26.81 26.80 26.82 26.84
X 0 50 100 50 0
X² 0 2500 10000 2500 0
Y 0 0.00 -0.01 0.00 0
Tinggi total patok 26.81 26.81 26.81 26.83 26.84
cekung tinggi totalnya bertambah
cekung tinggi totalnya bertambah
cekung tinggi totalnya bertambah
Perhitungan Derajat lengkung cembung
b. 1.
P2-P4
Dik :
100
q1 q2 PPV LV
= = = =
0.12 0.07 24.63 200
% % m m
A
= = =
q1
-
0.07
*
100
*
100
* 800 * 800
LV
= = =
EV
= = =
Y
= = =
= = =
24.51 24.63 24.70
0.12 0.06 LV A 200 0.06 347368.42 A 0.06 0.01 A 200 0.06 200 0.06
* * * * * 40000
q2
200
X² LV X² 200 X²
P10 - P12
Dik :
100
P2 P3 P4
100
R
2.
Tinggi Stasiun
q1 q2 PPV LV
= = = =
0.49 -0.26 26.18 200
% % m m
Tinggi Stasiun
A
= = =
q1
-
-0.26
*
100
*
100
p10 p11 p12
100
R
= = =
0.49 0.74 LV A 200 0.74 26870.95
q2
= = =
25.70 26.18 25.93
EV
= = =
Y
= = =
3.
* 800 * 800
0.74 0.19 A 200 0.74 200 0.74
* * * * * 40000
LV 200
X² LV X² 200 X²
p18 -p20
Dik :
100
A
q1 q2 PPV LV
= = = =
0.12 0.01 29.00 300
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.01
*
100
*
100
* 800 * 800
LV
P18 P19 P20
200
R
= = =
EV
= = =
Y
= = =
0.12 0.11 LV A 300 0.11 278858.89 A 0.11 0.04 A 200 0.11 200 0.11
* * * * * 60000
q2
300
X² LV X² 300 X²
= = =
28.86 29.00 28.50
JARAK LANGSUNG 300 350 400 450 500
Tinggi stasiun 24.51 24.57 24.63 24.67 24.70
X 0 50 100 50 0
X² 0 2500 10000 2500 0
Y 0.00 0.00 0.01 0.00 0.00
Tinggi total patok 24.51 24.57 24.62 24.66 24.70
JARAK LANGSUNG 3000 3050 3100 3150 3200
Tinggi stasiun 25.70 25.94 26.18 26.05 25.93
X 0 50 100 50 0
X² 0 2500 10000 2500 0
Y 0 0.05 0.19 0.05 0
Tinggi total patok 25.70 25.89 26.00 26.01 25.93
JARAK LANGSUNG 3360 3410 3460 3560 3660
Tinggi stasiun 28.86 28.93 29.00 28.75 28.50
X 0 50 100 100 0
X² 0 2500 10000 10000 0
Y 0 0.00 0.02 0.02 0.00
Tinggi total patok 28.86 28.92 28.98 28.73 28.50
cembung nilai total patok berkurang
cembung nilai total patok berkurang
cembung nilai total patok berkurang
Jarak Penyinaran Lampu Kendaraan Rumus : L
=
2
S
P0 P2
= 18.00 = 20.10
A
=
q2
-
q1
=
P0
-(
-
200
h
+
Tan α
A Dik :
penye :
= 18.00
q2
= -0.012 = 0.50
q1
P0 50 - ( 18.00 50
+ +
A
= 0.500 = 0.512
- -0.012
Lv
=
2
S
-
200
100
=
2
S
-
200
100
=
2
S
-
200
= 2 S = 246.8 2 = 123.4
-
246.8
100 S
0.6 ) 0.6 )
( 0.6
+ 0.50 ( 0.6 + 0.50 * 0.50
tan α 0.017 0.617
)
4. Volume Timbunan dan Galian a. timbunan atau cekung 1. Patok (P0 - P2) Dik: L = 100 T = 19.03 = 0.02 b = 7 penye: 1 A = 2 1 = 2 = 0.625 = 1.250 V = A = 1.250 = 8.75 2. Patok (P21 - P23) Dik: L = 100 T = 21.42 = 0.05 b = 7 penye: 1 A = 2 1 = 2 = 1.37 = 2.75 V = A = 2.75 = 19.22 3. Patok (P32 - P34) Dik: L = 50 T = 18.65 = 0.03 b = 7 penye: 1 A = 2
= -
50 19.00
*
L
*
T
*
50
*
0.02
*
2
* *
b 7
= -
50 21.36 4.706 32.94
*
L
*
T
*
50
*
0.05
*
2
* *
b 7
= -
25 18.63
*
L
*
T
=
V
= = = = =
1 2 0.36 0.71 A 0.71 4.97
*
25
*
2
* *
b 7
*
0.03
a. Galian atau cembung 1. Patok (P3 - P5) Dik: L = 100 = T = 21.50 = 0.11 b = 7 penye: 1 A = * 2 1 = * 2 = 2.6786 * = 5.36 V = A * = 5.36 * = 37.5 2. Patok (P41 - P43) Dik: L = 50 = T = 20.41 = 0.08 b = 7 penye: 1 A = * 2 1 = * 2 = 1.0243 * = 2.05 V = A * = 2.05 * = 14.34 3. Patok (P49 - P51) Dik: L = 100 T = 19.80 = 0.04 b = 7 penye: 1 A = 2
50 21.39
L
*
T
50
*
0.11
L
*
T
25
*
0.08
*
T
2 b 7
25 20.33
2 b 7
= -
50 19.76
*
L
9.41 65.9
=
V
= = = = =
1 2 1.00 2.01 A 2.01 14.046
*
50
*
2
* *
b 7
*
0.04
Tabel 3.1 Rekapitulasi Perhitungan PERHITUNGAN TINGGI TITIK STASIUN
Pn = Kt
PERHITUNGAN BEDA TINGGI
+ (x/y) . Ik
Dimana : Pn = Tinggi titik Stasiun ke - n Kt = Tinggi Kontur Terendah x = Jarak stasiun dengan kontur terendah y = Jarak antara kontur terendah dan tertinggi Ik = Nilai interval kontur
No Stasiun (Pn)
Jarak STA (m)
Lengkung Horisontal R (m) Sudut (p) L(m)
P0
KM = (BT/Jarak stasiun) x 100 %
Dimana : BT = Beda Tinggi (m) Pn akhir = Tinggi Stasiun Akhir (m) Pn awal = Tinggi Stasiun awal (m)
Dimana : KM = Kemiringan (%) BT = Beda Tinggi (m)
Kt
X
Y
Interval
Jarak Langsung (m)
23.50
0
0.2
0.5
200
23.90
0.4
0.6
0.5
300
Kontur
200 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 P21 P22 P23 P24 P25 P26
100 100 100
480
60
100 200 100 100 100
500
60
250
100 100 100 200 200 100 100
520
45
260
100 100 200 200 100 100 100 100 200
530
45
24.10
0.9
1.1
0.5
400
24.20
1.3
1.5
0.5
500
24.30
0.8
1
0.5
600
240
100
265
24.35
0.2
0.4
0.5
700
24.40
0.7
0.9
0.5
900
24.60
1.1
1.3
0.5
1000
24.75
1.6
1.8
0.5
1100
24.80
2
2.2
0.5
1200
25.25
1.7
1.9
0.5
1300
25.75
1.3
1.5
0.5
1400
25.55
0.6
0.8
0.5
1500
25.65
1.2
1.4
0.5
1700
25.80
1.9
2.1
0.5
1900
25.90
2.3
2.5
0.5
2000
26.00
2.8
3
0.5
2100
26.05
2.4
2.6
0.5
2200
26.10
1.5
1.7
0.5
2300
26.20
2.1
2.3
0.5
2500
26.20
2.6
2.8
0.5
2700
26.35
2.2
2.4
0.5
2800
26.40
1.7
2.1
0.5
2900
26.40
1.5
1.7
0.5
3000
26.45
0.6
0.8
0.5
3100
26.75
0.1
0.3
0.5
3300
27.75
0
0.2
0.5
3500
200 P27
27.65
0
0.1
0.5
3600
27.00
0
0.2
0.5
3600
100 P28
PERHITUNGAN KEMIRIGAN
BT = Pn akhir - Pn awal
Tinggi Stasiun (m)
Kemiringan (%)
Jarak Sta 100 Beda Tinggi (m) Keterangan
23.50
0.37
0.73
24.23
0.28
0.28
24.51
0.12
0.12
24.63
0.07
0.07
24.70
-0.10
-0.10
24.60
0.19
0.19
24.79
0.12
0.23
25.02
0.17
0.17
25.19
0.06
0.06
25.25
0.44
0.44
25.70
0.49
0.49
26.18
-0.26
-0.26
25.93
0.15
0.15
26.08
0.09
0.17
26.25
0.05
0.11
26.36
0.11
0.11
26.47
0.04
0.04
26.51
0.03
0.03
26.54
0.12
0.12
26.66
0.00
0.01
26.66
0.07
0.14
26.81
0.00
0.00
26.80
0.04
0.04
26.84
-0.02
-0.02
26.83
0.09
0.09
26.92
0.42
0.83
27.75
-0.05
-0.10
-0.65 0.00
-0.65 0.00
27.65 27.00
tikungan 1 Kelandaian (%)
24.70
- 24.23 240
x
100
0.19 (datar)
%
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 480 m tikungan 2 Kelandaian (%)
26.18
- 25.02 250
x
100
0.46 (datar)
%
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 500 m tikungan 3 Kelandaian (%)
26.54
- 26.36 260
x
100
0.07 (datar)
%
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 520 m tikungan 4 Kelandaian (%)
26.84
- 26.80 265
x
100
0.01 (datar)
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 530 m
%
90 Km/Jam 480 60 o 0.4 (perubahan kecepatan dari 0,1 - 0,4)
=
Fm
=
0.00125
=
0.35
=
1432.39 Rc
=
1432.39 480
=
2.98
Nilai e Ls
= =
0.081 80
Ls min
=
0.022
Vr3 Rc.C
-
x =
0.022
729000 192
-
Maka D
x
x
90
(digunakan Untuk interpolasi)
=
84 -
=
34
50
Syarat
:
Ls min
<
34
<
80
=
Ls
x x
90 Rc
=
80 3.14
x x
90 480
θs
π
=
+
Ls
5
Syarat
2θ
� 10
θc
�
< <
60
(Aman)
= =
-
2θs
60
-
=
60
-
=
50
o
> >
0 0
2
x 10
Syarat 50
θc Lc
=
θc
Syarat Lc 422
P
x
2
50 360
x
2
=
422
m
≥ ≥
20 20
=
Ls2 6 xθs Rc
=
=
K
(Aman)
π
x
360
(Aman)
-
6400 2880
-
=
2.22
-
=
0.55
m
=
Ls
=
80
=
80
x
Rc
480
1.67
Ls3
θs
-
40 x Rc 2 512000 9216000 -
40
= TS
=
(
=
440
=
θs
+ -
0.55
x
0.87
+
m ( Rc + P ) Cos 1/2 B
)
480
480.55 0.87
=
75
m
=
LC
+
=
422
+
=
490
< <
2 Ts 879
θc
-
+ Cos 1/2 x
=
Kelas jalan IIA �
+k
480
=
(
490
)
480.55
=
Syarat L
( Rc + P ) tan 1/2 B
=
Es
L
40
0.55 60
-
- 480
2 Ls 68
(Aman)
Vr Rc
= =
90 480
km/jam m
60 80
o
LS
= = =
5
o
n dari 0,1 - 0,4) 0.24 2.98 3.00
-
2.50 2.50
=
0.48 0.50 0.081
2.727
x
Vr x e C
2.727
x
7.2615 0.4
=
5
Rc
3.14
(1- cos
x
480
Sin
5
)
0.0035
Rc Sin
480
tan 1/2
60
+
40
40
RC
480
km/jam
= Lc =
50 o 422
P = K =
0.55 40
Ts =
440
e 0.081 e 0.010
-
0.071 0.071 0.071
Tikungan 1 perencanaan untuk kelas II A Diketahui : Vr = Rc = � = C
90 Km/Jam 500 60 o 0.4 (perubahan kecepatan dari 0,1 - 0,4)
=
Fm
=
0.00125
=
0.35
=
1432.39 Rc
=
1432.39 500
=
2.86
Nilai e Ls
= =
0.078 80
Ls min
=
0.022x
Vr3 Rc.C
-
2.727
x
Vr x e C
=
0.022x
729000 200
-
2.727
x
7.046604 0.4
Maka D
x
90
0.24
(digunakan Untuk interpolasi)
=
80 -
=
32
48
Syarat
:
Ls min
<
32
<
80
θs
=
Ls
x x
90 Rc
=
80 3.14
x x
90 500
=
+
Ls
π
4.586
Syarat
2θ
9.172
θc
< <
�
=
�
=
60
(Aman) -
60
-
2θs 2
=
60
-
=
50.828
o
θc
> >
0 0
Lc
=
θc
x
2
=
50.828 360
x
2
=
443
Syarat Lc 443
≥ ≥
20 20
P
=
Ls2
x
4.586
9.17197
Syarat 50.828
360
6 x Rc
K
(Aman)
π x x
Rc
3.14
x
500
Sin
4.586
m
(Aman)
-
=
6400 3000
-
=
2.13
-
=
0.53
m
=
Ls
Rc
(1- cosθs )
500
0.0032
1.60
Ls3 40 x Rc
-
2
=
80
512000 10000000
=
80
-
40
-
Rc Sin θs 500
= TS
= = = =
Es
L
Syarat L 508
40 ( Rc + P ) tan 1/2 B ( 500
-
459.7
m
=
0.53 )
+
500.53
+k
0.87
( Rc + P ) Cos 1/2 B
=
( 500 + Cos 1/2 x
=
500.53 0.87
=
78
m
=
LC
+
=
443
+
=
508
< <
2 Ts 919.4
x
tan 1/2
+
40
LS
θs
+
40
- RC 0.53 ) - 500 60
- 500
2 Ls 64
(Aman)
Kelas jalan IIA Vr Rc �
60
= =
90 500
km/jam m
= =
60 80
o
=
4.586
o
θc
= Lc =
50.828 o 443
P = K =
0.53 40
Ts =
459.7
2.86 3.00
-
2.50 2.50
0.36 0.50 0.078
=
e 0.081
=
e 0.010
-
0.071 0.071 0.071
PERENCANAAN TIPE TIKUNGAN S-S Diketahui
Vr = Rc = β=
90 520 45
km/jam m
π=
3.14
o
SILAHKAN DI ISI DATA DIATAS θs = = = Ls =
1 2 1 2 22.50 θs
x
β
x
45
o
x
π
x
Rc
x
520
90 22.50
x
408
m
Ls 6
^ x
2 Rc
-
Rc
(
1
-
cos
=
408 6
^ x
2 520
-
520
(
1
-
cos
=
14
k=
Ls
-
Ls 40
^ x
3 Rc
^
2
-
Rc
x
sin
=
408
-
408 40
^ x
3 520
^
2
-
520
x
sin
=
203
= = p=
3.14 90
θs
22.50 )
Ts =
(
Rc
+
P
)
x
tan
θs
+
=
(
520
+
14
)
x
tan
22.50
+
=
424
Rc
+
P
)
-
Rc
Es =
(
)
k 203
Es =
= =
cos (
θs
520 cos
+
14 22.50
TS 424
> >
LS 408
)
-
Rc
-
520
58
SYARAT
AMAN
RC Maka di peroleh e tabel e
= = =
sin
θs
sin
22.50
203
520 0.07 0.07
PERENCANAAN TIPE TIKUNGAN S-S Diketahui
Vr = Rc = β=
90 530 45
km/jam m
π=
3.14
o
SILAHKAN DI ISI DATA DIATAS θs = = = Ls =
1 2 1 2 22.50 θs
x
β
x
45
o
x
π
x
Rc
x
530
90 22.50
x
416
m
Ls 6
^ x
2 Rc
-
Rc
(
1
-
cos
=
416 6
^ x
2 530
-
530
(
1
-
cos
=
14
k=
Ls
-
Ls 40
^ x
3 Rc
^
2
-
Rc
x
sin
=
416
-
416 40
^ x
3 530
^
2
-
530
x
sin
=
207
= = p=
3.14 90
θs
22.50 )
Ts =
(
Rc
+
P
)
x
tan
θs
+
=
(
530
+
14
)
x
tan
22.50
+
=
432
Rc
+
P
)
-
Rc
Es =
(
)
k 207
Es =
= =
cos (
θs
530 cos
+
14 22.50
TS 432
> >
LS 416
)
-
Rc
-
530
59
SYARAT
AMAN
RC Maka di peroleh e tabel e
= = =
sin
θs
sin
22.50
207
530 0.07 0.07
PANJANG " Ls " UNTUK VARIASI KECEPATAN D (°) 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.50 3.00 3.50 4.00 4.50 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 15.00 16.00 17.00 18.00
R (m) 5730 2865 1910 1432 1146 955 819 716 573 477 409 358 318 286 239 205 179 159 143 130 119 110 102 95 90 84 80
V = 40 km/jam e Ls LN 35 LN 35 LN 35 LN 35 LN 35 LP 35 LP 35 LP 35 LP 35 LP 35 0.023 35 0.026 35 0.029 35 0.032 35 0.038 35 0.043 35 0.049 35 0.053 35 0.056 35 0.062 35 0.066 35 0.070 35 0.073 35 0.077 35 0.080 35 0.082 35 0.085 40
V = 50 km/jam e Ls LN 30 LN 30 LN 30 LP 30 LP 30 LP 30 LP 30 0.021 40 0.026 40 0.030 40 0.035 40 0.039 40 0.044 40 0.048 40 0.055 40 0.062 40 0.068 40 0.074 40 0.079 40 0.084 40 0.088 40 0.091 40 0.094 50 0.097 50 0.098 50 0.100 50 0.100 50
V = 60 km/jam e Ls LN 50 LN 50 LP 50 LP 50 LP 50 0.023 50 0.026 50 0.029 50 0.036 50 0.042 50 0.048 50 0.054 50 0.059 50 0.064 50 0.072 50 0.080 50 0.086 50 0.091 60 0.095 60 0.098 60 0.099 60 Dmaks = 12,78
V = 70 km/jam e Ls LN 60 LP 60 LP 60 0.021 60 0.025 60 0.030 60 0.035 60 0.039 60 0.047 60 0.055 60 0.062 60 0.068 60 0.074 60 0.079 60 0.088 60 0.094 60 0.098 70 0.099 70 0.100 70 Dmaks = 9,13
V = 80 km/jam e Ls LN 70 LP 70 0.020 70 0.027 70 0.033 70 0.038 70 0.044 70 0.049 70 0.059 70 0.068 70 0.076 70 0.082 70 0.088 70 0.093 70 0.099 70 0.100 80 Dmaks = 6,82
V = 90 km/jam e Ls LN 80 LP 80 0.025 80 0.033 80 0.040 80 0.047 80 0.054 80 0.060 80 0.071 80 0.081 80 0.089 80 0.095 80 0.098 80 0.100 90 Dmaks = 5,12
V = 100 km/jam e LP 0.021 0.031 0.040 0.049 0.057 0.065 0.072 0.085 0.094 0.099 0.100 Dmaks = 3,91
19.00 20.00 21.00 22.00 23.00 24.00 25.00 26.00 27.00 28.00 29.00 30.00
75 72 69 65 62 60 57 55 53 51 49 48
0.087 40 0.090 40 0.092 45 0.093 45 0.095 45 0.096 45 0.097 45 0.098 50 0.099 50 0.0996 50 0.0999 50 Dmaks = 29,90
Dmaks = 18,85
V = 100 km/jam Ls 90 90 90 90 90 90 90 90 90 90 90 90 Dmaks = 3,91
V = 120 km/jam e Ls LP 100 0.030 100 0.043 100 0.055 100 0.066 100 0.076 100 0.090 100 0.096 100 0.100 100 Dmaks = 2,40
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN A. Tikungan 1 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
90 480
=
km/ Jam
x
Vr2 127
x Rc
=
8100 60960
-
=
0.005
= =
0.24 90
+
-
Fm
0.24
0.128
e < e tabel 0.005
<
Landai Relatif
=
(
0.02
+
=
(
0.02
+
=
e maks
0.081
< <
0.081 90
127
Vr2 x
Rc
127
8100 x
480
=
Syarat e maks 0.005
e tabel vr
x
3.5 )
x
3.5
-
Fm
-
0.1275
0.0039
=
=
(Aman)
0.005
e tabel 0.081
(Aman)
)
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN B. Tikungan 2 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
90 500
=
km/ Jam
x
Vr2 127
x Rc
=
8100 63500
-
=
0.000
= =
0.24 90
+
-
Fm
0.24
0.128
e < e tabel 0.000 Landai Relatif
<
0.02
+
(
0.02
+
=
=
=
< <
e tabel vr 0.078 90
x
3.5 )
x
3.5
-
Fm
-
0.1275
0.0038
127
Vr2 x
Rc
127
8100 x
500
=
Syarat e maks 0.000
(Aman)
= ( =
e maks
0.078
0.000
e tabel 0.078
(Aman)
)
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN C. Tikungan 3 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
90 520
=
km/ Jam
x
Vr2 127
x Rc
=
8100 66040
-
=
-0.005
= =
0.24 90
+
-
Fm
0.24
0.1275
e < e tabel -0.005 Landai Relatif
)
<
0.02
+
(
0.02
+
=
=
=
< <
e tabel vr 0.071 90
x
x
0.0035
127
Vr2 x
RC
127
8100 x
520
=
Syarat e maks -0.005
(Aman)
= ( =
e maks
0.071
-0.005
e tabel 0.071
(Aman)
-
-
2. PERHITUNGAN SUPERELEVASI PADA TIKUNGAN D. Tikungan 4 diketahui : Vr Rc
= =
Maka Fm
=
-0.00125 x VR +
=
-0.00125
=
0.128
e
Syarat
=
3.5
km/ Jam
=
8100 67310
-
=
-0.007
-
0.1275
e < e tabel -0.007 Landai Relatif
)
<
0.02
(
0.02
=
e maks
0.071
= (
=
Fm
90
Vr2 x Rc
=
0.24
x
127
= 3.5 )
90 530
vr
90 0.0035
=
Vr2 127
=
0.1275
8100 127
= Syarat e maks -0.007
< <
-0.007
e tabel 0.071
+
0.24
Fm
(Aman) +
e tabel vr
+
0.071 90
Vr2 x
RC
8100 x
530
(Aman)
x
3.5 )
x
3.5
-
Fm
-
0.1275
)
pelebaran pada tikungan Rumus : B
= n x ( B' + C ) + ( n - 1 ) x Td + z
B n B' Td C P z A L
= = = = = = = = =
Keterangan : Lebar perkerasan jalan pada tikungan jumlah jalur lalu lintas lebar lintas truk pada tikungan lebar melintang akibat tanjakan depan (m) kebebasan samping ( 0,4 - 0,8 ) jarak roda depan dengan roda belakang ( 3,4 m ) Lebar tambahan karena kelalaian dalam mengemudi (m) Jarak ujung mobil dengan ban depan (0,9 m) Lebar mobil (2,1)
P A L
Tikungan 1 Diketahui : Rc Vr
= =
480 90
m km/ jam
B'
=
L
=
2.1
+ (Rc
=
2.1
=
2.1
=
2.1
=
2.112042
Td
= √ Rc
2
- √ Rc2 - P2)
+ ( 480
-
+ ( 480
-
+
( 480
-
11.56
)
x
3.40
+
√230400
- √230388.4 )
+ 0.012042
479.988 )
+ n ( 2 x P x A ) - Rc
= √ 230400
+
= √230415.4 =
480.016 -
=
0.016041 m
2
x
480 480
( 2
0.9
) -
480
Z
B
=
0.105
x Rc
Vr
=
0.105 x √ 480
90
=
9.45 21.909
=
0.431
= n x ( B' + C ) + ( n - 1 ) x Td + z =
x ( 2.112
2
= 5.024084 + 0.016041 =
0.4 )
+ +
0.431
m <
7m
+ (
2
-
x
0.016
+
5.471 Karena B =
5.471
Maka tidak perlu di adakan pelebaran jalan
Tikungan 2 Diketahui : Rc Vr
= =
500 90
m km/ jam
B'
=
L
=
2.1
+ (Rc
=
2.1
=
2.1
=
2.1
=
2.11156
Td
1 )
= √ Rc √
2
- √ Rc2 - P2)
+ ( 500
-
+ ( 500
-
+
+
( 500
√250000
- √249988.4 )
0.01156
499.9884 )
+ n ( 2 x P x A ) - Rc (
-
11.56
)
= √ 250000
+
2
= √250015.4 -
Z
B
=
500.0154 -
=
0.0154 m
500
0.105
x Rc
Vr
=
0.105 x √ 500
90
9.45 22.361
=
0.423
( 2
x
3.40
+
2
-
0.9
) -
500
500
=
=
x
= n x ( B' + C ) + ( n - 1 ) x Td + z = = =
x ( 2.112
2
5.02312 +
0.4 )
+
0.0154
+
0.423
5.461
m <
7m
+ (
1 )
x
0.015
+
5.461 Karena B =
Maka tidak perlu di adakan pelebaran jalan
Tikungan 3 Diketahui : Rc Vr
= =
520 90
m km/ jam
B'
=
L
=
2.1
+ (Rc
=
2.1
=
2.1
=
2.1
- √ Rc2 - P2)
+ ( 520
-
+ ( 520
-
+
( 520
+ 0.011116
√270400
- √270388.4 )
519.9889 )
-
11.56
)
=
Td
2.111116
= √ Rc
2
+ n ( 2 x P x A ) - Rc
= √ 270400
+
2
= √270415.4 -
Z
B
=
520.0148 -
=
0.014807 m
520
0.105
x Rc
Vr
=
0.105 x √ 520
90
9.45 22.804
=
0.414
( 2
x
3.40
+
2
-
0.9
) -
520
520
=
=
x
= n x ( B' + C ) + ( n - 1 ) x Td + z =
x ( 2.111
2
= 5.022231 + 0.014807 =
+
0.4 )
+
0.414
m <
7m
+ (
1 )
x
0.015
+
5.451 Karena B =
5.451
Maka tidak perlu di adakan pelebaran jalan
Tikungan 4 Diketahui : Rc Vr
= =
530 90
m km/ jam
B'
=
L
+ (Rc (
- √ Rc2 - P2)
)
=
2.1
=
2.1
=
2.1
=
2.1
=
Td
+ ( 530
-
+ ( 530
-
+
+
= √ Rc
2
)
x
0.00
+
0
2
-
1 )
- √ 280900 ) 530
0
)
+ n ( 2 x P x A ) - Rc +
= √ 280900 -
B
0
2.1
= √ 280900
Z
( 530
-
√280900
=
530 -
=
0 m
2 530
0.105
x Rc
Vr
=
0.105 x √ 530
90
9.45 23.022
=
0.410
( 2
) -
530
530
=
=
x
= n x ( B' + C ) + ( n - 1 ) x Td + z =
2
=
5
=
5.410 Karena B =
x ( 2.100
+
+
0.4 )
0
+
0.410
5.410
m <
7m
+ (
x
0.000
+
Maka tidak perlu di adakan pelebaran jalan
3.4 0.9 2.1
0.431
0.423
0.414
0.410
Perhitungan Landai Relatif pada lengkung Horisontal
Tikungan 1 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
480 60 80 0.081 5.471
=
e x B Ls
=
8.1
=
0.552
8.1 %
x 80
Jarak Pandang Lengkung Horisontal R' = Rc - B = 480 = 474.529 m θ = =
S
m
1 2 1 2
=
30.0
=
�
5.5
5.471
� x
60
2 x ∏ x R'
360
x
=
60 360
x
=
496.673 m
2
= S x ( 1 - Cos θ )
x
=
496.673 x ( 1 - cos 30.0 )
=
496.673
x
=
66.542
m
0.13
3.14
x 474.529
Tikungan 2 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
500 60 80 0.078 5.461
=
e x B Ls
=
7.8
=
0.534
7.8 %
x 80
Jarak Pandang Lengkung Horisontal R' = Rc - B = 500 = 494.539 m θ = =
S
m
1 2 1 2
=
30.0
=
�
5.5
5.461
� x
60
2 x ∏ x R'
360
x
=
60 360
x
=
517.617 m
2
= S x ( 1 - Cos θ )
x
=
517.617 x ( 1 - cos 30.0 )
=
517.617
x
=
69.348
m
0.13
3.14
x 494.539
Tikungan 3 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
520 45 408.20 0.071 5.451
=
e x B Ls
=
7.1
=
0.095
7.1 %
x 408.2
Jarak Pandang Lengkung Horisontal R' = Rc - B = 520 = 514.549 m θ = =
S
m
1 2 1 2
=
22.5
=
�
5.5
5.451
� x
45
2 x ∏ x R'
360
x
=
45 360
x
=
403.921 m
2
= S x ( 1 - Cos θ )
x
=
403.921 x ( 1 - cos 22.5 )
=
403.921
x
=
30.747
m
0.08
3.14
x 514.549
Tikungan 4 Diketahui : Rc �
Ls e B Landai Relatif I/m
= = = = =
530 45 416.05 0.071 % 5.410
=
e x B Ls
=
7.1
=
0.092
x 416.05
Jarak Pandang Lengkung Horisontal R' = Rc - B = 530 = 524.590 m θ = =
S
m
1 2 1 2
=
22.5
=
�
7.1
5.4
5.410
� x
45
2 x ∏ x R'
360
x
=
45 360
x
=
411.803 m
2
= S x ( 1 - Cos θ )
x
=
411.803 x ( 1 - cos 22.5 )
=
411.803
x
=
31.347
m
0.08
3.14
x 524.590
PERHITUNGAN JARAK PANDANG PADA LENGKUNG HORIZONTAL 1.
TIKUNGAN 1 Dik :
Rc β Ls B
= = = =
480 60 80 5.471
m Π = 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
R'
=
Rc
-
=
480
-
=
477.264
penye : -
-
θ
= = =
-
S
2.
m
*
β
*
60
=
β * 360 60 * 360 499.5366 m
= = = =
S *( 499.5366 * ( 499.5366 * 66.925
= =
-
1 2 1 2 30
B 2 5.471 2
2
*
Π
*
R'
2
*
3.14
*
477.264
1 1 0.134
cos θ ) cos
30 )
TIKUNGAN 2 Dik :
Rc β Ls B
= = = =
500 60 80 5.461
m Π 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
penye : -
-
R'
θ
=
Rc
-
=
500
-
=
497.269
= = =
-
S
3.
m
*
β
*
60
⁰
=
β * 360 60 * 360 520.4753 m
= = = =
S *( 520.4753 * ( 520.4753 * 69.73047 m
= =
-
1 2 1 2 30
B 2 5.461 2
2
*
Π
*
R'
2
*
3.14
*
497.269
1 1 0.134
cos θ ) cos
30 )
TIKUNGAN 3 Dik :
Rc β Ls B
= = = =
520 45 80 5.451
m Π 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
R'
=
Rc
-
=
520
-
=
517.274
penye : -
-
θ
=
1
*
B 2 5.451 2
β
-
θ
= = =
-
S
4
m
*
β
*
45
⁰
=
β * 360 45 * 360 406.0603 m
= = = =
S *( 406.0603 * ( 406.0603 * 30.9095 m
= =
-
2 1 2 22.5
2
*
Π
*
R'
2
*
3.14
*
517.274
1 1 0.0761
cos θ ) cos
22.5 )
TIKUNGAN 4 Dik :
Rc β Ls B
= = = =
530 45 80 5.410
m Π 3.14 ⁰ m m (diperoleh dari perhitungan pelebaran jalan)
R'
=
Rc
-
=
530
-
=
527.295
penye : -
-
θ
= = =
-
S
= =
1 2 1 2 22.5
β 360 45 360
B 2 5.410 2
*
β
*
45
*
2
*
Π
*
R'
*
2
*
3.14
*
527.295
⁰
-
m
=
413.9264 m
= = = =
S *( 413.9264 * ( 413.9264 * 31.50827 m
1 1 0.0761
cos θ ) cos
22.5 )
Jarak Pandang Henti Rumus Dh Dp
= =
Dr
=
Dh Dp Dr t Fm
= = = = =
Dh + Dr 0.278
x Vr x t
Vr2 254 x Fm
keterangan Jarak Pandang Henti Jarak dari saat melihat rintangan sampai menginjak pedal rem Jarak pengereman waktu siap ( 2,5 detik ) koefisien gesek minimum antara ban dan muka jalan dalam arah memanjang jalan
Daerah Datar Diketahui Vr t Fm
: = = =
90 2.5 0.285
km/ jam detik
Dp
= = =
0.278 0.278 62.55
x x m
Dr
=
Vr2 254 x Fm
= 254
Dh
Vr 90
8100 x 0.285
=
111.894 m
= = =
Dp + Dr 62.55 + 111.894 174.444 m Daerah Bukit
Diketahui Vr t Fm
: = = =
70 2.5 0.300
km/ jam detik
x x
t 2.5
Dp
= = =
Dr
=
0.278 0.278 48.65
254
Dh
= = =
Vr 70
x x
t 2.5
x x
t 2.5
Vr2 254 x Fm
=
=
x x m
4900 x
0.3
64.304 m Dp + Dr 48.65 + 64.304 112.954 m Daerah Gunung
Diketahui Vr t Fm
: = = =
50 2.5 0.330
km/ jam detik
Dp
= = =
0.278 0.278 34.75
x x m
Dr
=
Vr2 254 x Fm
= 254 = Dh
= = =
Vr 50
2500 x
0.33
29.826 m Dp + Dr 34.75 + 29.826 64.576 m
jarak pandang menyiap rumus D D1=+ D2 + D3 + D4 D1 = 0.278 x t1 x
(
T1 x Vr - m + a - t1 2 D2 = 0.278 x Vr + t2 D3 = 30-100 meter D4 = 2 d2 3
)
Daerah datar diketahui Vr = m =
90 15
km/ jam
data dari tabel jp 1 di peroleh t1 t2 a
= = =
4.72 11.36 2.412
D1
=
0.278
x
4.72
=
1.31216
x
73.846
=
96.90
=
0.278
x
Vr
x
t2
=
0.278
x
90
x
11.36
=
284.2272
m
D2
D3
=
60
D4
=
2 3
=
2
detik detik km/ jam
m
(
90
- 15
(di peroleh dari tabel jp 2) d2
284
+
2 - 4.72 2
)
3 =
189.485
284 m
Jadi D = D1 + D2 + D3 + d4 D
=
96.90
+
=
630.610 m
284.2
+
60
+
189.485
Daerah Bukit diketahui Vr = m =
70 15
detik
data dari tabel jp 1 di peroleh t1 t2 a
= = =
4.2 10.4 2.34
D1
=
0.278
x
=
1.1676
x
=
63.13
=
0.278
x
Vr
x
t2
=
0.278
x
70
x
10.4
=
202.384
m
D2
detik detik km/ jam 4.2
=
40
D4
=
2 3
d2
=
2 3
202
=
134.923
m
70
- 15
54.07
m
D3
(
(di peroleh dari tabel jp 2)
Jadi D = D1 + D2 + D3 + d4
+
2 - 4.2 2
)
D
=
63.13
+
=
440.439 m
202.4
+
40
+
134.923
Daerah gunung diketahui Vr = m =
50 15
km/ jam
data dari tabel jp 1 di peroleh t1 t2 a
= = =
3.68 9.44 2.268
D1
=
0.278
x
3.68
=
1.023
x
34.294
=
35.08
=
0.28
x
Vr
x
t2
=
0.28
x
50
x
9.44
=
131.22
m
D2
detik detik km/ jam
m
D3
=
60
D4
=
2 3
d2
=
2 3
131
=
87.477
m
(
50
- 15
+
2 - 3.68 2
)
(di peroleh dari tabel jp 2)
Jadi D = D1 + D2 + D3 + d4 D
=
35.08
+
131.2
+
60
+
87.477
=
313.777 m
Derajat lengkung maksimum Tikungan 1 Rumus : Dmax
=
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max = f max = Vr =
0.005 0.128 90
Maka Dmax
=
=
Tikungan 2 Rumus : Dmax
=
181913.53 (0.005374
+ 8100
0.1275 )
2.984
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max = f max = Vr =
0.000 0.128 90
Maka Dmax
=
=
Tikungan 3 Rumus : Dmax
=
181913.53 (5.91E-05 2.865
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max =
+ 8100
-0.005
0.1275 )
f max Vr
= =
0.128 90
Maka Dmax
=
=
Tikungan 4 Rumus : Dmax
=
181913.53 (-0.00485
+ 8100
0.1275 )
2.755
181912.53 ( e max + F max ) Vr2
Tikungan 1 diketahui : e max = f max = Vr =
-0.007 0.128 90
Maka Dmax
=
=
181913.53 (-0.00716 2.703
+ 8100
0.1275 )
Perhitungan Derajat lengkung cekung
a.
1. P6-p8
Dik :
200
q1 q2 PPV LV
= = = =
0.23 0.17 25.02 300
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.17
*
100
*
100
* 800 * 800
LV
p6 p7 p8
= = =
24.79 25.02 25.19
100
R
= = =
EV
= = =
Y
= = =
0.23 0.06 LV A 300 0.06 477551 A 0.06 0.02 A 200 0.06 200 0.06
* * * * * 60000
q2
300
X² LV X² 300 X²
2. P14-P16
Dik :
200
q1 q2 PPV LV
= = = =
0.11 0.11 26.36 300
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.11
*
100
*
100
P14 P15 P16
100
R
= = =
0.11 0.00 LV A 300 0.00 31500000
q2
= = =
26.25 26.36 26.47
EV
= = =
Y
= = =
3
* 800 * 800
0.00 0.00 A 200 0.00 200 0.00
* * * * * 60000
LV 300
X² LV X² 300 X²
P21-P23
Dik :
100
A
q1 q2 PPV LV
= = = =
0.00 0.04 26.80 200
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.04
*
100
*
100
* 800 * 800
LV
P21 P22 P23
100
R
= = =
EV
= = =
Y
= = =
0.00 -0.04 LV A 200 -0.04 -500175 A -0.04 -0.01 A 200 -0.04 200 -0.04
* * * * * 40000
q2
200
X² LV X² 200 X²
= = =
26.81 26.80 26.84
JARAK LANGSUNG 0 100 200 250 300
Tinggi stasiun 24.79 24.91 25.02 25.11 25.19
X 0 100 200 50 0
X² 0 10000 40000 2500 0
Y 0.00 0.01 0.04 0.00 0.00
Tinggi total patok 24.79 24.90 24.98 25.11 25.19
JARAK LANGSUNG 1550 1650 1750 1800 1850
Tinggi stasiun 26.25 26.31 26.36 26.41 26.47
X 0 100 200 50 0
X² 0 10000 40000 2500 0
Y 0 0.00 0.00 0.00 0
Tinggi total patok 26.25 26.31 26.36 26.41 26.47
JARAK LANGSUNG 0 50 100 150 200
Tinggi stasiun 26.81 26.81 26.80 26.82 26.84
X 0 50 100 50 0
X² 0 2500 10000 2500 0
Y 0 0.00 -0.01 0.00 0
Tinggi total patok 26.81 26.81 26.81 26.83 26.84
cekung tinggi totalnya bertambah
cekung tinggi totalnya bertambah
cekung tinggi totalnya bertambah
Perhitungan Derajat lengkung cembung
b. 1.
P2-P4
Dik :
100
q1 q2 PPV LV
= = = =
0.12 0.07 24.63 200
% % m m
A
= = =
q1
-
0.07
*
100
*
100
* 800 * 800
LV
= = =
EV
= = =
Y
= = =
= = =
24.51 24.63 24.70
0.12 0.06 LV A 200 0.06 347368.42 A 0.06 0.01 A 200 0.06 200 0.06
* * * * * 40000
q2
200
X² LV X² 200 X²
P10 - P12
Dik :
100
P2 P3 P4
100
R
2.
Tinggi Stasiun
q1 q2 PPV LV
= = = =
0.49 -0.26 26.18 200
% % m m
Tinggi Stasiun
A
= = =
q1
-
-0.26
*
100
*
100
p10 p11 p12
100
R
= = =
0.49 0.74 LV A 200 0.74 26870.95
q2
= = =
25.70 26.18 25.93
EV
= = =
Y
= = =
3.
* 800 * 800
0.74 0.19 A 200 0.74 200 0.74
* * * * * 40000
LV 200
X² LV X² 200 X²
p18 -p20
Dik :
100
A
q1 q2 PPV LV
= = = =
0.12 0.01 29.00 300
% % m m
Tinggi Stasiun
A
= = =
q1
-
0.01
*
100
*
100
* 800 * 800
LV
P18 P19 P20
200
R
= = =
EV
= = =
Y
= = =
0.12 0.11 LV A 300 0.11 278858.89 A 0.11 0.04 A 200 0.11 200 0.11
* * * * * 60000
q2
300
X² LV X² 300 X²
= = =
28.86 29.00 28.50
JARAK LANGSUNG 300 350 400 450 500
Tinggi stasiun 24.51 24.57 24.63 24.67 24.70
X 0 50 100 50 0
X² 0 2500 10000 2500 0
Y 0.00 0.00 0.01 0.00 0.00
Tinggi total patok 24.51 24.57 24.62 24.66 24.70
JARAK LANGSUNG 3000 3050 3100 3150 3200
Tinggi stasiun 25.70 25.94 26.18 26.05 25.93
X 0 50 100 50 0
X² 0 2500 10000 2500 0
Y 0 0.05 0.19 0.05 0
Tinggi total patok 25.70 25.89 26.00 26.01 25.93
JARAK LANGSUNG 3360 3410 3460 3560 3660
Tinggi stasiun 28.86 28.93 29.00 28.75 28.50
X 0 50 100 100 0
X² 0 2500 10000 10000 0
Y 0 0.00 0.02 0.02 0.00
Tinggi total patok 28.86 28.92 28.98 28.73 28.50
cembung nilai total patok berkurang
cembung nilai total patok berkurang
cembung nilai total patok berkurang
Jarak Penyinaran Lampu Kendaraan Rumus : L
=
2
S
P0 P2
= 18.00 = 20.10
A
=
q2
-
q1
=
P0
-(
-
200
h
+
Tan α
A Dik :
penye :
= 18.00
q2
= -0.012 = 0.50
q1
P0 50 - ( 18.00 50
+ +
A
= 0.500 = 0.512
- -0.012
Lv
=
2
S
-
200
100
=
2
S
-
200
100
=
2
S
-
200
= 2 S = 246.8 2 = 123.4
-
246.8
100 S
0.6 ) 0.6 )
( 0.6
+ 0.50 ( 0.6 + 0.50 * 0.50
tan α 0.017 0.617
)
4. Volume Timbunan dan Galian a. timbunan atau cekung 1. Patok (P0 - P2) Dik: L = 100 T = 19.03 = 0.02 b = 7 penye: 1 A = 2 1 = 2 = 0.625 = 1.250 V = A = 1.250 = 8.75 2. Patok (P21 - P23) Dik: L = 100 T = 21.42 = 0.05 b = 7 penye: 1 A = 2 1 = 2 = 1.37 = 2.75 V = A = 2.75 = 19.22 3. Patok (P32 - P34) Dik: L = 50 T = 18.65 = 0.03 b = 7 penye: 1 A = 2
= -
50 19.00
*
L
*
T
*
50
*
0.02
*
2
* *
b 7
= -
50 21.36 4.706 32.94
*
L
*
T
*
50
*
0.05
*
2
* *
b 7
= -
25 18.63
*
L
*
T
=
V
= = = = =
1 2 0.36 0.71 A 0.71 4.97
*
25
*
2
* *
b 7
*
0.03
a. Galian atau cembung 1. Patok (P3 - P5) Dik: L = 100 = T = 21.50 = 0.11 b = 7 penye: 1 A = * 2 1 = * 2 = 2.6786 * = 5.36 V = A * = 5.36 * = 37.5 2. Patok (P41 - P43) Dik: L = 50 = T = 20.41 = 0.08 b = 7 penye: 1 A = * 2 1 = * 2 = 1.0243 * = 2.05 V = A * = 2.05 * = 14.34 3. Patok (P49 - P51) Dik: L = 100 T = 19.80 = 0.04 b = 7 penye: 1 A = 2
50 21.39
L
*
T
50
*
0.11
L
*
T
25
*
0.08
*
T
2 b 7
25 20.33
2 b 7
= -
50 19.76
*
L
9.41 65.9
=
V
= = = = =
1 2 1.00 2.01 A 2.01 14.046
*
50
*
2
* *
b 7
*
0.04
Tabel 3.1 Rekapitulasi Perhitungan PERHITUNGAN TINGGI TITIK STASIUN
Pn = Kt
PERHITUNGAN BEDA TINGGI
+ (x/y) . Ik
Dimana : Pn = Tinggi titik Stasiun ke - n Kt = Tinggi Kontur Terendah x = Jarak stasiun dengan kontur terendah y = Jarak antara kontur terendah dan tertinggi Ik = Nilai interval kontur
No Stasiun (Pn)
Jarak STA (m)
Lengkung Horisontal R (m) Sudut (p) L(m)
P0
KM = (BT/Jarak stasiun) x 100 %
Dimana : BT = Beda Tinggi (m) Pn akhir = Tinggi Stasiun Akhir (m) Pn awal = Tinggi Stasiun awal (m)
Dimana : KM = Kemiringan (%) BT = Beda Tinggi (m)
Kt
X
Y
Interval
Jarak Langsung (m)
23.50
0
0.2
0.5
200
23.90
0.4
0.6
0.5
300
Kontur
200 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 P21 P22 P23 P24 P25 P26
100 100 100
480
60
100 200 100 100 100
500
60
250
100 100 100 200 200 100 100
520
45
260
100 100 200 200 100 100 100 100 200
530
45
24.10
0.9
1.1
0.5
400
24.20
1.3
1.5
0.5
500
24.30
0.8
1
0.5
600
240
100
265
24.35
0.2
0.4
0.5
700
24.40
0.7
0.9
0.5
900
24.60
1.1
1.3
0.5
1000
24.75
1.6
1.8
0.5
1100
24.80
2
2.2
0.5
1200
25.25
1.7
1.9
0.5
1300
25.75
1.3
1.5
0.5
1400
25.55
0.6
0.8
0.5
1500
25.65
1.2
1.4
0.5
1700
25.80
1.9
2.1
0.5
1900
25.90
2.3
2.5
0.5
2000
26.00
2.8
3
0.5
2100
26.05
2.4
2.6
0.5
2200
26.10
1.5
1.7
0.5
2300
26.20
2.1
2.3
0.5
2500
26.20
2.6
2.8
0.5
2700
26.35
2.2
2.4
0.5
2800
26.40
1.7
2.1
0.5
2900
26.40
1.5
1.7
0.5
3000
26.45
0.6
0.8
0.5
3100
26.75
0.1
0.3
0.5
3300
27.75
0
0.2
0.5
3500
200 P27
27.65
0
0.1
0.5
3600
27.00
0
0.2
0.5
3600
100 P28
PERHITUNGAN KEMIRIGAN
BT = Pn akhir - Pn awal
Tinggi Stasiun (m)
Kemiringan (%)
Jarak Sta 100 Beda Tinggi (m) Keterangan
23.50
0.37
0.73
24.23
0.28
0.28
24.51
0.12
0.12
24.63
0.07
0.07
24.70
-0.10
-0.10
24.60
0.19
0.19
24.79
0.12
0.23
25.02
0.17
0.17
25.19
0.06
0.06
25.25
0.44
0.44
25.70
0.49
0.49
26.18
-0.26
-0.26
25.93
0.15
0.15
26.08
0.09
0.17
26.25
0.05
0.11
26.36
0.11
0.11
26.47
0.04
0.04
26.51
0.03
0.03
26.54
0.12
0.12
26.66
0.00
0.01
26.66
0.07
0.14
26.81
0.00
0.00
26.80
0.04
0.04
26.84
-0.02
-0.02
26.83
0.09
0.09
26.92
0.42
0.83
27.75
-0.05
-0.10
-0.65 0.00
-0.65 0.00
27.65 27.00
tikungan 1 Kelandaian (%)
24.70
- 24.23 240
x
100
0.19 (datar)
%
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 480 m tikungan 2 Kelandaian (%)
26.18
- 25.02 250
x
100
0.46 (datar)
%
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 500 m tikungan 3 Kelandaian (%)
26.54
- 26.36 260
x
100
0.07 (datar)
%
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 520 m tikungan 4 Kelandaian (%)
26.84
- 26.80 265
x
100
0.01 (datar)
jalan kelas II A (lihat tabel perencanaan geometrik) kecepatan rencana : 90 Km/Jam Jari -Jari rencana minimum : 530 m
%
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