Download - BAB III FIX FIX FIX.docx
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4. Melakukan pembidikan kea rah rambu ukur pada titik II, III, BM2, BM4, dan
BM5 dengan membaca benang atas, benang bawah, sudut vertikal, sudut
horizontal biasa, dan sudut horizontal luar biasa.
5. Melakukan penembakan ke arah rambu ukur pada titik-titik penyebaran di
sekitar titik I sebanyak 25 titik, dan mendapatkan nilai pembacaan benang atas,
benang bawah, sudut horinzontal, dan sudut vertikal.
6. Memindahkan alat (Theodolith) ke titik II dan melakukan penyettingan alat
dengan benar serta mengukur tinggi alat, kemudian melakukan pembidikan ke
arah rambu ukur pada titik I dan III, dengan membaca benang atas, benang
bawah, sudut vertikal, sudut horizontal biasa, dan sudut horizontal luar biasa.
7. Melakukan penembakan ke arah rambu ukur pada titik-titik penyebaran di
sekitar titik II sebanyak 25 titik, dan mendapatkan nilai pembacaan benang
atas, benang bawah, sudut horizontal, dan sudut vertikal.
8. Memindahkan alat ke titik III dan melakukan langkah yang sama seperti pada
titik II.
3.4 Perhitungan Data
3.4.1 Perhitungan Titik Acuan
a. Data Lapangan
1. Titik I – BM2
Ba = 1,64 m
Bb = 1,55 m
Bt = 1,595 m
Sudut horizontal biasa = 90° 18' 53"
Sudut horizontal luar biasa = 270° 19' 30"
Sudut vertikal = 94° 52' 36"
2. Titik I – BM3
Ba = 2,576 m
Bb = 2,469 m
Bt = 2,532 m
Sudut horizontal biasa = 128° 11' 35"
Sudut horizontal luar biasa = 308° 12' 07"
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Sudut vertikal = 91° 57' 22"
b. Perhitungan Azimuth
Sudut Horizontal < 180°, Azimuth= Sudut Horizontal + 180°
Sudut Horizontal > 180°, Azimuth= Sudut Horizontal + 180°.............3.1
αBM2 = 90° 18' 53" + 180° = 270° 18' 53"
αBM3 = 128° 11' 35" + 180° = 308° 11' 35"
c. Perhitungan Jarak
1. Y = ba – bb........................................................................................ 3.2
Y titik ikat 2 = 1,64-1,55 = 0,09 m
Y titik ikat 3 = 2,576-2,469 = 0,107 m
2. Heling (h) = 90° - sudut vertikal........................................................3.3
h = 90° - 94° 52' 36"= -4° 52' 36"
h = 90° - 91° 57' 22"= -1° 57' 22"
3. Jarak (D) = A . Y . cos²h....................................................................3.4
A = 100
D titik ikat 3 = 100 . 0,09 . cos² (-4° 52' 36") = 8,935 m
D titik ikat 4 = 100 . 0,107 . cos² (-1° 57' 22") = 10,688 m
ΣD = 24,475 m
d. Perhitungan beda tinggi
1. V = D tan h........................................................................................3.5
V titik ikat 2 = 8,935 tan (-4° 52' 36") = -0,762 m
V titik ikat 4 = 10,688 tan (-1° 57' 22") = -0,365 m
2. ∆h = V + tinggi alat – bt....................................................................3.6
∆h titik ikat 2 = -0,762 + 1,403– (1,64+1,55
2¿ = -0,954 m
∆h titik ikat 4 = -0,365 + 1,403 – ( 2,576+2,469
2¿= -1,485 m
Σ∆h = -2,439 m
3. Elevasi BM2 = 351,080
Elevasi BM4 = 350,53
Elevasi Titik I= Elevasi BM + ∆h.....................................................3.7
Elevasi titik ikat 1-3 = 351,080 – (-0,954) = 350,126
Elevasi titik ikat 1-4 = 349,910 – (-1,485) = 349,045
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4. Selisih Elevasi Titik I
Elevasi Titik I (Titik Ikat 2) Elevasi Titik Ikat 2……….…….….3.8
= 350,126 – 349,045
= 1,080
5. KoreksiElevasi=Drata−rataΣ Drata−ratax
fh......................................................3.9
Koreksi Elevasi titik ikat 2 = 8,93519,623 x – (1,080) = -0,555
Koreksi Elevasi titik ikat 4 = 10,68819,623 x – (1,080) = +0,664
6. Elevasi terkoreksi = Elevasi titik I (besar) – koreksi elevasi
Elevasi titik I (kecil) + koreksi elevasi...........................................3.10
Elevasi terkoreksi titik ikat 3 = 350,126 - 0,555= 349,57
Elevasi terkoreksi titik ikat 4 = 349,396 + 0,664= 349,71
e. Perhitungan koordinat di titik I
1. D sin α……………………………………………………………3.11
D sin α titik ikat 2 = 8,935 sin 270°18’53” = -8,935 m
D sin α titik ikat 4 = 10,688 sin 308°11’35” = -8,399 m
2. D cos α…………………………………….…………………..…3.12
D cos α titik ikat 2 = 8,935 cos 270°18’53” = 0,049 m
D cos α titik ikat 4 = 10,688 sin 308°11’35” = 6,608 m
3. Koordinat Titik TI BM 2 X = 436315,246
Koordinat Titik TI BM 4 X = 436315,212
Koordinat Titik X=Koordinat BM X+ D sin α…………………3.13
Titik ikat 2 = 436315,246 + (-8,935) = 436306,311
Titik ikat 4 = 436315,212 + (-8,399) = 436306,812
Selisih koordinat = -0,501
4. Koordinat Titik TI BM 3 Y = 9150041,074
Koordinat Titik TI BM 3 Y = 9150034,563
Koordinat Titik Y = Koordinat BM Y+D cos α……………...….3.14
Titik ikat 3 = 9150041,074 + (0,049) = 9150041,123
Titik ikat 4 = 9150034,563 + (6,608) = 9150041,717
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Selisih koordinat = -0,594
5. Koreksi Koordinat X
Koreksi Koordinat =Drata−rataΣ Drata−rata x selisih koordinat...…...……
3.15
Titik ikat 2 = 8,93519,623 x 0,501 = 0,228195
Titik ikat 4 = 10,68819,623 x 0,501 = -0,27296
6. Koreksi Koordinat Y
Koreksi Koordinat= Drata−rataΣ Drata−rata x selisih koordinat………….…
3.16
Titik ikat 2 = 8,93519,623 x 0,594 = 0,02193
Titik ikat 4 = 10,68819,623x 0,594= -0,0262
7. Koordinat Terkoreksi Titik X
Koordinat Titik X = Koordinat Titik X – Koreksi Koordinat
(Terbesar)
Koordinat Titik X = Koordinat Titik X + Koreksi Koordinat
(Terkecil)
Nilai Elevasi Harus Sama………………………..………...……3.17
Titik ikat 2 = 436306,311 + 0,228195 = 436306,539
Titik ikat 4 = 436306,812 – 0,27296 = 436306,539
8. Koordinat Terkoreksi Titik Y
Koordinat Titik Y = Koordinat Titik Y – Koreksi Koordinat
(Terbesar)
Koordinat Titik Y = Koordinat Titik Y + Koreksi Koordinat
(Terkecil)
Nilai Elevasi Harus Sama…...…………….…………….……3.18
Titik ikat 2 = 9150041,123 + 0,02193= 9150041,145
Titik ikat 4 = 9150041,717 - 0,0262 = 9150041,145
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3.4.2 Perhitungan Poligon Tertutup
I. Perhitungan Sudut Dalam
a. Sudut dalam biasa dan luar biasa
Sudut dalam biasa
B = B besar – B kecil apabila sudutnya ≥ 180° maka sudut dalam
biasanya:
B= 360°- (B besar – B kecil)...........................................................3.19
Sudut dalam luar biasa
LB = LB besar – LB kecil apabila sudutnya ≥ 180° maka sudut dalam
luar biasanya:
LB= 360° - (LB besar – LB kecil)...................................................3.20
Perhitungan sudut dalam di titik poligon 1
B = 216° 12' 40" - 175° 52' 00" = 40° 20' 40"
LB = 336° 13' 29" - 335° 53' 01" = 40° 20' 28"
Perhitungan sudut dalam di titik poligon 2
B = 00° 00' 00" - 225° 54' 46" = 34° 05' 14"
LB = 180° 00' 00" - 145° 54' 45" = 34° 05' 15"
Perhitungan sudut dalam di titik poligon 3
B = 00° 00' 00" - 254° 34’ 44" = 105° 25' 16"
LB = 180° 00' 00" - 74° 34' 20" = 105° 25' 40"
b. Sudut Dalam Rata – rata
Σɵ = B+LB
2
.......................................................................................3.21
Titik I
ɵ1 = 40 °20 ' 40 + 40° 20' 28
2 = 40° 20' 34"
Titik II
ɵ2 =34 °05 ' 14 + 34° 05' 15
2 = 34° 05' 14,5"
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Titik III
ɵ3 =105° 25' 16 + 105° 2 5' 40
2 = 105° 25' 28"
ΣSudut dalam rata – rata = 179° 51' 16,5"
c. Koreksi Tiap Sudut
Koreksi = (-1) xΣsudut dalamrata 2−(n−2 ) 180°
n
........................3.22
n = banyaknya poligon
Koreksi = (-1) x179 °51 ' 16,5 - left (3-2 right ) 180°} over {3 ¿ = 00°
02' 54,5"
Σkoreksi = 00° 08' 43,5"
d. Sudut Dalam Terkoreksi
Sudut dalam terkoreksi = sudut dalam rata – rata + Koreksi ..........3.23
Titik I
ɵ1 = 40° 20' 34" + (00° 02' 54,5") = 40° 23' 28,5"
Titik II
ɵ2 = 34° 05' 14,5" + (00° 02' 54,5") = 34° 08' 09"
Titik III
ɵ3 = 105° 25' 28" + (00° 02' 54,5") = 105° 28' 22,5"
Σɵ terkoreksi = 180° 00' 00"
II. Perhitungan Azimuth
α ± 180° ± ɵ...........................................................................................3.24
α 12 = azimuth awal = 175° 52' 00"
α 23 = α 12 ± 180° ± ɵ2 = 175° 52' 00" – 180° - 34° 08' 09" = 321° 43' 51"
α 31 = α 23± 180° ± ɵ3 = 321° 43' 51" + 180° – 105° 28' 22,5" = 36°15'
28,5"
α 12 = α 31± 180° ± ɵ1 = 36°15' 28,5" – 180° + 40° 23' 28,5" = 175° 52' 00"
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III. Perhitungan Jarak
1. Y= ba – bb.......................................................................................3.25
I arah III = 1,97 – 1,73 = 0,240 m
II = 1,39 – 0,97 = 0,420 m
II arah I = 1,88 – 1,46 = 0,420 m
III = 1,519 – 1,24 = 0,279 m
III arah II= 1,271 – 0,994 = 0,277 m
I = 2,024 – 1,778 = 0,246 m
2. Heling (h) = 90° - sudut vertikal.....................................................3.26
I arah III = 90° – 97°09'37° = -07°09'37"
II = 90° – 94°12'28" = -04°12'28"
II arah I = 90° – 85°46'35" = 04°13'25"
III = 90° – 91°23'58" = -01°23'58"
III arah II = 90° – 89°03'37" = 00°56'23"
I = 90° – 80°30'54" = 09°29'06"
3. D = A . Y . cos²h..............................................................................3.27
A = 100 Y = ba – bb
Heling (h) = 90° - sudut vertikal
I arah III = 100 . 0,240 . cos²(-07°09'37") = 23,627 m
II = 100 . 0,420 . cos²(-04°12'28") = 41,774 m
II arah I = 100 . 0,420 . cos²(04°13'25") = 41,772 m
III = 100 . 0,279 . cos²(-01°23'58") = 27,883 m
III arah II = 100 . 0,277 . cos²(00°56'23") = 27,693 m
I = 100 . 0,246 . cos²(00°56'23") = 23,932 m
I arah III = 100 . 0,240 . cos²(-07°09'37") = 23,627 m
4. Jarak rata – rata (ΣD) :
ΣD=(41,774+41,772
2 )+(27,883+27,693
2 )+¿)
= 41,773 + 27,778 + 23,78 = 93,331 m
IV. Perhitungan Beda Tinggi
1. V = D tan h.....................................................................................3.28
I arah III = 23,627 tan (-07°09'37") = -2,698 m
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II = 41,774 tan (-04°12'28") = -3,073 m
II arah I = 41,772 tan (04°13'25") = 3,085 m
III = 27,883 tan (-01°23'58") = -0,681 m
III arah II = 27,693 tan (00°56'23") = 0,454 m
I = 23,932 tan (09°29'06") = 3,998 m
2. ∆h = V + ti – bt................................................................................3.29
I arah III = -2,698 + 1,403 – 1,85 = -3,415 m
II = -3,073 + 1,403 – 1,18 = -0,637 m
II arah I = 3,085 + 1,404 – 1,67 = 2,819 m
III = -0,681 + 1,404 – 1,379= -0,657 m
III arah II = 0,454 + 1,336 – 1,132 = 0,658 m
I = 3,998 + 1,336 – 1,901 = 3,433 m
I arah III = -2,698 + 1,403 – 1,85 = -3,415 m
3. ∆h rata – rata
I – II = −2,85+2,819
2 = -0,016 m
II – III = −0,657+0,658
2 = 0,001 m
III – I = 3,433+(−3,415)
2= 0,009 m
Σ∆h rata – rata = -0,006 m
4. Koreksi
k = Drata−rataΣDrata−rata
x (-Σ∆hrata-rata)......................................................3.30
I – II = 41,77393,331 x (0,006) = 0,002685
II – III = 27,77893,331 x (0,006) = 0,001786
III – I =23,78
93,331 x (0,006) = 0,001592
5. ∆h terkoreksi
∆hrata-rata + koreksi............................................................................3.31
I – II = -0,016 + 0,002685 = -0,013 m
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II – III = 0,001 + 0,001786 = 0,002 m
III – I = 0,009 + 0,001592 = 0,011 m
6. Elevasi
I = 349,5704
II = 349,5704 + (-0,013) = 349,557
III = 349,557 + (0,002) = 349,560
I = 349,560 + 0,011 = 349,570
V. Perhitungan Koordinat Poligon
1. D sin α……………..………..…………………………………3.32
Titik I – II
D sin α = 41,773 sin 175°52'00" = 3,012 m
Titik II – III
D sin α = 27,778 sin 321°43'51" = -17,213 m
Titik III – I
D sin α = 23,780 sin 36°15'28,5" = 14,064 m
2. D cos α…………………………….………………………………3.33
Titik I – II
D sin α = 41,773 cos 175°52'00" = -41,664 m
Titik II – III
D sin α = 27,778 cos 321°43'51" = 21,817 m
Titik III – I
D sin α = 23,780 cos 36°15'28,5" = 19,175 m
3. FxijD ij
ΣD . (-Σfx).......................................................................................3.34
-Σfx = ΣD sin α
Fx1−2 = 41,77393,341 . (0,136) = 0,061
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Fx2−3 = 27,78893,341 . (0,136) = 0,040
Fx3−1 = 23,78
93,341 . (0,136) = 0,035
4. Fy ijD ij
ΣD . (-
Σfy)........................................................................................3.35
-Σfy = ΣD cos α
Fx1−2 = 41,77393,341 . (0,673) = 0,301
Fx2−3 = 27,78893,341 . (0,673) = 0,2003
Fx3−1 = 23,78
93,341 . (0,673) = 0,171
5. D sin α terkoreksi = D sin α + Fxij……………………….………..3.36
D sin α terkoreksi 1-2 = 3,012 + (0,061) = 3,072
D sin α terkoreksi 2-3 = -17,211 + 0,040 = -17,1702
D sin α terkoreksi 3-1 = 14,064 + 0,035 = 14,098
6. D cos α terkoreksi = D cos α + Fyij……………….…………..3.37
D cos α terkoreksi 1-2 = -41,664 + 0,301 = -41,363
D cos α terkoreksi 2-3 = 21,817 + 0,2003 = 22,0169
D cos α terkoreksi 3-1 = 19,175 + 0,171 = 19,346
7. Koordinat X = Koordinat terkoreksi X + D sin α terkoreks….…3.38
Koordinat X1= 436306,5394
Koordinat X1-2 = 436306,5394 + 3,072 = 436309,6112
Koordinat X2-3 = 436309,6112 +(-17,1702) = 436292,441
Koordinat X3-1 = 436292,441 + 14,098 = 436306,5394
Koordinat Y = Koordinat terkoreksi Y + D cos α terkoreksi……..3.39
Koordinat Y1= 9150041,145
Koordinat Y1-2 = 9150041,145 + (-41,363) = 9149999,782
Koordinat Y2-3 = 9149999,782 + 22,0169 = 9150021,799
Koordinat Y3-1 = 9150021,799+ 19,346 = 9150041,145
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3.5 Perhitungan Titik Detail Penyebaran
1. Perhitungan jarak optis
D = A (ba-bb) . cos²h............................................................................3.40
D = Jarak optis ba = Benang atas
A = Konstanta bb = Benang bawah
h = Heling (90° - Arah vertikal)
a. Perhitungan jarak optis pada kedudukan I
D1 = 100 (1,302 – 1,066). cos2( 90o-90o 51’ 58”) = 23,596 m
D2 = 100 (1,845 – 1,581) . cos²(90° - 91° 45' 37") = 26,375 m
D3 = 100 (1,114 – 0,945) . cos²(90° - 91° 45' 46") = 16,884 m
D4 = 100 (3,081 – 2,939) . cos²(90° - 89° 05' 44") = 14,197 m
D5 = 100 (1,325 – 1,160) . cos²(90° - 93° 27' 20") = 16,440 m
D6 = 100 (0,755 – 0,574) . cos²(90° - 91° 07' 06") = 18,093 m
D7 = 100 (1,833 – 1,746) . cos²(90° - 85° 28' 40") = 8,6460 m
D8 = 100 (1,438 – 1,400) . cos²(90° - 84° 54' 46") = 3,7810 m
D9 = 100 (1,800 – 1,778) . cos²(90° - 85° 21' 34") = 2,1856 m
D10 = 100 (1,911 – 1,845) . cos²(90° - 85° 23' 46") = 6,5575 m
D11 = 100 (2,367 – 2,545). cos²(90° - 89° 25' 48") = 9,1991 m
D12 = 100 (1,744 – 1,682) . cos²(90° - 91° 56' 05") = 6,1929 m
D13 = 100 (2,715 – 2,595) . cos²(90° - 91° 22' 29") = 11,993 m
D14 = 100 (2,854 – 2,716) . cos²(90° - 92° 22' 39") = 13,776 m
D15 = 100 (2,485 – 2,355) . cos²(90° - 97° 02' 10") = 12,805 m
D16 = 100 (2,383 – 2,243) . cos²(90° - 95° 47' 15") = 13,858 m
D17 = 100 (1,686 – 1,534) . cos²(90° - 95° 10' 51") = 15,076 m
D18 = 100 (3,35 – 3,244 ) . cos²(90° - 95° 13' 46") = 10,512 m
D19 = 100 (2,40 – 2,22 ) . cos²(90° - 95° 14' 07") = 17,850 m
D20 = 100 (3,21 – 3,08 ) . cos²(90° - 94° 58' 28") = 12,902 m
b. Perhitungan jarak optis pada kedudukan II
D1 = 100 (1,06 – 0,54) . cos²(90° - 87° 40' 03") = 51,914 m
D2 = 100 (1,542 – 2,05) . cos²(90° - 86° 33' 59") = 50,618 m
D3 = 100 (1,699 – 1,423) . cos²(90° - 89° 10' 40") = 27,593 m
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D4 = 100 (1,433 – 1,220) . cos²(90° - 89° 41' 51") = 21,299 m
D5 = 100 (1,42 – 1,305) . cos²(90° - 89° 55' 22") = 11,500 m
D6 = 100 (1,635 – 1,478) . cos²(90° - 89° 07' 57") = 15,696 m
D7 = 100 (1,466 – 1,396) . cos²(90° - 89° 07' 57") = 6,9980 m
D8 = 100 (1,442 – 1,328) . cos²(90° - 89° 41' 31") = 11,399 m
D9 = 100 (1,779 – 1,625) . cos²(90° - 88° 18' 08") = 15,387 m
D10 = 100 (1,462 – 1,286) . cos²(90° - 89° 28' 05") = 17,599 m
D11 = 100 (0,885 – 0,722) . cos²(90° - 87° 53' 16") = 16,278 m
D12 = 100 (1,39 – 1,20) . cos²(90° - 84° 57' 25") = 18,853 m
D13 = 100 (2,449 – 2,265) . cos²(90° - 90° 43' 19") = 18,397 m
D14 = 100 (1,374 – 1,273) . cos²(90° - 98° 43' 19") = 9,8680 m
D15 = 100 (4,248 – 4,136) . cos²(90° - 88° 13' 34") = 11,189 m
D16 = 100 (1,92 – 1,785) . cos²(90° - 99° 50' 14") = 13,104 m
D17 = 100 (1,276 – 1,137) . cos²(90° - 92° 46' 36") = 13,867 m
D18 = 100 (2,535 – 2,435) . cos²(90° - 102° 37' 05") = 9,5228 m
D19 = 100 (2,418 – 2,303) . cos²(90° - 90° 44' 54") = 11,498 m
D20 = 100 (2,58 – 2,536) . cos²(90° - 94° 27' 23") = 4,3734 m
c. Perhitungan jarak optis pada kedudukan III
D1 = 100 (2,372 – 2,12 ) . cos²(90° - 88° 14' 15") = 25,176 m
D2 = 100 (1,29 – 1,075) . cos²(90° - 91° 08' 19") = 21,471 m
D3 = 100 (1,644 – 1,411) . cos²(90° - 91° 08' 19") = 23,291 m
D4 = 100 (0,856 – 0,696) . cos²(90° - 91° 41' 07") = 15,986 m
D5 = 100 (2,75 – 2,551) . cos²(90° - 89° 22' 52") = 19,898 m
D6 = 100 (1,75 – 1,585) . cos²(90° - 91° 18' 41") = 16,491 m
D7 = 100 (3,149 – 2,959) . cos²(90° - 89° 18' 36") = 18,997 m
D8 = 100 (1,553 – 1,41) . cos²(90° - 95° 43' 42") = 14,158 m
D9 = 100 (1,71 – 1,54) . cos²(90° - 80° 24' 46") = 16,528 m
D10 = 100 (1,785 – 1,711) . cos²(90° - 94° 01' 06") = 7,3637 m
D11 = 100 (0,99 – 0,84) . cos²(90° - 86° 01' 00") = 14,928 m
D12 = 100 (1,283 – 1,15) . cos²(90° - 88° 28' 46") = 13,291 m
D13 = 100 (1,564 – 1,429) . cos²(90° - 87° 05' 13") = 13,465 m
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D14 = 100 (1,755 – 1,659) . cos²(90° - 97° 19' 16") = 9,4441 m
D15 = 100 (0,743 – 0,661) . cos²(90° - 93° 23' 29") = 8,1713 m
D16 = 100 (1,381 – 1,29) . cos²(90° - 95° 06' 59") = 9,0276 m
D17 = 100 (0,966 – 0,916) . cos²(90° - 92° 17' 08") = 4,9920 m
D18 = 100 (1,00 – 0,885) . cos²(90° - 81° 30' 25") = 11,249 m
D19 = 100 (2,169 – 2,084) . cos²(90° - 89° 48' 18") = 8,4999 m
D20 = 100 (1,413 – 1,202) . cos²(90° - 81° 58' 21") = 20,688 m
2. Perhitungan beda tinggi (∆h)
∆h = D tan h + ti – bt...........................................................................3.41
∆h = Beda tinggi ti = Tinggi alat
D = Jarak optis bt = Benang tengah
heling = Heling (90° - sudut vertikal)
a. Perhitungan beda tinggi pada kedudukan I
∆h1 = 1,403 + 23,593. tan (-0,866) – 1,184 = -0,138 m
∆h2 = 1,403 + 26,375. tan (-1,760) – 1,713 = -1,121 m
∆h3 = 1,403 + 16,884. tan (-1,763) – 1,030 = -0,146 m
∆h4 = 1,403 + 14,146. tan (0,904 ) – 3,010 = -1,383 m
∆h5 = 1,403 + 16,440. tan (-3,456) – 1,243 = -0,832 m
∆h6 = 1,403 + 18,093. tan (-1,118) – 0,665 = 0,385 m
∆h7 = 1,403 + 8,646. tan (4,522) – 1,790 = 0,297 m
∆h8 = 1,403 + 3,770. tan (5,087) – 1,419 = 0,320 m
∆h9 = 1,403 + 2,186. tan (4,641) – 1,798 = -0,209 m
∆h10 = 1,403 + 6,557. tan (4,604) – 1,878 = 0,053 m
∆h11 = 1,403 + 9,199. tan (0,570) – 2,591 = -1,096 m
∆h12 = 1,403 + 6,193. tan (-1,935) – 1,713 = -0,519 m
∆h13 = 1,403 + 11,993. tan (-1,375) – 2,655 = -1,540 m
∆h14 = 1,403 + 13,776. tan (-2,378) – 2,785 = -1,954 m
∆h15 = 1,403 + 12,805. tan (-7,036) – 2,420 = -2,597 m
∆h16 = 1,403 + 13,858. tan (-5,787) – 2,313 = -2,315 m
∆h17 = 1,403 + 15,076. tan (-5,181) – 1,610 = -1,574 m
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∆h18 = 1,403 + 10,512. tan (-5,229) – 3,297 = -2,856 m
∆h19 = 1,403 + 17,850. tan (-5,235) – 2,310 = -2,543 m
∆h20 = 1,403 + 12,902. tan (-4,974) – 3,145 = -2,865 m
b. Perhitungan beda tinggi pada kedudukan II
∆h1 = 1,404 + 51,914. tan (2,333) – 0,800 = 2,719 m
∆h2 = 1,404 + 50,618. tan (3,434) – 1,786 = 2,645 m
∆h3 = 1,404 + 27,594. tan (0,822) – 1,561 = 0,239 m
∆h4 = 1,404 + 21,299. tan (0,302) – 1,327 = 0,190 m
∆h5 = 1,404 + 11,500. tan (0,077) – 1,363 = 0,057 m
∆h6 = 1,404 + 15,696 . tan (0,874) – 1,557 = 0,087 m
∆h7 = 1,404 + 6,998. tan (0,868) – 1,431 = 0,079 m
∆h8 = 1,404 + 11,400. tan (0,308) – 1,385 = 0,080 m
∆h9 = 1,404 + 15,386. tan (1,698) – 1,702 = 0,158 m
∆h10 =1,404 + 17,598. tan (0,532) – 1,374 = 0,193 m
∆h11 = 1,404 + 16,278. tan (2,112) – 0,804 = 1,201 m
∆h12 = 1,404 + 18,853. tan (5,043) – 1,295 = 1,773 m
∆h13 = 1,404 + 18,397. tan (-0,721) – 2,357 = -1,185 m
∆h14 = 1,404 + 9,868. tan (-8,722) – 1,324 = -1,433 m
∆h15 = 1,404 + 11,189. tan (1,774) – 4,192 = -2,441 m
∆h16 = 1,404 + 13,106. tan (-9,837) – 1,853 = -2,721 m
∆h17 = 1,404 + 13,867. tan (-2,777) – 1,207 = -0,475 m
∆h18 = 1,404 + 9,523. tan (-12,618) – 2,485 = -3,213 m
∆h19 = 1,404 + 11,498. tan (-0,748) – 2,361 = -1,107 m
∆h20 = 1,404 + 4,373. tan (-4,456) – 2,558 = -1,495 m
c. Perhitungan beda tinggi pada kedudukan III
∆h1 = 1,336 + 25,176. tan (1,763) – 2,246 = -0,135 m
∆h2 = 1,336 + 21,471. tan (2,106) – 1,183 = 0,943 m
∆h3 = 1,336 + 23,291. tan (-1,139) – 1,528 = -0,645 m
∆h4 = 1,336 + 15,986. tan (-1,685) – 0,776 = 0,090 m
∆h5 = 1,336 + 19,898. tan (0,619) – 2,651 = -1,100 m
40
∆h6 = 1,336 + 16,491. tan (-1,311) – 1,668 = -0,709 m
∆h7 = 1,336 + 18,997. tan (0,690) – 3,054 = -1,489 m
∆h8 = 1,336 + 14,158. tan (-5,728) – 1,482 = -1,566 m
∆h9 =1,336 + 16,528. tan (9,857) – 1,625 = 2,503 m
∆h10 = 1,336 + 7,364. tan (-4,018) – 1,748 = -0,929 m
∆h11 = 1,336 + 14,928. tan (3,983) – 0,915 = 1,460 m
∆h12 = 1,336 + 13,291 . tan (1,521) – 1,217 = 0,472 m
∆h13 = 1,336 + 13,465. tan (2,913) – 1,497 = 0,525 m
∆h14 = 1,336 + 9,444. tan (-7,321) – 1,707 = -1,584 m
∆h15 = 1,336 + 8,171. tan (-3,391) – 0,702 = 0,150 m
∆h16 = 1,336 + 9,028. tan (-5,116) – 1,336 = -,808 m
∆h17 = 1,336 + 4,992. tan (-2,286) – 0,941 = 0,196 m
∆h18 = 1,336 + 11,249. tan (8,493) – 0,943 = 2,073 m
∆h19 = 1,336 + 8,500. tan (0,195) – 2,127 = -0,762 m
∆h20 = 1,336 + 20,689. tan (8,028) – 1,308 = 2,946 m
3. Perhitungan Elevasi Detail
Elevasi + ∆h.........................................................................................3.42
a. Perhitungan elevasi detail pada kedudukan I
ED 1 = 349,570 + (-0,138) = 349,433 m
ED 2 = 349,570 + (-1,121) = 348,450 m
ED 3 = 349,570 + (-0,146) = 349,424 m
ED 4 = 349,570 + (-1,383) = 348,188 m
ED 5 = 349,570 + (-0,832) = 348,738 m
ED 6 = 349,570 + 0,385 =349,956 m
ED 7 = 349,570 + 0,297 = 349,686 m
ED 8 = 349,570 + 0,320 = 349,890 m
ED 9 = 349,570 + (-0,209) = 349,362 m
ED 10 = 349,570 + 0,053 = 349,623 m
ED 11 = 349,570 + (-1,096) = 348,474 m
ED 12 = 349,570 + (-0,519) = 349,051 m
ED 13 = 349,570 + (-1,540) = 348,031 m
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ED 14 = 349,570 + (-1,594) = 347,616 m
ED 15 = 349,570 + (-2,597) = 346,973 m
ED 16 = 349,570 + (-2,315) = 347,256 m
ED 17 = 349,570 + (-1,574) = 347,996 m
ED 18 = 349,570 + (-2,856) = 346,714 m
ED 19 = 349,570 + (-2,543) = 347,028 m
ED 20 = 349,570 + (-2,865) = 346,705 m
b. Perhitungan elevasi detail pada kedudukan II
ED 1 = 349,557 + 2,719 = 352,276 m
ED 2 = 349,557 + 2,645 = 352,202 m
ED 3 = 349,557 + 0,239 = 349,796 m
ED 4 = 349,557 + 0,190 = 349,747 m
ED 5 = 349,557 + 0,057 = 349,614 m
ED 6 = 349,557 + 0,087 = 349,644 m
ED 7 = 349,557 + 0,079 = 349,636 m
ED 8 = 349,557 + 0,080 = 349,637 m
ED 9 = 349,557 + 0.158 = 349,715 m
ED 10 = 349,557 + 0,193 = 349,750 m
ED 11 = 349,557 + 1,201 = 350,758 m
ED 12 = 349,557 + 1,773 = 351,330 m
ED 13 = 349,557 + (-1,185) = 348,372 m
ED 14 = 349,557 +(-1,433) = 348,124 m
ED 15 = 349,557 +(-2,441) = 347,116 m
ED 16 = 349,557 + (-2,721) = 346,836 m
ED 17 = 349,557 + (-0,475) = 349,082 m
ED 18 = 349,557 + (-3,213) = 346,344 m
ED 19 = 349,557 + (-1,107) = 348,450 m
ED 20 = 349,557 + (-1,495) = 348,062 m
c. Perhitungan elevasi detail pada kedudukan III
42
ED 1 = 349,560 + (-0,135) = 349,425 m
ED 2 = 349,560 + (0.943) = 350,503 m
ED 3 = 349,560 + (-0,654) = 348,906 m
ED 4 = 349,560 + (0,090) = 349,650 m
ED 5 = 349,560 + (-1,100) = 348,460 m
ED 6 = 349,560 + (-0,709) = 348,851 m
ED 7 = 349,560 + (-1,489) = 348,071 m
ED 8 = 349,560 + (-1,566) = 347,994 m
ED 9 = 349,560 + 2,503 = 352,063 m
ED 10 = 349,560 + (-0,929) = 348,631 m
ED 11 = 349,560 + 1,460 = 351,020 m
ED 12 = 349,560 + 0,472 = 350,032 m
ED 13 = 349,560 + 0,525 = 350,085 m
ED 14 = 349,560 + (-1,584) = 347,976 m
ED 15 = 349,560 + 0,150 = 349,710 m
ED 16 = 349,560 + (-0,808) = 348,752 m
ED 17 = 349,560 + 0,196 = 349,756 m
ED 18 = 349,560 + 2,073 = 351,633 m
ED 19 = 349,560 + (-0,762) = 348,798 m
ED 20 = 349,560 + 2,946 = 352,506 m
3.6 Perhitungan Galian dan Timbunan
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Tabel 3.1 Koordinat, Elevasi, dan ∆h Poligon TertutupTitik Koordinat Elevasi ∆h
X Y
I 436306.539 9150041.145 349,570 +0,01
II 436309.611 9149999.782 349,557 -0,003
III 436292.441 9150021.799 349,560 0
Gambar 3.1 Sketsa Poligon
Untuk mendapatkan jarak antara titik polygon I dan III dapat dilihat pada
Gambar 3.2 kemudian menggunakan persamaan berikut:
Gambar 3. 2 Sketsa Panjang L
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L = √ (X II−X I )2+(Y I−Y II )
2
= √ (436309,611−436306,539 )2+(9150041,145−9149999,782 )2
= √ (3,072 )2+ (41,363 )2
= √1720,335
= 41,447 m
Perhitungan jarak I dan P:
∆H I
∆ H II¿ L−XX
0,010,003 =
41,447−XX
0,01 X= −0,003 X+0,124
0,013X= 0,124
X = 9,54 m
Perhitungan jarak titik II dan P :
L – X = 41,447 m – 9,54 m = 10,5177 m
Menentukan koordinat titik P :
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Gambar 3.3 Posisi titik P terhadap titik I dan III
Xa
= LX II−X I
9,54a
= 41,477436309,611−436306,539
41,477a = 29,307
a = 0,707 m
b =√X2−a2=√9,542−0,7072
¿√91,0116−0,4998
¿√90,5118
¿9,514m
XP = XII – a = 436309,611 m – 0,707 m = 436308,904 m
YP = YII – b = 9149999,782 m – 9,514 m = 9149990,268 m
Jadi, koordinattitik P adalah (436308,904 ; 9149990,268)
3.6.1 Timbunan
46
Gambar 3.4 Sket Timbunan
L = 12 {(XII . YIII + XIII . YP + XP . YII) – (XIII . YII + XP . YIII + XII . YP)}
= 12 {(436309,611 . 9150021,799 + 436292,441 . 9149990,268 +
436308,904 . 9149999,782) – (436292,441 . 9149999,782 +
436308,904 . 9150021,799 + 436309,611 . 9149990,268)}
= 126,444 m2
V = 13 . L . ∆HII
= 13 . 126,444 m2 . 0,003 m
= 12,939 m3
3.6.2 Galian
47
Gambar 3.5 Sket Galian
L = 12 {(XIII . YI + XI . YP + XP . YIII) – (XI . YIII + XP . YI + XIII . YP)}
= 12 {(436292,441 . 9150041,145 + 436306,539 . 9149990,268 +
436308,904 . 9150021,799) – (436306,539 . 9150021,799 +
436308,904 . 9150041,145 + 436292,441 . 9149990,268)}
= 122,411 m2
V = 13 . L . ∆HIII
= 13 . 126,444 m2 . 0,01 m
= 12,241 m3
3.7 Pembahasan
Poligon tertutup merupakan serangkaian titik yang dihubungkan dengan
garis lurus yang membentuk suatu bidang dimana titik awal dan titik akhir
mempunyai koordinat yang sama.Perhitungan sudut dalam pada sudut dalam
terkoreksi (Σɵ terkoreksi) harus didapat 180° 00' 00", supaya membentuk suatu
segitiga dan dapat kembali pada titik awal, karena jumlah sudut dalam suatu
segitiga adalah 180°. Pada perhitungan azimuth, didapat berdasarkan rumus
(azimuth awal α 12) = pembacaan theodolit α 12 sebesar 175° 52' 00", maka
pada perhitungan akhir harus sama.
Perhitungan jarak pada titik 1 arah 3 dan 3 arah 1 harus sama besar atau
jika ada selisih tidak terlalu besar, salah satu penyebabnya adalah kurangnya
48
ketelitian dalam membaca rambu. Pada perhitungan beda tinggi (∆H), sudut
elevasi awal dan akhir harus sama besar. Jika berbeda, titik awal dan titik
akhir tidak mempunyai koordinat yang sama. Begitu pula pada perhitungan
koordinat poligon di koordinat x dan koordinat y harus didapat hasil yang
sama besar pada titik awal dan titik akhir.
3.8 Kesimpulan
Dalam bab ini, praktikum mengenai poligon tertutup yang kami lakukan
menggunakan alat theodolit. Pengertian poligon tertutup adalah serangkaian
titik yang dihubungkan dengan garis lurus yang membentuk suatu bidang,
dimana titik awal dan titik akhir memiki koordinat yang sama dan
membentuk suatu bidang datar di lapangan.
Dalam praktikum poligon tertutup bertujuan untuk mencari sudut
azimuth, jarak optis, beda tinggi, heling dan koordinat titik. Setelah semua
data yang diperlukan dalam praktikum ini terkumpul maka akan diolah dalam
perhitungan dan hasil akhir dari praktikum ini akan dihasilkan suatu peta
situasi atau peta kontur.