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"


B = 00° 00' 00" - 225° 54' 46" = 34° 05' 14"

LB = 180° 00' 00" - 145° 54' 45" = 34° 05' 15"


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

38

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

39

∆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

41

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

43

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

44

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 :

45

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.

bab iii fix fix fix.docx

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