CONTOH KASUS POLIGON TERTUTUP

CONTOH KASUS POLIGON TERTUTUPUntuk menentukan kerangka suatu proyek bangunan dilakukan dengan cara polygon tertutup pada 5 titik P, Q , R , S dan T , s dst t . .

S T drs dtp R P p r dqr Q dpq

. q

Diketahui : P (1500,000 m , 1200,000 m) sebagai titik awal dan titik akhir

pq = azimuth PQ = azimuth awal dan azimuth akhir = 248 23 42Diukur : Sudut-sudut hasil ukuran : p = 266 09 21

q = 218 16 50

r = 262 51 20 s = 256 44 21

t = 255 58 16

Jarak-jarak hasil ukuran : dpq = 728,142 m

dqr = 696,992 m

drs = 756,509 m

dst = 984,109 m

dtp = 778,819 m Ditanya : Hitung koordinat (posisi) titik-titik Q , R , S dan T jika ketelitian sudut = 10 nJawab : 1. Menghitung kesalahan sudut

= p + q + r + s + t = 266 09 21 + 218 16 50 + 262 51 20 + 256 44 21 + 255 58 16 = 1260 00 08

= (n + 2 ) x 180 = (5 + 2) x 180 = 7 x 180 = 1260

Kesalahan sudut ( f ) = 08 f = 8

10 n = 10 5 = 10x 2,24 = 22,4 f < 10 n hasil pengukuran sudut dapat diterima / memenuhi syarat geometris / sesuai spesifikasi teknis

2. Menghitung koreksi sudut Kesalahan f = 8 hrs dikoreksikan secara merata ke semua sudut hasil ukuran , jadi koreksi tiap sudut = f /n = 8/ 5 = 1,6 : p = p 1,6 q = q 1,6

r = r 1,6 krn (p + q + r + s + t) > {(n + 2) x 180 } , maka koreksinya ( - ) s = s 1,6

t = t 1,6

3. Menghitung sudut terkoreksi

p = 266 09 21 - 1,6 = 266 09 19,4 q = 218 16 50 - 1,6 = 218 16 48,4

r = 262 51 20 - 1,6 = 262 51 18,4 s = 256 44 21 - 1,6 = 256 44 19,4

t = 255 58 16 - 1,6 = 255 58 14,4

4. Menghitung Azimuth tiap sisi poligon () pq = 248 23 42

qr = pq + q - 180 = 248 23 42 + 218 16 48,4 - 180 = 286 40 30,4 rs = qr + r - 180 = 286 40 30,4 + 262 51 18,4 - 180 = 369 31 48,8 - 360

st = rs + s - 180 = 9 31 48,8 + 256 44 19,4 - 180 = 86 16 8,2 tp = st + t - 180 = 86 16 8,2 + 255 58 14,4 - 180 = 162 14 22,6

pq = tp + p - 180 = 162 14 22,6 + 266 09 19,4 - 180 = 248 23 425. Menghitung jumlah jarak ( d )

d = d pq + d qr + d rs + d st + d tp

= 728,142 m + 696,992 m + 756,509 m + 984,109 m + 778,819 m

= 3944,571 m

6. Menghitung (d sin ) : d pq sin pq = 728,142 m x sin 248 23 42 = - 676,986 m

d qr sin qr = 696,992 m x sin 286 40 30,4 = - 667,681557 m

d rs sin rs = 756,509 m x sin 9 31 48,8 = 125,253551 m

d st sin st = 984,109 m x sin 86 16 8,2 = 982,023175 m

d tp sin tp = 778,819 m x sin 162 14 22,6 = 237,568598 m

+ d sin = fx = 0,177 m7. Menghitung koreksi absis :

fx1 = (dpq / d) x fx = (728,142 / 3944,571) x 0,178 m = 0,033 m

fx2 = (dqr / d) x fx = (696,992 / 3944,571) x 0,178 m = 0,031 m

fx3 = (drs / d) x fx = (756,509 / 3944,571) x 0,178 m = 0,034 m

fx4 = (dst / d) x fx = (984,109 / 3944,571) x 0,178 m = 0,044 m

fx5 = (dtp / d) x fx = (778,.819 / 3944,571) x 0,178 m = 0,035 m

= 0,177 m8. Menghitung absis :

Karena fx > 0 , maka koreksi absis negatip

dpq sin pq = dpq sin pq - fx1 = - 676,986 0,033 = - 677,019 m

dqr sin qr = dqr sin qr - fx2 = - 667,682 - 0,031 = - 667,713 m

drs sin rs = drs sin rs - fx3 = 125,254 - 0,034 = 125,220 m

dst sin st = dst sin st - fx4 = 982,023 - 0,044 = 981,979 m

dtp sin tp = dtp sin tp - fx5 = 237,568 - 0,035 = 237,533 m

9. Perhitungan X

Xp = 1500,000 m

Xq = Xp + dpq sin pq = 1500,000 + (- 677,019) = 822,981 m Xr = Xq + dqr sin qr = 822,981 + (- 667,713) = 155,268 m

Xs = Xr + drs sin rs = 155,268 + 125,220 = 280,488 m

Xt = Xs + dst sin st = 280,488 + 981,979 = 1262,467 m

Xp = Xt + dtp sin tp = 1262,467 + 237,533 = 1500,000 m

10. Menghitung (d cos ) :

dpq cos pq = 728,142 m x cos 248 23 42 = - 268,10603 m dqr cos qr = 696,992 m x cos 286 40 30,4 = 199,99797 m drs cos rs = 756,509 m x cos 9 31 48,8 = 746,06797 m

dst cos st = 984,109 m x cos 86 16 8,2 = 64,03911 m dtp cos tp = 778,819 m x cos 162 14 22,6 = - 741,70088 m + d cos = fy = 0,298 m11. Menghitung koreksi ordinat :

fy1 = (dpq / d) x fy = (728,142 / 3944,571) x 0,298 m = 0,05501 m

fy2 = (dqr / d) x fy = (696,992 / 3944,571) x 0,298 m = 0,05265 m

fy3 = (drs / d) x fy = (756,509 / 3944,571) x 0,298 m = 0,05715 m

fy4 = (dst / d) x fy = (984,109 / 3944,571) x 0,298 m = 0,07435 m

fy5 = (dtp / d) x fy = (778,819 / 3944,571) x 0,298 m = 0,05884 m

= 0,298 m

12. Menghitung Ordinat :

Karena fy > 0 , maka koreksi ordinat negatip

dpq cos pq = dpq cos pq - fy1 = - 268,10603 0,05501 = - 268,16104 m

dqr cos qr = dqr cos qr - fy2 = 199,99797 0,05265 = 199,94532 m

drs cos rs = drs cos rs - fy3 = 746,06797 - 0,05715 = 746,01082 m

dst cos st = dst cos st - fy4 = 64,03911 0,07435 = 63,96474 m

dtp cos tp = dtp cos tp - fy5 = - 741,70088 0,05884 = - 741,75972 m

13. Perhitungan Y :

Yp = 1200,000 m

Yq = Yp + dpq cos pq = 1200,000 + (- 268,16104 ) = 931,839 m

Yr = Yq + dqr cos qr = 931,83896 + 199,94532 = 1131,784 m

Ys = Yr + drs cos rs = 1131,784 + 746,01082 = 1877,795 m

Yt = Ys + dst cos st = 1877,795 + 63,96474 = 1941,760 m

Yp = Yt + dtp cos tp = 1941,760 + (- 741,75972) = 1200,000 m