1. CONTOH KASUS POLIGON TERTUTUP
Untuk menentukan kerangka suatu proyek bangunan dilakukan dengan cara polygon tertutup pada 5
titik P, Q , R , S dan T ,
βs dst βt
. .
S T
drs dtp
R P βp
βr
dqr Q dpq
.
Βq
Diketahui : P (1500,000 m , 1200,000 m) sebagai titik awal dan titik akhir
α pq = azimuth PQ = azimuth awal dan azimuth akhir = 248° 23′ 42″
Diukur : Sudut-sudut hasil ukuran : βp = 266° 09′ 21″
βq = 218° 16′ 50″
βr = 262° 51′ 20″
βs = 256° 44′ 21″
βt = 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″ √ n
Jawab :
1. Menghitung kesalahan sudut
∑ β = βp + βq + βr + βs + βt
= 266° 09′ 21″ + 218° 16′ 50″ + 262° 51′ 20″ + 256° 44′ 21″ + 255° 58′ 16″ = 1260° 00′ 08″
∑ β = (n + 2 ) x 180° = (5 + 2) x 180° = 7 x 180° = 1260°
Kesalahan sudut ( f β ) = 08″

2. f β = 8″
10″√ n = 10″ √ 5 = 10″x 2,24 = 22,4″
fβ < 10″ √ n → hasil pengukuran sudut dapat diterima / memenuhi syarat geometris / sesuai
spesifikasi teknis
2. Menghitung koreksi sudut
Kesalahan f β = 8″ hrs dikoreksikan secara merata ke semua sudut hasil ukuran , jadi koreksi tiap
sudut = f β/n = 8″/ 5 = 1,6″ :
βp = βp ± 1,6″
βq = βq ± 1,6″
βr = βr ± 1,6″ krn (βp + βq + βr + βs + βt) > {(n + 2) x 180° } , maka koreksinya ( - )
βs = βs ± 1,6″
βt = βt ± 1,6″
3. Menghitung sudut terkoreksi
βp = 266° 09′ 21″ - 1,6″ = 266° 09′ 19,4″
βq = 218° 16′ 50″ - 1,6″ = 218° 16′ 48,4″
βr = 262° 51′ 20″ - 1,6″ = 262° 51′ 18,4″
βs = 256° 44′ 21″ - 1,6″ = 256° 44′ 19,4″
βt = 255° 58′ 16″ - 1,6″ = 255° 58′ 14,4″
4. Menghitung Azimuth tiap sisi poligon (α)
α pq = 248° 23′ 42″
α qr = α pq + βq - 180° = 248° 23′ 42″ + 218° 16′ 48,4″ - 180° = 286° 40′ 30,4″
α rs = α qr + βr - 180° = 286° 40′ 30,4″ + 262° 51′ 18,4″ - 180° = 369° 31′ 48,8″ - 360°
α st = α rs + βs - 180° = 9° 31′ 48,8″ + 256° 44′ 19,4″ - 180° = 86° 16′ 8,2″
α tp = α st + βt - 180° = 86° 16′ 8,2″ + 255° 58′ 14,4″ - 180° = 162° 14′ 22,6″
α pq = α tp + βp - 180° = 162° 14′ 22,6″ + 266° 09′ 19,4″ - 180° = 248° 23′ 42″
5. Menghitung jumlah jarak ( ∑ d )

3. ∑ 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 α ) :
d pq sin α pq = 728,142 m x sin 248° 23′ 42″ = - 676,986 m
d qr sin α qr = 696,992 m x sin 286° 40′ 30,4″ = - 667,681557 m
d rs sin α rs = 756,509 m x sin 9° 31′ 48,8″ = 125,253551 m
d st sin α st = 984,109 m x sin 86° 16′ 8,2″ = 982,023175 m
d tp sin α tp = 778,819 m x sin 162° 14′ 22,6″ = 237,568598 m
+
∑ d sin α = fx = 0,177 m
7. 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 m
8. Menghitung absis :
Karena fx > 0 , maka koreksi absis negatip
dpq sin α pq = dpq sin α pq - fx1 = - 676,986 – 0,033 = - 677,019 m
dqr sin α qr = dqr sin α qr - fx2 = - 667,682 - 0,031 = - 667,713 m
drs sin α rs = drs sin α rs - fx3 = 125,254 - 0,034 = 125,220 m
dst sin α st = dst sin α st - fx4 = 982,023 - 0,044 = 981,979 m
dtp sin α tp = dtp sin α tp - fx5 = 237,568 - 0,035 = 237,533 m
9. Perhitungan X
Xp = 1500,000 m
Xq = Xp + dpq sin α pq = 1500,000 + (- 677,019) = 822,981 m
Xr = Xq + dqr sin α qr = 822,981 + (- 667,713) = 155,268 m

4. Xs = Xr + drs sin α rs = 155,268 + 125,220 = 280,488 m
Xt = Xs + dst sin α st = 280,488 + 981,979 = 1262,467 m
Xp = Xt + dtp sin α tp = 1262,467 + 237,533 = 1500,000 m
10. Menghitung (d cos α ) :
dpq cos α pq = 728,142 m x cos 248° 23′ 42″ = - 268,10603 m
dqr cos α qr = 696,992 m x cos 286° 40′ 30,4″ = 199,99797 m
drs cos α rs = 756,509 m x cos 9° 31′ 48,8″ = 746,06797 m
dst cos α st = 984,109 m x cos 86° 16′ 8,2″ = 64,03911 m
dtp cos α tp = 778,819 m x cos 162° 14′ 22,6″ = - 741,70088 m
+
∑ d cos α = fy = 0,298 m
11. 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 α pq = dpq cos α pq - fy1 = - 268,10603 – 0,05501 = - 268,16104 m
dqr cos α qr = dqr cos α qr - fy2 = 199,99797 – 0,05265 = 199,94532 m
drs cos α rs = drs cos α rs - fy3 = 746,06797 - 0,05715 = 746,01082 m
dst cos α st = dst cos α st - fy4 = 64,03911 – 0,07435 = 63,96474 m
dtp cos α tp = dtp cos α tp - fy5 = - 741,70088 – 0,05884 = - 741,75972 m
13. Perhitungan Y :
Yp = 1200,000 m
Yq = Yp + dpq cos α pq = 1200,000 + (- 268,16104 ) = 931,839 m

5. Yr = Yq + dqr cos α qr = 931,83896 + 199,94532 = 1131,784 m
Ys = Yr + drs cos α rs = 1131,784 + 746,01082 = 1877,795 m
Yt = Ys + dst cos α st = 1877,795 + 63,96474 = 1941,760 m
Yp = Yt + dtp cos α tp = 1941,760 + (- 741,75972) = 1200,000 m

6. Yr = Yq + dqr cos α qr = 931,83896 + 199,94532 = 1131,784 m
Ys = Yr + drs cos α rs = 1131,784 + 746,01082 = 1877,795 m
Yt = Ys + dst cos α st = 1877,795 + 63,96474 = 1941,760 m
Yp = Yt + dtp cos α tp = 1941,760 + (- 741,75972) = 1200,000 m