Contoh kasus poligon tertutup

Contoh kasus poligon tertutup

• 1.
  CONTOH KASUS POLIGONTERTUTUP 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