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SISTEM KOORDINAT

SISTEM KOORDINAT. Proyeksi. Koordinat X ( Bujur ). Koordinat Y ( Lintang ). X = 106 o 50’30” BT Y = 6 o 20’00 LS. X = 106 o 50’30” BT Y = 6 o 20’00 LS. 1 o = 60’ (1 Derajat = 60 Menit ) 1’ = 60” (1 Menit = 60 Detik ) 1 o = 60’x60’ = 360” (1 Derajat = 3600 Detik ) JADI

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SISTEM KOORDINAT

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  1. SISTEM KOORDINAT

  2. Proyeksi

  3. Koordinat X (Bujur) Koordinat Y (Lintang)

  4. X = 106o50’30” BTY = 6o20’00 LS

  5. X = 106o50’30” BTY = 6o20’00 LS • 1o = 60’ (1 Derajat = 60 Menit) • 1’= 60” (1 Menit = 60 Detik) • 1o = 60’x60’ = 360” (1 Derajat = 3600 Detik) JADI • 50’ = 50/60 = 0.833o • (50 Menit = 0.833 Derajat) • 30” = 30/3600 = 0.00833o • (30 Detik = 0.00833 Derajat) • 50’30” = 0.833 + 0.00833 = 84133 o X = 106,8413 o (Desimal Degree)

  6. SISTEM PROYEKSI UTM • Universal Transverse Mercator (UTM) • Untukmengetahui zone UTM • wilayah yang akanditransformasikan • digunakanrumus : • GarisBujur/6 + 30 = ZONE* * Hasilnilai zone selaludibulatkankeatas (zone 49,1 ~ zone 50)

  7. X = 106o50’30” BTY = 6o20’00 LS Zona Utara Lintang 0 Zona Selatan Bujur Barat (BT) Bujur Timur (BT) 106/6 +30 = 17,66 + 30 = 47,66 ~ 48 Y = 6o20’00 LS Maka Zona 48 South/Selatan

  8. Sistem Koordinat TM3 • Penggunaansistemkoordinat yang dianggapakuratyaitu Transverse Mercator 3 yang lebihdikenalsebagaiSistemKoordinat TM3. • Sistemkoordinatinimemodifikasisistemkoordinat yang sudahadasebelumnyayaitu UTM WGS 1984, • DengancaramembagiSistemProyeksi UTM 6oke 3o • Sehingga dalam satu zona UTM 48 selatan misalnya, terdiri dari 2 zona TM3, yaitu TM3 zona 48.1 dan TM3 zona 48.2.

  9. 48 North UTM 0O 48 South 102O 108O 114O zone 48 50 49 1 Zone = 6 Derajat 48.1 N 48.2 N TM3 0O 48.1 S 48.2 S 105O 108O 102O 106O30’ 103O30’

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