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      Distance from Altus to Saint-Andre-les-Vergers

      The shortest distance (air line) between Altus and Saint-Andre-les-Vergers is 5,020.68 mi (8,080.01 km).

      How far is Altus from Saint-Andre-les-Vergers

      Altus is located in Oklahoma, United States within 34° 39' 23.4" N -100° 41' 38.04" W (34.6565, -99.3061) coordinates. The local time in Altus is 04:07 (27.01.2025)

      Saint-Andre-les-Vergers is located in Aube, France within 48° 16' 46.92" N 4° 3' 14.04" E (48.2797, 4.0539) coordinates. The local time in Saint-Andre-les-Vergers is 11:07 (27.01.2025)

      The calculated flying distance from Saint-Andre-les-Vergers to Saint-Andre-les-Vergers is 5,020.68 miles which is equal to 8,080.01 km.

      Altus, Oklahoma, United States

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      Altus to Tampa (Florida)1,094.51 mi (1,761.45 km)
      Altus to Washington (District of Columbia)1,264.47 mi (2,034.96 km)
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      Altus to Brooklyn (New York)1,441.78 mi (2,320.32 km)
      Altus to New York (New York)1,443.51 mi (2,323.11 km)
      Altus to Queens (New York)1,450.68 mi (2,334.64 km)
      Altus to Seattle (Washington)1,486.08 mi (2,391.62 km)
      Altus to Boston (Massachusetts)1,606.61 mi (2,585.58 km)
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      Saint-Andre-les-Vergers, Aube, France

      Related Distances to Saint-Andre-les-Vergers