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      Distance from Waterbury to Saint-Andre-les-Vergers

      The shortest distance (air line) between Waterbury and Saint-Andre-les-Vergers is 3,639.13 mi (5,856.62 km).

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

      Waterbury is located in Connecticut, United States within 41° 33' 29.52" N -74° 57' 50.04" W (41.5582, -73.0361) coordinates. The local time in Waterbury is 11:53 (17.03.2025)

      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 17:53 (17.03.2025)

      The calculated flying distance from Saint-Andre-les-Vergers to Saint-Andre-les-Vergers is 3,639.13 miles which is equal to 5,856.62 km.

      Waterbury, Connecticut, United States

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      Saint-Andre-les-Vergers, Aube, France

      Related Distances to Saint-Andre-les-Vergers