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      Distance from Aleksandrow Kujawski to Saint-Andre-les-Vergers

      The shortest distance (air line) between Aleksandrow Kujawski and Saint-Andre-les-Vergers is 714.79 mi (1,150.35 km).

      How far is Aleksandrow Kujawski from Saint-Andre-les-Vergers

      Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 11:41 (08.05.2025)

      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 11:41 (08.05.2025)

      The calculated flying distance from Saint-Andre-les-Vergers to Saint-Andre-les-Vergers is 714.79 miles which is equal to 1,150.35 km.

      Aleksandrow Kujawski, Włocławski, Poland

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      Aleksandrow Kujawski to Gdynia112.35 mi (180.81 km)
      Aleksandrow Kujawski to Wroclaw141.04 mi (226.98 km)
      Aleksandrow Kujawski to Czestochowa144.62 mi (232.74 km)
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      Aleksandrow Kujawski to Zielona Gora149.55 mi (240.68 km)
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      Aleksandrow Kujawski to Szczecin175.97 mi (283.20 km)
      Aleksandrow Kujawski to Zabrze177.90 mi (286.30 km)
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      Aleksandrow Kujawski to Sosnowiec180.18 mi (289.97 km)
      Aleksandrow Kujawski to Katowice180.87 mi (291.08 km)
      Aleksandrow Kujawski to Bialystok185.92 mi (299.22 km)
      Aleksandrow Kujawski to Rybnik193.10 mi (310.77 km)
      Aleksandrow Kujawski to Lublin199.20 mi (320.59 km)
      Aleksandrow Kujawski to Krakow201.74 mi (324.67 km)
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      Saint-Andre-les-Vergers, Aube, France

      Related Distances to Saint-Andre-les-Vergers