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      Distance from Lake Oswego to San Agustin de Guadalix

      The shortest distance (air line) between Lake Oswego and San Agustin de Guadalix is 5,399.39 mi (8,689.49 km).

      How far is Lake Oswego from San Agustin de Guadalix

      Lake Oswego is located in Oregon, United States within 45° 24' 46.44" N -123° 17' 58.56" W (45.4129, -122.7004) coordinates. The local time in Lake Oswego is 02:39 (31.08.2025)

      San Agustin de Guadalix is located in Madrid, Spain within 40° 40' 41.16" N -4° 23' 6" W (40.6781, -3.6150) coordinates. The local time in San Agustin de Guadalix is 11:39 (31.08.2025)

      The calculated flying distance from San Agustin de Guadalix to San Agustin de Guadalix is 5,399.39 miles which is equal to 8,689.49 km.

      Lake Oswego, Oregon, United States

      Related Distances from Lake Oswego

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      Lake Oswego to Baltimore (Maryland)2,357.37 mi (3,793.82 km)
      Lake Oswego to Philadelphia (Pennsylvania)2,407.85 mi (3,875.05 km)
      Lake Oswego to New York (New York)2,445.96 mi (3,936.39 km)
      Lake Oswego to Brooklyn (New York)2,446.04 mi (3,936.52 km)
      Lake Oswego to Queens (New York)2,450.46 mi (3,943.64 km)
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      San Agustin de Guadalix, Madrid, Spain

      Related Distances to San Agustin de Guadalix