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      Distance from Seymour to Sint-Pieters-Leeuw

      The shortest distance (air line) between Seymour and Sint-Pieters-Leeuw is 4,210.16 mi (6,775.60 km).

      How far is Seymour from Sint-Pieters-Leeuw

      Seymour is located in Indiana, United States within 38° 56' 51.36" N -86° 6' 32.04" W (38.9476, -85.8911) coordinates. The local time in Seymour is 23:32 (26.04.2025)

      Sint-Pieters-Leeuw is located in Arr. Halle-Vilvoorde, Belgium within 50° 46' 59.88" N 4° 15' 0" E (50.7833, 4.2500) coordinates. The local time in Sint-Pieters-Leeuw is 05:32 (27.04.2025)

      The calculated flying distance from Sint-Pieters-Leeuw to Sint-Pieters-Leeuw is 4,210.16 miles which is equal to 6,775.60 km.

      Seymour, Indiana, United States

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      Sint-Pieters-Leeuw, Arr. Halle-Vilvoorde, Belgium

      Related Distances to Sint-Pieters-Leeuw