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      Distance from Reynoldsburg to Sint-Pieters-Leeuw

      The shortest distance (air line) between Reynoldsburg and Sint-Pieters-Leeuw is 4,041.63 mi (6,504.38 km).

      How far is Reynoldsburg from Sint-Pieters-Leeuw

      Reynoldsburg is located in Ohio, United States within 39° 57' 31.68" N -83° 12' 20.52" W (39.9588, -82.7943) coordinates. The local time in Reynoldsburg is 17:46 (24.02.2025)

      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 23:46 (24.02.2025)

      The calculated flying distance from Sint-Pieters-Leeuw to Sint-Pieters-Leeuw is 4,041.63 miles which is equal to 6,504.38 km.

      Reynoldsburg, Ohio, United States

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

      Related Distances to Sint-Pieters-Leeuw