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      Distance from Oxon Hill to Sint-Pieters-Leeuw

      The shortest distance (air line) between Oxon Hill and Sint-Pieters-Leeuw is 3,862.76 mi (6,216.50 km).

      How far is Oxon Hill from Sint-Pieters-Leeuw

      Oxon Hill is located in Maryland, United States within 38° 47' 18.24" N -77° 1' 38.28" W (38.7884, -76.9727) coordinates. The local time in Oxon Hill is 02:45 (01.08.2025)

      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 08:45 (01.08.2025)

      The calculated flying distance from Sint-Pieters-Leeuw to Sint-Pieters-Leeuw is 3,862.76 miles which is equal to 6,216.50 km.

      Oxon Hill, Maryland, United States

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      Oxon Hill to Phoenix (Arizona)1,978.82 mi (3,184.60 km)
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      Oxon Hill to Brooklyn (New York)205.79 mi (331.19 km)
      Oxon Hill to New York (New York)208.69 mi (335.85 km)
      Oxon Hill to Queens (New York)216.30 mi (348.11 km)
      Oxon Hill to Boston (Massachusetts)393.57 mi (633.39 km)
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      Sint-Pieters-Leeuw, Arr. Halle-Vilvoorde, Belgium

      Related Distances to Sint-Pieters-Leeuw