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      Distance from San Angelo to Sint-Pieters-Leeuw

      The shortest distance (air line) between San Angelo and Sint-Pieters-Leeuw is 5,147.36 mi (8,283.87 km).

      How far is San Angelo from Sint-Pieters-Leeuw

      San Angelo is located in Texas, United States within 31° 26' 32.64" N -101° 32' 58.56" W (31.4424, -100.4504) coordinates. The local time in San Angelo is 19:55 (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 02:55 (25.02.2025)

      The calculated flying distance from Sint-Pieters-Leeuw to Sint-Pieters-Leeuw is 5,147.36 miles which is equal to 8,283.87 km.

      San Angelo, Texas, United States

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      San Angelo to Minneapolis (Minnesota)1,011.31 mi (1,627.55 km)
      San Angelo to Los Angeles (California)1,057.88 mi (1,702.49 km)
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      San Angelo to Washington (District of Columbia)1,415.32 mi (2,277.74 km)
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      San Angelo to Portland (Oregon)1,536.74 mi (2,473.14 km)
      San Angelo to Brooklyn (New York)1,603.30 mi (2,580.26 km)
      San Angelo to Seattle (Washington)1,604.36 mi (2,581.97 km)
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      Sint-Pieters-Leeuw, Arr. Halle-Vilvoorde, Belgium

      Related Distances to Sint-Pieters-Leeuw