Distance from Austintown to Sint-Pieters-Leeuw
The shortest distance (air line) between Austintown and Sint-Pieters-Leeuw is 3,908.95 mi (6,290.84 km).
How far is Austintown from Sint-Pieters-Leeuw
Austintown is located in Ohio, United States within 41° 5' 35.52" N -81° 15' 34.2" W (41.0932, -80.7405) coordinates. The local time in Austintown is 06:39 (23.11.2024)
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 12:39 (23.11.2024)
The calculated flying distance from Sint-Pieters-Leeuw to Sint-Pieters-Leeuw is 3,908.95 miles which is equal to 6,290.84 km.
Austintown, Ohio, United States
Related Distances from Austintown
Sint-Pieters-Leeuw, Arr. Halle-Vilvoorde, Belgium