Distance from Lexington Park to Sint-Andries
The shortest distance (air line) between Lexington Park and Sint-Andries is 3,811.98 mi (6,134.78 km).
How far is Lexington Park from Sint-Andries
Lexington Park is located in Maryland, United States within 38° 15' 10.08" N -77° 33' 27.36" W (38.2528, -76.4424) coordinates. The local time in Lexington Park is 20:27 (23.01.2025)
Sint-Andries is located in Arr. Brugge, Belgium within 51° 12' 0" N 3° 10' 59.88" E (51.2000, 3.1833) coordinates. The local time in Sint-Andries is 02:27 (24.01.2025)
The calculated flying distance from Sint-Andries to Sint-Andries is 3,811.98 miles which is equal to 6,134.78 km.
Lexington Park, Maryland, United States
Related Distances from Lexington Park
Sint-Andries, Arr. Brugge, Belgium