Distance from Sint-Oedenrode to Lexington Park
The shortest distance (air line) between Sint-Oedenrode and Lexington Park is 3,895.29 mi (6,268.86 km).
How far is Sint-Oedenrode from Lexington Park
Sint-Oedenrode is located in Noordoost-Noord-Brabant, Netherlands within 51° 33' 48.96" N 5° 27' 38.88" E (51.5636, 5.4608) coordinates. The local time in Sint-Oedenrode is 11:12 (13.12.2025)
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 05:12 (13.12.2025)
The calculated flying distance from Lexington Park to Lexington Park is 3,895.29 miles which is equal to 6,268.86 km.
Sint-Oedenrode, Noordoost-Noord-Brabant, Netherlands
Related Distances from Sint-Oedenrode
Lexington Park, Maryland, United States