Distance from Sint-Oedenrode to Lexington
The shortest distance (air line) between Sint-Oedenrode and Lexington is 4,221.11 mi (6,793.23 km).
How far is Sint-Oedenrode from Lexington
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 is located in Kentucky, United States within 38° 2' 32.28" N -85° 32' 28.68" W (38.0423, -84.4587) coordinates. The local time in Lexington is 05:12 (13.12.2025)
The calculated flying distance from Lexington to Lexington is 4,221.11 miles which is equal to 6,793.23 km.
Sint-Oedenrode, Noordoost-Noord-Brabant, Netherlands
Related Distances from Sint-Oedenrode
Lexington, Kentucky, United States