Distance from Lukovica to Jefferson Valley-Yorktown
The shortest distance (air line) between Lukovica and Jefferson Valley-Yorktown is 4,191.21 mi (6,745.10 km).
How far is Lukovica from Jefferson Valley-Yorktown
Lukovica is located in Osrednjeslovenska, Slovenia within 46° 10' 6.96" N 14° 41' 21.12" E (46.1686, 14.6892) coordinates. The local time in Lukovica is 01:38 (17.12.2025)
Jefferson Valley-Yorktown is located in New York, United States within 41° 19' 4.8" N -74° 11' 57.12" W (41.3180, -73.8008) coordinates. The local time in Jefferson Valley-Yorktown is 19:38 (16.12.2025)
The calculated flying distance from Jefferson Valley-Yorktown to Jefferson Valley-Yorktown is 4,191.21 miles which is equal to 6,745.10 km.
Lukovica, Osrednjeslovenska, Slovenia
Related Distances from Lukovica
Jefferson Valley-Yorktown, New York, United States