Distance from Jefferson Valley-Yorktown to Begijnendijk
The shortest distance (air line) between Jefferson Valley-Yorktown and Begijnendijk is 3,637.49 mi (5,853.97 km).
How far is Jefferson Valley-Yorktown from Begijnendijk
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 17:12 (22.07.2025)
Begijnendijk is located in Arr. Leuven, Belgium within 51° 1' 6.96" N 4° 47' 6" E (51.0186, 4.7850) coordinates. The local time in Begijnendijk is 23:12 (22.07.2025)
The calculated flying distance from Begijnendijk to Begijnendijk is 3,637.49 miles which is equal to 5,853.97 km.
Jefferson Valley-Yorktown, New York, United States
Related Distances from Jefferson Valley-Yorktown
Begijnendijk, Arr. Leuven, Belgium