Distance from Jefferson Valley-Yorktown to Zonnebeke
The shortest distance (air line) between Jefferson Valley-Yorktown and Zonnebeke is 3,568.48 mi (5,742.91 km).
How far is Jefferson Valley-Yorktown from Zonnebeke
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 17:32 (31.07.2025)
Zonnebeke is located in Arr. Ieper, Belgium within 50° 52' 0.12" N 2° 58' 59.88" E (50.8667, 2.9833) coordinates. The local time in Zonnebeke is 23:32 (31.07.2025)
The calculated flying distance from Zonnebeke to Zonnebeke is 3,568.48 miles which is equal to 5,742.91 km.
Jefferson Valley-Yorktown, New York, United States
Related Distances from Jefferson Valley-Yorktown
Zonnebeke, Arr. Ieper, Belgium