Distance from Waterbury to Sveti Ivan Zelina
The shortest distance (air line) between Waterbury and Sveti Ivan Zelina is 4,221.00 mi (6,793.05 km).
How far is Waterbury from Sveti Ivan Zelina
Waterbury is located in Connecticut, United States within 41° 33' 29.52" N -74° 57' 50.04" W (41.5582, -73.0361) coordinates. The local time in Waterbury is 05:48 (10.03.2025)
Sveti Ivan Zelina is located in Zagrebačka županija, Croatia within 45° 57' 34.56" N 16° 14' 35.16" E (45.9596, 16.2431) coordinates. The local time in Sveti Ivan Zelina is 11:48 (10.03.2025)
The calculated flying distance from Sveti Ivan Zelina to Sveti Ivan Zelina is 4,221.00 miles which is equal to 6,793.05 km.
Waterbury, Connecticut, United States
Related Distances from Waterbury
Sveti Ivan Zelina, Zagrebačka županija, Croatia