Distance from Brunswick to Sveti Ivan Zelina
The shortest distance (air line) between Brunswick and Sveti Ivan Zelina is 4,586.60 mi (7,381.42 km).
How far is Brunswick from Sveti Ivan Zelina
Brunswick is located in Ohio, United States within 41° 14' 47.4" N -82° 10' 48.72" W (41.2465, -81.8198) coordinates. The local time in Brunswick is 21:36 (27.11.2024)
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 03:36 (28.11.2024)
The calculated flying distance from Sveti Ivan Zelina to Sveti Ivan Zelina is 4,586.60 miles which is equal to 7,381.42 km.
Brunswick, Ohio, United States
Related Distances from Brunswick
Sveti Ivan Zelina, Zagrebačka županija, Croatia