Distance from Lake Oswego to Sveti Ivan Zelina
The shortest distance (air line) between Lake Oswego and Sveti Ivan Zelina is 5,646.47 mi (9,087.11 km).
How far is Lake Oswego from Sveti Ivan Zelina
Lake Oswego is located in Oregon, United States within 45° 24' 46.44" N -123° 17' 58.56" W (45.4129, -122.7004) coordinates. The local time in Lake Oswego is 22:35 (11.06.2025)
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 07:35 (12.06.2025)
The calculated flying distance from Sveti Ivan Zelina to Sveti Ivan Zelina is 5,646.47 miles which is equal to 9,087.11 km.
Lake Oswego, Oregon, United States
Related Distances from Lake Oswego
Sveti Ivan Zelina, Zagrebačka županija, Croatia