Distance between Belgrade and Waterbury

The distance between Belgrade (Serbia) and Waterbury (Connecticut, USA) is ca. 7.140,7 km (air line).

Belgrade

Latidute: 44.787198
Longitude: 20.457274

Waterbury

Koordinaten: 41°33′22″N 73°2′29″W
Latidute: 41.55611111
Longitude: -73.04138889

Distance between Belgrade and other cities and places worldwide

Distance between Belgrade and other major cities