Distance between Ouagadougou and Waterbury

The distance between Ouagadougou (Burkina Faso) and Waterbury (Connecticut, USA) is ca. 7.565,7 km (air line).

Ouagadougou

Latidute: 12.36571
Longitude: -1.533881

Waterbury

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

Distance between Ouagadougou and other cities and places worldwide

Distance between Ouagadougou and other major cities