Distance from Waukegan to Saint-Genis-Pouilly
The shortest distance (air line) between Waukegan and Saint-Genis-Pouilly is 4,361.43 mi (7,019.04 km).
How far is Waukegan from Saint-Genis-Pouilly
Waukegan is located in Illinois, United States within 42° 22' 11.28" N -88° 7' 42.24" W (42.3698, -87.8716) coordinates. The local time in Waukegan is 05:46 (04.03.2025)
Saint-Genis-Pouilly is located in Ain, France within 46° 14' 35.88" N 6° 1' 17.04" E (46.2433, 6.0214) coordinates. The local time in Saint-Genis-Pouilly is 12:46 (04.03.2025)
The calculated flying distance from Saint-Genis-Pouilly to Saint-Genis-Pouilly is 4,361.43 miles which is equal to 7,019.04 km.
Waukegan, Illinois, United States
Related Distances from Waukegan
Saint-Genis-Pouilly, Ain, France