Distance from Waukegan to San Lazzaro di Savena
The shortest distance (air line) between Waukegan and San Lazzaro di Savena is 4,644.74 mi (7,474.98 km).
How far is Waukegan from San Lazzaro di Savena
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 07:17 (11.11.2025)
San Lazzaro di Savena is located in Bologna, Italy within 44° 28' 17.76" N 11° 24' 17.64" E (44.4716, 11.4049) coordinates. The local time in San Lazzaro di Savena is 14:17 (11.11.2025)
The calculated flying distance from San Lazzaro di Savena to San Lazzaro di Savena is 4,644.74 miles which is equal to 7,474.98 km.
Waukegan, Illinois, United States
Related Distances from Waukegan
San Lazzaro di Savena, Bologna, Italy