Distance from Aleksandrow Lodzki to Severn
The shortest distance (air line) between Aleksandrow Lodzki and Severn is 4,388.22 mi (7,062.16 km).
How far is Aleksandrow Lodzki from Severn
Aleksandrow Lodzki is located in Łódzki, Poland within 51° 49' 9.84" N 19° 18' 14.04" E (51.8194, 19.3039) coordinates. The local time in Aleksandrow Lodzki is 13:29 (16.12.2025)
Severn is located in Maryland, United States within 39° 8' 7.8" N -77° 18' 15.84" W (39.1355, -76.6956) coordinates. The local time in Severn is 07:29 (16.12.2025)
The calculated flying distance from Severn to Severn is 4,388.22 miles which is equal to 7,062.16 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Severn, Maryland, United States