Distance from San Severino Marche to Aleksandrow Lodzki
The shortest distance (air line) between San Severino Marche and Aleksandrow Lodzki is 658.23 mi (1,059.32 km).
How far is San Severino Marche from Aleksandrow Lodzki
San Severino Marche is located in Macerata, Italy within 43° 13' 44.04" N 13° 10' 37.56" E (43.2289, 13.1771) coordinates. The local time in San Severino Marche is 13:48 (07.02.2025)
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 13:48 (07.02.2025)
The calculated flying distance from Aleksandrow Lodzki to Aleksandrow Lodzki is 658.23 miles which is equal to 1,059.32 km.
San Severino Marche, Macerata, Italy
Related Distances from San Severino Marche
Aleksandrow Lodzki, Łódzki, Poland