Distance from San Severino Marche to Aleksandrow Kujawski
The shortest distance (air line) between San Severino Marche and Aleksandrow Kujawski is 713.10 mi (1,147.63 km).
How far is San Severino Marche from Aleksandrow Kujawski
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 13:37 (07.02.2025)
Aleksandrow Kujawski is located in Włocławski, Poland within 52° 52' 36.12" N 18° 41' 36.96" E (52.8767, 18.6936) coordinates. The local time in Aleksandrow Kujawski is 13:37 (07.02.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 713.10 miles which is equal to 1,147.63 km.
San Severino Marche, Macerata, Italy
Related Distances from San Severino Marche
Aleksandrow Kujawski, Włocławski, Poland