Distance from Siemiatycze to Alexandroupoli
The shortest distance (air line) between Siemiatycze and Alexandroupoli is 812.32 mi (1,307.30 km).
How far is Siemiatycze from Alexandroupoli
Siemiatycze is located in Łomżyński, Poland within 52° 25' 37.92" N 22° 51' 45" E (52.4272, 22.8625) coordinates. The local time in Siemiatycze is 23:00 (23.02.2025)
Alexandroupoli is located in Evros, Greece within 40° 51' 0" N 25° 52' 0.12" E (40.8500, 25.8667) coordinates. The local time in Alexandroupoli is 00:00 (24.02.2025)
The calculated flying distance from Alexandroupoli to Alexandroupoli is 812.32 miles which is equal to 1,307.30 km.
Siemiatycze, Łomżyński, Poland