Distance from Cheat Lake to Aleksandrow Kujawski
The shortest distance (air line) between Cheat Lake and Aleksandrow Kujawski is 4,415.28 mi (7,105.70 km).
How far is Cheat Lake from Aleksandrow Kujawski
Cheat Lake is located in West Virginia, United States within 39° 40' 1.2" N -80° 8' 36.96" W (39.6670, -79.8564) coordinates. The local time in Cheat Lake is 15:55 (19.05.2025)
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 21:55 (19.05.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 4,415.28 miles which is equal to 7,105.70 km.
Cheat Lake, West Virginia, United States
Related Distances from Cheat Lake
Aleksandrow Kujawski, Włocławski, Poland