Distance from Aleksandrow Kujawski to Lacey
The shortest distance (air line) between Aleksandrow Kujawski and Lacey is 5,169.84 mi (8,320.05 km).
How far is Aleksandrow Kujawski from Lacey
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 00:12 (12.12.2025)
Lacey is located in Washington, United States within 47° 2' 46.32" N -123° 12' 24.12" W (47.0462, -122.7933) coordinates. The local time in Lacey is 15:12 (11.12.2025)
The calculated flying distance from Lacey to Lacey is 5,169.84 miles which is equal to 8,320.05 km.
Aleksandrow Kujawski, Włocławski, Poland
Related Distances from Aleksandrow Kujawski
Lacey, Washington, United States