Distance from Aleksandrow Kujawski to Deer Park
The shortest distance (air line) between Aleksandrow Kujawski and Deer Park is 5,488.74 mi (8,833.27 km).
How far is Aleksandrow Kujawski from Deer Park
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 03:16 (10.12.2025)
Deer Park is located in Texas, United States within 29° 41' 23.28" N -96° 53' 5.64" W (29.6898, -95.1151) coordinates. The local time in Deer Park is 20:16 (09.12.2025)
The calculated flying distance from Deer Park to Deer Park is 5,488.74 miles which is equal to 8,833.27 km.
Aleksandrow Kujawski, Włocławski, Poland
Related Distances from Aleksandrow Kujawski
Deer Park, Texas, United States