Distance from Ponce to Aleksandrow Kujawski
The shortest distance (air line) between Ponce and Aleksandrow Kujawski is 5,039.39 mi (8,110.11 km).
How far is Ponce from Aleksandrow Kujawski
Ponce is located in Puerto Rico, United States within 18° 0' 45.72" N -67° 22' 43.68" W (18.0127, -66.6212) coordinates. The local time in Ponce is 11:53 (12.11.2025)
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 16:53 (12.11.2025)
The calculated flying distance from Aleksandrow Kujawski to Aleksandrow Kujawski is 5,039.39 miles which is equal to 8,110.11 km.
Ponce, Puerto Rico, United States
Related Distances from Ponce
Aleksandrow Kujawski, Włocławski, Poland