Distance from Aleksandrow Lodzki to Sapulpa
The shortest distance (air line) between Aleksandrow Lodzki and Sapulpa is 5,227.80 mi (8,413.34 km).
How far is Aleksandrow Lodzki from Sapulpa
Aleksandrow Lodzki is located in Łódzki, Poland within 51° 49' 9.84" N 19° 18' 14.04" E (51.8194, 19.3039) coordinates. The local time in Aleksandrow Lodzki is 11:18 (16.12.2025)
Sapulpa is located in Oklahoma, United States within 36° 0' 32.76" N -97° 53' 58.92" W (36.0091, -96.1003) coordinates. The local time in Sapulpa is 04:18 (16.12.2025)
The calculated flying distance from Sapulpa to Sapulpa is 5,227.80 miles which is equal to 8,413.34 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Sapulpa, Oklahoma, United States