Distance from Aleksandrow Lodzki to San Angelo
The shortest distance (air line) between Aleksandrow Lodzki and San Angelo is 5,629.07 mi (9,059.12 km).
How far is Aleksandrow Lodzki from San Angelo
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 03:40 (15.12.2025)
San Angelo is located in Texas, United States within 31° 26' 32.64" N -101° 32' 58.56" W (31.4424, -100.4504) coordinates. The local time in San Angelo is 20:40 (14.12.2025)
The calculated flying distance from San Angelo to San Angelo is 5,629.07 miles which is equal to 9,059.12 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
San Angelo, Texas, United States