Distance from Sokolow Podlaski to San Angelo
The shortest distance (air line) between Sokolow Podlaski and San Angelo is 5,694.06 mi (9,163.70 km).
How far is Sokolow Podlaski from San Angelo
Sokolow Podlaski is located in Siedlecki, Poland within 52° 23' 60" N 22° 15' 0" E (52.4000, 22.2500) coordinates. The local time in Sokolow Podlaski is 03:55 (24.08.2025)
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 20:55 (23.08.2025)
The calculated flying distance from San Angelo to San Angelo is 5,694.06 miles which is equal to 9,163.70 km.
Sokolow Podlaski, Siedlecki, Poland
Related Distances from Sokolow Podlaski
San Angelo, Texas, United States