Distance from Janow Lubelski to San Angelo
The shortest distance (air line) between Janow Lubelski and San Angelo is 5,779.63 mi (9,301.41 km).
How far is Janow Lubelski from San Angelo
Janow Lubelski is located in Puławski, Poland within 50° 43' 0.12" N 22° 25' 0.12" E (50.7167, 22.4167) coordinates. The local time in Janow Lubelski is 09:21 (21.12.2025)
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 02:21 (21.12.2025)
The calculated flying distance from San Angelo to San Angelo is 5,779.63 miles which is equal to 9,301.41 km.
Janow Lubelski, Puławski, Poland
Related Distances from Janow Lubelski
San Angelo, Texas, United States