Distance from Faribault to San Angelo
The shortest distance (air line) between Faribault and San Angelo is 969.67 mi (1,560.54 km).
How far is Faribault from San Angelo
Faribault is located in Minnesota, United States within 44° 17' 58.56" N -94° 43' 15.96" W (44.2996, -93.2789) coordinates. The local time in Faribault is 08:15 (06.10.2025)
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 08:15 (06.10.2025)
The calculated flying distance from San Angelo to San Angelo is 969.67 miles which is equal to 1,560.54 km.
Faribault, Minnesota, United States
Related Distances from Faribault
San Angelo, Texas, United States