Distance from Burleson to San Angelo
Distance from Burleson to San Angelo is 318 kilometers. This air travel distance is equal to 198 miles.
The air travel (bird fly) shortest distance between Burleson and San Angelo is 318 km or 198 miles.
If you travel with an airplane (which has average speed of 560 miles) from Burleson to San Angelo, It takes 0.35 hours to arrive.
Burleson
Burleson is located in United States.
| GPS Coordinates (DMS) | 32° 32´ 31.4880'' N 97° 19´ 15.0600'' W |
|---|---|
| Latitude | 32.54208 |
| Longitude | -97.32085 |
| Altitude | 220 m |
| Country | United States |
Burleson Distances to Cities
San Angelo
San Angelo is located in United States.
| GPS Coordinates | 31° 27´ 49.5720'' N 100° 26´ 13.3440'' W |
|---|---|
| Latitude | 31.46377 |
| Longitude | -100.43704 |
| Altitude | 564 m |
| Country | United States |
San Angelo Distances to Cities
| San Angelo | Distance |
|---|---|
| Distance from San Angelo to Eagle | 1,948 km |
| Distance from San Angelo to West Odessa | 200 km |
| Distance from San Angelo to Moscow | 2,210 km |
| Distance from San Angelo to Twin Falls | 1,749 km |
| Distance from San Angelo to Albany Or | 2,455 km |