Distance from Watauga to San Angelo
Distance from Watauga to San Angelo is 338 kilometers. This air travel distance is equal to 210 miles.
The air travel (bird fly) shortest distance between Watauga and San Angelo is 338 km or 210 miles.
If you travel with an airplane (which has average speed of 560 miles) from Watauga to San Angelo, It takes 0.38 hours to arrive.
Watauga
Watauga is located in United States.
| GPS Coordinates (DMS) | 32° 51´ 28.4760'' N 97° 15´ 17.0640'' W |
|---|---|
| Latitude | 32.85791 |
| Longitude | -97.25474 |
| Altitude | 188 m |
| Country | United States |
Watauga Distances to Cities
| Watauga | Distance |
|---|---|
| Distance from Watauga to Plum | 1,775 km |
| Distance from Watauga to Weatherford | 52 km |
| Distance from Watauga to Parker | 1,000 km |
| Distance from Watauga to Hobbs | 551 km |
| Distance from Watauga to Amarillo | 497 km |
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 |