Distance from Harker Heights to San Angelo
Distance from Harker Heights to San Angelo is 268 kilometers. This air travel distance is equal to 167 miles.
The air travel (bird fly) shortest distance between Harker Heights and San Angelo is 268 km or 167 miles.
If you travel with an airplane (which has average speed of 560 miles) from Harker Heights to San Angelo, It takes 0.3 hours to arrive.
Harker Heights
Harker Heights is located in United States.
| GPS Coordinates (DMS) | 31° 5´ 0.6360'' N 97° 39´ 35.0640'' W |
|---|---|
| Latitude | 31.08351 |
| Longitude | -97.65974 |
| Altitude | 236 m |
| Country | United States |
Harker Heights 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 |