Distance from Cedar Park to San Angelo
Distance from Cedar Park to San Angelo is 272 kilometers. This air travel distance is equal to 169 miles.
The air travel (bird fly) shortest distance between Cedar Park and San Angelo is 272 km or 169 miles.
If you travel with an airplane (which has average speed of 560 miles) from Cedar Park to San Angelo, It takes 0.3 hours to arrive.
Cedar Park
Cedar Park is located in United States.
| GPS Coordinates (DMS) | 30° 30´ 18.7200'' N 97° 49´ 13.0440'' W |
|---|---|
| Latitude | 30.50520 |
| Longitude | -97.82029 |
| Altitude | 278 m |
| Country | United States |
Cedar Park 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 |