Distance from Lincoln to San Angelo
Distance from Lincoln to San Angelo is 1,090 kilometers. This air travel distance is equal to 677 miles.
The air travel (bird fly) shortest distance between Lincoln and San Angelo is 1,090 km= 677 miles.
If you travel with an airplane (which has average speed of 560 miles) from Lincoln to San Angelo, It takes 1.21 hours to arrive.
Lincoln
Lincoln is located in United States.
| GPS Coordinates (DMS) | 40° 47´ 60.0000'' N 96° 40´ 1.0560'' W |
|---|---|
| Latitude | 40.80000 |
| Longitude | -96.66696 |
| Altitude | 366 m |
| Country | United States |
Lincoln Distances to Cities
| Lincoln | Distance |
|---|---|
| Distance from Lincoln Ne to Las Vegas | 1,687 km |
| Distance from Lincoln Ne to Los Angeles | 2,045 km |
| Distance from Lincoln Ne to San Francisco | 2,239 km |
| Distance from Lincoln Ne to Sacramento | 2,137 km |
| Distance from Lincoln Ne to Denver | 717 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 |