Distance from Illinois to San Angelo
Distance from Illinois to San Angelo is 1,421 kilometers. This air travel distance is equal to 883 miles.
The air travel (bird fly) shortest distance between Illinois and San Angelo is 1,421 km= 883 miles.
If you travel with an airplane (which has average speed of 560 miles) from Illinois to San Angelo, It takes 1.58 hours to arrive.
Illinois
Illinois is located in United States.
GPS Coordinates (DMS) | 40° 37´ 59.2320'' N 89° 23´ 54.7080'' W |
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Latitude | 40.63312 |
Longitude | -89.39853 |
Altitude | 252 m |
Country | United States |
Illinois Distances to Cities
Illinois | Distance |
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Distance from Illinois to California | 2,633 km |
Distance from Illinois to New York | 1,300 km |
Distance from Illinois to Michigan | 514 km |
Distance from Illinois to Ohio | 550 km |
Distance from Illinois to Wisconsin | 354 km |
San Angelo
San Angelo is located in United States.
GPS Coordinates | 31° 27´ 49.5720'' N 100° 26´ 13.3440'' W |
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Latitude | 31.46377 |
Longitude | -100.43704 |
Altitude | 564 m |
Country | United States |
San Angelo Distances to Cities
San Angelo | Distance |
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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 |