Distance from Jefferson City to Sandy Hills
Distance from Jefferson City to Sandy Hills is 1,702 kilometers. This air travel distance is equal to 1,058 miles.
The air travel (bird fly) shortest distance between Jefferson City and Sandy Hills is 1,702 km= 1,058 miles.
If you travel with an airplane (which has average speed of 560 miles) from Jefferson City to Sandy Hills, It takes 1.89 hours to arrive.
Jefferson City
Jefferson City is located in United States.
| GPS Coordinates (DMS) | 38° 34´ 36.1200'' N 92° 10´ 24.6720'' W |
|---|---|
| Latitude | 38.57670 |
| Longitude | -92.17352 |
| Altitude | 195 m |
| Country | United States |
Jefferson City Distances to Cities
| Jefferson City | Distance |
|---|---|
| Distance from Jefferson City to Sacramento | 2,544 km |
| Distance from Jefferson City to Boston | 1,834 km |
| Distance from Jefferson City to San Francisco | 2,640 km |
| Distance from Jefferson City to Los Angeles | 2,386 km |
| Distance from Jefferson City to Dallas | 767 km |
Sandy Hills
Sandy Hills is located in United States.
| GPS Coordinates | 40° 34´ 51.8160'' N 111° 51´ 2.7720'' W |
|---|---|
| Latitude | 40.58106 |
| Longitude | -111.85077 |
| Altitude | 1459 m |
| Country | United States |