Distance from Jefferson City to Sandusky
Distance from Jefferson City to Sandusky is 868 kilometers. This air travel distance is equal to 539 miles.
The air travel (bird fly) shortest distance between Jefferson City and Sandusky is 868 km or 539 miles.
If you travel with an airplane (which has average speed of 560 miles) from Jefferson City to Sandusky, It takes 0.96 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 |
Sandusky
Sandusky is located in United States.
| GPS Coordinates | 41° 26´ 56.1840'' N 82° 42´ 28.6560'' W |
|---|---|
| Latitude | 41.44894 |
| Longitude | -82.70796 |
| Altitude | 183 m |
| Country | United States |
Sandusky Distances to Cities
| Sandusky | Distance |
|---|---|
| Distance from Sandusky to South Euclid | 100 km |
| Distance from Sandusky to Waukesha | 488 km |
| Distance from Sandusky to Pittsburgh | 254 km |
| Distance from Sandusky to Tucson | 2,702 km |
| Distance from Sandusky to Janesville | 540 km |