Distance from Slantsy to Jacksonville Beach
Distance from Slantsy to Jacksonville Beach is 8,183 kilometers. This air travel distance is equal to 5,085 miles.
The air travel (bird fly) shortest distance between Slantsy and Jacksonville Beach is 8,183 km or 5,085 miles.
If you travel with an airplane (which has average speed of 560 miles) from Slantsy to Jacksonville Beach, It takes 9.08 hours to arrive.
Slantsy
Slantsy is located in Russia.
| GPS Coordinates (DMS) | 59° 7´ 5.4120'' N 28° 5´ 28.9320'' E |
|---|---|
| Latitude | 59.11817 |
| Longitude | 28.09137 |
| Altitude | 41 m |
| Country | Russia |
Slantsy Distances to Cities
| Slantsy | Distance |
|---|---|
| Distance from Slantsy to League City | 8,885 km |
| Distance from Slantsy to Jacksonville Beach | 8,183 km |
| Distance from Slantsy to Tysons Corner | 7,169 km |
| Distance from Slantsy to Cahokia | 7,785 km |
| Distance from Slantsy to Lompoc | 9,215 km |
Jacksonville Beach
Jacksonville Beach is located in United States.
| GPS Coordinates | 30° 17´ 40.8840'' N 81° 23´ 35.3040'' W |
|---|---|
| Latitude | 30.29469 |
| Longitude | -81.39314 |
| Altitude | 3 m |
| Country | United States |