Distance from Swakopmund to Jeffersonville
Distance from Swakopmund to Jeffersonville is 12,404 kilometers. This air travel distance is equal to 7,707 miles.
The air travel (bird fly) shortest distance between Swakopmund and Jeffersonville is 12,404 km or 7,707 miles.
If you travel with an airplane (which has average speed of 560 miles) from Swakopmund to Jeffersonville, It takes 13.76 hours to arrive.
Swakopmund
Swakopmund is located in Namibia.
| GPS Coordinates (DMS) | 22° 40´ 59.9880'' S 14° 31´ 59.9880'' E |
|---|---|
| Latitude | -22.68333 |
| Longitude | 14.53333 |
| Altitude | 14 m |
| Country | Namibia |
Swakopmund Distances to Cities
| Swakopmund | Distance |
|---|---|
| Distance from Swakopmund to Waipahu | 19,249 km |
| Distance from Swakopmund to Desert Hot Springs | 15,113 km |
| Distance from Swakopmund to Baton Rouge | 12,703 km |
| Distance from Swakopmund to Sunrise Manor | 14,972 km |
| Distance from Swakopmund to Elizabeth City | 11,535 km |
Jeffersonville
Jeffersonville is located in United States.
| GPS Coordinates | 38° 16´ 39.2520'' N 85° 44´ 13.8480'' W |
|---|---|
| Latitude | 38.27757 |
| Longitude | -85.73718 |
| Altitude | 138 m |
| Country | United States |