Distance between Swakopmund and Jacksonville

The distance between Swakopmund (Namibia) and Jacksonville (Florida, USA) is ca. 11.818,1 km (air line).

Swakopmund

Latidute: -22.67756818
Longitude: 14.52633535

Jacksonville

Latidute: 30.332184
Longitude: -81.655648

Distance between Swakopmund and other cities and places worldwide

Distance between Swakopmund and other major cities