Distance from Martinsburg to Six-Fours-les-Plages
The shortest distance (air line) between Martinsburg and Six-Fours-les-Plages is 4,167.75 mi (6,707.35 km).
How far is Martinsburg from Six-Fours-les-Plages
Martinsburg is located in West Virginia, United States within 39° 27' 29.52" N -78° 1' 20.64" W (39.4582, -77.9776) coordinates. The local time in Martinsburg is 04:50 (16.12.2025)
Six-Fours-les-Plages is located in Var, France within 43° 6' 3.24" N 5° 49' 12" E (43.1009, 5.8200) coordinates. The local time in Six-Fours-les-Plages is 10:50 (16.12.2025)
The calculated flying distance from Six-Fours-les-Plages to Six-Fours-les-Plages is 4,167.75 miles which is equal to 6,707.35 km.
Martinsburg, West Virginia, United States
Related Distances from Martinsburg
Six-Fours-les-Plages, Var, France