Distance from Lake Charles to Bad Sassendorf
The shortest distance (air line) between Lake Charles and Bad Sassendorf is 5,061.12 mi (8,145.08 km).
How far is Lake Charles from Bad Sassendorf
Lake Charles is located in Louisiana, United States within 30° 12' 3.6" N -94° 47' 20.04" W (30.2010, -93.2111) coordinates. The local time in Lake Charles is 18:45 (24.02.2025)
Bad Sassendorf is located in Soest, Germany within 51° 34' 59.16" N 8° 10' 0.12" E (51.5831, 8.1667) coordinates. The local time in Bad Sassendorf is 01:45 (25.02.2025)
The calculated flying distance from Bad Sassendorf to Bad Sassendorf is 5,061.12 miles which is equal to 8,145.08 km.
Lake Charles, Louisiana, United States
Related Distances from Lake Charles
Bad Sassendorf, Soest, Germany