Distance from Oxon Hill to Bad Sassendorf
The shortest distance (air line) between Oxon Hill and Bad Sassendorf is 3,998.32 mi (6,434.67 km).
How far is Oxon Hill from Bad Sassendorf
Oxon Hill is located in Maryland, United States within 38° 47' 18.24" N -77° 1' 38.28" W (38.7884, -76.9727) coordinates. The local time in Oxon Hill is 01:17 (25.02.2025)
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 07:17 (25.02.2025)
The calculated flying distance from Bad Sassendorf to Bad Sassendorf is 3,998.32 miles which is equal to 6,434.67 km.
Oxon Hill, Maryland, United States
Related Distances from Oxon Hill
Bad Sassendorf, Soest, Germany