Distance from Oxon Hill to Bad Sooden-Allendorf
The shortest distance (air line) between Oxon Hill and Bad Sooden-Allendorf is 4,078.19 mi (6,563.22 km).
How far is Oxon Hill from Bad Sooden-Allendorf
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 23:21 (13.01.2026)
Bad Sooden-Allendorf is located in Werra-Meißner-Kreis, Germany within 51° 16' 59.88" N 9° 58' 59.88" E (51.2833, 9.9833) coordinates. The local time in Bad Sooden-Allendorf is 05:21 (14.01.2026)
The calculated flying distance from Bad Sooden-Allendorf to Bad Sooden-Allendorf is 4,078.19 miles which is equal to 6,563.22 km.
Oxon Hill, Maryland, United States
Related Distances from Oxon Hill
Bad Sooden-Allendorf, Werra-Meißner-Kreis, Germany