Distance from Oxon Hill to Noordwijkerhout
The shortest distance (air line) between Oxon Hill and Noordwijkerhout is 3,835.82 mi (6,173.15 km).
How far is Oxon Hill from Noordwijkerhout
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 05:55 (13.03.2025)
Noordwijkerhout is located in Agglomeratie Leiden en Bollenstreek, Netherlands within 52° 16' 0.12" N 4° 30' 0" E (52.2667, 4.5000) coordinates. The local time in Noordwijkerhout is 11:55 (13.03.2025)
The calculated flying distance from Noordwijkerhout to Noordwijkerhout is 3,835.82 miles which is equal to 6,173.15 km.
Oxon Hill, Maryland, United States
Related Distances from Oxon Hill
Noordwijkerhout, Agglomeratie Leiden en Bollenstreek, Netherlands