Distance from Seven Hills to Begijnendijk
The shortest distance (air line) between Seven Hills and Begijnendijk is 3,945.92 mi (6,350.35 km).
How far is Seven Hills from Begijnendijk
Seven Hills is located in Ohio, United States within 41° 22' 49.08" N -82° 19' 35.04" W (41.3803, -81.6736) coordinates. The local time in Seven Hills is 11:57 (06.04.2025)
Begijnendijk is located in Arr. Leuven, Belgium within 51° 1' 6.96" N 4° 47' 6" E (51.0186, 4.7850) coordinates. The local time in Begijnendijk is 17:57 (06.04.2025)
The calculated flying distance from Begijnendijk to Begijnendijk is 3,945.92 miles which is equal to 6,350.35 km.
Seven Hills, Ohio, United States
Related Distances from Seven Hills
Begijnendijk, Arr. Leuven, Belgium