Distance from Koekelberg to Seven Hills
The shortest distance (air line) between Koekelberg and Seven Hills is 3,933.10 mi (6,329.72 km).
How far is Koekelberg from Seven Hills
Koekelberg is located in Arr. de Bruxelles-Capitale/Arr. Brussel-Hoofdstad, Belgium within 50° 52' 0.12" N 4° 19' 59.88" E (50.8667, 4.3333) coordinates. The local time in Koekelberg is 10:17 (22.12.2025)
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 04:17 (22.12.2025)
The calculated flying distance from Seven Hills to Seven Hills is 3,933.10 miles which is equal to 6,329.72 km.
Koekelberg, Arr. de Bruxelles-Capitale/Arr. Brussel-Hoofdstad, Belgium
Related Distances from Koekelberg
Seven Hills, Ohio, United States