Distance from Koekelberg to Sandusky
The shortest distance (air line) between Koekelberg and Sandusky is 3,970.02 mi (6,389.13 km).
How far is Koekelberg from Sandusky
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 10:34 (22.12.2025)
Sandusky is located in Ohio, United States within 41° 26' 48.48" N -83° 17' 51.36" W (41.4468, -82.7024) coordinates. The local time in Sandusky is 04:34 (22.12.2025)
The calculated flying distance from Sandusky to Sandusky is 3,970.02 miles which is equal to 6,389.13 km.
Koekelberg, Arr. de Bruxelles-Capitale/Arr. Brussel-Hoofdstad, Belgium