Distance from Secaucus to Diepenbeek
The shortest distance (air line) between Secaucus and Diepenbeek is 3,700.09 mi (5,954.72 km).
How far is Secaucus from Diepenbeek
Secaucus is located in New Jersey, United States within 40° 46' 51.6" N -75° 56' 2.76" W (40.7810, -74.0659) coordinates. The local time in Secaucus is 16:22 (24.02.2025)
Diepenbeek is located in Arr. Hasselt, Belgium within 50° 54' 25.92" N 5° 25' 3" E (50.9072, 5.4175) coordinates. The local time in Diepenbeek is 22:22 (24.02.2025)
The calculated flying distance from Diepenbeek to Diepenbeek is 3,700.09 miles which is equal to 5,954.72 km.
Secaucus, New Jersey, United States
Related Distances from Secaucus
Diepenbeek, Arr. Hasselt, Belgium