Distance from Florida Ridge to Scherpenheuvel
The shortest distance (air line) between Florida Ridge and Scherpenheuvel is 4,567.67 mi (7,350.95 km).
How far is Florida Ridge from Scherpenheuvel
Florida Ridge is located in Florida, United States within 27° 34' 49.8" N -81° 36' 54.72" W (27.5805, -80.3848) coordinates. The local time in Florida Ridge is 08:45 (02.02.2025)
Scherpenheuvel is located in Arr. Leuven, Belgium within 51° 0' 37.08" N 4° 58' 22.08" E (51.0103, 4.9728) coordinates. The local time in Scherpenheuvel is 14:45 (02.02.2025)
The calculated flying distance from Scherpenheuvel to Scherpenheuvel is 4,567.67 miles which is equal to 7,350.95 km.
Florida Ridge, Florida, United States
Related Distances from Florida Ridge
Scherpenheuvel, Arr. Leuven, Belgium