Distance from Ruskin to Sint-Andries
The shortest distance (air line) between Ruskin and Sint-Andries is 4,568.25 mi (7,351.90 km).
How far is Ruskin from Sint-Andries
Ruskin is located in Florida, United States within 27° 42' 23.4" N -83° 34' 44.76" W (27.7065, -82.4209) coordinates. The local time in Ruskin is 00:32 (25.02.2025)
Sint-Andries is located in Arr. Brugge, Belgium within 51° 12' 0" N 3° 10' 59.88" E (51.2000, 3.1833) coordinates. The local time in Sint-Andries is 06:32 (25.02.2025)
The calculated flying distance from Sint-Andries to Sint-Andries is 4,568.25 miles which is equal to 7,351.90 km.
Ruskin, Florida, United States
Related Distances from Ruskin
Sint-Andries, Arr. Brugge, Belgium