Distance from Peekskill to Sint-Andries
The shortest distance (air line) between Peekskill and Sint-Andries is 3,574.49 mi (5,752.58 km).
How far is Peekskill from Sint-Andries
Peekskill is located in New York, United States within 41° 17' 18.24" N -74° 4' 38.28" W (41.2884, -73.9227) coordinates. The local time in Peekskill is 15:57 (20.04.2025)
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 21:57 (20.04.2025)
The calculated flying distance from Sint-Andries to Sint-Andries is 3,574.49 miles which is equal to 5,752.58 km.
Peekskill, New York, United States
Related Distances from Peekskill
Sint-Andries, Arr. Brugge, Belgium