Distance from Peekskill to Sfantu-Gheorghe
The shortest distance (air line) between Peekskill and Sfantu-Gheorghe is 4,652.54 mi (7,487.54 km).
How far is Peekskill from Sfantu-Gheorghe
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 19:26 (20.05.2025)
Sfantu-Gheorghe is located in Covasna, Romania within 45° 51' 48.96" N 25° 47' 15" E (45.8636, 25.7875) coordinates. The local time in Sfantu-Gheorghe is 02:26 (21.05.2025)
The calculated flying distance from Sfantu-Gheorghe to Sfantu-Gheorghe is 4,652.54 miles which is equal to 7,487.54 km.
Peekskill, New York, United States
Related Distances from Peekskill
Sfantu-Gheorghe, Covasna, Romania