Distance from Chestnut Ridge to Aleksandrow Lodzki
The shortest distance (air line) between Chestnut Ridge and Aleksandrow Lodzki is 4,194.50 mi (6,750.40 km).
How far is Chestnut Ridge from Aleksandrow Lodzki
Chestnut Ridge is located in New York, United States within 41° 4' 58.44" N -75° 56' 41.64" W (41.0829, -74.0551) coordinates. The local time in Chestnut Ridge is 16:00 (02.01.2026)
Aleksandrow Lodzki is located in Łódzki, Poland within 51° 49' 9.84" N 19° 18' 14.04" E (51.8194, 19.3039) coordinates. The local time in Aleksandrow Lodzki is 22:00 (02.01.2026)
The calculated flying distance from Aleksandrow Lodzki to Aleksandrow Lodzki is 4,194.50 miles which is equal to 6,750.40 km.
Chestnut Ridge, New York, United States
Related Distances from Chestnut Ridge
Aleksandrow Lodzki, Łódzki, Poland