Distance from Aleksandrow Lodzki to Vineland
The shortest distance (air line) between Aleksandrow Lodzki and Vineland is 4,309.01 mi (6,934.69 km).
How far is Aleksandrow Lodzki from Vineland
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 15:42 (09.01.2026)
Vineland is located in New Jersey, United States within 39° 27' 55.08" N -75° 0' 6.84" W (39.4653, -74.9981) coordinates. The local time in Vineland is 09:42 (09.01.2026)
The calculated flying distance from Vineland to Vineland is 4,309.01 miles which is equal to 6,934.69 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Vineland, New Jersey, United States