Distance from Aleksandrow Lodzki to Shepherdsville
The shortest distance (air line) between Aleksandrow Lodzki and Shepherdsville is 4,770.66 mi (7,677.64 km).
How far is Aleksandrow Lodzki from Shepherdsville
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 05:15 (18.12.2025)
Shepherdsville is located in Kentucky, United States within 37° 58' 52.68" N -86° 17' 57.48" W (37.9813, -85.7007) coordinates. The local time in Shepherdsville is 23:15 (17.12.2025)
The calculated flying distance from Shepherdsville to Shepherdsville is 4,770.66 miles which is equal to 7,677.64 km.
Aleksandrow Lodzki, Łódzki, Poland
Related Distances from Aleksandrow Lodzki
Shepherdsville, Kentucky, United States