Distance from Seymour to Janow Lubelski
The shortest distance (air line) between Seymour and Janow Lubelski is 4,879.17 mi (7,852.26 km).
How far is Seymour from Janow Lubelski
Seymour is located in Indiana, United States within 38° 56' 51.36" N -86° 6' 32.04" W (38.9476, -85.8911) coordinates. The local time in Seymour is 23:18 (02.01.2026)
Janow Lubelski is located in Puławski, Poland within 50° 43' 0.12" N 22° 25' 0.12" E (50.7167, 22.4167) coordinates. The local time in Janow Lubelski is 05:18 (03.01.2026)
The calculated flying distance from Janow Lubelski to Janow Lubelski is 4,879.17 miles which is equal to 7,852.26 km.
Seymour, Indiana, United States
Related Distances from Seymour
Janow Lubelski, Puławski, Poland