Distance from Seymour to Sannicolau Mare
The shortest distance (air line) between Seymour and Sannicolau Mare is 5,014.80 mi (8,070.54 km).
How far is Seymour from Sannicolau Mare
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 05:08 (07.01.2026)
Sannicolau Mare is located in Timiş, Romania within 46° 3' 48.96" N 20° 36' 45" E (46.0636, 20.6125) coordinates. The local time in Sannicolau Mare is 12:08 (07.01.2026)
The calculated flying distance from Sannicolau Mare to Sannicolau Mare is 5,014.80 miles which is equal to 8,070.54 km.
Seymour, Indiana, United States
Related Distances from Seymour
Sannicolau Mare, Timiş, Romania