Distance from Lakeville to Sannicolau Mare
The shortest distance (air line) between Lakeville and Sannicolau Mare is 4,984.33 mi (8,021.50 km).
How far is Lakeville from Sannicolau Mare
Lakeville is located in Minnesota, United States within 44° 40' 38.64" N -94° 44' 52.8" W (44.6774, -93.2520) coordinates. The local time in Lakeville is 03:50 (10.03.2025)
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 11:50 (10.03.2025)
The calculated flying distance from Sannicolau Mare to Sannicolau Mare is 4,984.33 miles which is equal to 8,021.50 km.
Lakeville, Minnesota, United States
Related Distances from Lakeville
Sannicolau Mare, Timiş, Romania