Distance from Lakeville to Bad Sassendorf
The shortest distance (air line) between Lakeville and Bad Sassendorf is 4,310.86 mi (6,937.66 km).
How far is Lakeville from Bad Sassendorf
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 06:43 (11.02.2025)
Bad Sassendorf is located in Soest, Germany within 51° 34' 59.16" N 8° 10' 0.12" E (51.5831, 8.1667) coordinates. The local time in Bad Sassendorf is 13:43 (11.02.2025)
The calculated flying distance from Bad Sassendorf to Bad Sassendorf is 4,310.86 miles which is equal to 6,937.66 km.
Lakeville, Minnesota, United States
Related Distances from Lakeville
Bad Sassendorf, Soest, Germany