Distance from Seymour to Buggenhout
The shortest distance (air line) between Seymour and Buggenhout is 4,201.40 mi (6,761.50 km).
How far is Seymour from Buggenhout
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 21:36 (24.02.2025)
Buggenhout is located in Arr. Dendermonde, Belgium within 51° 0' 0" N 4° 12' 0" E (51.0000, 4.2000) coordinates. The local time in Buggenhout is 03:36 (25.02.2025)
The calculated flying distance from Buggenhout to Buggenhout is 4,201.40 miles which is equal to 6,761.50 km.
Seymour, Indiana, United States
Related Distances from Seymour
Buggenhout, Arr. Dendermonde, Belgium