Distance from Lake Mary to Buggenhout
The shortest distance (air line) between Lake Mary and Buggenhout is 4,516.23 mi (7,268.18 km).
How far is Lake Mary from Buggenhout
Lake Mary is located in Florida, United States within 28° 45' 33.12" N -82° 39' 50.4" W (28.7592, -81.3360) coordinates. The local time in Lake Mary is 22:13 (22.01.2025)
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 04:13 (23.01.2025)
The calculated flying distance from Buggenhout to Buggenhout is 4,516.23 miles which is equal to 7,268.18 km.
Lake Mary, Florida, United States
Related Distances from Lake Mary
Buggenhout, Arr. Dendermonde, Belgium