Distance from Cesky Tesin to Bloomingdale
The shortest distance (air line) between Cesky Tesin and Bloomingdale is 5,222.31 mi (8,404.50 km).
How far is Cesky Tesin from Bloomingdale
Cesky Tesin is located in Moravskoslezský kraj, Czechia within 49° 44' 45.96" N 18° 37' 33.96" E (49.7461, 18.6261) coordinates. The local time in Cesky Tesin is 02:45 (05.10.2025)
Bloomingdale is located in Florida, United States within 27° 52' 42.24" N -83° 44' 15.36" W (27.8784, -82.2624) coordinates. The local time in Bloomingdale is 20:45 (04.10.2025)
The calculated flying distance from Bloomingdale to Bloomingdale is 5,222.31 miles which is equal to 8,404.50 km.
Cesky Tesin, Moravskoslezský kraj, Czechia
Related Distances from Cesky Tesin
Bloomingdale, Florida, United States