12,822 km
7,967 mi14h 45min
Tiempo Est.1475 kg
Huella de CarbonoN/A
Zona HorariaMapa de Ruta
Sobre esta ruta
The direct flight distance between Aeropuerto Internacional Netaji Subhas Chandra Bose (Calcuta) (CCU) and Aeropuerto Internacional de Cleveland-Hopkins (Cleveland) (CLE) is approximately 12,822 kilometers (7,967 miles). An average commercial jet flying non-stop at a cruising speed of 900 km/h requires about 14h 45min of total flight time, including 30 minutes for taxiing, takeoff, and landing procedures. Travelers flying this route will cross time zones, experiencing an official time difference of N/A between the departure and arrival locations. Estimated carbon emissions for an economy class passenger on this route are roughly 1,475 kg of CO2. Use our interactive flight range map and route calculator to plan connections, compare airlines, and analyze flight paths between CCU and CLE.
Detalles del Aeropuerto
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