Air Seychelles HM5908 Flight Status Today

International flight Air Seychelles HM5908 takes off from Abu Dhabi (AUH) United Arab Emirates to New York (JFK) United States. It's operated by Air Seychelles. The plane leaves Abu Dhabi International Airport at 02:50 Asia/Dubai. The flight is expected to land at John F. Kennedy International Airport at 09:10 America/New_York. The flight will last about 14 hours 20 minutes.

HM5908 - SEY5908 Air Seychelles (HM)
AUH Abu Dhabi United Arab Emirates
Flight AUH to JFK
JFK New York United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 30-September-2018
Departure Abu Dhabi International Airport
IATA: AUHICAO: OMAA
Scheduled 02:50
3
Arrival John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 09:10
4
HM5908 Detail
  • Type: International Flight
  • Flight Duration: 14 hours 20 minutes
  • Flight Distance: 11052 kms / 6867 miles
Airline
  • Air Seychelles
  • IATA: HM
  • ICAO: SEY
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 498
  • Freight capacity: 15.3 tons
  • Passenger classes: First Class, Economy, Business Class
More Detail
  • Aircraft: Airbus A380-800 Passenger
  • Callsign: SEY5908
  • Departure Timezone: Asia/Dubai
  • Arrival Timezone: America/New_York
  • Current Time in Abu Dhabi: Sunday 2026-01-18 09:02 AM
  • Current Time in New York: Sunday 2026-01-18 00:02 AM

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Air Seychelles Flight HM5908 FAQs

What is the scheduled flight duration for Air Seychelles HM5908 flight?

On average, nonstop flight takes 14 hour(s) 20 minutes, with the flight distance of 11052 km (6867 miles).

What type of aircraft is used for the Air Seychelles HM5908 flight?

All Air Seychelles HM5908 flights are operated using Airbus A380-800 Passenger aircraft.

Which terminal the flight Air Seychelles HM5908 is arriving at?

Flight Air Seychelles HM5908 arrives in John F. Kennedy International Airport at Terminal 4.