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Kalitta Air K49705 Flight Status Today
Domestic flight Kalitta Air K49705 takes off from New York (JFK) United States to Oscoda (OSC) United States. It's operated by Kalitta Air. The plane leaves John F. Kennedy International Airport at 07:03 20 March 2026 America/New_York. The flight is expected to land at Wurtsmith Air Force Base at 15:01 20 March 2026 America/New_York. The flight will last about 9 hours 5 minutes.
Current status of K49705 is Unknown
- Type: Domestic Flight
- Flight Duration: 9 hours 5 minutes
- Flight Distance: 894 kms / 555 miles
- Kalitta Air
- IATA: K4
- ICAO: CKS
- Operating Days: Friday
- Service type: N/A
- Seats: N/A
- Freight capacity: N/A
- Passenger classes: N/A
- Aircraft: Boeing 747-4B5F
- Callsign: F-KFNT1
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in New York: Friday 2026-03-20 04:59 AM
- Current Time in Oscoda: Friday 2026-03-20 04:59 AM
Kalitta Air Flight K49705 FAQs
What is the scheduled flight duration for Kalitta Air K49705 flight?
On average, nonstop flight takes 9 hour(s) 5 minutes, with the flight distance of 894 km (555 miles).
What is the current status of Kalitta Air K49705 flight?
The current status of flight Kalitta Air K49705 is Unknown.
What type of aircraft is used for the Kalitta Air K49705 flight?
All Kalitta Air K49705 flights are operated using Boeing 747-4B5F aircraft.
When does the flight Kalitta Air K49705 arrive?
Flight Kalitta Air K49705 arrives at 15:01.
Which terminal the flight Kalitta Air K49705 is arriving at?
Flight Kalitta Air K49705 arrives in Wurtsmith Air Force Base at Terminal -.
How many Kalitta Air K49705 flights are operated a week?
1 flights per week. The Flight Kalitta Air K49705 is operated on Friday.
