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 14:30 America/New_York. The flight is expected to land at Wurtsmith Air Force Base at 15:52 America/New_York. The flight will last about 49 minutes.

K49705 - CKS9705 Kalitta Air (K4)
JFK New York United States
Flight JFK to OSC
OSC Oscoda United States
Contact Airlines It may not operate on the date requested
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 14:30
None None
Arrival Wurtsmith Air Force Base
IATA: OSCICAO: KOSC
Scheduled 15:52
None None
K49705 Detail
  • Type: Domestic Flight
  • Flight Duration: 49 minutes
  • Flight Distance: 894 kms / 555 miles
Airline
  • Kalitta Air
  • IATA: K4
  • ICAO: CKS
  • Operating Days: Friday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 747-4B5F
  • Callsign: F-KFNT1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in New York: Sunday 2026-01-18 13:14 PM
  • Current Time in Oscoda: Sunday 2026-01-18 13:14 PM

Flight routes similar to Kalitta Air K49705

Airline Flight no Departure Arrival
Kalitta Air K49716
Kalitta Air 		18/01/2026
13:15 		1 hours 2 minutes
14:37

Kalitta Air Flight K49705 FAQs

What is the scheduled flight duration for Kalitta Air K49705 flight?

On average, nonstop flight takes 49 minutes, with the flight distance of 894 km (555 miles).

What type of aircraft is used for the Kalitta Air K49705 flight?

All Kalitta Air K49705 flights are operated using Boeing 747-4B5F aircraft.

How many Kalitta Air K49705 flights are operated a week?

1 flights per week. The Flight Kalitta Air K49705 is operated on Friday.