Kalitta Air K49873 Flight Status Today

Domestic flight Kalitta Air K49873 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 04:29 America/New_York. The flight is expected to land at Wurtsmith Air Force Base at None America/New_York. The flight will last about 1 hours 28 minutes.

K49873 - CKS9873 Kalitta Air (K4)
JFK New York United States
Flight JFK to OSC
OSC Oscoda United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 04:29
None None
Arrival Wurtsmith Air Force Base
IATA: OSCICAO: KOSC
Scheduled None
None None
K49873 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 28 minutes
  • Flight Distance: 894 kms / 555 miles
Airline
  • Kalitta Air
  • IATA: K4
  • ICAO: CKS
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 747-4B5F
  • Callsign: CKS9873
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in New York: Sunday 2026-01-18 18:25 PM
  • Current Time in Oscoda: Sunday 2026-01-18 18:25 PM

Flight routes similar to Kalitta Air K49873

Airline Flight no Departure Arrival
Kalitta Air K49705
Kalitta Air 		18/01/2026
14:30 		49 minutes
15:52
Kalitta Air K49716
Kalitta Air 		18/01/2026
13:15 		1 hours 2 minutes
14:37

Kalitta Air Flight K49873 FAQs

What is the scheduled flight duration for Kalitta Air K49873 flight?

On average, nonstop flight takes 1 hour(s) 28 minutes, with the flight distance of 894 km (555 miles).

What type of aircraft is used for the Kalitta Air K49873 flight?

All Kalitta Air K49873 flights are operated using Boeing 747-4B5F aircraft.