Allegiant Air G49398 Flight Status Today

Domestic flight Allegiant Air G49398 takes off from Knoxville (TYS) United States to Orlando (SFB) United States. It's operated by Allegiant Air. The plane leaves McGhee Tyson Airport at 10:30 America/New_York. The flight is expected to land at Orlando Sanford International Airport at 11:55 America/New_York. The flight will last about 51 minutes.

G49398 - AAY9398 Allegiant Air (G4)
TYS Knoxville United States
Flight TYS to SFB
SFB Orlando United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 10-November-2023
Departure McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 10:30
None 2
Arrival Orlando Sanford International Airport
IATA: SFBICAO: KSFB
Scheduled 11:55
None None
G49398 Detail
  • Type: Domestic Flight
  • Flight Duration: 51 minutes
  • Flight Distance: 821 kms / 510 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • 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: Airbus A320-214
  • Callsign: AAY9398
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Monday 2026-01-19 07:00 AM
  • Current Time in Orlando: Monday 2026-01-19 07:00 AM

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Allegiant Air Flight G49398 FAQs

What is the scheduled flight duration for Allegiant Air G49398 flight?

On average, nonstop flight takes 51 minutes, with the flight distance of 821 km (510 miles).

What type of aircraft is used for the Allegiant Air G49398 flight?

All Allegiant Air G49398 flights are operated using Airbus A320-214 aircraft.