Allegiant Air G46908 flight status now

G46908 - AAY6908 Allegiant Air (G4)
OKC Oklahoma City United States
Flight OKC to VPS
VPS Fort Walton Beach United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure Will Rogers World Airport
IATA: OKCICAO: KOKC
Scheduled 14:25
None None
Arrival Destin Fort Walton Beach Airport
IATA: VPSICAO: KVPS
Scheduled None
None None
G46908 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 36 minutes
  • Flight Distance: 1166 kms / 725 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: AAY6908
  • Departure Timezone: America/Chicago
  • Arrival Timezone: America/Chicago
  • Current Time in Oklahoma City: Thursday 2025-11-20 19:28 PM
  • Current Time in Fort Walton Beach: Thursday 2025-11-20 19:28 PM

Domestic flight Allegiant Air G46908 takes off from Oklahoma City (OKC) United States to Fort Walton Beach (VPS) United States. It's operated by Allegiant Air. The plane leaves Will Rogers World Airport at 14:25 America/Chicago. The flight is expected to land at Destin Fort Walton Beach Airport at None America/Chicago. The flight will last about 1 hours 36 minutes.

Frequently asked questions, answered

How long is the flight from Oklahoma City to Fort Walton Beach?

On average, nonstop flight takes 1 hour(s) 36 minutes, with the flight distance of 1166 km (725 miles).

What type of aircraft is used for the Allegiant Air G46908 flight?

All Allegiant Air G46908 flights are operated using Airbus A320-214 aircraft.