Allegiant Air G41453 Flight Status from Fort Walton Beach to Oklahoma City, United States today

Domestic flight Allegiant Air G41453 from Fort Walton Beach (VPS) United States to Oklahoma City (OKC) United States operated by Allegiant Air. Scheduled time of departure from Destin Fort Walton Beach Airport is 16:12 America/Chicago and scheduled time of arrival in Will Rogers World Airport is 18:25 America/Chicago. The duration of the flight is 2 hours 13 minutes.

G41453 - AAY1453 Allegiant Air (G4)
VPS Fort Walton Beach United States
Flight VPS to OKC
OKC Oklahoma City United States
Contact Airlines It may not operate on the date requested
Departure Destin Fort Walton Beach Airport
IATA: VPSICAO: KVPS
Scheduled 16:12
None None
Arrival Will Rogers World Airport
IATA: OKCICAO: KOKC
Scheduled 18:25
None 22
G41453 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 13 minutes
  • Flight Distance: 1166 kms / 725 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Friday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-KROG1
  • Departure Timezone: America/Chicago
  • Arrival Timezone: America/Chicago
  • Current Time in Fort Walton Beach: Thursday 2024-05-02 12:42 PM
  • Current Time in Oklahoma City: Thursday 2024-05-02 12:42 PM

Allegiant Air G41453 flight status on other days

Frequently asked questions, answered

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

On average, nonstop flight takes 2 hour(s) 13 minutes, with the flight distance of 1166 km (725 miles).

What type of aircraft is used for the Allegiant Air G41453 flight?

All Allegiant Air G41453 flights are operated using Airbus A320 aircraft.

How many Allegiant Air G41453 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41453 is operated on Friday.