Allegiant Air G44804 Flight Status Today

Domestic flight Allegiant Air G44804 takes off from Indianapolis (IND) United States to Cleveland (CLE) United States. It's operated by Allegiant Air. The plane leaves Indianapolis International Airport at 18:30 America/Indiana/Indianapolis. The flight is expected to land at Cleveland Hopkins International Airport at 19:29 America/New_York. The flight will last about 59 minutes.

G44804 - AAY4804 Allegiant Air (G4)
IND Indianapolis United States
Flight IND to CLE
CLE Cleveland United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 25-September-2020
Departure Indianapolis International Airport
IATA: INDICAO: KIND
Scheduled 18:30
Arrival Cleveland Hopkins International Airport
IATA: CLEICAO: KCLE
Scheduled 19:29
G44804 Detail
  • Type: Domestic Flight
  • Flight Duration: 59 minutes
  • Flight Distance: 422 kms / 262 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 156
  • Freight capacity: 6.7 tons
  • Passenger classes: First Class, Economy, Business Class
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-KLUK3
  • Departure Timezone: America/Indiana/Indianapolis
  • Arrival Timezone: America/New_York
  • Current Time in Indianapolis: Wednesday 2026-01-21 07:29 AM
  • Current Time in Cleveland: Wednesday 2026-01-21 07:29 AM

Flight routes similar to Allegiant Air G44804

Airline Flight no Departure Arrival
FedEx FX1731
FedEx 		21/01/2026
04:46 		1 hours 8 minutes
05:38
American Airlines AA9694
American Airlines 		21/01/2026
16:00 		1 hours 3 minutes
17:15

Allegiant Air Flight G44804 FAQs

What is the scheduled flight duration for Allegiant Air G44804 flight?

On average, nonstop flight takes 59 minutes, with the flight distance of 422 km (262 miles).

What type of aircraft is used for the Allegiant Air G44804 flight?

All Allegiant Air G44804 flights are operated using Airbus A320 aircraft.