Allegiant Air G43053 Flight Status Today

Domestic flight Allegiant Air G43053 takes off from Middletown (MDT) United States to Jacksonville (JAX) United States. It's operated by Allegiant Air. The plane leaves Harrisburg International Airport at 18:09 America/New_York. The flight is expected to land at Jacksonville International Airport at 20:10 America/New_York. The flight will last about 2 hours 1 minutes.

G43053 - AAY3053 Allegiant Air (G4)
MDT Middletown United States
Flight MDT to JAX
JAX Jacksonville United States
Contact Airlines It may not operate on the date requested
Departure Harrisburg International Airport
IATA: MDTICAO: KMDT
Scheduled 18:09
None B6
Arrival Jacksonville International Airport
IATA: JAXICAO: KJAX
Scheduled 20:10
None C6
G43053 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 1 minutes
  • Flight Distance: 1166 kms / 724 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Wednesday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-KCTY2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Middletown: Wednesday 2026-01-21 09:03 AM
  • Current Time in Jacksonville: Wednesday 2026-01-21 09:03 AM

Allegiant Air Flight G43053 FAQs

What is the scheduled flight duration for Allegiant Air G43053 flight?

On average, nonstop flight takes 2 hour(s) 1 minutes, with the flight distance of 1166 km (724 miles).

What type of aircraft is used for the Allegiant Air G43053 flight?

All Allegiant Air G43053 flights are operated using Airbus A320 aircraft.

How many Allegiant Air G43053 flights are operated a week?

1 flights per week. The Flight Allegiant Air G43053 is operated on Wednesday.