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Allegiant Air G43053 Flight Status
Current status of G43053 is Scheduled On time Expected Depart in 12 hours, 58 minutes
- Type: Domestic Flight
- Flight Duration: 1 hours 37 minutes
- Flight Distance: 1166 kms / 724 miles
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Wednesday
- Service type: N/A
- Seats: N/A
- Freight capacity: N/A
- Passenger classes: N/A
- Aircraft: Airbus A320
- Callsign: F-KCTY2
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Middletown: Friday 2025-12-12 07:37 AM
- Current Time in Jacksonville: Friday 2025-12-12 07:37 AM
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 20:35 12 December 2025 America/New_York. The flight is expected to land at Jacksonville International Airport at 22:12 12 December 2025 America/New_York. The flight will last about 1 hours 37 minutes.
Similar flight route from Middletown (MDT) to Jacksonville (JAX)
| Airline | Flight no | Departure | Arrival |
|---|---|---|---|
|
G43052 Allegiant Air |
12/12/2025 18:15 |
2 hours 2 minutes 20:17 |
Frequently asked questions, answered
What is the scheduled flight duration for Allegiant Air G43053 flight?
On average, nonstop flight takes 1 hour(s) 37 minutes, with the flight distance of 1166 km (724 miles).
What is the current status of Allegiant Air G43053 flight?
The current status of flight Allegiant Air G43053 is Scheduled On time.
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.
