Check your Flight status online
Allegiant Air G41078 Flight Status Today
Domestic flight Allegiant Air G41078 takes off from Middletown (MDT) United States to Saint Petersburg (PIE) United States. It's operated by Allegiant Air. The plane leaves Harrisburg International Airport at 16:54 13 April 2026 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 19:17 13 April 2026 America/New_York. The flight will last about 2 hours 23 minutes.
Current status of G41078 is Scheduled On time Expected Depart in 2 hours, 50 minutes
- Type: Domestic Flight
- Flight Duration: 2 hours 23 minutes
- Flight Distance: 1468 kms / 912 miles
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Monday, Friday, Saturday
- Service type: N/A
- Seats: N/A
- Freight capacity: N/A
- Passenger classes: N/A
- Aircraft: Boeing 737MAX 8 Passenger
- Callsign: F-KATL3
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Middletown: Monday 2026-04-13 14:04 PM
- Current Time in Saint Petersburg: Monday 2026-04-13 14:04 PM
Allegiant Air Flight G41078 FAQs
What is the scheduled flight duration for Allegiant Air G41078 flight?
On average, nonstop flight takes 2 hour(s) 23 minutes, with the flight distance of 1468 km (912 miles).
What is the current status of Allegiant Air G41078 flight?
The current status of flight Allegiant Air G41078 is Scheduled On time.
What type of aircraft is used for the Allegiant Air G41078 flight?
All Allegiant Air G41078 flights are operated using Boeing 737MAX 8 Passenger aircraft.
How many Allegiant Air G41078 flights are operated a week?
3 flights per week. The Flight Allegiant Air G41078 is operated on Monday, Friday, Saturday.
