Allegiant Air G41041 Flight Status Today

Domestic flight Allegiant Air G41041 takes off from Saint Petersburg (PIE) United States to Middletown (MDT) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 17:13 America/New_York. The flight is expected to land at Harrisburg International Airport at 19:34 America/New_York. The flight will last about 2 hours 23 minutes.

G41041 - AAY1041 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to MDT
MDT Middletown United States
Contact Airlines It may not operate on the date requested
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 17:13
None 10
Arrival Harrisburg International Airport
IATA: MDTICAO: KMDT
Scheduled 19:34
None B6
G41041 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 23 minutes
  • Flight Distance: 1468 kms / 912 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Monday, Wednesday, Friday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A319
  • Callsign: F-KALX1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Wednesday 2025-12-31 23:14 PM
  • Current Time in Middletown: Wednesday 2025-12-31 23:14 PM

Flight routes similar to Allegiant Air G41041

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None
19:30

Allegiant Air Flight G41041 FAQs

What is the scheduled flight duration for Allegiant Air G41041 flight?

On average, nonstop flight takes 2 hour(s) 23 minutes, with the flight distance of 1468 km (912 miles).

What type of aircraft is used for the Allegiant Air G41041 flight?

All Allegiant Air G41041 flights are operated using Airbus A319 aircraft.

How many Allegiant Air G41041 flights are operated a week?

3 flights per week. The Flight Allegiant Air G41041 is operated on Monday, Wednesday, Friday.