Allegiant Air G41048 Flight Status Today

Domestic flight Allegiant Air G41048 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 15:27 America/New_York. The flight is expected to land at Harrisburg International Airport at 17:47 America/New_York. The flight will last about 2 hours 19 minutes.

G41048 - AAY1048 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 15:27
None 5
Arrival Harrisburg International Airport
IATA: MDTICAO: KMDT
Scheduled 17:47
None B6
G41048 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 19 minutes
  • Flight Distance: 1468 kms / 912 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-KDNL2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Thursday 2026-01-22 15:49 PM
  • Current Time in Middletown: Thursday 2026-01-22 15:49 PM

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19:30

Allegiant Air Flight G41048 FAQs

What is the scheduled flight duration for Allegiant Air G41048 flight?

On average, nonstop flight takes 2 hour(s) 19 minutes, with the flight distance of 1468 km (912 miles).

What type of aircraft is used for the Allegiant Air G41048 flight?

All Allegiant Air G41048 flights are operated using Airbus A320 aircraft.

How many Allegiant Air G41048 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41048 is operated on Sunday.