Allegiant Air G41040 Flight Status Today

Domestic flight Allegiant Air G41040 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 20:20 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 22:48 America/New_York. The flight will last about 2 hours 28 minutes.

G41040 - AAY1040 Allegiant Air (G4)
MDT Middletown United States
Flight MDT to PIE
PIE Saint Petersburg United States
Contact Airlines It may not operate on the date requested
Departure Harrisburg International Airport
IATA: MDTICAO: KMDT
Scheduled 20:20
None None
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 22:48
None None
G41040 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 28 minutes
  • Flight Distance: 1468 kms / 912 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Saturday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 737MAX 8 Passenger
  • Callsign: F-KRDU1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Middletown: Monday 2026-02-09 14:30 PM
  • Current Time in Saint Petersburg: Monday 2026-02-09 14:30 PM

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None
22:44

Allegiant Air Flight G41040 FAQs

What is the scheduled flight duration for Allegiant Air G41040 flight?

On average, nonstop flight takes 2 hour(s) 28 minutes, with the flight distance of 1468 km (912 miles).

What type of aircraft is used for the Allegiant Air G41040 flight?

All Allegiant Air G41040 flights are operated using Boeing 737MAX 8 Passenger aircraft.

How many Allegiant Air G41040 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41040 is operated on Saturday.