Allegiant Air G44847 Flight Status Today

Domestic flight Allegiant Air G44847 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 15:40 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 18:08 America/New_York. The flight will last about 2 hours 28 minutes.

G44847 - AAY4847 Allegiant Air (G4)
MDT Middletown United States
Flight MDT to PIE
PIE Saint Petersburg United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure Harrisburg International Airport
IATA: MDTICAO: KMDT
Scheduled 15:40
None None
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 18:08
None None
G44847 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: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 177
  • Freight capacity: 9.4 tons
  • Passenger classes: First Class, Economy, Business Class
More Detail
  • Aircraft: Airbus A319
  • Callsign: AAY847
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Middletown: Monday 2026-01-19 19:44 PM
  • Current Time in Saint Petersburg: Monday 2026-01-19 19:44 PM

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None
22:44

Allegiant Air Flight G44847 FAQs

What is the scheduled flight duration for Allegiant Air G44847 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 G44847 flight?

All Allegiant Air G44847 flights are operated using Airbus A319 aircraft.