Allegiant Air G42598 Flight Status Today

Domestic flight Allegiant Air G42598 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 14:41 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 17:14 America/New_York. The flight will last about 2 hours 31 minutes.

G42598 - AAY2598 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 14:41
None B6
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 17:14
None 7
G42598 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 31 minutes
  • Flight Distance: 1468 kms / 912 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Monday, Thursday, Friday, Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 737MAX 8 Passenger
  • Callsign: F-KBID1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Middletown: Monday 2026-01-19 19:45 PM
  • Current Time in Saint Petersburg: Monday 2026-01-19 19:45 PM

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Allegiant Air Flight G42598 FAQs

What is the scheduled flight duration for Allegiant Air G42598 flight?

On average, nonstop flight takes 2 hour(s) 31 minutes, with the flight distance of 1468 km (912 miles).

What type of aircraft is used for the Allegiant Air G42598 flight?

All Allegiant Air G42598 flights are operated using Boeing 737MAX 8 Passenger aircraft.

How many Allegiant Air G42598 flights are operated a week?

4 flights per week. The Flight Allegiant Air G42598 is operated on Monday, Thursday, Friday, Sunday.