Allegiant Air G45613 Flight Status Today

Domestic flight Allegiant Air G45613 takes off from Saint Petersburg (PIE) United States to Nashville (BNA) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 15:00 America/New_York. The flight is expected to land at Nashville International Airport at 15:47 America/Chicago. The flight will last about 1 hours 4 minutes.

G45613 - AAY5613 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to BNA
BNA Nashville United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 30-November-2021
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 15:00
None 9
Arrival Nashville International Airport
IATA: BNAICAO: KBNA
Scheduled 15:47
None None
G45613 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 4 minutes
  • Flight Distance: 985 kms / 612 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A319
  • Callsign: AAY5613
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/Chicago
  • Current Time in Saint Petersburg: Monday 2026-04-13 18:34 PM
  • Current Time in Nashville: Monday 2026-04-13 17:34 PM

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Allegiant Air Flight G45613 FAQs

What is the scheduled flight duration for Allegiant Air G45613 flight?

On average, nonstop flight takes 1 hour(s) 4 minutes, with the flight distance of 985 km (612 miles).

What type of aircraft is used for the Allegiant Air G45613 flight?

All Allegiant Air G45613 flights are operated using Airbus A319 aircraft.