Allegiant Air G45452 Flight Status Today

Domestic flight Allegiant Air G45452 takes off from Knoxville (TYS) United States to Saint Petersburg (PIE) United States. It's operated by Allegiant Air. The plane leaves McGhee Tyson Airport at 13:30 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 15:05 America/New_York. The flight will last about 30 minutes.

G45452 - AAY5452 Allegiant Air (G4)
TYS Knoxville United States
Flight TYS to PIE
PIE Saint Petersburg United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 03-August-2021
Departure McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 13:30
None None
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 15:05
None None
G45452 Detail
  • Type: Domestic Flight
  • Flight Duration: 30 minutes
  • Flight Distance: 884 kms / 549 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: None
  • Callsign: None
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Sunday 2026-05-03 15:33 PM
  • Current Time in Saint Petersburg: Sunday 2026-05-03 15:33 PM

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Allegiant Air Flight G45452 FAQs

What is the scheduled flight duration for Allegiant Air G45452 flight?

On average, nonstop flight takes 30 minutes, with the flight distance of 884 km (549 miles).