Allegiant Air G45429 Flight Status Today

Domestic flight Allegiant Air G45429 takes off from Saint Petersburg (PIE) United States to Knoxville (TYS) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 09:45 America/New_York. The flight is expected to land at McGhee Tyson Airport at 11:20 America/New_York. The flight will last about 1 hours 24 minutes.

G45429 - AAY5429 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to TYS
TYS Knoxville United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 30-June-2021
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 09:45
None None
Arrival McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 11:20
None None
G45429 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 24 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 Saint Petersburg: Wednesday 2025-12-31 11:14 AM
  • Current Time in Knoxville: Wednesday 2025-12-31 11:14 AM

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Allegiant Air Flight G45429 FAQs

What is the scheduled flight duration for Allegiant Air G45429 flight?

On average, nonstop flight takes 1 hour(s) 24 minutes, with the flight distance of 884 km (549 miles).