Allegiant Air G43038 Flight Status : Live Tracking & Updates

Domestic flight Allegiant Air G43038 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 19:00 America/New_York. The flight is expected to land at McGhee Tyson Airport at 20:40 America/New_York. The flight will last about 1 hours 31 minutes.

G43038 - AAY3038 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to TYS
TYS Knoxville United States
Contact Airlines It may not operate on the date requested
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 19:00
None 2
Arrival McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 20:40
None 4
G43038 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 31 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Thursday, Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: F-KCRG1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Monday 2026-05-25 20:52 PM
  • Current Time in Knoxville: Monday 2026-05-25 20:52 PM

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Allegiant Air Flight G43038 FAQs

What is the scheduled flight duration for Allegiant Air G43038 flight?

On average, nonstop flight takes 1 hour(s) 31 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G43038 flight?

All Allegiant Air G43038 flights are operated using Airbus A320 aircraft.

How many Allegiant Air G43038 flights are operated a week?

2 flights per week. The Flight Allegiant Air G43038 is operated on Thursday, Sunday.