Allegiant Air G43647 Flight Status Today

Domestic flight Allegiant Air G43647 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 21:01 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 22:42 America/New_York. The flight will last about 1 hours 41 minutes.

G43647 - AAY3647 Allegiant Air (G4)
TYS Knoxville United States
Flight TYS to PIE
PIE Saint Petersburg United States
Contact Airlines It may not operate on the date requested
Departure McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 21:01
None 4
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 22:42
None None
G43647 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 41 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Tuesday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320
  • Callsign: None
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Sunday 2026-04-12 23:07 PM
  • Current Time in Saint Petersburg: Sunday 2026-04-12 23:07 PM

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Allegiant Air Flight G43647 FAQs

What is the scheduled flight duration for Allegiant Air G43647 flight?

On average, nonstop flight takes 1 hour(s) 41 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G43647 flight?

All Allegiant Air G43647 flights are operated using Airbus A320 aircraft.

How many Allegiant Air G43647 flights are operated a week?

1 flights per week. The Flight Allegiant Air G43647 is operated on Tuesday.