Allegiant Air G46107 Flight Status Today

Domestic flight Allegiant Air G46107 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 09:48 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 11:23 America/New_York. The flight will last about 57 minutes.

G46107 - AAY6107 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 12-October-2021
Departure McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 09:48
None 6
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 11:23
None 5
G46107 Detail
  • Type: Domestic Flight
  • Flight Duration: 57 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: Airbus A320
  • Callsign: AAY6107
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Monday 2026-03-23 15:45 PM
  • Current Time in Saint Petersburg: Monday 2026-03-23 15:45 PM

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Allegiant Air Flight G46107 FAQs

What is the scheduled flight duration for Allegiant Air G46107 flight?

On average, nonstop flight takes 57 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G46107 flight?

All Allegiant Air G46107 flights are operated using Airbus A320 aircraft.