Allegiant Air G41423 Flight Status Today

Domestic flight Allegiant Air G41423 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 08:53 America/New_York. The flight is expected to land at McGhee Tyson Airport at 10:35 America/New_York. The flight will last about 1 hours 42 minutes.

G41423 - AAY1423 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 08:53
None 10
Arrival McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 10:35
None 4
G41423 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 42 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320 (sharklets)
  • Callsign: F-KMQJ1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Friday 2026-05-01 12:17 PM
  • Current Time in Knoxville: Friday 2026-05-01 12:17 PM

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Allegiant Air Flight G41423 FAQs

What is the scheduled flight duration for Allegiant Air G41423 flight?

On average, nonstop flight takes 1 hour(s) 42 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G41423 flight?

All Allegiant Air G41423 flights are operated using Airbus A320 (sharklets) aircraft.

How many Allegiant Air G41423 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41423 is operated on Sunday.