Allegiant Air G4410 Flight Status Today

Domestic flight Allegiant Air G4410 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 15:48 America/New_York. The flight is expected to land at McGhee Tyson Airport at 17:25 America/New_York. The flight will last about 1 hours 37 minutes.

G4410 - AAY410 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 15:48
None 3
Arrival McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 17:25
None 2
G4410 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 37 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Friday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320 (sharklets)
  • Callsign: F-KLGC2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Wednesday 2025-12-31 05:01 AM
  • Current Time in Knoxville: Wednesday 2025-12-31 05:01 AM

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Allegiant Air Flight G4410 FAQs

What is the scheduled flight duration for Allegiant Air G4410 flight?

On average, nonstop flight takes 1 hour(s) 37 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G4410 flight?

All Allegiant Air G4410 flights are operated using Airbus A320 (sharklets) aircraft.

How many Allegiant Air G4410 flights are operated a week?

1 flights per week. The Flight Allegiant Air G4410 is operated on Friday.