Allegiant Air G41410 Flight Status Today

Domestic flight Allegiant Air G41410 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 08:30 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 10:18 America/New_York. The flight will last about 1 hours 40 minutes.

G41410 - AAY1410 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 08:30
None 4
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 10:18
None 3
G41410 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 40 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Wednesday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A320 (sharklets)
  • Callsign: F-KFLL2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Monday 2026-02-09 00:38 AM
  • Current Time in Saint Petersburg: Monday 2026-02-09 00:38 AM

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Allegiant Air Flight G41410 FAQs

What is the scheduled flight duration for Allegiant Air G41410 flight?

On average, nonstop flight takes 1 hour(s) 40 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G41410 flight?

All Allegiant Air G41410 flights are operated using Airbus A320 (sharklets) aircraft.

How many Allegiant Air G41410 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41410 is operated on Wednesday.