Allegiant Air G42709 Flight Status Today

Domestic flight Allegiant Air G42709 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 06:20 America/New_York. The flight is expected to land at McGhee Tyson Airport at 07:57 America/New_York. The flight will last about 1 hours 35 minutes.

G42709 - AAY2709 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 06:20
None 11
Arrival McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 07:57
None 2
G42709 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 35 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Thursday, Friday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A319
  • Callsign: F-KTLH2
  • 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 G42709 FAQs

What is the scheduled flight duration for Allegiant Air G42709 flight?

On average, nonstop flight takes 1 hour(s) 35 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G42709 flight?

All Allegiant Air G42709 flights are operated using Airbus A319 aircraft.

How many Allegiant Air G42709 flights are operated a week?

2 flights per week. The Flight Allegiant Air G42709 is operated on Thursday, Friday.