Allegiant Air G41428 Flight Status : Live Tracking & Updates

Domestic flight Allegiant Air G41428 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 19:33 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 21:12 America/New_York. The flight will last about 1 hours 39 minutes.

G41428 - AAY1428 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 19:33
None 2
Arrival St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 21:12
None None
G41428 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 39 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Thursday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 737MAX 8 Passenger
  • Callsign: F-KPGD2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Tuesday 2026-05-05 21:10 PM
  • Current Time in Saint Petersburg: Tuesday 2026-05-05 21:10 PM

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Allegiant Air Flight G41428 FAQs

What is the scheduled flight duration for Allegiant Air G41428 flight?

On average, nonstop flight takes 1 hour(s) 39 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G41428 flight?

All Allegiant Air G41428 flights are operated using Boeing 737MAX 8 Passenger aircraft.

How many Allegiant Air G41428 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41428 is operated on Thursday.