Allegiant Air G4973 Flight Status Today

Domestic flight Allegiant Air G4973 takes off from Saint Petersburg (PIE) United States to Akron (CAK) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 07:10 America/New_York. The flight is expected to land at Akron-Canton Airport at 09:30 America/New_York. The flight will last about 2 hours 15 minutes.

G4973 - AAY973 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to CAK
CAK Akron United States
Contact Airlines It may not operate on the date requested
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 07:10
None 11
Arrival Akron-Canton Airport
IATA: CAKICAO: KCAK
Scheduled 09:30
None 5
G4973 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 15 minutes
  • Flight Distance: 1447 kms / 899 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Tuesday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Airbus A319
  • Callsign: F-KMCF1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Thursday 2026-04-30 21:03 PM
  • Current Time in Akron: Thursday 2026-04-30 21:03 PM

Allegiant Air Flight G4973 FAQs

What is the scheduled flight duration for Allegiant Air G4973 flight?

On average, nonstop flight takes 2 hour(s) 15 minutes, with the flight distance of 1447 km (899 miles).

What type of aircraft is used for the Allegiant Air G4973 flight?

All Allegiant Air G4973 flights are operated using Airbus A319 aircraft.

How many Allegiant Air G4973 flights are operated a week?

1 flights per week. The Flight Allegiant Air G4973 is operated on Tuesday.