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Allegiant Air G4974 Flight Status Today
Domestic flight Allegiant Air G4974 takes off from Albany (ALB) United States to Saint Petersburg (PIE) United States. It's operated by Allegiant Air. The plane leaves Albany International Airport at 09:37 19 January 2026 America/New_York. The flight is expected to land at St. Petersburg-Clearwater International Airport at 13:07 19 January 2026 America/New_York. The flight will last about 3 hours 30 minutes.
Current status of G4974 is Scheduled On time Expected Depart in 5 hours, 14 minutes
- Type: Domestic Flight
- Flight Duration: 3 hours 30 minutes
- Flight Distance: 1831 kms / 1138 miles
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Monday, Wednesday, Friday
- Service type: N/A
- Seats: N/A
- Freight capacity: N/A
- Passenger classes: N/A
- Aircraft: Airbus A318 / A319 / A320 / A321
- Callsign: F-KTLH2
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Albany: Monday 2026-01-19 04:23 AM
- Current Time in Saint Petersburg: Monday 2026-01-19 04:23 AM
Flight routes similar to Allegiant Air G4974
| Airline | Flight no | Departure | Arrival |
|---|---|---|---|
|
G43667 Allegiant Air |
19/01/2026 20:00 |
3 hours 4 minutes 23:04 |
Allegiant Air Flight G4974 FAQs
What is the scheduled flight duration for Allegiant Air G4974 flight?
On average, nonstop flight takes 3 hour(s) 30 minutes, with the flight distance of 1831 km (1138 miles).
What is the current status of Allegiant Air G4974 flight?
The current status of flight Allegiant Air G4974 is Scheduled On time.
What type of aircraft is used for the Allegiant Air G4974 flight?
All Allegiant Air G4974 flights are operated using Airbus A318 / A319 / A320 / A321 aircraft.
How many Allegiant Air G4974 flights are operated a week?
3 flights per week. The Flight Allegiant Air G4974 is operated on Monday, Wednesday, Friday.
