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Allegiant Air G4983 Flight Status Today
Domestic flight Allegiant Air G4983 takes off from Saint Petersburg (PIE) United States to Bloomington (BMI) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 11:26 01 March 2026 America/New_York. The flight is expected to land at Central Illinois Regional Airport at 14:17 01 March 2026 America/Chicago. The flight will last about 2 hours 39 minutes.
Current status of G4983 is Arrived On time
- Type: Domestic Flight
- Flight Duration: 2 hours 39 minutes
- Flight Distance: 1507 kms / 936 miles
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Thursday, Sunday
- Service type: N/A
- Seats: N/A
- Freight capacity: N/A
- Passenger classes: N/A
- Aircraft: Boeing 737MAX 8 Passenger
- Callsign: F-KBOW3
- Departure Timezone: America/New_York
- Arrival Timezone: America/Chicago
- Current Time in Saint Petersburg: Sunday 2026-03-01 23:19 PM
- Current Time in Bloomington: Sunday 2026-03-01 22:19 PM
Flight routes similar to Allegiant Air G4983
| Airline | Flight no | Departure | Arrival |
|---|---|---|---|
|
G42676 Allegiant Air |
01/03/2026 17:19 |
2 hours 27 minutes 18:46 |
Allegiant Air Flight G4983 FAQs
What is the scheduled flight duration for Allegiant Air G4983 flight?
On average, nonstop flight takes 2 hour(s) 39 minutes, with the flight distance of 1507 km (936 miles).
What is the current status of Allegiant Air G4983 flight?
The current status of flight Allegiant Air G4983 is Arrived On time.
What type of aircraft is used for the Allegiant Air G4983 flight?
All Allegiant Air G4983 flights are operated using Boeing 737MAX 8 Passenger aircraft.
How many Allegiant Air G4983 flights are operated a week?
2 flights per week. The Flight Allegiant Air G4983 is operated on Thursday, Sunday.
