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Allegiant Air G41088 Flight Status Today
Domestic flight Allegiant Air G41088 takes off from Saint Petersburg (PIE) United States to Clarksburg (CKB) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 16:18 23 March 2026 America/New_York. The flight is expected to land at Benedum Airport at 18:00 23 March 2026 America/New_York. The flight will last about 1 hours 42 minutes.
Current status of G41088 is Scheduled On time Expected Depart in 7 hours, 36 minutes
- Type: Domestic Flight
- Flight Duration: 1 hours 42 minutes
- Flight Distance: 1283 kms / 797 miles
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Saturday
- Service type: N/A
- Seats: N/A
- Freight capacity: N/A
- Passenger classes: N/A
- Aircraft: Airbus A319
- Callsign: F-KGNV1
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Saint Petersburg: Monday 2026-03-23 08:42 AM
- Current Time in Clarksburg: Monday 2026-03-23 08:42 AM
Allegiant Air Flight G41088 FAQs
What is the scheduled flight duration for Allegiant Air G41088 flight?
On average, nonstop flight takes 1 hour(s) 42 minutes, with the flight distance of 1283 km (797 miles).
What is the current status of Allegiant Air G41088 flight?
The current status of flight Allegiant Air G41088 is Scheduled On time.
What type of aircraft is used for the Allegiant Air G41088 flight?
All Allegiant Air G41088 flights are operated using Airbus A319 aircraft.
How many Allegiant Air G41088 flights are operated a week?
1 flights per week. The Flight Allegiant Air G41088 is operated on Saturday.
