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Allegiant Air G41047 Flight Status Today
Domestic flight Allegiant Air G41047 takes off from Saint Petersburg (PIE) United States to Louisville (SDF) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 08:00 America/New_York. The flight is expected to land at Louisville International Airport-Standiford Field at 10:07 America/New_York. The flight will last about 2 hours 3 minutes.
G41047 - AAY1047
Allegiant Air (G4)
PIE
Saint Petersburg United States
SDF
Louisville United States
Contact Airlines
It may not operate on the date requested
Scheduled
08:00
None
11
Scheduled
10:07
None
A10
G41047 Detail
- Type: Domestic Flight
- Flight Duration: 2 hours 3 minutes
- Flight Distance: 1173 kms / 729 miles
Airline
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Saturday, Sunday
Passenger Services
- Service type: None
- Seats: None
- Freight capacity: None
- Passenger classes: None
More Detail
- Aircraft: Airbus A320
- Callsign: F-KTLH2
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Saint Petersburg: Wednesday 2026-02-11 07:15 AM
- Current Time in Louisville: Wednesday 2026-02-11 07:15 AM
Allegiant Air Flight G41047 FAQs
What is the scheduled flight duration for Allegiant Air G41047 flight?
On average, nonstop flight takes 2 hour(s) 3 minutes, with the flight distance of 1173 km (729 miles).
What type of aircraft is used for the Allegiant Air G41047 flight?
All Allegiant Air G41047 flights are operated using Airbus A320 aircraft.
How many Allegiant Air G41047 flights are operated a week?
2 flights per week. The Flight Allegiant Air G41047 is operated on Saturday, Sunday.