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Allegiant Air G42958 Flight Status Today
Domestic flight Allegiant Air G42958 takes off from Knoxville (TYS) United States to Orlando (SFB) United States. It's operated by Allegiant Air. The plane leaves McGhee Tyson Airport at 22:04 America/New_York. The flight is expected to land at Orlando Sanford International Airport at 23:39 America/New_York. The flight will last about 1 hours 43 minutes.
G42958 - AAY2958
Allegiant Air (G4)
TYS
Knoxville United States
SFB
Orlando United States
Contact Airlines
It may not operate on the date requested
G42958 Detail
- Type: Domestic Flight
- Flight Duration: 1 hours 43 minutes
- Flight Distance: 821 kms / 510 miles
Airline
- Allegiant Air
- IATA: G4
- ICAO: AAY
- Operating Days: Wednesday
Passenger Services
- Service type: None
- Seats: None
- Freight capacity: None
- Passenger classes: None
More Detail
- Aircraft: Boeing 737MAX 8 Passenger
- Callsign: F-KCHS2
- Departure Timezone: America/New_York
- Arrival Timezone: America/New_York
- Current Time in Knoxville: Monday 2026-01-19 05:12 AM
- Current Time in Orlando: Monday 2026-01-19 05:12 AM
Flight routes similar to Allegiant Air G42958
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G41420 Allegiant Air |
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G43033 Allegiant Air |
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G43043 Allegiant Air |
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G41418 Allegiant Air |
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G41434 Allegiant Air |
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Allegiant Air Flight G42958 FAQs
What is the scheduled flight duration for Allegiant Air G42958 flight?
On average, nonstop flight takes 1 hour(s) 43 minutes, with the flight distance of 821 km (510 miles).
What type of aircraft is used for the Allegiant Air G42958 flight?
All Allegiant Air G42958 flights are operated using Boeing 737MAX 8 Passenger aircraft.
How many Allegiant Air G42958 flights are operated a week?
1 flights per week. The Flight Allegiant Air G42958 is operated on Wednesday.
