Allegiant Air G41410 Flight Status Today

Domestic flight Allegiant Air G41410 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 17:46 America/New_York. The flight is expected to land at Orlando Sanford International Airport at 19:21 America/New_York. The flight will last about 1 hours 35 minutes.

G41410 - AAY1410 Allegiant Air (G4)
TYS Knoxville United States
Flight TYS to SFB
SFB Orlando United States
Contact Airlines It may not operate on the date requested
Departure McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 17:46
None 4
Arrival Orlando Sanford International Airport
IATA: SFBICAO: KSFB
Scheduled 19:21
None 11
G41410 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 35 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: Airbus A320 (sharklets)
  • Callsign: F-KFLL2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Knoxville: Sunday 2026-03-22 19:34 PM
  • Current Time in Orlando: Sunday 2026-03-22 19:34 PM

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Allegiant Air Flight G41410 FAQs

What is the scheduled flight duration for Allegiant Air G41410 flight?

On average, nonstop flight takes 1 hour(s) 35 minutes, with the flight distance of 821 km (510 miles).

What type of aircraft is used for the Allegiant Air G41410 flight?

All Allegiant Air G41410 flights are operated using Airbus A320 (sharklets) aircraft.

How many Allegiant Air G41410 flights are operated a week?

1 flights per week. The Flight Allegiant Air G41410 is operated on Wednesday.