Swift Air WQ438 Flight Status Today

Domestic flight Swift Air WQ438 takes off from Columbus (CSG) United States to Alexandria (AEX) United States. It's operated by Swift Air. The plane leaves Columbus Airport at 18:15 America/New_York. The flight is expected to land at Alexandria International Airport at 18:32 America/Chicago. The flight will last about 4 hours 10 minutes.

WQ438 - SWQ438 Swift Air (WQ)
CSG Columbus United States
Flight CSG to AEX
AEX Alexandria United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 16-March-2017
Departure Columbus Airport
IATA: CSGICAO: KCSG
Scheduled 18:15
Arrival Alexandria International Airport
IATA: AEXICAO: KAEX
Scheduled 18:32
WQ438 Detail
  • Type: Domestic Flight
  • Flight Duration: 4 hours 10 minutes
  • Flight Distance: 731 kms / 454 miles
Airline
  • Swift Air
  • IATA: WQ
  • ICAO: SWQ
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft:
  • Callsign: None
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/Chicago
  • Current Time in Columbus: Monday 2026-01-19 15:44 PM
  • Current Time in Alexandria: Monday 2026-01-19 14:44 PM

Swift Air Flight WQ438 FAQs

What is the scheduled flight duration for Swift Air WQ438 flight?

On average, nonstop flight takes 4 hour(s) 10 minutes, with the flight distance of 731 km (454 miles).