Southwest Airlines WN6807 Flight Status Today

Domestic flight Southwest Airlines WN6807 takes off from Atlanta (ATL) United States to Columbus (CMH) United States. It's operated by Southwest Airlines. The plane leaves Hartsfield-Jackson Atlanta International Airport at 08:15 America/New_York. The flight is expected to land at John Glenn Columbus International Airport at 09:50 America/New_York. The flight will last about 1 hours 35 minutes.

WN6807 - SWA6807 Southwest Airlines (WN)
ATL Atlanta United States
Flight ATL to CMH
CMH Columbus United States
Contact Airlines It may not operate on the date requested
Departure Hartsfield-Jackson Atlanta International Airport
IATA: ATLICAO: KATL
Scheduled 08:15
N None
Arrival John Glenn Columbus International Airport
IATA: CMHICAO: KCMH
Scheduled 09:50
None None
WN6807 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 35 minutes
  • Flight Distance: 719 kms / 447 miles
Airline
  • Southwest Airlines
  • IATA: WN
  • ICAO: SWA
  • Operating Days: Sunday
Passenger Services
  • Service type: Normal passenger
  • Seats: 175
  • Freight capacity: 8.7 tons
  • Passenger classes: Economy, Premium Economy
More Detail
  • Aircraft: Boeing 737-800 (winglets) Passenger/BBJ2
  • Callsign: F-KOKC2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Atlanta: Sunday 2026-01-18 20:51 PM
  • Current Time in Columbus: Sunday 2026-01-18 20:51 PM

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Southwest Airlines Flight WN6807 FAQs

What is the scheduled flight duration for Southwest Airlines WN6807 flight?

On average, nonstop flight takes 1 hour(s) 35 minutes, with the flight distance of 719 km (447 miles).

What type of aircraft is used for the Southwest Airlines WN6807 flight?

All Southwest Airlines WN6807 flights are operated using Boeing 737-800 (winglets) Passenger/BBJ2 aircraft.

How many Southwest Airlines WN6807 flights are operated a week?

1 flights per week. The Flight Southwest Airlines WN6807 is operated on Sunday.