Amapola HP6224 Flight Status Today

Domestic flight Amapola HP6224 takes off from Oklahoma City (OKC) United States to Houston (IAH) United States. It's operated by Amapola. The plane leaves Will Rogers World Airport at 12:34 America/Chicago. The flight is expected to land at George Bush Intercontinental Airport at 14:05 America/Chicago. The flight will last about 1 hours 31 minutes.

HP6224 - APF6224 Amapola (HP)
OKC Oklahoma City United States
Flight OKC to IAH
IAH Houston United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure Will Rogers World Airport
IATA: OKCICAO: KOKC
Scheduled 12:34
None None
Arrival George Bush Intercontinental Airport
IATA: IAHICAO: KIAH
Scheduled 14:05
None None
HP6224 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 31 minutes
  • Flight Distance: 636 kms / 395 miles
Airline
  • Amapola
  • IATA: HP
  • ICAO: APF
  • 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: None
  • Callsign: None
  • Departure Timezone: America/Chicago
  • Arrival Timezone: America/Chicago
  • Current Time in Oklahoma City: Monday 2026-01-19 06:55 AM
  • Current Time in Houston: Monday 2026-01-19 06:55 AM

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Amapola Flight HP6224 FAQs

What is the scheduled flight duration for Amapola HP6224 flight?

On average, nonstop flight takes 1 hour(s) 31 minutes, with the flight distance of 636 km (395 miles).