Amapola HP6048 Flight Status Today

Domestic flight Amapola HP6048 takes off from Houston (IAH) United States to Oklahoma City (OKC) United States. It's operated by Amapola. The plane leaves George Bush Intercontinental Airport at 21:00 America/Chicago. The flight is expected to land at Will Rogers World Airport at 22:35 America/Chicago. The flight will last about 1 hours 35 minutes.

HP6048 - APF6048 Amapola (HP)
IAH Houston United States
Flight IAH to OKC
OKC Oklahoma City United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure George Bush Intercontinental Airport
IATA: IAHICAO: KIAH
Scheduled 21:00
None None
Arrival Will Rogers World Airport
IATA: OKCICAO: KOKC
Scheduled 22:35
None None
HP6048 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 35 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 Houston: Tuesday 2026-03-03 16:49 PM
  • Current Time in Oklahoma City: Tuesday 2026-03-03 16:49 PM

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Amapola Flight HP6048 FAQs

What is the scheduled flight duration for Amapola HP6048 flight?

On average, nonstop flight takes 1 hour(s) 35 minutes, with the flight distance of 636 km (395 miles).