Amapola HP6258 Flight Status

HP6258 - APF6258 Amapola (HP)
IAH Houston United States
Flight IAH to CLE
CLE Cleveland United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure George Bush Intercontinental Airport
IATA: IAHICAO: KIAH
Scheduled 10:45
None None
Arrival Cleveland Hopkins International Airport
IATA: CLEICAO: KCLE
Scheduled 14:11
None None
HP6258 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 26 minutes
  • Flight Distance: 1757 kms / 1092 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/New_York
  • Current Time in Houston: Thursday 2025-12-11 14:42 PM
  • Current Time in Cleveland: Thursday 2025-12-11 15:42 PM

Domestic flight Amapola HP6258 takes off from Houston (IAH) United States to Cleveland (CLE) United States. It's operated by Amapola. The plane leaves George Bush Intercontinental Airport at 10:45 America/Chicago. The flight is expected to land at Cleveland Hopkins International Airport at 14:11 America/New_York. The flight will last about 2 hours 26 minutes.

Similar flight route from Houston (IAH) to Cleveland (CLE)

Airline Flight no Departure Arrival
United Airlines UA544
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United Airlines UA1934
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Air New Zealand NZ6413
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Air New Zealand NZ6411
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Emirates EK6445
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11/12/2025
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22:10
Air New Zealand NZ6410
Air New Zealand
11/12/2025
07:20
2 hours 44 minutes
11:04
Air New Zealand NZ6412
Air New Zealand
11/12/2025
18:50
2 hours 48 minutes
22:38
Air New Zealand NZ6414
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11/12/2025
13:58
2 hours 56 minutes
17:54

Frequently asked questions, answered

What is the scheduled flight duration for Amapola HP6258 flight?

On average, nonstop flight takes 2 hour(s) 26 minutes, with the flight distance of 1757 km (1092 miles).