JetSuiteX XE4973 Flight Status Today

Domestic flight JetSuiteX XE4973 takes off from Oklahoma City (OKC) United States to Houston (IAH) United States. It's operated by JSX. The plane leaves Will Rogers World Airport at 13:35 America/Chicago. The flight is expected to land at George Bush Intercontinental Airport at None America/Chicago. The flight will last about 1 hours 9 minutes.

XE4973 - JSX4973 JSX (XE)
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 13:35
None None
Arrival George Bush Intercontinental Airport
IATA: IAHICAO: KIAH
Scheduled None
None None
XE4973 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 9 minutes
  • Flight Distance: 636 kms / 395 miles
Airline
  • JSX
  • IATA: XE
  • ICAO: JSX
  • 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: Embraer ERJ-145LR
  • Callsign: UCA4973
  • Departure Timezone: America/Chicago
  • Arrival Timezone: America/Chicago
  • Current Time in Oklahoma City: Saturday 2026-05-02 09:10 AM
  • Current Time in Houston: Saturday 2026-05-02 09:10 AM

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JetSuiteX Flight XE4973 FAQs

What is the scheduled flight duration for JetSuiteX XE4973 flight?

On average, nonstop flight takes 1 hour(s) 9 minutes, with the flight distance of 636 km (395 miles).

What type of aircraft is used for the JetSuiteX XE4973 flight?

All JetSuiteX XE4973 flights are operated using Embraer ERJ-145LR aircraft.