Delta DL4758 Flight Status Today

Domestic flight Delta DL4758 takes off from New York (JFK) United States to Cleveland (CLE) United States. It's operated by Delta Airlines. The plane leaves John F. Kennedy International Airport at 15:00 America/New_York. The flight is expected to land at Cleveland Hopkins International Airport at 17:09 America/New_York. The flight will last about 1 hours 17 minutes.

DL4758 - DAL4758 Delta Airlines (DL)
JFK New York United States
Flight JFK to CLE
CLE Cleveland United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 30-April-2022
Departure John F. Kennedy International Airport
IATA: JFKICAO: KJFK
Scheduled 15:00
4 A15
Arrival Cleveland Hopkins International Airport
IATA: CLEICAO: KCLE
Scheduled 17:09
TERM B6
DL4758 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 17 minutes
  • Flight Distance: 682 kms / 424 miles
Airline
  • Delta Airlines
  • IATA: DL
  • ICAO: DAL
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 69
  • Freight capacity: 1.9 tons
  • Passenger classes: First Class, Economy, Premium Economy
More Detail
  • Aircraft: Mitsubishi CRJ-900LR
  • Callsign: F-KHGR1
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in New York: Sunday 2026-01-18 09:10 AM
  • Current Time in Cleveland: Sunday 2026-01-18 09:10 AM

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Delta Flight DL4758 FAQs

What is the scheduled flight duration for Delta DL4758 flight?

On average, nonstop flight takes 1 hour(s) 17 minutes, with the flight distance of 682 km (424 miles).

What type of aircraft is used for the Delta DL4758 flight?

All Delta DL4758 flights are operated using Mitsubishi CRJ-900LR aircraft.

Which terminal the flight Delta DL4758 is arriving at?

Flight Delta DL4758 arrives in Cleveland Hopkins International Airport at Terminal TERM.