Allegiant Air G4978 Flight Status Today

Domestic flight Allegiant Air G4978 takes off from Saint Petersburg (PIE) United States to Cedar Rapids (CID) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 06:30 America/New_York. The flight is expected to land at The Eastern Iowa Airport at 08:26 America/Chicago. The flight will last about 2 hours 55 minutes.

G4978 - AAY978 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to CID
CID Cedar Rapids United States
Contact Airlines It may not operate on the date requested
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 06:30
None 9
Arrival The Eastern Iowa Airport
IATA: CIDICAO: KCID
Scheduled 08:26
None 1
G4978 Detail
  • Type: Domestic Flight
  • Flight Duration: 2 hours 55 minutes
  • Flight Distance: 1753 kms / 1089 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: Monday, Thursday, Friday, Sunday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 737MAX 8 Passenger
  • Callsign: F-KTPA2
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/Chicago
  • Current Time in Saint Petersburg: Thursday 2026-02-12 08:19 AM
  • Current Time in Cedar Rapids: Thursday 2026-02-12 07:19 AM

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Allegiant Air Flight G4978 FAQs

What is the scheduled flight duration for Allegiant Air G4978 flight?

On average, nonstop flight takes 2 hour(s) 55 minutes, with the flight distance of 1753 km (1089 miles).

What type of aircraft is used for the Allegiant Air G4978 flight?

All Allegiant Air G4978 flights are operated using Boeing 737MAX 8 Passenger aircraft.

How many Allegiant Air G4978 flights are operated a week?

4 flights per week. The Flight Allegiant Air G4978 is operated on Monday, Thursday, Friday, Sunday.