Flight G4 1075: Roanoke → St. Petersburg-Clearwater
Allegiant Air4 October 2026LandedUpdated 47 min ago, official airport data
Flight G4 1075, 4 October 2026
All flights Roanoke → St. Petersburg-ClearwaterLanded
Flight time: 2 h 04 min
Data from the official airport boards. Times are local to each airport.
- Flight time
- 2 h 04 min
- Distance
- ≈ 1,077 km
- Days of the week
Punctuality of flight G4 1075
The last 60 days of boards hold fewer than 5 finished flights with a known departure time: too few to count punctuality.
Flight G4 1075: schedule and upcoming dates
- Days it flew in the last 60 days
- Usual departure time
- 14:17
- Usual arrival time
- 16:21
Dates already on the boards
- Sun 4 OctDeparture 14:17, Arrival 16:21Landed
Route of flight G4 1075
Departure airport: Roanoke (ROA). Arrival airport: St. Petersburg-Clearwater (PIE). Scheduled flight time: about 2 h 04 min. Distance between the airports: about 1,077 km in a straight line.
See also
Questions about flight G4 1075
Is flight G4 1075 on time today?
For 4 October 2026 the airport board shows flight G4 1075 as “Landed”. Departure: Roanoke, 14:26. Arrival: St. Petersburg-Clearwater, 16:21.
Which terminal and gate does flight G4 1075 use?
Roanoke (ROA): Gate 6. Gates are announced shortly before boarding and can change: check the screens at the airport.
How often is flight G4 1075 delayed?
The last 60 days of boards hold fewer than 5 finished flights with a known departure time: too few to count punctuality.
How long does flight G4 1075 take?
About 2 h 04 min by the schedule. The airports are 1077 km apart.
On which days does flight G4 1075 operate?
In the last 60 days it was on the boards on: Sunday.