AA7258 — Flight status: Guayaquil

American Airlines AA7258

Guayaquil — Galapagos Islands

José Joaquín de Olmedo International Airport (GYE)		Seymour Airport (GPS)
Usual departure time		Usual arrival time
12:00		12:57

Operating days: Fri. Usual times and days are based on observations in 2018–2020. The schedule may have changed.

Flight status AA7258

Flight American Airlines AA7258 (Guayaquil — Galapagos Islands) running from the airport José Joaquín de Olmedo International Airport (GYE), to the airport Seymour Airport (GPS). The flight is operated by the family's aircraft Airbus A319. Flight distance is 1170 km / 631 miles, and the average flight length is 1 h. 57 min..

Common questions about flight AA7258

What time does flight AA7258 depart?

Scheduled departure is 12:00 and arrival 12:57, local time at each airport. Operating days: Fri.

How long does flight AA7258 take?

On average 1 h 57 min. The airports are 1170 km / 631 miles apart.

Which aircraft operates flight AA7258?

The flight is operated by the family's aircraft Airbus A319.

Low Price Calendar Guayaquil — Galapagos Islands

Types of aircraft flying AA7258 American Airlines: Guayaquil — Galapagos Islands

Airbus A319

100%

This section contains information about the types of aircraft and flight numbers that perform the flight. Based on observations in 2018–2020.

Other flights on the route Guayaquil — Galapagos Islands

Flight No.	Departure time*	Arrival time*	Aircraft
CM3452	10:40	11:30	Airbus A319
AA7254	10:20	11:20	Airbus A319
CM3450	08:10	09:00	Embraer 190

* Local time

All data is taken from open sources. The administration of the portal is not responsible for the accuracy of the information. For exact flight information, see the airline's website or the airport's website.