Information about the train from Auburn to Portland. The train is one of the safest modes of transportation in existence, and offers a comfortable atmosphere for you to relax on your journey from Auburn to Portland. Best of all, getting from Auburn to Portland is budget-friendly, with train tickets starting at just $403. This is an estimate, so please contact the train ticket seller directly for precise information.
According to our database, the best route to go from Auburn to Portland is through Montgomery and Carbondale. Take a look at the route we propose for this trip. We show you all available schedules for each train trip.
You can find the most accurate information for the route Auburn AL - Montgomery AL by clicking here
Montgomery, AL (MGM)
Carbondale, IL (CDL)
25h 26m
$189
Amtrak
Carbondale Station
Portland Station
56h 54m
$214
Amtrak
Carbondale Station
Portland Station
52h 40m
$214
Amtrak
If you want to get cheap train tickets from Auburn to Portland we recommend that you book in advance as the best Amtrak tickets sell out fast.The price of the ticket is usually $403.Remember that you must also add the cost of the journey: Auburn -> Montgomery
The first trains for the route we propose are:
Auburn -> Montgomery: -
Montgomery -> Carbondale: 01:45
Carbondale -> Portland: 03:16
The companies that can help you are: Amtrak.
Each company has its rules and depending on the ticket, price, and offer different refund policies apply. We recommend that you contact the company where you bought the ticket to get a solution.
In Goticket we try to give a solution for every trip the user asks. Sometimes is not possible but we do our best to group different train direct routes to in order to create a multi-step journey that satisfies the user. This is the route we recommend::
1.- Auburn -> Montgomery
2.- Montgomery -> Carbondale
3.- Carbondale -> Portland
The approximate distance between the two places is 4035 km. With the route we propose, it will take approximately 78h 6m (plus the time it takes to you to make: Auburn -> Montgomery).