| Train Ticket Price: | $289 |
|---|---|
| Avg. Train Duration: | 58h 31m |
| Train Companies: | Amtrak |
| Daily Trains: | 2 |
| Trains depart from: | Oregon City |
| Trains arrive in: | Newton |
Information about the train from Oregon City to Newton. 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 Oregon City to Newton. Best of all, getting from Oregon City to Newton is budget-friendly, with train tickets starting at just $289. 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 Oregon City to Newton is through Portland. Take a look at the route we propose for this trip. We show you all available schedules for each train trip.
Oregon City Station
Portland Station
41m
$5
Amtrak
Oregon City Station
Portland Station
41m
$5
Amtrak
Portland, OR - Union Station (PDX)
Newton, KS (NEW)
57h 50m
$548
Amtrak
If you want to get cheap train tickets from Oregon City to Newton we recommend that you book in advance as the best Amtrak tickets sell out fast.The cheapest ticket is usually $289 and the most expensive one to go to Newton is approximately $553. .
The first train leaves at 10:54 from Oregon City and costs $5 while the last one arriving at Newton costs $548 and it is at 02:15.
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.- Oregon City -> Portland
2.- Portland -> Newton
The approximate distance between the two places is 2596 km. With the route we propose, it will take approximately 58h 31m.