| Departure: | Cleveland, OH |
| Arrival: | Harpers Ferry, WV |
| Fastest route: | 9h 37min |
| Distance: | 475km |
| Cheapest route: | $65 |
| Transfers: | 0 |
| Train companies: | Amtrak |
One Passenger / One Trip
1:54am
Cleveland, OH
Cleveland
11:31am
Harpers Ferry, WV
Harpers Ferry
9h 37min
Amtrak
Capitol Limited
$65
The Cleveland - Harpers Ferry route has approximately 1 frequencies and its minimum duration is around 9h 37min. It is important you book your ticket in advance to avoid running out, since $65 tickets tend to run out quickly.
The distance between Cleveland and Harpers Ferry is around 475 kilometers and bus companies that can help you in your journey are: Amtrak.
The train journey may vary depending on the stops. The minimum duration is usually around 9h 37min to cover 475 kilometers.
According to our data, the cheapest ticket costs $65 and leaves Cleveland. You will not have to do any transfers, the trip will go direct to Harpers Ferry.
The last train leaves at 1:54am from Cleveland and arrives at 11:31am at Harpers Ferry. It will take 9h 37min, its price is $65 and the number of changes will be 0.
Yes, there are direct train routes, their duration is usually around 9h 37min and the price is $65.