Departure: | Harpers Ferry, WV |
Arrival: | Cleveland, OH |
Fastest route: | 9h 37min |
Distance: | 475km |
Cheapest route: | $52 |
Transfers: | Between 0 and 1 |
Train companies: | Amtrak |
One Passenger / One Trip
5:16pm
Harpers Ferry, WV
Harpers Ferry
2:53am
Cleveland, OH
Cleveland
9h 37min
Amtrak
Capitol Limited
$52
5:16pm
Harpers Ferry, WV
Harpers Ferry
2:53am
Cleveland, OH
Cleveland
9h 37min
+
$57
5:16pm
Harpers Ferry, WV
Harpers Ferry
7:17pm
Cumberland, MD
Cumberland
2h 1min
Amtrak
Capitol Limited
$15
0h 7min layover
7:24pm
Cumberland, MD
Cumberland
2:53am
Cleveland, OH
Cleveland
7h 29min
Amtrak
Capitol Limited
$42
The Harpers Ferry - Cleveland route has approximately 2 frequencies and its minimum duration is around 9h 37min. It is important you book your ticket in advance to avoid running out, since $52 tickets tend to run out quickly.
The distance between Harpers Ferry and Cleveland 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 $52 and leaves Harpers Ferry. You will not have to do any transfers, the trip will go direct to Cleveland.
The last train leaves at 5:16pm from Harpers Ferry and arrives at 2:53am at Cleveland. It will take 9h 37min, its price is $57 and the number of changes will be 1.
Yes, there are direct train routes, their duration is usually around 9h 37min and the price is $52.