Departure: | Harpers Ferry, WV |
Arrival: | Cleveland, OH |
Fastest route: | 8h 37min |
Distance: | 475km |
Cheapest route: | $83 |
Transfers: | Between 0 and 1 |
Train companies: | Amtrak |
One Passenger / One Trip
5:16pm
Harpers Ferry, WV
Harpers Ferry
3:53am
Cleveland, OH
Cleveland
8h 37min
Amtrak
Capitol Limited
$83
5:16pm
Harpers Ferry, WV
Harpers Ferry
3:53am
Cleveland, OH
Cleveland
8h 37min
+
$91
5:16pm
Harpers Ferry, WV
Harpers Ferry
7:17pm
Cumberland, MD
Cumberland
2h 1min
Amtrak
Capitol Limited
$24
0h 7min layover
7:24pm
Cumberland, MD
Cumberland
3:53am
Cleveland, OH
Cleveland
6h 29min
Amtrak
Capitol Limited
$67
The Harpers Ferry - Cleveland route has approximately 2 frequencies and its minimum duration is around 8h 37min. It is important you book your ticket in advance to avoid running out, since $83 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 8h 37min to cover 475 kilometers.
According to our data, the cheapest ticket costs $83 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 3:53am at Cleveland. It will take 8h 37min, its price is $91 and the number of changes will be 1.
Yes, there are direct train routes, their duration is usually around 8h 37min and the price is $83.