Departure: | Cleveland, OH |
Arrival: | Somerdale, NJ |
Fastest route: | 14h 4min |
Distance: | 680km |
Cheapest route: | $78.25 |
Transfers: | 2 |
Train companies: | Amtrak |
One Passenger / One Trip
1:54am
Cleveland, OH
Cleveland
3:58pm
Somerdale, NJ
Lindenwold - NJ Transit Station
14h 4min
++
$78.25
1:54am
Cleveland, OH
Cleveland
5:05am
Pittsburgh, PA
Pittsburgh
3h 11min
Amtrak
Capitol Limited
$23
2h 25min layover
7:30am
Pittsburgh, PA
Pittsburgh
2:59pm
Philadelphia, PA
Philadelphia - 30th Street Station
7h 29min
Amtrak
Pennsylvanian
$50
0h 20min layover
3:19pm
Philadelphia, PA
Philadelphia - 30th Street Station
3:58pm
Somerdale, NJ
Lindenwold - NJ Transit Station
0h 39min
Amtrak
NJ Transit Train
$5.25
The Cleveland - Somerdale route has approximately 1 frequencies and its minimum duration is around 14h 4min. It is important you book your ticket in advance to avoid running out, since $78.25 tickets tend to run out quickly.
The distance between Cleveland and Somerdale is around 680 kilometers and bus companies that can help you in your journey are: Amtrak.
Remember that the number of transfers to be made will be at least 2 so in some cases you should book the tickets separately.
The train journey may vary depending on the stops. The minimum duration is usually around 14h 4min to cover 680 kilometers.
According to our data, the cheapest ticket costs $78.25 and leaves Cleveland. If you decide to make this journey you will have to do 2 stops before reaching Lindenwold - NJ Transit Station.
The last train leaves at 1:54am from Cleveland and arrives at 3:58pm at Lindenwold - NJ Transit Station. It will take 14h 4min, its price is $78.25 and the number of changes will be 2.
We do not have direct routes in our database. The minimum number of transfers will be 2 transshipments and the total duration of the trip will be approximately 14h 4min