Departure: | Old Saybrook, CT |
Arrival: | Fort Edward, NY |
Fastest route: | 6h 43min |
Distance: | 276km |
Cheapest route: | $102 |
Transfers: | 2 |
Train companies: | Amtrak |
One Passenger / One Trip
11:31am
Old Saybrook, CT
Old Saybrook
6:14pm
Fort Edward, NY
Fort Edward-Glens Falls
6h 43min
++
$102
11:31am
Old Saybrook, CT
Old Saybrook
1:48pm
New York, NY
New York - Penn Station
2h 17min
Amtrak
Northeast Regional
$38
0h 32min layover
2:20pm
New York, NY
New York - Penn Station
4:50pm
Rensselaer, NY
Albany-Rensselaer
2h 30min
Amtrak
Ethan Allen Express
$44
0h 15min layover
5:05pm
Rensselaer, NY
Albany-Rensselaer
6:14pm
Fort Edward, NY
Fort Edward-Glens Falls
1h 9min
Amtrak
Ethan Allen Express
$20
The Old Saybrook - Fort Edward route has approximately 1 frequencies and its minimum duration is around 6h 43min. It is important you book your ticket in advance to avoid running out, since $102 tickets tend to run out quickly.
The distance between Old Saybrook and Fort Edward is around 276 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 6h 43min to cover 276 kilometers.
According to our data, the cheapest ticket costs $102 and leaves Old Saybrook. If you decide to make this journey you will have to do 2 stops before reaching Fort Edward-Glens Falls.
The last train leaves at 11:31am from Old Saybrook and arrives at 6:14pm at Fort Edward-Glens Falls. It will take 6h 43min, its price is $102 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 6h 43min