Distance between cities › USA › Vermont › Forest Dale, VT › Distance from Forest Dale to Ashland

Distance from Forest Dale, VT to Ashland, WV

Distance

812 mi

Time

13 hours 30 mins

Gas Cost

$78 - $152

There are 620.80 miles from Forest Dale to Ashland in southwest direction and 812 miles (1,306.79 kilometers) by car, following the I-81 S route.

Forest Dale and Ashland are 13 hours 30 mins far apart, if you drive non-stop .

This is the fastest route from Forest Dale, VT to Ashland, WV. The halfway point is Penbrook, PA.

Forest Dale, VT and Ashland, WV are in the same time zone (EDT). Current time in both locations is 11:58 pm.

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Gas Consumption and Emissions

A car with a fuel efficiency of MPG will need 32.61 gallons of gas to cover the route between Forest Dale, VT and Ashland, WV.

The estimated cost of gas to go from Forest Dale to Ashland is $110.09.

During the route, an average car will release 638.91 pounds of CO₂ to the atmosphere. The carbon footprint would be 0.79 pounds of CO₂ per mile.

Average USA gas price used for calculation is $3.38 per gallon of regular gas. Price last updated on April 17, 2024.
Read more about our CO₂ calculation.

Best Hotels In or Near Ashland, WV

Do you have where to stay when you arrive to Ashland, WV? Check out our hotel recommendations:

Halfway Point Between Forest Dale, VT and Ashland, WV

If you want to meet halfway between Forest Dale, VT and Ashland, WV or just make a stop in the middle of your trip, the exact coordinates of the halfway point of this route are 40.299938 and -76.839371, or 40º 17' 59.7768" N, 76º 50' 21.7356" W. This location is 406.01 miles away from Forest Dale, VT and Ashland, WV and it would take approximately 6 hours 45 mins to reach the halfway point from both locations.

Closest City or Town to Halfway Point

The closest town to the halfway point is Penbrook, PA, situated 404 miles from Forest Dale, VT and 412 miles from Ashland, WV. It would take 7 hours 1 min to go from Forest Dale to Penbrook and 6 hours 39 mins to go from Ashland to Penbrook.