Distance between cities › USA › Oklahoma › Jay, OK › Distance from Jay to Ace Book Hotels

Distance from Jay, OK to Ace, TX

Distance

491 mi

Time

9 hours 27 mins

Gas Cost

$46 - $89

There are 406.61 miles from Jay to Ace in south direction and 491 miles (790.19 kilometers) by car, following the US-64 and US-69 route.

Jay and Ace are 9 hours 27 mins far apart, if you drive non-stop .

This is the fastest route from Jay, OK to Ace, TX. The halfway point is Pattonville, TX.

Jay, OK and Ace, TX are in the same time zone (CDT). Current time in both locations is 2:19 am.

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Lowest Price Rental Cars in Jay, OK

Planning on renting a car to go from Jay, OK to Ace, TX? Here there are some offers to rent a car in Jay, OK

Small

From $14

Intermediate

From $18

Full-size

From $24

SUVs

From $28

Vans

From $40

Compare rental car prices in Jay »

Gas Consumption and Emissions

A car with a fuel efficiency of MPG will need 19.72 gallons of gas to cover the route between Jay, OK and Ace, TX.

The estimated cost of gas to go from Jay to Ace is $64.32.

During the route, an average car will release 386.33 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 $3.26 per gallon of regular gas. Price last updated on March 12, 2026.
Read more about our CO₂ calculation.

Halfway Point Between Jay, OK and Ace, TX

If you want to meet halfway between Jay, OK and Ace, TX or just make a stop in the middle of your trip, the exact coordinates of the halfway point of this route are 33.621094 and -95.452446, or 33º 37' 15.9384" N, 95º 27' 8.8056" W. This location is 245.50 miles away from Jay, OK and Ace, TX and it would take approximately 4 hours 43 mins to reach the halfway point from both locations.

Closest City or Town to Halfway Point

The closest town to the halfway point is Pattonville, TX, situated 248 miles from Jay, OK and 238 miles from Ace, TX. It would take 4 hours 18 mins to go from Jay to Pattonville and 4 hours 42 mins to go from Ace to Pattonville.