Ready to head out on vacation after the long pandemic-era drought — but not looking to go broke in the process?
Online airport parking platform ParkSleepFly has ranked the most — and least — affordable choices out of 51 U.S. destinations for vacationers, weighing six metrics ranging from the average price of a beer to the typical cost of a night’s accommodations.
The capital of Oklahoma, the aptly named Oklahoma City, came it at No. 1, ParkSleepFly found, with a score of 8.58 out of a possible 10. Thanks to great deals like beers for as little as $3, $11.50 restaurant meals and $106 for a hotel room, “Oklahoma City tourists could extend their vacations almost three times longer than if visiting New York for exactly the same price,” said ParkSleepFly CEO Martin Jones.
“Despite not being the top of most people’s bucket lists, Oklahoma City is a great place to visit,” Jones added, noting that rodeos, horseback riding and roping and herding cattle are all popular local activities, among other, more obviously urban pastimes. “Known for its cowboy culture, the city is a must-visit for tourists wanting to experience the Old West.”
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Speaking of the Old West, three cities in Texas — San Antonio, Houston and Fort Worth — made the top 10 for affordability, ranking fifth, sixth and seventh, respectively. The other affordable destinations (in descending order of affordability) included Indianapolis, Indiana, at No. 2; Tucson, Arizona (3); Memphis, Tennessee (4); Louisville, Kentucky (8); Orlando, Florida (9); and Raleigh, North Carolina (10).
What cities should you avoid to steer clear of high vacation costs? No surprises there, said Jones. The five most expensive U.S. destinations included, at No. 1, New York City (in general), and, at No. 4, its constituent borough of Brooklyn. Their affordability scores were just 2.56 and 3.76, respectively, out of 10.
