HOBOKEN - The U.S. Department of Defense (DOD) has announced the scheduled July 4th flyover of New York City and the Hudson River, as part of its "salute to the Great Cities of the American Revolution."
Reportedly led by the U.S. Air Force Thunderbirds, the flyover will also consist of F-15 and F-22 fighters U.S. Marine Corps F-35 fighters, as well as USAF B-1, B-2, and B-52 bombers.
According to the DOD, "The flyovers provide an opportunity for DOD to demonstrate the capabilities and professionalism of the United States Armed Forces. Flying hours are a sunk cost for the Department of Defense, and these aircraft and crews would be using these hours for proficiency and training at other locations if they were not conducting these flyovers."
July 4th is Independence Day in America, celebrating the date on which the Declaration of Independence was signed in 1776.
"DOD is proud to help celebrate the nation’s 244th birthday. We are grateful for our nation’s support as we defend our country, 24 hours a day, 7 days a week, 365 days a year."
The aircraft are scheduled to fly south along the Hudson River toward the Statue of Liberty at approximately 5:00 p.m. on Saturday, July 4. Current weather forecasts call for clear skies.
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