Roy E. Halsey rescued Roger B. Turrell from burning, Middletown, Ohio, February 7, 1963. A single-engine airplane piloted by Turrell, 35, lawyer, and with Halsey, 33, accounting supervisor, as the only passenger, crashed at take-off, falling from a height of 15 feet into a parking lot where it stopped with its left side against an unoccupied automobile. Of the plane’s two fuel tanks containing 40 gallons of gasoline, the one in the left wing was ruptured, and the escaping fuel caught fire. Halsey, who had sustained minor injuries, left through the plane’s right door and ran 20 feet. Noting that Turrell still was in the aircraft, on which flames eight to 10 feet high burned on the left wing and the outside of the cabin, Halsey ran back to the open door. Turrell, who had sustained rib fractures, was trying vainly to unfasten his seat belt as flames extended into the cabin through the broken left window. Halsey stepped into the plane and released Turrell’s seat belt. He also beat out flames on Turrell’s coat, sustaining burns to his hand. Halsey drew Turrell to the doorway, stepped out, and pulled him from the cabin. Turrell was too weak to stand alone. As Halsey aided him away from the plane, the flames reached the other fuel tank, causing the fire to spread more rapidly over the aircraft. Firemen put out the flames. Halsey and Turrell were hospitalized for their injuries and burns. Both recovered.
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