Matches in DBpedia 2016-04 for { ?s ?p "The Tybee Island B-47 crash was an incident on February 5, 1958, in which the United States Air Force lost a 7,600-pound (3,400 kg) Mark 15 nuclear bomb in the waters off Tybee Island near Savannah, Georgia, United States. During a practice exercise, an F-86 fighter plane collided with the B-47 bomber carrying the bomb. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned."@en }
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- 1958_Tybee_Island_mid-air_collision comment "The Tybee Island B-47 crash was an incident on February 5, 1958, in which the United States Air Force lost a 7,600-pound (3,400 kg) Mark 15 nuclear bomb in the waters off Tybee Island near Savannah, Georgia, United States. During a practice exercise, an F-86 fighter plane collided with the B-47 bomber carrying the bomb. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned.".
- Q1514690 comment "The Tybee Island B-47 crash was an incident on February 5, 1958, in which the United States Air Force lost a 7,600-pound (3,400 kg) Mark 15 nuclear bomb in the waters off Tybee Island near Savannah, Georgia, United States. During a practice exercise, an F-86 fighter plane collided with the B-47 bomber carrying the bomb. To protect the aircrew from a possible detonation in the event of a crash, the bomb was jettisoned.".