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. Following several unsuccessful searches, the
bomb was presumed lost somewhere in Wassaw Sound off the shores of Tybee

The B-47
bomber was on a simulated combat mission from Homestead Air Force Base in
Florida. It was carrying a single 7,600-pound (3,400 kg) bomb. At about 2:00
AM, an F-86 fighter collided with the B-47. The F-86 crashed, after the pilot
ejected from the plane. The damaged B-47 remained airborne, plummeting 18,000
feet (5,500 m) from flight level 380 (38,000 feet (12,000 m)) when Major
Richardson regained flight control.

The crew
requested permission to jettison the bomb, in order to reduce weight and
prevent the bomb from exploding during an emergency landing. Permission was
granted, and the bomb was jettisoned at 7,200 feet (2,200 m) while the bomber
was traveling at about 200 knots (370 km/h). The crew did not see an explosion
when the bomb struck the sea. They managed to land the B-47 safely at the
nearest base, Hunter Air Force Base. The pilot, Colonel Howard Richardson, was
awarded the Distinguished Flying Cross after this incident.

Declassified Documents

Aircraft Mishap Report, 5 February 1958
[154 Pages, 64.7MB]

Bir cevap yazın

E-posta hesabınız yayımlanmayacak. Gerekli alanlar * ile işaretlenmişlerdir