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2007-04-13 04:20:20 · 4 answers · asked by dwijesh91 1 in Science & Mathematics Engineering

4 answers

It depends on the yield and the height of detonation. There is an optimum height for every yield that maximizes blast damage.

Yield scales as the 3rd-root. As a reference, for an 1 kT blast (from optimmum height) there is 10 psi overpressure at appriximately 1100 feet from ground-zero.

Scaling that up to a 1 Megaton (factor of 1000, 3rd-root of 1000 = 10), there will be 10 psi overpressure at 11,000 feet from ground-zero.

10 psi overpressure is enough to destroy most factories, residences, office buildings, etc. with F5 tornado-force wind. 5 psi overpressure will severly damage brick and wood structures -- that range is out to about 13,000 feet for a 1 Mt device.

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2007-04-13 04:40:06 · answer #1 · answered by tlbs101 7 · 0 0

depends on the size of the bomb, what altitude it is detonated at, and what level of effect you consider the "blast radius"

A 1 megaton bomb at ground level will make a 1000 ft wide crater, with a 1.7 miles of near-total destructon radiating from center. Some heavily reinforced buildings may survive on the edges.

8 miles out, there's still damage, but not much.

Airbursts tend to leave a wider path of destruction, but less concentrated in the center, and a logarythmic increase in damage will occur with increasing power. I.e. a 2 megaton blast won't be twice as damaging

2007-04-13 11:34:21 · answer #2 · answered by Anonymous · 0 0

It depends on which one. The largest ever detonated had a fireball radius of 4.6 km (2.85 miles). In comparison the A bomb drooped on Nagasaki had a fireball radius of 0.2 km.

2007-04-13 11:38:59 · answer #3 · answered by joatman71 3 · 0 0

nothing like a 60 mega-ton bomb

but interesting question

I think you already have some good answers... I give the same general figures

2007-04-13 14:02:46 · answer #4 · answered by TEST 1 · 0 0

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