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2006-08-21 06:02:09 · 15 answers · asked by shdw471@sbcglobal.net 2 in Politics & Government Military

15 answers

it depends on the height of the bomb in the air when detonation occurs. Ground level covers around 1.5 to 3 miles in diameter. An air burst at 10 thousand feet will cover a diameter of 25 plus miles. That would be around 625 square miles of destruction. and death.

2006-08-21 06:46:46 · answer #1 · answered by Anonymous · 0 0

There are a number of different ways to measure effects. You must consider blast effects, radiation effects and electro-magnetic pulse effets.

These effects can be selected for depending on how you 'tune' the weapon on yield and altitude.

A high altitude detonation will create a wider spread EMP and knock out electronics in a wider footprint than a lower detonation.

A low altitude detonation will create a wider spread of fallout than a high altitude detonation. There will be a greater amount of ejecta from the creation of a crater and surface matter swept up in the explosion.

Increasing yield will increase these effects at a similar altitude.

If you 'tune' the altitude of detonation of any explosive to the yield you can get an interaction between the reflected blast wave from the ground and the blast wave from the weapon, creating an amplified wave form. The leads to greater over-pressurisation wave. This gives the weapon a much greater 'knock-down' power. This works for both conventional and nuclear weapons.

For a nuclear weapon this coverage can be as wide as global for fallout and EMP and as local as a tuned theatre weapon taking out a division of amour.

2006-08-21 13:40:55 · answer #2 · answered by Anonymous · 0 0

Depends on two factors.

1) Yield of the weapon. Pretty self explanatory.

2) Altitude of the blast. A blast right against the surface of the earth will have the smallest radius since a lot of the explosion will be absorbed into the ground; a high-altitude blast covers the most, but with very little surface damage, although the EMP will fry every electronic device that is currently turned on in a very, very large area; a low altitude blast (altitude determined as a function of the warhead) will cause the best compromise in surface damage and blast area.

Hope it helps.

2006-08-21 13:14:28 · answer #3 · answered by Anonymous · 0 0

It depends on the warhead type and yield (size).

It also depends on whether you're talking about the area that is instantly disintegrated, the radius that is just flash-fried, the radius of the concussion wave, the area affected by radioactive fallout, etc.

A one-kiloton blast levels an area up to one kilometer in diameter, and causes second degree burns to twice that range.

A 20-megaton blast levels an area up to 15-20 kilometers in radius, charbroils anything within 30-40 kilometers, and causes radiation burns out to over 50 kilometers in radius.

2006-08-21 13:28:33 · answer #4 · answered by coragryph 7 · 0 0

Depends on the amount of fusionable material, but an average estimate is that it will level everything in a 8 mile radius, and cause severe damage out to 12 miles.

2006-08-21 13:10:40 · answer #5 · answered by Black Sabbath 6 · 0 0

It depends on a few factors:
1: Size of warhead
2: height above ground.

Scary thought: the largest nukes can be put miles away from cities and still wreak havock.

2006-08-21 13:17:27 · answer #6 · answered by Anonymous · 0 0

Depends upon the size of the weapon and altitude of detonation.

2006-08-21 13:21:12 · answer #7 · answered by Anonymous · 0 0

It is always a ration between the power of the nuclear warhead and the altitude of detonation.

2006-08-21 13:10:21 · answer #8 · answered by ColdWarrior 3 · 0 0

I don't know in detail, but I know that they've got them small enough to limit to a neighborhood, and probabaly even a city block. Of course on the large scale, it's probabaly in the 100's of sq miles.

2006-08-21 13:13:53 · answer #9 · answered by Padrecero 1 · 0 0

It depends greatly upon the size of the weapon.

2006-08-21 13:13:21 · answer #10 · answered by Anonymous · 0 0

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