Talk:Effects of nuclear explosions

Question?
How big is a average nuclear blast? And how far can the radiation travel from impact?

(Federal111 19:42, 31 January 2007 (UTC))


 * There is no "average". A pure fission blast can range from 10 tons to 500 kt.  A blast with a fusion stage (and possibly a third fission stage) can vary from 100 kt to 100 Mt.  The Prompt radiation intensity (gamma and neutrons) are generally halved for every 162 m of air, so prompt radiation is difficult to find more than a mile out where intensity is 2^-10 (one millionth) that it close to the bomb.SkoreKeep (talk) 04:00, 7 December 2013 (UTC)

Comments and Suggestions
The Flash Burn topic in this article should be a totally independent article because it is a phenomenon that many people are not familiar with yet it is a strong topic by itself.

Machpovii3...Holla Atcha Bezzle, Kid! 15:51, 9 February 2007 (UTC)

Layout
This article seems to start with the history, then got into some applications without first describing what a nuclear explosion is concisely. May I add this as a first paragraph and move the history to the history section?

"Surviving A Nuclear Blast, A Historical Perspective" Paradigmbuff 18:44, Jan 31, 2005 (UTC)

Is anyone aware of the source of Image:Blast positive and negative.gif? 119 00:35, 6 Mar 2005 (UTC)

Units and blast absorption due to destructive work
Is "kt" or "kT" the correct spelling? In German Wikipedia it has been widely accepted to use "T" for "tons of TNT equivalent" to distinguish from the mass unit "t". In other words, 1 kt TNT equivalent is the same as 1 kT. In this article, on the other hand, the lower case is used. But which of them is correct?--SiriusB 14:41, 15 Apr 2005 (UTC)


 * Dr William G. Penney used "kT" in his article on the nuclear explosive yields at Hiroshima and Nagasaki, Proc. Roy. Soc. London, 1970. Penney's paper is cited in Glasstone & Dolan (ENW 1977), although they only use it for the source of the yields of Hiroshima and Nagasaki.  Penney had issues with the 1962/4 edition of Glasstone, and these are ignored.  The British manual "Nuclear Weapons" (H.M. Stationery Office, 1974) uses "KT", but most sources use "kt".


 * Incidentally, Penney reproduces British nuclear test data and disputes the blast wave height-of-burst curves. Penney found that the 'peaking' effect in the Mach region for air bursts is due to the heating of the air just above the ground by the heat flash, and almost disappears if you measure the blast with sensors on poles 3 m high.  Penney also discredits Glasstone's dismissal of blast damage in reducing the blast pressure.  Accurate data on the crushing of empty petrol cans at Hiroshima by the blast showed that the overpressure decreased due to damage done to wooden houses.  (You can't cause mass destruction without using up a lot of energy, which causes an irreversible loss of blast pressure with distance.)  In a megaton detonation over a brick or concrete built city the loss of energy would reduce pressure ranges dramatically as the blast diverges outwards.  All the American data comes from tests in unobstructed deserts or Pacific atolls.

I discussed this by email with Dr Hal Brode, who did the original RAND Corp computer calculations of blast waves. His first response was the standard idea that the blast doesn't necessarily lose energy by doing work (causing destruction), since the debris will pick up some of the energy and carry it outward as flying bricks, panels and glass. However it is clear that the blast loses energy by the work done in breaking walls, which is irreversibly lost in warming up the rubble. If each house destroyed takes 1 % of the blast energy, then the energy after destroying 200 houses on a radial line outward from the explosion is down to just 100(0.99^200) = 13 % of what it would be over desert. This is valid for wood-frame houses. Brick and concrete buildings absorb far more energy per building destroyed, so in a modern city the blast pressure would fall very rapidly indeed. This is non-scalable, so it is most pronounced at high yields with large destruction radii computed for open terrain. Brode did concede, when presented with Penney's data, that this effect is not taken into account in American blast calculations at present. See http://glasstone.blogspot.com for further data. - Nigel Cook (edit by User:217.137.87.10)


 * This loss of energy may be less important due to the fact that blast waves move around obstacles and therefore the partition of blast which misses the building will replace the absorbed fraction within short distance behind the obstacle. This diffraction phenomenon is described in Effects of Nuclear Weapons, Chapter IV. The total blast energy will be reduced, of course, but by a much smaller amount, especially for high yields. I think, is does not really matter whether the obstacle is destroyed by the blast or not. Even if the object is fully resistant there will be loss due to turbulence behind the obstacle. The situation is somewhat different for surface bursts of low yields where the absorbed fraction is no longer negligible (consider a terrorist sub-kiloton or even conventional explosion in an urban area with huge buildings around it). Remember: Large bombs waste most of their energy into the atmosphere.--SiriusB 14:19, 19 December 2005 (UTC)

SiriusB, this is the standard fiction. The blast energy which diffracts back in is the incident blast energy minus the energy lost in causing destruction. The blast wave is always diverging, which is one of the reasons for the fall in overpressure with distance. Any sideways (non-radial) flow of energy to fill in areas where houses have been destroyed, reduces the energy somewhere else. You can't get something for nothing. If you have read the declassified book which is 1317 pages long, "Capabilities of Nuclear Weapons" by Philip J. Dolan of SRI, report DNA-EM-1 (Defence Nuclear Agency's Effects Manual number 1), you will see that this applies to forests. The blast diffracts around the tree trunks and fills in again afterwards. This was observed in forest stands at various tests, where the blast overpressure was measured on each side and found to be similar.

The blast wave cannot cause destruction without using energy, and this use of energy depletes the blast wave. The American manuals neglect the fact that energy used is lost from the blast. Visiting Hiroshima and Nagasaki, Penney recorded accurate measurements of damage effects on large objects that had been simply crushed or bent by the blast overpressure or by the blast wind pressure, respectively. At Hiroshima, a collapsed oil drum at 198 m and bent I-beams at 396 m from ground zero both implied a yield of 12-kt. But at 1,396 m data from the crushing of a blue print container indicated that the peak overpressure was down by 30%, due to damage caused, as compared to desert test data. At 1,737 m, damage to empty petrol cans showed a reduction in peak overpressure to 50%: ‘clear evidence that the blast was less that it would have been from an explosion over an open site.’

A similar pattern emerged at Nagasaki, with close-in effects indicating a yield of 22-kt and a 50% reduction in peak overpressure at 1,951 m as shown by empty petrol can damage: ‘clear evidence of reduction of blast by the damage caused…’ If each house destroyed in a radial line uses 1 % of the blast energy, then after an average of 200 houses in any radial line from ground zero outwards are destroyed, 87 % of the blast energy will have been lost in addition to the normal fall in blast pressure due to divergence in an unobstructed desert or Pacific ocean test. You can’t ‘have your cake and eat it’: either you get vast blast areas affected with no damage, or you get the energy being used to cause damage over a relatively limited area. The major effects at Hiroshima in the horizontal blast (Mach wave) zone from the air bursts were fires set off when the blast overturned paper screens, bamboo furniture, and such like on to charcoal cooking braziers being used in thousands of wooden houses to cook breakfast at 8.01 am. The heat flash can’t set wood alight directly, as proved in Nevada tests: it just scorches wood unless it is painted white. You need to have intermediaries like paper litter and trash in a line-of-sight from the fireball before you can get direct ignition, as proved by the clarity of ‘shadowing’ remaining afterwards (such as scorch protection of tarmac and dark paint by people who were flash burned). In general, each building will absorb a constant amount of energy from the blast wave (ranging from about 1 % for wood frame houses to about 5 % for brick or masonry buildings) despite varying overpressure, because more work is done on the building in causing destruction at higher pressures. At low pressures, the building just vibrates slightly. So the percentage of the blast energy incident on the building which is absorbed irreversibly in heating up the building is approximately constant, regardless of peak pressure. Hence, the energy loss in a city of uniform housing density is exponential with distance, and does not scale with weapon yield. Therefore, the reduction in damage distances is most pronounced at high yields.

-Nigel Cook 26 Dec 05


 * I do not know the source at the moment (I believe it was somethin in nuclearweaponarchive.org), but I remember that these extinction effect only applies on dense urban areas. But if you assume a large void area, e.g. a lake, within the Mach area of the blast (but with some area with buildings between the inner rim of the Mach zone and the void), than most of the initial blast strength will be restored at the outer rim of the void. And also the blast reduction in constantly built-up areas will be exponential only in the beginning and will reach a constant reduction factor (compared to a free flat terrain) after a certain distance. Of course, energy conservation also applies on blast waves, but diffraction allows redistribution of energy within a large fraction of the Mach stem. I will try to find the article again, but I believe that it also uses data from Hiroshima and Nagasaki as the main sources.


 * From my physical understanding this can be understood by the general mechanics of shock waves. The pressure inhomogenities behind the shock front can level because the flow in this region is sub-sonic with respect to the shock front (in contrast to the pre-shock region). The air volume will behave somewhat like a ballon where inhomogenities are also levelled after a certain relaxation time. And redistribution is easier the smaller the buildings are compared to a characteristic length scale of the blast (e.g. the length of the positive blast wave). If the obstacles are larger than this zone or the void behind the building is not much larger than the building, this relaxation cannot be effecient, and the blast will be strongly reduced. The same will happen if the height of the obstacles is so large that a large fraction of the total blast surface is affected and the absorbed energy is no longer negligible compared to the total blast energy (at least within the mach stem).


 * You can illustrate this phenomenon with the bang of a fired gun. Close to the gun the shot is by far louder in the direction of the shot. In larger distance the noise will become nearly isotropic (unless supersonic bullets are fired). Acoustical focusing over a large distance requires reflectors that are much wider than the typical wavelengths. For a nuclear blast of Hiroshima type or higher this length scale most probably exceeds the size of the tallest buildings. Thus the reduction of blast is only locally important. But "locally" means "everywere" in e densily built-up city.--SiriusB 13:34, 27 December 2005 (UTC)

The easiest way to deal with it is by energy use by the blast. The work energy used in pushing a wall distance x with force F is E = xF. Blast waves do diffract, but this doesn't violate conservation of energy. The problem with Glasstone and Dolan 1957-77 is that the book tries to dismiss the differences between a concrete city and a desert, without evidence. It is a cut down version of DNA-EM-1 which does contain sources. When you recognise that it was only in 1986 that they realised that gravity limits crater sizes in the megaton range to 1/4 power scaling (instead of 0.3 power scaling), you get an idea of the bureaucracy of the U.S. Government nuclear effects calculation business. The secrecy prevents a wide range of critical assessment, so fundamental new ideas are ignored, and errors can persist for decades. 172.189.174.108 14:49, 13 January 2006 (UTC)


 * Of course, energy conservation is not violated in blast effects. But the energy loss is, as far as I know, low compared with the amount of energy redistribution that can occur due to diffraction. Consider a city approximated by a more or less regular arrangement of tall concrete buildings. Without diffraction one expects the shock wave just above the roofs of the buildings to be similar to the flat area blast on the ground because there is no shielding. Maybe the shock reflectance of the ground is less than in flat terrain. Between the buildings, however, the shock strength is far less due to absorption. But due to diffraction some of the blast energy is diffracted into the gaps between the buildings an on the buildings themselves (destroying them). Thus, the shock strength in the free air above the roofs should also be reduced, but the strength that hits the buildings is somewhat higher than without diffraction. If this were not the case and if shock absorption were most efficient for huge blasts, this would contradict the observed destruction of dense forest areas in the Tunguska region. The total volume shielded by the buildings is nearly negligible compared to the total air volume engulfed by the blast wave. But as far as I anderstand shock wave dynamics, energy can be redistributed within large areas of the after-shock region because the flow with respect to the shock front is subsonic behind the front.--SiriusB 14:07, 15 January 2006 (UTC)

"The energy loss per square metre of diverging blast front is small for each building, 1% loss for destroying a wood frame house. So the blast reduction is only important for cities, not for isolated buildings on a desert.The American manuals neglect the fact that energy used is lost from the blast. Visiting Hiroshima and Nagasaki, Penney recorded accurate measurements of damage effects on large objects that had been simply crushed or bent by the blast overpressure or by the blast wind pressure, respectively. At Hiroshima, a collapsed oil drum at 198 m and bent I-beams at 396 m from ground zero both implied a yield of 12-kt. But at 1,396 m data from the crushing of a blue print container indicated that the peak overpressure was down by 30%, due to damage caused, as compared to desert test data. At 1,737 m, damage to empty petrol cans showed a reduction in peak overpressure to 50%: ‘clear evidence that the blast was less that it would have been from an explosion over an open site.’

"A similar pattern emerged at Nagasaki, with close-in effects indicating a yield of 22-kt and a 50% reduction in peak overpressure at 1,951 m as shown by empty petrol can damage: ‘clear evidence of reduction of blast by the damage caused…’ If each house destroyed in a radial line uses 1 % of the blast energy, then after 200 houses are destroyed, the blast will be down to just 0.99^200 = 0.13 of what it was before, so 87 % of the blast energy will have been lost in addition to the normal fall in blast pressure due to divergence in an unobstructed desert or Pacific ocean test. You can’t ‘have your cake and eat it’: either you get vast blast areas affected with no damage, or you get the energy being used to cause damage over a relatively limited area. The major effects at Hiroshima in the horizontal blast (Mach wave) zone from the air bursts were fires set off when the blast overturned paper screens, bamboo furniture, and such like on to charcoal cooking braziers being used in thousands of wooden houses to cook breakfast at 8.01 am." - http://glasstone.blogspot.com172.201.72.197 13:34, 30 January 2006 (UTC)


 * I have thought about the issue and hope, that the discussion could be revived after 3 1/2 years. It is clear that some of the blast is absorbed due to destrucive work. But I think that my central argument could have been misunderstood: The absorption takes place within the volume between the ground and the characteristic height of the obstacles (buildings, trees etc.). This volume is fairly constant since trees as well as buildings have a natural height limit that is far lower than the height of the air body available for blast development. Let's assume an average building height of 100 metres within a circular city area of 1 km radius. The corresponding "urban volume" is therefore about 0.314 km³ (pi*r^2*h). However, the total volume available to a low-altitude air burst above the city, given here by a hemisphere with radius of 1 km, is 4/3*pi*r³ = 4.19 km³, more than 13 times the absorbing volume. Since, according to me knowledge about shock wave physics, the Mach number is <1 behind the shock front, pressure can level within the shock hemisphere and thus re-distribute energy. This takes a while, of course, and therefore it is clear that high-rise urban areas will create a shadow effect. This, however, will not extend radially to infinity like the shadow created by a light-emitting point source but will level out after having traveled some "relaxation" or "levelling" distance. Below the line, the absorption in the case mentioned above will only be about 0.075% of the total blast energy. However, the reduction of the shock reflectance may be of much greater importance, but only up to 50% of the total blast yield, as discussed further below.


 * I don't know how large this distance might be, but according to the statements about diffraction in Glasstone & Dolan (1977), it should be of the order of several times the size of the obstruting object (i.e. the average building height), and thus be of the order of one kilometre for a Hiroshima-type bomb. This fits to the observations in Hiroshima and Nagasaki you mentioned. I'm also not sure about the role of the Mach stem height. The Mach enhancement of the blast may be more or less nullified since the typical Mach stem height for a Hiroshima-type air burst is just of the same order as the building height. But I guess that the absoption effect will be not or less significant several kilometres behind the absorbing.


 * There is another argument against the strict yield-independent exponential decay figure: In 1908, the putative meteor air burst over the Tunguska region levelled the (AFAIC very dense) forests within a radius of about 20-30 km. This is just the distance within the fully reflective BLAST.EXE model gives a wind speed of up hurricane force, required for levelling large forest areas. Therefore, the exponential decay law does not apply unless one assumes an extremely small absorption coefficient of forests. The latter can easily be neglected by the every-day experience that strong winds are not or far less felt in a forest compared to the free area. In other words: Forests *do* absorb wind (and probably shock waves) very efficiently, but the total blast volume affected by the Tunguska blast (blast area times about 10 kilometres of height) was *far* larger than the volume occupied by the forests. BTW it does not really matter whether this was really a meteor air burst or a methane eruption according to Prof. Wolfgang Kundt, since the cause of a blast becomes more and more unimportant with growing distance, and thus can both be compared to nuclear blast effects.


 * A very straight-forward consideration gives a similar answer: If a completely absorbing urban or forest area is assumed this absorption will only affect the urban or forest volume mentioned above. However, the air volume *above* the roof or tree tops will be completely uneffacted if the blast is completely absorbed below. I.e., it will behave just as a free-air burst with no reflection. In any case of less than total absorption there will be a more or less significant blast enhancement due to reflection. If the blast leaves the forest (or urban) area the shadow will smear out after a while, and then behave like an ordinary blast since the left-behind obstructive area becomes more and more unimportant with growing distance. In 10 km distance or so from a 20 kt air bust in 500 m there will probably no significant difference whether there is (or has been...) a three-kilometre-sized city below the burst point or not. But *within* the city, it makes a great difference.--SiriusB (talk) 10:20, 25 June 2009 (UTC)

Hmmm... So the blast obeys inverse cube law. But you just use the overpressure to calculate the damage but what about the time? Larger explosions take more time and that may expose stuff to overpressure for a longer duration and bingo! it suddenly obeys Y^0.4 or something law. —Preceding unsigned comment added by 91.99.165.174 (talk) 11:54, 4 April 2010 (UTC)

Blast Damage
The high temperatures and pressures cause gas to move outward radially in a thin, dense shell called "the hydrodynamic front." The front acts like a piston that pushes against and compresses the surrounding medium to make a spherically expanding shock wave. At first, this shock wave is inside the surface of the developing fireball, which is created in a volume of air by the X-rays. However, within a fraction of a second the dense shock front obscures the fireball, making the characteristic double pulse of light seen from a nuclear detonation.

Much of the destruction caused by a nuclear explosion is due to blast effects. Most buildings, except reinforced or blast-resistant structures, will suffer moderate to severe damage when subjected to overpressures of only 35.5 kilopascals (kPa) (5 lbf/in&sup2; or 0.35 atm).

The blast wind may exceed several hundred kilometers per hour. The range for blast effects increases with the explosive yield of the weapon. In a typical air burst, these values of overpressure and wind velocity noted above will prevail at a range of 0.7 km for 1 kiloton of TNT (kt) yield; 3.2 km for 100 kt; and 15.0 km for 10 megatons (Mt).

Thermal radiation
Nuclear weapons emit large amounts of electromagnetic radiation as visible, infrared, and ultraviolet light. The chief hazards are burns and eye injuries. On clear days, these injuries can occur well beyond blast ranges. The light is so powerful that it can start fires that spread rapidly in the debris left by a blast. The range of thermal effects increases markedly with weapon yield.

In Hiroshima, a tremendous fire storm developed within 20 minutes after detonation. A fire storm has gale force winds blowing in towards the center of the fire from all points of the compass. It is not, however, a phenomenon peculiar to nuclear explosions, having been observed frequently in large forest fires and following incendiary raids during World War II.

Electromagnetic pulse (EMP)
Gamma rays from a nuclear explosion produce high energy electrons through Compton scattering. These electrons are captured in the earth's magnetic field, at altitudes between twenty and forty kilometers, where they resonate. The oscillating electric current produces a coherent EMP (electromagnetic pulse) which lasts about 1 millisecond. Secondary effects may last for more than a second. The pulse is powerful enough so that long metal objects (such as cables) act as antennas and generate high voltages when the pulse passes. These voltages, and the associated high currents, can destroy unshielded electronics and even many wires.

The largest-yield nuclear devices are designed for this use. An air burst at the right altitude could produce continent-wide effects.

Ionizing radiation
About 5% of the energy released in a nuclear air burst is in the form of neutrons, gamma rays, alpha particles, and electrons moving at incredible speeds. The neutrons result almost exclusively from the fission and fusion reactions, while the initial gamma radiation includes that arising from these reactions as well as that resulting from the decay of short-lived fission products. The neutron radiation serves to transmute the surrounding matter, often rendering it radioactive.

Nuclear fallout



 * Main article: Nuclear fallout

The residual radioactive contamination hazard from a nuclear explosion is in the form of radioactive fallout and neutron-induced activity. Residual ionizing radiation arises from:


 * Fission products. These are intermediate weight isotopes which are formed when a heavy uranium or plutonium nucleus is split in a fission reaction. There are over 300 different fission products that may result from a fission reaction. Many of these are radioactive with widely differing half-lives. Some are very short, i.e., fractions of a second, while a few are long enough that the materials can be a hazard for months or years. Their principal mode of decay is by the emission of beta and gamma radiation. Approximately 60 grams of fission products are formed per kiloton of yield (14 g/TJ). The estimated activity of this quantity of fission products 1 minute after detonation is equal to that of 1.1 &times; 1021 Bq (30 gigagrams of radium) in equilibrium with its decay products.
 * Unfissioned nuclear material. Nuclear weapons are relatively inefficient in their use of fissionable material, and much of the uranium and plutonium is dispersed by the explosion without undergoing fission. Such unfissioned nuclear material decays slowly by the emission of alpha particles and is of relatively minor importance.
 * Neutron-induced activity. If atomic nuclei capture neutrons when exposed to a flux of neutron radiation, they will, as a rule, become radioactive (neutron-induced activity) and then decay by emission of beta and gamma radiation over an extended period. Neutrons emitted as part of the initial nuclear radiation will cause activation of the weapon residues. In addition, atoms of environmental material, such as soil, air, and water, may be activated, depending on their composition and distance from the burst. For example, a small area around ground zero may become hazardous as a result of exposure of the minerals in the soil to initial neutron radiation. This is due principally to neutron capture by various elements, such as sodium, manganese, aluminum and silicon in the soil. This is a negligible hazard because of the limited area involved.

In an explosion near the surface large amounts of earth or water will be vaporized by the heat of the fireball and drawn up into the radioactive cloud. This material will become radioactive when it condenses, mixed with fission products and other radiocontaminants that have become neutron-activated. The larger particles will settle back to the earth's surface near ground zero (depending on wind and weather conditions of course) within 24 hours, while fine particles will rise to the stratosphere and be distributed globally over the course of weeks or months.

Severe local fallout contamination can extend far beyond the blast and thermal effects, particularly in the case of high yield surface detonations. In detonations near a water surface, the particles tend to be lighter and smaller and produce less local fallout but will extend over a greater area. The particles contain mostly sea salts with some water; these can have a cloud seeding effect causing local rainout and areas of high local fallout.

The radiobiological hazard of worldwide fallout is essentially a long-term one due to the potential accumulation of long-lived radioisotopes, such as strontium-90 and caesium-137, in the body as a result of ingestion of foods incorporating these radioactive materials. Chemically, both isotopes are recognized as similar to calcium and deposited in bone structure throughout the body. These highly-radioactive substances then interfere with white blood cell production, which is a prime effect of radiation sickness. The hazard of worldwide fallout is much less serious than the hazards which are associated with local fallout.

Blast and thermal injuries in many cases will far outnumber radiation injuries. However, radiation effects are considerably more complex and varied than are blast or thermal effects and are subject to considerable misunderstanding. A wide range of biological changes may follow the irradiation of animals, ranging from rapid death following high doses of penetrating whole-body radiation to essentially normal lives for a variable period of time until the development of delayed radiation effects, in a portion of the exposed population, following low dose exposures.



Plasma Ring at nuclear explosions
At some photographs of nuclear explosions, as http://nuclearweaponarchive.org/Usa/Tests/Ukgrable2.jpg, a plasma ring at the mushroom of the atomic bomb can be seen. How is it generated? Describe it in a chapter!

Lightning at nuclear explosions
Does lightning occur in the mushroom clouds of atomic bombs? Describe it in a chapter!

The first nuclear test lightning photographed was 5 lightning flashes at tens of milliseconds around the fireball of the 10.4 Mt Mike test in 1952, caused by the effect of gamma radiation ionising the air and making it conductive. Still pictures of this and similar lightning during the 14.8 Mt Bravo test in 1954 have been published in Journal of Geophysical Research and elsewhere since they were declassified around 1970. The lightning bolts range from about 900 m to 1.2 km from ground zero and are vertical but slightly curved to follow the shape of the hemispherical fireball. The radial electric field of the EMP at those distances was estimated to be on the order of 100 kV/m. Natural lightning strokes have recently been explained as initiated by cosmic radiation hitting air in low density air at high altitude, which sets off an avalanche of ionisation, like the Geiger counter effect, but with the electric field caused by the natural difference in electric potential of about 100 v/m between the ground and the ionosphere. If any lightning occurred in the mushroom cloud minutes after a test, it would be unrelated to ionisation of air because of the large amount of radioactive decay which occurs. All the lightning caused by nuclear explosions is over within 0.1 second. - Nigel Cook

Strange formula in spanish article
In the spanish article there is a strange formula that seems to be related to the blast magnitude. It is
 * $$p=\frac{6Y}{R^{3}}$$

where Y is the yield in megatons, R the range in kilometres and p the magnitude in psi. But this formula gives horribly wrong results (by several orders of magnitude). For example, for 1 MT and 6.2 km it yields 0.025 psi, 1/200th of the true value (5 psi, see table). But since I do not understand Spanish I cannot exclude that there is a different meaning of this formula. The article is labelled as a featured article. Does anybody understand the spanish article?--SiriusB 20:09, 7 January 2006 (UTC)

The correct formula for peak overpressure is at http://glasstone.blogspot.com/2006/03/physical-understanding-of-blast-wave.html:

$$p={\frac{3.55W^{1/3}}{R}}+{\frac{2.00W^{2/3}}{R^{2}}}+{\frac{0.387W}{R^{3}}}$$

kilopascals. Note that 1 psi = 6.9 kilopascals (kPa). This formula holds good for sea-level air bursts with a total yield of W kilotons and a distance from detonation point of R kilometres. Source: it is based on the curve in Philip J. Dolan's manual Capabilities of Nuclear Weapons DNA-EM-1, 1981, and the same curve is in Glasstone & Dolan, The Effects of Nuclear Weapons, U.S. Department of Defense, 1977, Fig. 3.72. For a surface burst, the blast is hemispherical and if the ground was a perfect reflector the blast would have twice the energy per unit area as in an air burst, so the factor 2 would need to be inserted into the yield W in the formula above. However, the data in DNA-EM-1 and Glasstone and Dolan for surface bursts suggests that due to loss of fireball energy in surface bursts due to ground shock, cratering, and melting surface material to create fallout, the true reflection factor is 1.68, not 2. Hence the peak overpressure in a surface burst is roughly:

$$p={\frac{4.22W^{1/3}}{R}}+{\frac{2.83W^{2/3}}{R^{2}}}+{\frac{0.650W}{R^{3}}}$$

kilopascals. This formula is consistent with 70 accurate measurements at 7 American surface burst tests conducted in the Nevada and at Eniwetok and Bikini Atolls from 1951-4 (Sugar, Mike, Bravo, Romeo, Union, Yankee and Nectar), for scaled distances equivalent to 55-300 m from a 1 kiloton burst. Reference: see the original plots of nuclear test data in reports like WT-602 and WT-934   Nigel 172.212.202.81 16:28, 31 March 2006 (UTC)

13:36, 30 January 2006 (UTC)


 * Thanks, interesting. I have got another formula which is said to be accurate. Given the slant range R in metres the overpressure of an 1 kiloton free-air burst (i.e. in an ideal atmosphere without density gradient and without a solid surface limiting the shock propagation) is

\begin{matrix}\mathit{OP}_\mathrm{DNA}&=&\frac{3.04\times10^{11}}{R^3} +\frac{1.13\times10^9}{R^2}\\&&\\ &&+\quad\frac{7.9\times 10^6}{R\sqrt{\ln \left(\frac{R}{445.42}+3\exp\left(-\frac{1}{3}\sqrt{\frac{R}{445.42}}\right)\right)}} \end{matrix} $$ (source: the DOS program BLAST, 1984). This formula is taken from the German Wikipedia article Formelsammlung Kernwaffenexplosion together with formulae which describe the shock reflection and Mach stem formation (in German, but mathmatics is the same ;-)). It uses the full reflection model. But an interpolation between the free-air case and the full-reflection should give approximately reliable answers.--SiriusB 19:23, 3 April 2006 (UTC)


 * One question: Could you please give a reference for the formula you have posted? I did not find anything like it in the original PDFs.--SiriusB 13:55, 5 April 2006 (UTC)

About the first equation, the Spanish article described it above it as: "the formula that measures the statistical damage produced by the wave on a colectivity of individuals that are in the zone of the explosion based on the overpressure produced by the shock wave" (rough translation). Is this an attempt to describe the damage to people? Certainly a raw number cannot describe physical injuries. If anyone (who knows Spanish better than I) can help, please do. CharonM72 06:10, 4 September 2006 (UTC)

How big?
In layman's terms, how much damage does a nuclear explosion yield, in terms of physical damage?

How big would a crater be after an explosion of a 10 kT bomb? 100?

How big is the "size" of the base of the mushroom cloud of a 60 mT nuke? Like Tsar Bomba?

This would just be interesting stuff to know, and I am having trouble finding it other places.

Peer Review
There is a peer review now occurring for this page, please join in and contribute! This page could be a featured article, if we just put our minds to it, so if you see anything you can help fix as they review it, jump in! Judgesurreal777 01:41, 12 March 2006 (UTC)
 * I'm reviewing all the scientific-quality nuclear weapons effects data in the public domain at http://glasstone.blogspot.com - feel free to use data and links to reports from there. Nigel 172.212.202.81 16:01, 31 March 2006 (UTC)

"Summary of the effects" table
The "Summary of the effects" table is poorly formatted. After the yield/height of burst the following rows of the table are quite confusing. What do the numbers mean? My assumption is that for 1 MT/2.0 km warhead the "Urban areas almost completely levelled (20 PSI)" value is in kilometers meaning that everything in 2.4 km radius is pretty much obliterated. -Khokkanen 13:23, 5 April 2006 (UTC)

Urban areas completely leveled: @1kT => 0.2km and @1MT => 2.4km but lethal dose @1kT => 0.8km while @1MT => 2.3km This seems strange. At 1kT, the area with a lethal dose has a radius four times as big as the area completely leveled. But at 1MT, the radius gets smaller?! How can the lethal dose radius be smaller than the completely leveled radius? Beau (talk) 17:36, 24 July 2009 (UTC)


 * The two measures don't depend on each other. Lethal dose comes from the near-instantaneous flash of radiation from the nuclear reaction, and destruction comes a bit later from the force of pressure. Binksternet (talk) 18:15, 24 July 2009 (UTC)

Title
What's wrong with just "nuclear explosion" (which is currently a redirect)? Fredrik Johansson 21:44, 6 September 2006 (UTC)
 * The article name was changed for accuracy purposes; see the peer review on the discussion page. Judgesurreal777 01:23, 7 September 2006 (UTC)


 * All I see is that the article has to be called "effects of" because it doesn't deal with the causes of nuclear explosions. The problem is that the current lead section is misleading: it claims to be about the concept of nuclear explosions when it is in fact just about a specific property of nuclear explosions. We should definitely have a non-redirecting article called nuclear explosion, but I think modifying this article would be better than creating a separate article with a "main article" link pointing here, since the effects are the most interesting property of nuclear explosions. Fredrik Johansson 12:30, 7 September 2006 (UTC)

Also, most articles linking here go via the redirect.

No one is providing a good reason not to move. Who supports? Fredrik Johansson 17:27, 11 October 2006 (UTC)


 * It was moved here because of the articles content. It's true, the lead needs rewriting, so lets rewrite it. Judgesurreal777 00:16, 12 October 2006 (UTC)


 * The article already is about more than just "effects of". With your approach, we'd have to throw out content and create a separate article for nuclear explosion. But that article would have to include much of the content of this page, thereby making this page essentially redundant, or utself be a fairly short article that points here, which is worse for readers since they'll have to visit two pages to find what they are looking for. Fredrik Johansson 12:04, 12 October 2006 (UTC)


 * This is what the peer review said when it was called Nuclear Explosion : "This article apparently isn't attempting to treat the causes of a nuclear explosion, e.g. the design of an atomic weapon or the related physics. So it might be worth renaming it to something like Effects of nuclear explosions, or any name that better reflects exactly what this article is about. Another point: I'd like to see a little more on the applications, which are mentioned in the lead but not again. Christopher Parham (talk) 03:44, 13 March 2006 (UTC)

GA Re-Review and In-line citations
Note: This article has a very small number of in-line citations for an article of its size and currently would not pass criteria 2b. Members of the WikiProject Good articles are in the process of doing a re-review of current Good Article listings to ensure compliance with the standards of the Good Article Criteria. (Discussion of the changes and re-review can be found here). A significant change to the GA criteria is the mandatory use of some sort of in-line citation (In accordance to WP:CITE) to be used in order for an article to pass the verification and reference criteria. Currently this article does not include in-line citations. It is recommended that the article's editors take a look at the inclusion of in-line citations as well as how the article stacks up against the rest of the Good Article criteria. GA reviewers will give you at least a week's time from the date of this notice to work on the in-line citations before doing a full re-review and deciding if the article still merits being considered a Good Article or would need to be de-listed. If you have any questions, please don't hesitate to contact us on the Good Article project talk page or you may contact me personally. On behalf of the Good Articles Project, I want to thank you for all the time and effort that you have put into working on this article and improving the overall quality of the Wikipedia project. Agne 20:23, 25 September 2006 (UTC)
 * Thank you much. Although I am not an expert on this topic, I have given it some attention and will work on improving it :) Judgesurreal777 22:46, 25 September 2006 (UTC)

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This article's GA status has been revoked because it fails criterion 2. b. of 'What is a Good Article?', which states;


 * (b) the citation of its sources using inline citations is required (this criterion is disputed by editors on Physics and Mathematics pages who have proposed a subject-specific guideline on citation, as well as some other editors &mdash; see talk page).

LuciferMorgan 21:22, 18 November 2006 (UTC)

Sources for energy distribution
Most sources I have studied mention 50% to 60% for blast and 35% to 45% for the thermal radiation (e.g. Carey Sublette's Nuclear Weapon FAQ). The significantly wider ranges given at the top of the article should be substantiated by sources. However, the wider range for the thermal fraction would (at least partially) explain the longer duration of the glowing and larger radius of the fireball with respect to the cube root scaling-law (Glasstone & Dolan 1977 state that the fireball light curve vs. time scales with about W0.4 instead of W1/3 like the blast range) which, under the assumption of equal effective surface temperatures, would violate the conservation of energy.--SiriusB 18:42, 18 October 2007 (UTC)

Long-term effects
I am wondering about the radiation levels persisting at ground zero, owing to the fallout. If these effects are negligible in the long term, then this should be stated in the section on "Indirect Effects". What is needed are some ball-park numbers about the rate at which cities such as Hiroshima can recover from the "effects of nuclear explosions". Currently, this article discusses the effects with not enough reference to a long-term timeline. Mebden (talk) 13:43, 18 November 2007 (UTC)


 * They don't recover "per se". Japan is a tiny island considering its 125 million population and most of the middle is made up of big mountains. They simply had no suitable empty space to relocate either Hiroshima or Nagasaki. Chernobyl, however, gives a good insight of what kind of habitation denial a three-phase (Pt-239 or U-235 to Li-D to U-238 or Th-232) salted nuke would cause. That is really a terror-only weapon, way too powerful for any logically imaginable military aim in war. 91.83.4.250 (talk) 20:13, 12 April 2008 (UTC)

Nukes are terrible, about the only kind of weapon whose effect is UNDERestimated by the public.
1., The use of imperial PSI in a summary table which has kilometer basis looks really stupid. Please go fully metricated!

2., Data I have collected from books:


 * 20kt-class 560m alt. low airburst (e.g Nagasaki)
 * 20kt everbody dies outside of shelter up to 800 meters distance from gound zero
 * 20kt overpressure at 1km is 0.7kg/cm2 and at 1.5km it is 0.4kg/cm2
 * 20kt notable civilian building damage up to 18km, brick and wood structures collapse up to 2.3km
 * 20kt wall and roof damage mostly beyond repair up to 2.75km (minimum 0.14kg/cm2 overpressure)
 * 20kt 30-40pcs of these will level 15-20 thousand square km2 and cause 30 Roentgen/hour pollution
 * 20kt Armoured personnel carriers crushed up to 500-600 meters (1kg/cm2 overpressure)
 * 20kt Battle tanks are not killed beyond 500-600 meters, but badly contanimated up to 2kilometers
 * 20kt All railway carts thrown from track and crushed up to 1km range (up to 0.63kg/cm2 overpressure)


 * 50kt-class: Ten of these will entirely destroy a city of 2million people
 * 50kt up to 2500 meters everybody dies outside shelter due to third-degree burns


 * 100kt-class: All rebar concrete material civil / industry structures destroyed up to 900 meters
 * 100kt underwater throws up 1km diameter column of seawater, takes 10-12 seconds to fall back
 * 100kt creates seawaves at distance 540 meters that are still 54 meters tall from top to bottom
 * 100kt ground surface blast creates 20km diameter lethal contanimated area for up to two weeks


 * 1mt-class explosion: fireball diameter 1750m, 30x brighter than the Sun even at 100km distance
 * 1mt shockwave front reaches 19km in 50sec, shock speed is still 345m/sec (barely supersonic)
 * 1mt second degree burns up to 19km
 * 1mt rips and ignites most oil and gas tanks up to 14km (0.2-0.3kg/cm2 overpressure)
 * 1mt bad radioactive fallout up to 32km/350km against/alongside wind direction, width 64 kilometers
 * 1mt rebar concrete destroyed up to 2100 meters
 * 1mt surface explosion destroys an area of 15-20.000 square kilometers and pollute it with 30R/h
 * 1mt One such high altitude explosion at H=60km will EMP-blind radars in 400km diameter for 5 min.
 * 1mt One such high altitude explosion az H=100-200km will EMP blind early warning radars for 17 min.
 * 1mt Digs ground crater if explodes at 135 meters or less altitude
 * 1mt Underground explosion at H=-15 meter digs a crater 60meters deep with a diameter of 400 meters.
 * 1mt shallow underground explosion throws out ten million metric tons of soil
 * 1mt underwater explosion produces one hundred thousand (10**5) metric tons of steam
 * 1mt bombers on runway wrecked up to 5-6 kilometer distance (minimum 0.4kg/cm2 overpressure)
 * 1mt all airplanes seriously damaged on runway up to 8km (min. 0.14 to 0.28 kg/cm2 overpressure)
 * 1mt low altitude airburst destroys all moored or surfaced submarines up to 2.3 kilometers


 * several mt class: Rebar concrete and metal grid industrial structures badly damaged up to 5-7 km
 * several mt: aerials, pole mounted wires, light structures suffer medium bad damage up to 9-10 km
 * several mt: metal grid structure bridges suffer medium bad damage up to 5-7 km


 * 5mt-class: 420pcs of these were calculated to guarantee destruction of 1000 pcs Minuteman silos
 * 5mt Some 263pcs of these were calculated to destroy 71 major USA cities and 132 top military targets
 * 5mt Same 263pcs attack would isotope pollute 50% of CONUS territory beyond habitability for weeks


 * 10mt-class: One such bomb can destroy a city of 2million
 * 10mt Ships destroyed up to 4.5 kilometers distance
 * 10mt Two of these secures killing of one 21kg/cm2 overpressure resistant enemy underground missile silo at a 2.4 kilometers average error of aiming (e.g. early generation R7-based USSR attack ICBM).
 * 10mt One single-head 10MT ICBM 100% kills the 21kg/cm2 silo at 800 meters medium aiming error (e.g Minuteman-II class precision, which was exactly quoted as 900 meters CEP)


 * 100mt-class (hypothetical full yield version of Tsar bomba, never actually tested this high, it was "only" 57-58MT in year 1960)
 * 100mt theorised: low airburst or ground burst causes total destruction in a radius of 96 kilometers (overall almost 30,000 square kilometer area)

Overpressure effects in numbers:
 * 70kg/cm2 in shallow water immobilizes warships due to hull and screw crushing
 * 10kg/cm2 ship engine screw-driving axles and bearings much damaged, ships on the move hurt more
 * 0.5kg/cm2 loaded railways carts thrown away from track and crushed on impact
 * 0.4kg/cm2 aircraft on the ground destroyed
 * 0.28kg/cm2 railway carts damaged, potentially won't roll along without repair
 * 0.27kg/cm2 almost all trees fell in a forest
 * 0.07-0.14 kg/cm2 aircraft skin and spars suffer medium damaged, won't fly without serious repair 91.83.4.250 (talk) 20:03, 12 April 2008 (UTC)


 * I have added the SI values for the peak overpressures to the psi values. Please note that many nuclear weapon related sources use imperial units, thus they should also be given. Although simple conversion of units should not violate the non-OR rule, in my opinion, it is still an unaccuracy to omit the values in the given sources (e.g. nuclearweaponarchive.org that often uses psi). Please also note that kp/cm^2 is not an SI unit. Please use kilopascals (kPa) instead. 1 kpa = 98.0665 kp/cm^2. BTW, kg/cm^2 is not a pressure unit at all, it's surface density.


 * Furthermore, I have added estimates for the effects on railroad trains by using the cube root scaling (see Glasstone & Dolan 1977). Since overturning cars is not only a matter of dynamic pressure (in this case related to a distance where the static peak pressure is 62 kPa) but also of the exposure time (time that is needed to lift the cars over there center of mass rather than simply cracking walls), the values are to be taken with caution. Therefore the "≈".--SiriusB (talk) 14:35, 24 October 2008 (UTC)


 * "about the only kind of weapon whose effect is UNDERestimated by the public"


 * You are dead wrong about this. It is the one type of weapon that is overestimated by the public.  Ask an average American, for example, about "nucular" weapons and they will tell you that you might as well just kill yourself if you know one is about to explode because apparently they have a %100 mortality rate.  Also they believe that nuclear weapons have the ability to "destroy entire states" (with just one).  Even the Tsar Bomba wouldn't destroy the entire state of Rhode Island.  I seriously doubt there are any studies on this but I will look.  I am fairly confident that they will show that there is an extreme overestimation of nuclear weapon damage.  I might even do a small curiosity study.  Perhaps someone could help me look?JohnnyTopQuark (talk) 21:16, 6 February 2010 (UTC)


 * I'll agree with JohnnyTopQuark on this. One of the most heard comments after watching Hashimoto's simulation of testing video is "Why is the West Coast/Los Angeles/Las Vegas/Nevada/the west/the entire continent still inhabited" in the face of 900+ nuclear explosions?  Most non-technical people believe that 2 or 3 50 Mt weapons would wipe out all life on Earth, and possibly crack the planet open.  SkoreKeep (talk) 04:15, 7 December 2013 (UTC)

I would love to see any data re. this Pixelson (talk) 20:50, 12 September 2011 (UTC)

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Tubes??
"All American and Russian backup communications systems are tube (UK "valve") operated to ensure effective operation during or after a nuclear incident.[dubious – discuss]"

Where does this statement come from? I can find no references for it past or present. It seems highly unlikely as the military has methods to "harden" it assets against EMP. Are tubes even manufactured anymore? Wlmg (talk) 15:18, 19 October 2010 (UTC)

Jokes or what
Table claims that 1000 times more powerful nuke (1kT vs 1Mt) does 10 times more damage. I almost believed in that lol. — Preceding unsigned comment added by 78.56.60.210 (talk) 14:12, 2 May 2012 (UTC)

For the most part that is true, that is why modern nuclear arsenals don't included anything with a higher yield than 1Mt, as smaller ~300 kt weapons get the job done much better, it is also why in conventional munitions cluster bombs are preferred over big dumb massive bombs.

You should look up Nuclear weapon Scaling laws and familiarise yourself with them. Especially the cubed root blast scaling law.  Sc-em Scaling laws - look it up, the 1st document will help you understand.

'''The overpressure from a given weapon yield decreases inversely as the cube of the radius (because space is 3 dimensional not linear), while the area of destruction increases with the square of the radius. ''' Suitable scaling constants for the equation r_blast = Y^0.33 * constant_bl are:

where Y is in kilotons and range (r_blast) is in km, and * is multiply.

constant_bl_1_psi = 2.2

constant_bl_3_psi = 1.0

constant_bl_5_psi = 0.71

constant_bl_10_psi = 0.45

constant_bl_20_psi = 0.28

Plug in a few Y yields yourself and you'll soon find that big bombs are really a waste of energy, carpet bombing with lower yield weapons ~300 kt would be far more effective(that's why the nuclear powers the world over design use weapons of approximately this yield e.g W78. Boundarylayer (talk) 04:23, 3 May 2012 (UTC)

Atmosphere ignition
The way I heard it, the Manhattan Project scientists had not been worried about a nuclear reaction of two nitrogen atoms but starting a global chemical reaction between the atmosphere's nitrogen and oxygen. However, this could be a chemistry teachers' conspiracy to teach students about activation energy.


 * I understand this is one of the problems they worked on during the initial planning at Los Alamos (the Nuclear Primary time). Hans Bethe was assigned to look into it and reported in the negative.  A short paper by Teller, C. Marvin, and EJ Kopinski (http://www.fas.org/sgp/othergov/doe/lanl/docs1/00329010.pdf) was submitted in 1946, possibly as a CYA that some due diligence had been done on the question, and declassified in 1970.  That apparently didn't keep the physicists from joking about it at Trinity. SkoreKeep (talk) 04:36, 7 December 2013 (UTC)

Note
Possible quality reference and source of further details to expand article.. http://history.nasa.gov/conghand/nuclear.htm

at high altitudes, where the air density is low, more energy is released as ionizing gamma radiation and x-rays than an atmosphere-displacing shockwave.
Isn't this strictly incorrect? With a thinner athmosphere there's just not enough air to combust to form a fireball. Additionally the same amount of radiation is produced, it's just not absorbed as much by the thinner athmosphere and consequently creates a weaker shockwave and the radiation travels much farther. But I can't say I'm 100% sure about this. 79.223.156.40 (talk) 18:45, 2 April 2015 (UTC)


 * The nuclear fireball is not air combusting. Though there is some of that (at least oxidation of some nitrogen), the fireball is 99+% air simply heated to millions of degrees by absorption of soft x-rays and everything higher in the spectrum. If there is no air, the x-rays become radiation-at-a-distance, and the heat pulse (which is light and some infrared) is simply not present, because there is no fireball. Yes, the same amount of radiation is produced, but its not absorbed and re-emitted as light close to the explosion. SkoreKeep (talk) 06:52, 10 July 2015 (UTC)

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Wilson clouds
Watching footage of the Hiroshima explosion, I saw a white cloud that overtook the mushroom cloud and continued expanding outward horizontally while disappearing from around the mushroom cloud. After Googling the phenomenon, I found this article which explains the effect as the condensing of water vapor in the low pressure and therefor cooled wave following the shockwave. It also called them Wilson clouds. As this phenomenon is not mentioned in the article, I found it worth mentioning. — Preceding unsigned comment added by 50.5.104.20 (talk) 01:30, 16 February 2017 (UTC)

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Survivability
Just to let you know the section on Survivability needs to be checked, also it should be in Sv not rems as it isn't an SI unit. Some of the data is wrong, people have survived higher than 2 Sv, Burgowski (though not whole body dose) survived a thousand times this, Yokokawa survived 3 Sv. Also may want to not only focus on deterministic effects but look at stochastic as well. Khu kri  22:48, 27 February 2022 (UTC)

Inconsistency in “Survivability” section
The section states, "50% of the 460–600 rems group will die within one to three weeks. If the group is exposed to 600 to 1000 rems, 50% will die in one to three weeks." Note that the same range (50% in one to three weeks) is given for two different doses. Zeligf (talk) 04:35, 1 April 2024 (UTC)