Talk:Nuclear weapon yield

Estimating/measuring yield
I have added a bit on the calcualtions that G.I. Taylor used. I'm not sure what the US military does for their calcualtions, but G.I. Taylor is well known in chemical engineering for having accurately determined the yield of weapons that only pictures were issued of without yields to accompany. An example regarding the trinity test is provided. Sorry about some of my grammar, but I'm an engineer, not an english major. -MDL 01:25 08 March 2006

Added links for Trinity test and G. I. Taylor. Listed reference for G.I. Taylor's paper on the subject. -MDL 09:37 09 March 2006

MDL: please note that a physically-based alternative to G.I. Taylor's approach proves that the correct formula for the energy is

$$E={\frac {8\pi\,\rho\,{R}^5}{75(\gamma\ - 1){t}^{2}}}$$

and avoids the approximate integration that Taylor used: http://glasstone.blogspot.com/2006/03/analytical-mathematics-for-physical.html 172.212.100.165 11:24, 30 March 2006 (UTC)

I'm sure there are many such approaches. I used G.I. Taylor's because it is commonly referred to in high level engineering math courses, so I am familiar with it and its derivation. It also produces very nice results. -MDL

COMMENT on above
One problem is that the result given in the WP section can't actually be used alone to figure out the answer, because the value of c is not given.

"A good approximation of the yield of the Trinity test device was obtained from simple dimensional analysis by the British physicist G. I. Taylor. Taylor noted that the radius R of the blast should initially depend only on the energy E of the explosion, the time t after the detonation, and the density ρ of the air.  The only number having dimensions of length that can be constructed from these quantities is: $$R=c\left( {\frac{{E{t}}^{2}}{\rho}} \right)^{\frac {1} {5}}$$

Using the picture of the Trinity test shown here (which had been publicly released by the U.S. government and published in Life magazine), Taylor estimated that at t = 0.025 s the blast radius was 140 metres. Taking ρ to be 1 kg/m³ and solving for E, he obtained that the yield was about 22 kilotons of TNT (90 TJ). This very simple argument agrees within 10% with the official value of the bomb's yield, 20 ktonTNT, which at the time that Taylor published his result was considered highly-classified information. (See G. I. Taylor, Proc. Roy. Soc. London A 200, pp. 235-247 (1950).)"

COMMENT: Cute, but if you go through the equation above, you find from pure dimensional analysis that "c" (not the speed of light!) is a dimensionless constant "k", but you need to know what it is, to calculate E from R. In the equation above, for example, you can estimate E from R if you assume c is about 1, but that's a coincidence. If you have two sets of t and R (two timed stills of the blast, which I assume were available), you can show R^5/t^2 is constant, and from them you can estimte E*c^5 if you know density rho, but without a value for c you still don't know E without some further considerations about the heat capacity of air. Taylor worked by integrating the energy in air shockwaves to get a numerical esitmate of c as approximately 1. The link given shows that you can use gas dynamical estimate of the heat capacity of air to show that c^5 = 8π/[75(γ-1)] or c = [75(gamma-1)/(8pi)]1/5 which is close to 1. If you assume γ = 7/5 for a hot diatomic gas, then this gives c^5 = 0.84, or c = 0.965. That's really close to 1, all right, but again, totally coincidentally! In the paper it notes that a combination of dissociation and vibration returned gamma coincidentally at 5000 K back to the "room temp" gamma of 1.4, so c becomes 1.036. Again, quite coincidentally close to 1, from the dimensional analysis sense. Also, interestingly Taylor's classified work was done in 1941, and declasified along with application to the Trinity results in 1950. By then the Soviets had the atom bomb anyway, so there was no point in keeping all this a secret. I'll add some material to the main article. BTW, Taylor used a density for air of 1.25 kg/m^3, but for the Trinity conditions it just happened to be closer to 1.0 (altitude and so on). S B Harris 23:46, 13 March 2011 (UTC).

I'd like to cite the following paper on the dimensional analysis for this problem: UtzT (talk) 14:23, 2 June 2014 (UTC)

Modern methods of yield estimation
It would be really great if someone could add a section on how yield has been historically measured, and about some of the more difficult yield estimates (i.e. exactly how powerful the Trinity, Little Boy, and Nagasaki bombs were; whether the Tsar Bomba was 50 or 55 Mt, and disputed yields in relation to the India/Pakistan tests). --Fastfission 17:33, 17 November 2005 (UTC)
 * I added some details about the controversies and some possible methods to calculate yields, though I have no idea how they are normally calculated these days. --Fastfission 16:57, 22 November 2005 (UTC)

Fastfission: it is done by radiochemistry. You measure the number of fissions in your sample by measuring the amount of unfractionated fission products. (Until 1961 they used Mo-99, but then they changed to Nb-95 because is is more abundant and easier to measure accurately in fallout, while not being fractionated relative to actinides.) Say this tells you that your sample has 1000 fissions. You then measure the amounts of the heavy elements (uranium, plutonium, plus neutron capture products), and from the ratios and the amounts of material you put into the bomb, you can work out the fission efficiency of the bomb, i.e., the number of fissions per fissionable atom initially present. It gets more complicated naturally when fusion reactions are present, which is why you also need a way to measure the total yield. The first major attempt at radiochemical yield determination was at the 1952 Mike shot, but it failed because they didn't take account of the contamination of the fallout by uranium in the sea water taken up into the fireball (it was several times bigger than the test island!). Other early problems were due to measuring plutonium-239 in the fallout, which if U-238 is present, is always greater than what you put into the bomb, due to the reaction: U-238 + neutron -> U-239 (23 mins half life) -> Np-239 (56 hours half life) -> Pu-239. (The average amount of Pu-239 formed this way by neutron capture was ~0.5 atom/fission in the devices tested in Operation Castle according to declassified fallout data, see for example CF Miller, USNRDL-466: . Redwing data: ) 172.212.100.165 11:44, 30 March 2006 (UTC)

- Here is some text from my old web page that's no longer up. If somebody knows how to make a table with wikimarkup, they are welcome to add this to the yields section.

Estimated Yield of nuclear explosion from illumination time (in kilotons and megatons)

Illumination Time (seconds) Yield Less than 1 1 to 2 KT 1 2.5KT, 2 10KT, 3 22KT 4 40KT 5 60KT 6 90KT 7 125KT 8 160KT 9 200KT 10 250KT 12 325KT 14 475KT 16 700KT 20 1MT 24 1.5MT 27 2MT 40 5MT 55 10MT 75 20MT

Mytwocents 22:21, 19 December 2005 (UTC) The north korean nuke test is believed to have 1/60th the power of the hiroshima nuke.

Yield estimation and energy distribution
The shock propagation method described above only allows to estimate the mechanical (blast) fraction of the yield. But the energy distribution between blast, thermal radiation and nuclear radiation greatly depends on the weapon type and the total yield. Bombs with large yield-to-weight ratio tend to release more energy as heat while reducing the blast fraction (between 50 and 60 according to Carey Sublette's Nuclear Weapon Archive). In fact, this variance is even larger then, e.g., the uncertainties of the yield of the Tsar Bomba (50-57 MT). Could it be the case that the 50-MT-estimate addresses the blast equivalent while the 57 MT is more closely related to the total energy yield that could be estimated e.g. by the amount of radioactive particles or simply by assuming a typical blast-to-thermal relation extrapolated from earlier tests?--SiriusB (talk) 07:56, 11 April 2008 (UTC)

Am I reading this section correctly ?
"The United States claimed they had the capability of tipping a Titan II ICBM with a 35-megaton-of-TNT (150 PJ) fusion bomb. If this were the case, the yield to weight ratio would be about 9.5 megatons of TNT per metric ton (40 TJ/kg)."

The theoretical limit is stated as 25 TJ/kg in the previous passage. Are both of these numbers right? If so, should a sentence be added here to make this contradiction clear? What is the source of that theoretical ceiling ? I don't know much about this subject, but it seems like a secondary's output might be on the order of 500 TJ/kg just based on T-D fusion and very rough math ... Obviously, there needs to be a primary, but what assumptions are included about the primary's design in that limit calculation? Again I don't know much about this subject I got the 500 TJ/kg number by using a 14 MeV number for D-T and converting it to TJ/kg, perhaps, I have gotten it wrong or this is completely wrong way to think about it... Can anyone expand on this part?

Garick —Preceding unsigned comment added by 128.91.197.64 (talk) 16:00, 27 January 2009 (UTC)

There is definitely something wrong with the passage. —Preceding unsigned comment added by 99.231.50.118 (talk) 00:38, 19 April 2009 (UTC)

Blast efficiency to target
In a fictional series I watch, they say that a next-generation atomic weapon is really advanced because it has an "8% blast to target ratio". That is...actual "weapon yield" is the amount of energy released when the bomb detonates....but most of this energy is radiated outward; an explosion happens in all directions. The bigger the bomb, according to the inverse-square law, actually the less efficient it is (i.e. a 20 megaton bomb only has 10 times the radius of a 20 kiloton bomb, not a thousand times greater). Regardless, what is the term for this? That is, "the amount of blast yield that actually hits the target and it not lost by being directed upwards and outwards"? The idea was that these fictional next-generation bombs were able to "focus" a lot better, losing less of their energy to the environment but concentrating it all more efficiently. Answer on my talk page. --Vi Veri Veniversum Vivus Vici (talk) 20:32, 16 September 2008 (UTC)

Amazing.
Wiki never stops to amaze me.

Claims to be encyclopedia, but presents nearly all US warheads, and only one soviet test. As I said, Amazing. Mk-36 bomb had a 19 megaton yield.6 mt was clean version,but clean weapons apparently never been deployed.Source-Report of NSC AD HOC Working Group on the technical feasibility of a cessation of nuclear testing(Hans Bethe and others).That can be found at www.gwu.edu/~nsarchiv/NSAEBB/NSAEBB94/tb02.pdf.So that Hansen numbers wrong. —Preceding unsigned comment added by 94.242.180.237 (talk) 10:56, 26 June 2010 (UTC)

"Vela Incident?"
Does it strike anyone else as odd that an event that can't be confirmed to have actually occurred, let alone what parties were involved, and doesn't even have any hypothetical information for yield and name can be classified as "milestone?" I'm not really certain of Wikipedia etiquette with regards to editing, and I don't want to screw anything up, but maybe that line in the table should be removed? Either that or a significant amount of caveats added within the "significance" cell for the line. As it stands now, I feel the entry is misleading and misplaced. 84.229.55.39 (talk) 08:29, 29 July 2012 (UTC)


 * Considering the source of the entry, we should all probably take it with a huge grain of salt. Removing the line as you recommend sounds wise.Magneticlifeform (talk) 03:21, 13 December 2012 (UTC)

Maximum yield
Why is 6 kt/kg the theoretical maximum yield? D-T fusion alone would be 81kt/kg if 100% of the fuel is consumed, and. U-235 would be around 20kt/kg. Also this doesn't even account for the energy released when the surrounding atmosphere reacts with excess neutrons like a thermobaric device. 2600:8801:0:1530:485E:7697:6684:D96F (talk) 07:28, 3 February 2017 (UTC)


 * is against Wikipedia guidelines and plain rude. The comment above is correct, 6 kt/kg is not a theoretical limit, it is a practical limit. The article is mistaken. Deleting an on-topic comment that corrects a mistake in the article is detrimental to Wikipedia, inhibiting the correction of an article both technically and behaviorally. BrightRoundCircle (talk) 00:18, 4 February 2017 (UTC)

Seems like you were WP:WIKIHOUNDING the contributor because. BrightRoundCircle (talk) 00:49, 4 February 2017 (UTC)
 * In the case of fission bombs, the limit is what fraction can fission before the fireball is too big, and fission stops. That is slightly less obvious for fusion bombs, but I suspect is again that only a certain fraction reacts before it spreads out too much. Gah4 (talk) 05:23, 24 March 2021 (UTC)
 * In the case of fission bombs, the limit is what fraction can fission before the fireball is too big, and fission stops. That is slightly less obvious for fusion bombs, but I suspect is again that only a certain fraction reacts before it spreads out too much. Gah4 (talk) 05:23, 24 March 2021 (UTC)

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fraction yield
I came to this page trying to remember what fraction of fissile material actually fissions in a fission bomb. I know it is a little more for implosion than gun design. It might also be nice to know for fusion bombs. Gah4 (talk) 05:25, 24 March 2021 (UTC)

sigfigs
The article says: measurement of the energy released by TNT has always been problematic, and then quotes yields to three digits, though most are only two. Gah4 (talk) 13:12, 8 November 2021 (UTC)

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