Talk:Nuclear weapon yield

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[edit] 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

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: [1]. Redwing data: [2]) 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.

[edit] 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)