Talk:B41 nuclear bomb

message: The mark number (Mk) denotes the warhead, the B41 is the bomb which carried the warhead. The warhead was also used in Atlas ICBMs. If a merge were to occur I think it would make more sense to merge the delivery systems into the warhead. Anynobody 03:25, 25 July 2007 (UTC)
replies: text: No. The warheads are denoted with a W, e.g. W41. "Mk" was used to describe the bombs as a whole up until the mid-1960s. This warhead was not used on Atlas ICBMs. In fact, the ICBM version was cancelled in 1957. Atlas used the W49 and W38 warheads. Dziban303 (talk) 12:37, 28 July 2014 (UTC)
thread_title:  Mk41 and B41 have different meanings 

message: Has this device ever been tested? Drhex (talk) 12:49, 26 October 2011 (UTC)
replies: text: No 27.79.186.97 (talk) 13:00, 9 May 2023 (UTC)
thread_title:  Testing? 

message: The Mk 41 was of the usual long cylindrical shape and weighed 10,670 lb (4,840 kg). The nuclear fusion warhead was of the Teller-Ulam type and used a 40–100 kiloton implosion type nuclear fission primary fueled by HEU to trigger the lithium-6 deuteride fusion fuel. Between 500 and 1,000 kg of lithium deuteride was used and was contained in a cylinder of natural uranium with an inner casing of U-238.
This makes no sense at all, as natural uranium is essentially U-238. Even attempts to make pure U-238 (depeleted uranium or DU) still have removed only about half of the U-235. There is no role for "pure" U-238 here when you have natural uranium or DU. It makes more sense that this is a typo and the inner casing was actually U-235 to increase yield (U-235 is more efficiently fissionable with fast neutrons than U-238). I suspect that the boasted-of 35 Mt weapon for the Titan was one in which all the tamper was U-235. This is a lot of U-235 to be sure, but tons of HEU were in existence then, and still are, for submarine and other military power reactors, and it would have been available for something where very high efficiency was needed, as in an ICBM. Ultimately, of course, the MIRV concept killed the monster bombs. However, even now I wonder if the jackets of W-88 and other US warheads aren't HEU (U-235) or at least enriched with 235. SBHarris 23:12, 27 April 2012 (UTC)
replies: text: In absense of comment I've gone ahead and replaced the typo of U-238 with the intended U-235. SBHarris 01:24, 7 May 2012 (UTC)
replies: text: So you have "inferred" that the prior unsourced edit was correct except for a typo that replaced U235 with U238?  How do you know it is a typo rather than the original edit being a confused, uninformed mess? On the strength of this dubious assumption, you have taken it upon yourself to tell the whole world that the Mk41 Y1 had a HEU (of course, there is no such thing as "pure" U235 in any usable amount, but you know that) encased tertiary?  Fascinating.  It might be true.  But is there ANY available source that says so (other than you, of course)?  It is not an assumption we can make on the available information.  I'm changing it to "uranium."  You are very welcome to put the HEU back into the Mk41 Y1's tamper when there is a source for it more reliable than guessing. Criticality (talk) 23:59, 27 February 2016 (UTC)
replies: text: All: The U-238 is converted to Pu-239 by the neutrons from the fusing  lithium deuteride. Wikipeidans really should deliberate more before wading into technical articles. Greg L (talk) 04:10, 23 March 2017 (UTC)
thread_title:  Inner casing of what? 

message: Would a photo of the B41 replica at the Air Force Museum be helpful to the article? Craigbucher (talk) 21:08, 18 May 2015 (UTC)
thread_title:  Photos 

message: It says the weapon hat a free fall option.
I did some approximate calculations (http://keisan.casio.com/exec/system/1231475371) with the speed ad service ceiling specified by Wikipedia for the B-52 (1,047 km/h and 15,000m). The casing does not look particularly aerodynamic, compared to the Tallboy, so I chose an air resistance factor high enough so it only just failed to go supersonic in free fall. That gave a maximum drop time of some 72 seconds (15,000 metres to ground level). During that time, the bomber could fly about 21 km. Using Pythagoras, that puts the bomber just a little under 26 km from the epicentre at the time of explosion.
Now I don't have the math to quickly approximate exactly when the shock wave would catch up with the bomber let's say it was about another 90 seconds later, during which the bomber could fly another 26 km horizontally. Some more Pythagoras and we have a distance of about 49 at impact. IF the bomb is dropped from the plane's ceiling almost all the way to the ground.
So that should be enough for the plane to get away.
But what about an airburst?--Cancun771 (talk) 22:22, 16 January 2016 (UTC)
replies: text: Air-dropped thermonuclear weapons from day-one fall under parachutes. Greg L (talk) 03:55, 23 March 2017 (UTC)
thread_title:  Free fall = suicide mission? 

message: I seriously doubt there would have been between 500 and 1000 kg of lithium deuteride in a B41 like this version of the article stated. Lithium deuteride releases orders of magnitude more energy per gram than uranium and plutonium do upon fissioning.
A serious shortcoming with such a statement is the cited reference says no such thing. Also, such a value implies spectacular inefficiency, and U.S. bombs were—and are—exceedingly efficient. A single 6Li2H molecule fusing releases 22.4 MeV of energy per molecule, which means less than 10 kg of 6Li2H (at 100% fusion) could provide all 40 MT of yield all by itself without the need for a third stage; that's 1% of the amount claimed to be in the device. Even the Fat Man plutonium bomb was 25% efficient.
The general rule is one can get a leverage of about 100:1 per stage, so that's 20 kT for a single-stage fission bomb, 2 MT for a fission/fusion bomb, and 200 MT for a three-stage bomb. Even if one surmised that the extra 99% 6Li2H was required to produce the excess neutrons necessary to convert the U-238 third stage to Pu-239, you don't get the neutrons without all the nuclei first fusing. A total of 1000 kg of 6Li2H fusing is a continent-busting 4 gigatons of yield; that’s an impossible 20,000:1 leverage ratio from the first-stage trigger. Out of a mere two-stage device? No one would even need three-stage weapons unless they wanted to usher the age of the dinosaurs back in.
It's not cited and none if this passes the grin test of someone in a persistent vegetative state.
There seems to be a 1000X error here and that is explainable by a simple mixup in units: grams vs. kilograms. If there was 1000 grams and 50% of it fused, that would be a perfectly plausible 2MT before we get to the third stage. But that's just a hunch and I'm not changing a unit of measure on a hunch.
Maybe I'm wrong on my understanding or math. But until there is a (very) reliable source backing up statement like “500 and 1000 kg of lithium deuteride”, the article should remain silent on such detail. I've deleted it. Greg L (talk) 04:34, 23 March 2017 (UTC)
replies: text: You made some significant mistake in your calculations. Li6D produces about 60 kt yield per kg. If we assume 50% fission, that's a minimum of 166kg of Li6D, but if you look at Swords, Hansen describes most of these large weapons as being about 25 to 50% efficient in their fusion stage.Kylesenior (talk) 09:45, 26 March 2022 (UTC)

text: 22.3 MeV * 6*10^3 (Avogadro constant) = 2.12*10^12 Joules per mol, or 507 tons TNT per mol. A mole of Li6D weighs 8 grams, therefore Li6D produces 2.7*10^14 Joules per kg or 63.4 kt per kg.Kylesenior (talk) 09:53, 26 March 2022 (UTC)
thread_title:  Between 500 and 1,000 kg of of lithium deuteride?