Talk:High energy weapon design

Post your comments and discuss them here. —The preceding unsigned comment was added by Tmayes1965 (talk • contribs) on 05:53, 10 February 2006.

internal contradiction
"The high explosive hollow sphere will at least double the density of the fissionable pit, and it will also reduce its volume by at least 20%." This was added in the revision 09:20, 26 November 2005 by 68.39.174.238

if you reduce its volume by 20% you've increased its density by 20%. if you double the density you've reduced its volume by half. My guess is these were two different designs and both somehow made it into the text. It makes me wonder if the rest of the facts have been checked too, or if many more of the definitive-sounding comments should be phrased as (e.g.) "in one design, the high explosive hollow sphere doubles the density of the fissionable pit" rather than "the high explosive hollow sphere will at least double the density of the fissionable pit".

Correct me if i'm wrong. —The preceding unsigned comment was added by Kstinch (talk • contribs) on 05:44, 17 April 2006.

When the density is doubled, the volume decreases by just 20% Some implosion weapon designs only double the density of the pit. Other designs may increase the density up to 5 times —The preceding unsigned comment was added by Tmayes1965 (talk • contribs) on 17:34, 25 April 2006.

This is nonsense.
A little common sense needs to be applied here. If the density is doubled and weight remains constant then volume halves. The reference to a reduction in volume of 20% is not consistent with a doubling of density. Not on this planet, or any other. However ..... I may be able to offer an explanation for how this confusing figure of 20% came to appear.


 * Assume a solid sphere of 1dm radius.


 * Volume = 1³ x π x 1.333 = 4.188 dm³


 * At double the density the volume will halve to 2.094 dm³


 * Therefore: 3√(2.094 ÷ 1.333 ÷ π) = 0.794 dm = 79%


 * ERGO: Doubling the density and halving the volume has decreased the solid sphere RADIUS and DIAMETER by 20%

But this is becoming tedious. Get some help with maths. Brian.Burnell 01:56, 10 June 2006 (UTC)


 * If I'm reading his comments right, he's referring to peak density during implosion, and volume of the fissionable material prior to implosion. That doesn't mean his numbers are accurate; just explains the apparent discrepancy you noted. Or, I could be misreading the thread above. --Christopher Thomas 04:43, 10 June 2006 (UTC)


 * I don't see it as being that complicated. Its really quite simple. If a soft rubber ball is squeezed down to half its volume while its weight remains constant, its density must double, and the maths shows that the rubber ball would decrease in diameter by 20%. I genuinely don't understand why others seem to be having such a problem with this.


 * It does illustrate a useful lesson though, about the need for clear and concise language. If Tmayes had meant to say that when the density of a sphere doubles, its DIAMETER decreases by 20%, then he should have said that and phrased it clearly, and he would have been correct. All that his muddled references to a 20% reduction in volume have achieved is to confuse others and divert them from doing useful things. Brian.Burnell 13:58, 10 June 2006 (UTC)

People the formula for calculating the volume of a sphere is 1/3 * Pi * R ^ 3* Tmayes1965


 * No, it's $$\left ( \frac{4}{3} \right ) \pi r^3$$. Check sphere, or the nearest geometry textbook, for confirmation. --Christopher Thomas 20:32, 26 June 2006 (UTC)


 * The formula listed above by Christopher Thomas is absolutely correct. Tmayes is wrong. There are no excuses for a failure to check the most basic facts, and it is a sad reflection of falling standards in education. What is worse though is a stubborn refusal to accept help freely offered and in good faith.Brian.Burnell 13:27, 27 June 2006 (UTC)