Timerover51
SOC-14 5K
A.T. Mahan, C.W. Nimitz, and H. Nelson all seem to have felt reserves to be worthwhile
Don John of Austria at the Battle of Lepanto would also concur.
A.T. Mahan, C.W. Nimitz, and H. Nelson all seem to have felt reserves to be worthwhile
A.T. Mahan, C.W. Nimitz, and H. Nelson all seem to have felt reserves to be worthwhile
I disagree. All three, and many others writing on naval strategy recognized that navies don't hold ground (ocean) like an army would on land. Instead, they advocate that naval power be concentrated and a large fleet sent against the enemy in which that entire fleet participates in battle.
If you are looking at much higher yield weapons, a reasonable rule of thumb is 1 kiloton per pound of warhead weight, with very high yield weapons, say 5 or more megatons being more efficient than that.
Interesting... Are there any online links to references you can post so I can have a look?
Carlobrand said:Occurs to me that an interesting alternative is to divide the fleet into 3 groups: a screen, a line, and a reserve. Line ships can only fire mesons through their own screen and can only be targeted by mesons while their screen is intact.
That would only strengthen meson dominance.
That would only strengthen meson dominance.
With a screen meson ships can't be hit by missiles or PAs, so don't need armour. We would see ridiculous unarmoured riders with spinals, sceens and nothing more.
I see no advantage...
In naval warfare a ship unused is a ship wasted. Reserves are for land warfare.
Switching to metric, that rule of thumb works out to a bit over 2kt/kg. Surfing around the web it looks like 6kt/kg is the yield limit on larger weapons or about 20Mt/dton to translate to CT terms (based very roughly on the B-41).
But that is with today's public technology. Can anyone estimate the theoretical yield per kg and per volume of a warhead? I'm going to try, but I doubt my ability to get it right.
EDIT: gonna show the work for the translation to volume because I keep making mistakes
B-41
dimensions: 52" radius, 148" long (model as a cylinder)
weight: 4,839 kg
cu. in to dton: 854,331 (I use 14m^3 per dton)
6kg/kt*4839kg= ~29MT yield at today's "practical" limit
pi*52^2*148/854331 = ~1.5 dton size of a B-41
29MT/1.5dton = 20MT (rounded up since this is all really rough)
All I can say is that the 6 kiloton per kilogram is far too high.
But that is with today's public technology. Can anyone estimate the theoretical yield per kg and per volume of a warhead?
The main limiting factor is the blowback that occurs as the critical mass (which for plutonium, by the way, is around a mere 11kg at the bottom end IIRC) comes together and the nuclear explosion overpowers the assembly implosion, blasting the whole thing apart before only a tiny fraction of the critical mass has been converted to energy.
The 11 kilograms is for U-235 and is high, for P-239 you are looking at more like 5 kilograms. For plutonium, the yield is also going to be dependent on how much Plutonium-240 is in your material. A lot depends on what type of a reflector you are using, and more than that I cannot say.
Wonder what nuclear damper tech will do to nuclear warhead yields...
or gravitic compression in the same strength gravity field that is used to focus lasers...
They say they can now set the yield on a given nuclear warhead, just prior to launch.
To prevent overkill.
...I have long been of the thought that space combat will have to be consensual or forced at choke points such as planets, (L)stations and moons which means orbital velocities or risk mutual destruction.