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[Jura The idiot]

Right now I wonder about this: If Yamato had met

TG 59.7 - Fast Battleship Bombardment Group - created only 45/03/24
BB South Dakota/FF, New Jersey, Wisconsin, Missouri

what would have been Yamato's tactics if she was engaged at good visibility conditions at, let's say, 1000 hours ("no surprising night attack")? I mean if to "try to get them one by one" or "distribute fire" or "this would depend" ??

 

[chuck731]

Right now I wonder about this: If Yamato had met

TG 59.7 - Fast Battleship Bombardment Group - created only 45/03/24
BB South Dakota/FF, New Jersey, Wisconsin, Missouri

what would have been Yamato's tactics if she was engaged at good visibility conditions at, let's say, 1000 hours ("no surprising night attack")? I mean if to "try to get them one by one" or "distribute fire" or "this would depend" ??

That is an overwhelming force. Individually, the defensive qualities of the American battleships are about on par with the Vanguard, ie none of them has a real immunity zone against the Yamato. But the last 3 can pierce Yamato's side armor from 20kms. So they can deal serious or fatal damage to the Yamato from reasonable battle ranges, unlike the Vanguard. They also have combined weight of broadside almost 3.5 times that of Yamato. Yamato doesn't have enough fire to distribute over 4 opponents, unless you consider the 6" secondary guns, which is unlikely to seriously weaken any one of the 4 opponents before Yamato herself is overwhelmed. In any case, the statistical chance of Yamato making any decisive hits on more than one the American ships while the American ships try to rapidly close to 20kms is not all that high. Yamato is also too slow to run or to dictate the range or the direction of the engagement. So there is not much tactical options for the Yamato.

I think Yamato's best alternative is to pursue a steady course to be the best gunnery platform she can be while the American forces are trying to close to 20kms, in the hopes of knocking out one american ship with a lucky hit before the Americans can close to range where American guns can also inflict decisive damage. Once the America ships come to 20KM, Yamato ought to accept that Yamato can no logner win, and can only hope to out last the ammunition supply of the 4 American ships by maneuvering vigorously to avoid getting hit while using the threat of her own more powerful guns to keep Americans from closing too aggressively, hoping American doctrine does not call for the American force to sacrafice one of their own just to get the Yamato.

Yamato is an exceptionally maneuverable battleship in terms of having unusually tight turning radius and relatively low loss of speed during sharp turns. She could make herself a very tough target for long range gunnery if she is willing to sacrafice her own gunnery in the bargain by jinking constantly. There are examples, such as battle of Kommandorski island, to show WWII surface fire control really is fundamentally not up to the challenge of hitting from long range an enemy who is not trying to steady her own gunnery at all, but is instead manuevering violently to avoid getting hit. If she can keep the Americans from closing to much less than 20km, if her steering doesn't give out under the strain, then she has a chance of outlasting American ammunition supply without taking a crippling hit.

But If Yamato has to seriously fight rather than just evade, or if the Americans just press in closer and accept that Yamato might just get one of the American ships before she herself succumbs, then I think Yamato is doomed. In that case she should concentrate on one target in the hopes that it can deal a crippling or even fatal blow to at least one enemy battleship before she herself is overwhelmed.

 

[Jura The idiot]

chuck731 I went over your interesting text, have just a couple of quotes which I think confirm what you've said:

1. Yamato had 16" belt inclined at 20 degrees, which makes its belt more or less equivalent to an vertical 18" plate. This means Yamato's side belt is totally immune to 15"/42 1920lbs AP shell, or immune at all but point blank range (~5km) for any other shells fired by that gun.

Garzke & Dulin on Vanguard's "Armor Penetration Capabilities, 15-inch Gun" (p. 290) don't give anything from below 10000 yards (and this would be: 16.5" Vertical Plates)

2. Yamato had 8" deck, protected by a 1.5" fuse trigger weather deck. It can not be penetrated by any 15"/42 shell fired up to maximum elevation of 30 degrees. So if vanguard's 15" turrets were limited to 30 degrees, Yamato's citadel is totally immune from vertical penetration.

Garzke & Dulin on Vanguard's "Armor Penetration Capabilities, 15-inch Gun" (p. 290) don't give anything from above 30000 yards (and this would be: 5.7" Deck Plates)

3. Vanguard had 14" vertical belt. Yamato can penetrated it in optimal condition any any range out to slightly beyond 30km.

Garzke & Dulin say about Vanguard (p. 293) "The belt abreast the magazines was proof against 15-inch shells outside 15400 yards; abreast of the propulsion plant it was sufficient down to 17600 yards".

4. Vanguard had 5.25" deck, Yamato can penetrate it at somewhere between 20km and 30km, likely much closer to 20km than 30km. But let's be conservative and say 25kms.

Garzke & Dulin say about Vanguard (p. 293) "Horizontal protection ... was also considered proof against 15-inch shells inside 31000 yards."

Yamato's citadel has a colossal immunity zone from around 5km out to maximum range of vanguard's guns. Vanguard has no immunity zone whatsoever against Yamato. In fact there is a substantial zone of double jeopardy from <25km - >30km where Yamato's shells will penetrate both the deck and the belt on the vanguard.

that's just great

... British shells in the same situation would likely be kicked into a wild unpreditcable underwater trejectory by water impact, and explode prematurely before going very far.

Yamato's armor design anticipates other countries might try to achieve the same underwater performances with their heavy shells (which none actually did).

Campbell in section GREAT BRITAIN/PROJECTILES: "In 1944 it was decided to make changes in AP shell design, though these were not carried through. It was realised for the first time that the assumption that a shell which did not ricochet continued in a straight line on striking water was entirely false, a fact that might have been learnt at any time for the Shoeburyness ranges, and designs to give a long, straight underwater trajectory were put in hand, as they had been earlier in Japan."

...

Vanguard was designed at a time when the British had no inkling that the Japanese were building monster battleships 50% heavier than anything the British were contemplating. ...

On p. 281 Garzke & Dulin mention one of the factors in Vanguard's design was the following: "British intelligence had also determined that Japan was preparing to build two 32000-ton battlecruisers armed with 12-inch guns and designed for high speed."

There's one more weakness of the Vanguard I don't think you mentioned (sorry if I overlooked, your post is pretty long THE CONNING TOWER PROTECTION (p. 293 of Garzke & Dulin: "The Royal Navy persisted in providing only protection sufficient to resist fragmentation damage or at best hits from 5-inch shells, even in vital ship-control areas.") It'd seem those "vital ship-control areas" could've been taken down by precise fire of the 15.5 cm/60 (6.1") 3rd Year Type

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You see, I'm armchair admiral on active duty

 

[Jura The idiot]

...

Thats very advanced physics and actually, is the defeat mechanism of soviet heavy ERA Kontakt 5 proven against M829A1 which guillotine the APFSDS round and spalls the penetrating half.

Lezt I don't understand tanks but I noticed an interesting effect: At long ranges, the striking velocity of 15" shell INCREASED, look (Campbell, p. 27, Range and Elevation Data for 1938 lb Shell, 2575 f/s Muzzle Velocity)

range (in yards) * elevation (ROUNDED BY ME) * descent (ROUNDED BY ME) * time of flight (in s) * striking velocity (in f/s)
25000 * 16 deg. * 22 deg. * 39.86 * 1560
30000 * 21 deg. * 30 deg. * 51.27 * 1497
35000 * 28 deg. * 38 deg. * 64.59 * 1496
NOT YET ... BUT NOW:
36500 * 30 deg. * 41 deg. * 69.20 * 1507
(hope this "table" is readable; the same trend can be seen for a 4-crh 15" shell, and for 14")

I read a comment on this in some book, but I forgot in which one, and now it's tough to find that comment without Ctrl+F heheh

 

[Lezt]

Lezt I don't understand tanks but I noticed an interesting effect: At long ranges, the striking velocity of 15" shell INCREASED, look (Campbell, p. 27, Range and Elevation Data for 1938 lb Shell, 2575 f/s Muzzle Velocity)

range (in yards) * elevation (ROUNDED BY ME) * descent (ROUNDED BY ME) * time of flight (in s) * striking velocity (in f/s)
25000 * 16 deg. * 22 deg. * 39.86 * 1560
30000 * 21 deg. * 30 deg. * 51.27 * 1497
35000 * 28 deg. * 38 deg. * 64.59 * 1496
NOT YET ... BUT NOW:
36500 * 30 deg. * 41 deg. * 69.20 * 1507
(hope this "table" is readable; the same trend can be seen for a 4-crh 15" shell, and for 14")

I read a comment on this in some book, but I forgot in which one, and now it's tough to find that comment without Ctrl+F heheh

Thats pretty normal, the energy of a shell can only be eroded after it leaves the barrel (as nothing is propelling it). The path of the shell follows the Brachistochrone curve... which going back to my university years in advanced dynamic... is the path determined cyclic integration between potential energy and kinetic energy.

in simple terms, potential energy does transform into kinetic energy perfectly depending on the path which the projectile takes.

More.. descriptively in this case, since aerodynamic drag is is basically proportional to the velocity of the shell, and for simplicity sake, potential energy is directly proportional of the height the shell is reached above the point of fire.

The shell loses kinetic energy when it gains height; there is a sweet point (angle) where the shell losses velocity so quickly when gaining height (which is quasi regenerated on descend), that it losses less energy to aerodynamically drag (which is not regenerate-able and therefore lost for good). Therefore, the shell losses velocity quickly, but it regains it when it descends.

So this might seem odd, but the flight times should be higher than the lower angle shells (disproportionately so) as the "cruising velocity" is lower.

 

[Jura The idiot]

Thats pretty normal, the energy of a shell can only be eroded after it leaves the barrel (as nothing is propelling it). The path of the shell follows the Brachistochrone curve... which going back to my university years in advanced dynamic... is the path determined cyclic integration between potential energy and kinetic energy.

in simple terms, potential energy does transform into kinetic energy perfectly depending on the path which the projectile takes.

More.. descriptively in this case, since aerodynamic drag is is basically proportional to the velocity of the shell, and for simplicity sake, potential energy is directly proportional of the height the shell is reached above the point of fire.

The shell loses kinetic energy when it gains height; there is a sweet point (angle) where the shell losses velocity so quickly when gaining height (which is quasi regenerated on descend), that it losses less energy to aerodynamically drag (which is not regenerate-able and therefore lost for good). Therefore, the shell losses velocity quickly, but it regains it when it descends.

So this might seem odd, but the flight times should be higher than the lower angle shells (disproportionately so) as the "cruising velocity" is lower.

Lezt your explanations inspired this simple question of mine: What was the maximum height of, say, a 15" shell shot at 30 degrees of elevation with the muzzle velocity of 785 m/s? I mean the true value, not obtained neglecting atmospheric drag etc.! as

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just gave me 7852 m (but the time of flight was not completely off: 80 s instead of about 69 s I guessed from Campbell's table) A moment ago I quit a google search so ... thanks!

 

[chuck731]

So this might seem odd, but the flight times should be higher than the lower angle shells (disproportionately so) as the "cruising velocity" is lower.

This feature is exploited in modern land artillery to achieve what is called Multiple Round Simultaneous Impact (MRSI), where the same gun would fire 3-5 rounds in quick succession, using successively larger propellant charges and lower elevation each time, to enable the rounds fired sequentially to follow successively faster trajectory and thus arrive at the same target at the same time.

This is done because it is observed that an artillery barrage against an enemy army position does the most damage during the first 10 seconds, because after than the enemy would have taken cover. So MRSI maximizes the effectiveness of artillery by allowing a gun that could fire only one round every 10 seconds to put 5 rounds on target within 1-2 seconds.

 

[Jeff Head]

Okay, I will give you a different, and perhaps more realistic "what if" scenario.

Say at the Battle of Leyte Gulf the Japanese Task Forces had been a little differently arranged and one of the IJN Groups had consisted of the following:

Battleships Yamato and Musashi, and the two Tone class heavy cruisers, Tone and Chikuma.

Now, let's further postulate that on the evening of October 24th, 1944, that that above mentioned Task Force had received only minimal damage from carrier attack during the day, had received no surface action damage and was proceeding towards the US anchorage.

Let's further imagine that Halsey had taken the bait and gone off after the Japanese carriers but had formed up and dispatched TG 59.7 consisting of the battleships South Dakota, New Jersey, Wisconsin, and Missouri, that evening because he knew there was this Japanese Task Force out there and that TF 59.7 intercepted these Japanese ships before they could get to the Taffy Escort carriers the morning of October 25, 1944, and the two groups came together and engaged that morning.

What then? How would the two Yamato Class faired against the four US BBs

 

[Jura The idiot]

Okay, I will give you a different, and perhaps more realistic "what if" scenario.

Say at the Battle of Leyte Gulf the Japanese Task Forces had been a little differently arranged and one of the IJN Groups had consisted of the following:

Battleships Yamato and Musashi, and the two Tone class heavy cruisers, Tone and Chikuma.

Now, let's further postulate that on the evening of October 24th, 1944, that that above mentioned Task Force had received only minimal damage from carrier attack during the day, had received no surface action damage and was proceeding towards the US anchorage.

Let's further imagine that Halsey had taken the bait and gone off after the Japanese carriers but had formed up and dispatched TG 59.7 consisting of the battleships South Dakota, New Jersey, Wisconsin, and Missouri, that evening because he knew there was this Japanese Task Force out there and that TF 59.7 intercepted these Japanese ships before they could get to the Taffy Escort carriers the morning of October 25, 1944, and the two groups came together and engaged that morning.

What then? How would the two Yamato Class faired against the four US BBs

Jeff why don't you tell us what you think would have been the tactics of the American battleships ... I'm eager to hear this too ... please

 

[Rutim]

Jeff why don't you tell us what you think would have been the tactics of the American battleships ... I'm eager to hear this too ... please

Keep in mind that US Navy didn't have a clue what Yamato was carrying aboard. Before Samar and photos they got there they thought it was 16in guns as indicated in reports from the battle off Samar. No to even mention they surely didn't had a clue about the protection of those ships. Any kind of of miscalculation could cost a lot of lives.