Best World war two warships?

Hello Juha,

On Your Scharnhorst, looked good, even if as a bit conservative person I like more the 3*2 15in solution.

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I know what you think, but the design of a Battleship is one huge compromise.
All german designs of WWII suffer on a very large citadel and that is one of the reason for their poor horizontal protection and their very heavy armour compared to their limited protection. Bismack has an armour weight of 19006 ts and Iowa 19.400 ts. But Iowa has the much much better armoured scheme and one turrent less. German BS weren't compact like South Dakota, Richellieu or Yamato and I'm a huge fan of compact citadells so I prefer as less turrents as possible. If I would design a "new" Bismarck she will have two quad turrents one front and one on the heck. So i prefer a short citadel and this is a lot easier with two turrents than three.

The armour scheme looks good, probably fuel and 45mm torpedo bulkhead would have contained splinters from any shell deflected by sloped belt into torpedo defence system, what you think?

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You are right. But this is an other compromise. The outer inclined Belt of the italian BB's VV class is my favorite but no german concept. The inline inclined belt is on the poketbattleships and the heavy cruisers so it's ok. The fuel is under the machinery rooms. So if splinters make holes it is not good for sea keeping but for a good and compact armoured scheme to my opinion you need an inclined belt.

Hello Don
My favourites among late BBs are Iowas and SDs, Iowas partly because of speed and looks but SDs because of compactness and effectiveness of its design. Even if US torpedo protection in North Caroline, which itself was a bit different than that in SDs, didn't work as well as was predicted when hit. But that seems to be almost universal, all big navies seemed to have suffered from this phenomenon. And I notice the French influence on your Scharnhorst, IIRC the next French BB would has had one quad turret forward and one rear. My inclination towards 3 twin turrets is based on fact that then loss of one turret would have been less important and SDs were compact design even if they had 3 turrets. On SDs, one thing which I don't like was the internal belt, but as you wrote, designs are compromise.

On inclined belt, yes, they saved weight and all but British and Germans adopted it for their BBs in 30s nad 40s.

Juha

Hello Juha

And I notice the French influence on your Scharnhorst,

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Lol yes I'm a huge fan fan of french designs only their guns were very bad.

On inclined belt, yes, they saved weight and all but British and Germans adopted it for their BBs in 30s nad 40s.

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Sorry but i don't understand this. All BBs from german and the KGV class have no inclined belt.
And the KGV is very vulnerable against all 15inch and more guns. In our german naval forum we calculate with Natan Okuns formula about armoured schemes and guns that KGV is vulnerable against all european 15 inch till 20000m through her main belt. So a straight main belt can only function like the german concect with the slops behind the belt and this is to my opinion not efficient.
So were did the german and the british adopted the inclined belt for their BBs?

Hello Don
by writing "all but British and Germans adopted it " I tied to say that all others adopted it, only British and Germans used vertical belt in their 30s and 40s BBs.

Juha

One 11" shell is still enough to sink any of the Battleships deployed during the war, and considering how accurate and far reaching Sharnhorst's guns were that made her a very lethal opponent to even the heaviest of adversaries.

Soren said:

One 11" shell is still enough to sink any of the Battleships deployed during the war, and considering how accurate and far reaching Sharnhorst's guns were that made her a very lethal opponent to even the heaviest of adversaries.

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Not even the Germans believed that, which is why they were equipping them with 6 x 15in. Only a fluke shot would achieve what you say.

But they didn't Glider, and a penetration of 335mm at 15.1 km is sufficient to be very lethal against even a ship like the Iowa whose belt was 305mm thick.

In an engagement taking place at for example 12 km distance, the Sharnhorst with 9 gun firing 3.5 rounds a min will have a significantly higher chance of a hit than the opposing ship. And a hit at 12 km was going to be devastating if it hit directly on the broadside.

Hello Soren
remember that the belt in Iowas inclined, so that increased its effectiveness and how you thing that Sch could get so near to Iowa, Iowa was marginally faster and would be able to keep the distance long, which suited it.

Juha

ut they didn't Glider, and a penetration of 335mm at 15.1 km is sufficient to be very lethal against even a ship like the Iowa whose belt was 305mm thick.

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Sorry Soren but that is completlely nonsens.
335mm at 15km is for a straight belt not an inclined belt. Iowas Belt is as thick 420-450mm straight belt!

One 11" shell is still enough to sink any of the Battleships deployed during the war, and considering how accurate and far reaching Sharnhorst's guns were that made her a very lethal opponent to even the heaviest of adversaries.

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No!
SH could only fight at short ranges 15000-18000m against thin armoured vessels like Hood, Dunkerque, Repulse, Renown etc.. All other Ships are save. And the range of 15000-18000m is WWI range not the average range for WWII. SHs armoured scheme is for short ranges and her guns too because of their "little" punch compare to all other capital ships and that is her big problem.
For example KGV has 385mm straight belt and nothing behind, from all european BBs in the 30er the worst vertical protection but SH has no chance to penetrate this belt with her 11in guns.

Wait a minute, the Hood was destroyed after being hit by the Bismarck at a range of about 15km, so why was this uncommon ?

Also a round coming in from 15 km away is definitely not going in a straight line, so unless the Iowa's armour belt was inclined by some 45 or so degrees then how is it going to take a direct hit coming from the Sharnhorst without taking any damage ?

Soren said:

Wait a minute, the Hood was destroyed after being hit by the Bismarck at a range of about 15km, so why was this uncommon ?

Also a round coming in from 15 km away is definitely not going in a straight line, so unless the Iowa's armour belt was inclined by some 45 or so degrees then how is it going to take a direct hit coming from the Sharnhorst without taking any damage ?

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First of all and most obviously, the Hood wasn't hit by an 11in shell.
Secondly as you have pointed out the Scharnhorst hit a British Heavy Cruiser with at least two shells and she was able to continue with the battle
Thirdly the Graf Spee hit three fairly small British Cruisers with a number of shells and only the Exeter was knocked out after she was hit I think by four 11 in shells.
The question is, if RN cruisers can take a number of hits and continue the action, why should one hit be devistating to a BB which is designed to take hits from larger shells than that carried by the Scharnhorst?

The unanswered question is, if the 11in was so good then why were the germans trying to rearm the ships with 6 x 15in guns?

Glider,

I'm not saying that a single 28cm shell will always prove devastating, I'm saying that it 'could' prove devastating, just as well as a 38 or 40cm 'could' prove devastating. It all depends on where the hit is made.

The point however being that if the weapon in question is capable of defeating the opponents armour, then it is also a lethal weapon versus that opponent. It's that simple.

And as to why the the German wanted 38cm gun instead, well obviously because they would've been even better in some respects, namely armour penetration. So if the max gun elevation angle was to have remained the same 40 degrees as with the original 28cm guns on the SH, then the 38cm guns would've proven even more useful.

The problem with the Bismarck it seems was that its guns couldn't be elevated past 30 degrees, cause otherwise they actually reached out futher than most other guns.

Hello Soren
Quote:"Wait a minute, the Hood was destroyed after being hit by the Bismarck at a range of about 15km, so why was this uncommon ?"

most of us probably know this but as explanation, the British commander Holland wanted to decrease range as fast as possible because of the weak deck armour of Hood. Why Germans allowed the decrease, difficult to say but the were 2 RN heavy cruisers shadowing Germans and maybe Germans wanted keep them away of the battle as long as possible, or maybe Germans calculated that shorter range suited also to them or it is true that after the ships aproaching from SE were identified as capital ships Lütjens seemed to "paralyzied" and didn't gave any orders for a while or...

Quote: "Also a round coming in from 15 km away is definitely not going in a straight line, so unless the Iowa's armour belt was inclined by some 45 or so degrees then how is it going to take a direct hit coming from the Sharnhorst without taking any damage ?"

Ohm, have you looked the "imaginary Schanhorst" armour scheme, inclined armour was inclined so that its lower end was more inward, so the shell was coming "downhill" and the inclined armour, in Iowa's case to 19deg, combined to form the strike/hit angle.

Juha

Yes Juha, so that means that at 15km the 28cm shells will be hitting at a 29 degree angle on the Iowa's side armour, at which range the 28cm shell was expected to slam through 335mm of armour IIRC, some 30mm more than what the Iowa was packing. So that in my book would make the 28cm guns quite lethal against the ship.

Soren said:

Yes Juha, so that means that at 15km the 28cm shells will be hitting at a 29 degree angle on the Iowa's side armour, at which range the 28cm shell was expected to slam through 335mm of armour IIRC, some 30mm more than what the Iowa was packing. So that in my book would make the 28cm guns quite lethal against the ship.

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If the 11in didn't prove lethal in any of the actions it was used in, why should it be lethal against a BB?

The only change the Scharhorst would have is a fluke shot.

PS you can include the sinking of the Glorious in this. The escort of 2 small destroyers were sunk but closed to within torpedo range. 2 hours to sink 1 carrier and 2 destroyers with 2 Battlecruisers is hardly impressive.

In real life the angles could be some what off.

Computing the angle these penetrations based on a most favorable condition seems like asking for a lot of luck.

As in ships steaming on exact parallel course. instead of converging or diverging by even 5 or 10 degrees.

Ships being exactly abeam of each other instead of one ship leading or following the other. AS in ship "A" being 20,000yds away but 1,000yd ahead steaming a roughly parallel course. Yes they are broadside to broadside but the impact is not going to be 90 degress in horizontal plane.

Ships are rolling, THis averages out but means that any one shot could hit more sloped armour (or less) in the vertical plane.

The firing rate for the Scharhorst seems just a bit high. At what angle of elevation did the Scharnhorst achieve the 3.5rpm rate? While it might get to 3.5rpm in a short range blast fest at medium to long range I would think that things would low down a bit. at 16-18Degrees of elevation the Scharhorst is going to spend 14 seconds out of every minute just lowering and raising the guns from shooting postion to loading position and back.

While the Bismark's guns couldn't elevate past 30^ this may not have been that much of a disadvantage.
At that elevation they could reach 35,500 meters which is a good way beyond the longest hit scored in combat. At these extreme ranges a hit on a moving ship was almost pure luck. Time of flight for a German 15in shell was almost 14 seconds longer from 30,000meters to 35,000 meters. A 30kt ship could move about 700ft further in those 14seconds than the amount it moved for the shell to reach 30,000yds. Total time of flight to 35,000meters is given as 69.9 seconds. Obviously longer ranges involve even more flight time. Time of flight For the American 16"/50 was 79.96 sec to 36,580 meters.

This also points out another problem with firing at long range. How many salvos do you want in the air at one time before one lands so it can be spotted and corrections made?
Waiting for one salvo to land before firing the next means a very slow rate of fire and lots of opertunity for the target to change course/speed between shoots.
Shooting as quickly as possiable means much more rapid feed back but also means some salvoes will be wasted on empty ocean.
Firing "half salvoes" at rapid rate does give prompt feed back and save ammo but cuts down on the chances of a hit.
Long range naval gunnery was a bit like duck hunting with buckshot. You fire a group of projectiles at the projected future path of the target. the smaller the area the group covers the better the chances of a hit if the path has been correctly predicted. The more shells in the group the the better the chance of a hit. Too few shells in the group and the target could very well be centered but unhit. If the group is too tight and the path prediction isn't good enough the entire group misses.
THe longer the range the larger the area the group covers again lowering the chances of a hit.

only usefull info, that i've, it's the rate of elevation of Scharnhorst it's 8°/second and loading angle it's +2 so from 18° (already high for real battle) need 2 second for down and 2 second for up

...same you tell

Soren, Soren
As an engineer, you must have heard trigonometry. 335mm from 15km for 28cm is against vertical belt, and again Iowa's 12.1in on 0.875in belt was/is inclined 19deg, so it's thickness is just enough, at least in theory and using simply LOS analyze. Of course armour penetration isn't wholly exact science.

Juha

Juha said:

Soren, Soren
As an engineer, you must have heard trigonometry. 335mm from 15km for 28cm is against vertical belt, and again Iowa's 12.1in on 0.875in belt was/is inclined 19deg, so it's thickness is just enough, at least in theory and using simply LOS analyze. Of course armour penetration isn't wholly exact science.

Juha

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Ah, back to your good old snide remarks Juha. I'm glad to see how you've grown as a person these past few years

1st. The 28cm shells achieved a penetration performance of 335mm of armour at an impact angle of 30 degrees AFAIK, seeing that this was std. practice for German gunnery tests, heck US British too. Also AFAIK for naval gunnery tests the Germans even had their targets stand a side angle of 20 degree's on top of this to simulate the uncertainty of impact angles in actual combat.

2nd. Oh yes I know trigonometry quite well Juha, I believe most peolpe past the age of 12 do. And you said yourself that the Iowa's armour was inclined inwards by 19 degrees, right? Well then taking into consideration that at a range of 15km the 28cm shell is coming in at an angle of 10 degrees then that equates to an impact angle of exactly 29 degree's on the belt armour. In short, the Iowa's armour should be quite vulnerable at that range then.

These are the characteristics of the 28cm gun which I recently posted on a rel;ated thread in this forum:

Range
.
Elevation With 727.5 lbs. (330 kg) APC L4,4 Striking Velocity Angle of Fall
2.0 degrees 5,470 yards (5,000 m) 2,513 fps (766 mps) 2.5
4.3 degrees 10,940 yards (10,000 m) 2,139 fps (652 mps) 5.7
7.4 degrees 16,400 yards (15,000 m) 1,824 fps (556 mps) 10.3
11.3 degrees 21,870 yards (20,000 m) 1,578 fps (481 mps) 17.2
16.2 degrees 27,340 yards (25,000 m) 1,430 fps (436 mps) 25.7
22.0 degrees 32,810 yards (30,000 m) 1,371 fps (418 mps) 35.3
29.2 degrees 38,280 yards (35,000 m) 1,404 fps (428 mps) 44.0
38.2 degrees 43,740 yards (40,000 m) 1,509 fps (460 mps) 52.0

Armor Penetration with 727.5 lbs. (330 kg) APC L4,4 Shell
.
Range Side Armor Deck Armor
0 yards (0 m) 23.79" (604 mm) ---
8,640 yards (7,900 m) 18.09" (460 mm) 0.76" (19 mm)
16,514 yards (15,100 m) 13.18" (335 mm) 1.63" (41 mm)
20,013 yards (18,288 m) 11.47" (291 mm) 1.87" (48 mm)
30,000 yards (27,432 m) 8.08" (205 mm) 2.99" (76 mm)
Note: The above information is from "Battleships: Axis and Neutral Battleships in World War Two" for a muzzle velocity of 2,920 fps (890 mps) and is based upon the USN Empirical Formula for Armor Penetration.

.
Range Side Armor Deck Armor
10,936 yards (10,000 m) 13.70" (348 mm) ---
16,404 yards (15,000 m) 11.02" (280 mm) ---
21,872 yards (20,000 m) 8.86" (225 mm) ---
27,340 yards (25,000 m) 7.64" (194 mm) ---
Note: The above information is from "German Capital Ships of World War Two." The data is based upon the pre-war Krupp test shoots on their range in Meppen with L/4,4 APC projectiles using RPC/32 propellant against KC-type armor at an impact angle of 70 degrees.

I copied all of this data from the site German 28 cm/54.5 (11") SK C/34

The Iowas had the following armouring scheme

Main Belt 12.1 case hardened steel on ).875 STS (Special Tensile Steel)
Deck: 6 in and 1.5 weather deck, on 0.625 STS
Bulkheads: 11.3 in

The Nav weapons data shows that the Iowas deck protection scheme was more or less completely immune to damage at any practical eange. To penetrate the side armour, the 28cm round would need to close to somewhere between 12-15000 metres to ahave any hope of a main belt penetration. For a Battleship, this is extremely close range, essentially point blank range in fact.

In comparison the USN 16/50 Mk VII has the following characteristics:

Range
.
Elevation With 2,700 lbs. (1,224.7 kg) AP Mark 8 (new gun) With 1,900 lbs. (861.8 kg) HC Mark 13 (new gun)
10 degrees 17,650 yards (16,139 m) 18,200 yards (16,642 m)
15 degrees 23,900 yards (21,854 m) 24,100 yards (22,037 m)
20 degrees 29,000 yards (26,518 m) 28,800 yards (26,335 m)
25 degrees 33,300 yards (30,450 m) 32,700 yards (29,901 m)
30 degrees 36,700 yards (33,558 m) 36,000 yards (32,918 m)
35 degrees 39,500 yards (36,119 m) 38,650 yards (35,342 m)
40 degrees 41,430 yards (37,884 m) 40,600 yards (37,163 m)
45 degrees 42,345 yards (38,720 m) 41,622 yards (38,059 m)
Notes:
1) With reduced charges, the AP Mark 8 had a maximum range of 24,180 yards (22,110 m) while the HC Mark 13 had a maximum range of 27,350 yards (25,010 m).

2) Time of flight for AP Shell with MV = 2,500 fps (762 mps)
10,000 yards (9,140 m): 13.2 seconds
20,000 yards (18,290 m): 29.6 seconds
30,000 yards (27,430 m): 50.3 seconds
36,000 yards (32,920 m): 66.1 seconds
40,000 yards (36,580 m): 80.0 seconds

3) Time of flight for HC Shell with MV = 2,615 fps (797 mps)
10,000 yards (9,140 m): 13.1 seconds
20,000 yards (18,290 m): 30.3 seconds
30,000 yards (27,430 m): 53.2 seconds
35,000 yards (32,000 m): 70.3 seconds
39,500 yards (36,120 m): 86.0 seconds

4) The maximum range with the originally planned 2,240 lbs. (1,016 kg) AP Mark 5 was 47,000 yards (42,980 m). Muzzle velocity would have been 2,700 fps (823 mps) with a charge of 640 lbs. (290 kg) SPD.

5) At an "average gun" MV of 2,425 fps (1,739 mps), the maximum range with AP Mark 8 at a 40 degree elevation was 40,185 yards (36,745 m

Armor Penetration with 2,700 lbs. (1,224.5 kg) AP Mark 8
.
Range Side Armor Deck Armor Striking Velocity Angle of Fall
0 yards (0 m) 32.62" (829 mm) --- 2,500 fps (762 mps) 0
5,000 yards (4,572 m) 29.39" (747 mm) 0.67" (17 mm) 2,280 fps (695 mps) 2.5
10,000 yards (9,144 m) 26.16" (664 mm) 1.71" (43 mm) 2,074 fps (632 mps) 5.7
15,000 yards (13,716 m) 23.04" (585 mm) 2.79" (71 mm) 1,893 fps (577 mps) 9.8
20,000 yards (18,288 m) 20.04" (509 mm) 3.90" (99 mm) 1,740 fps (530 mps) 14.9
25,000 yards (22,860 m) 17.36" (441 mm) 5.17" (131 mm) 1,632 fps (497 mps) 21.1
30,000 yards (27,432 m) 14.97" (380 mm) 6.65" (169 mm) 1,567 fps (478 mps) 28.25
35,000 yards (32,004 m) 12.97" (329 mm) 8.48" (215 mm) 1,555 fps (474 mps) 36.0
40,000 yards (36,576 m) 11.02" (280 mm) 11.26" (286 mm) 1,607 fps (490 mps) 45.47
42,345 yards (38,720 m) 9.51" (241 mm) 14.05" (357 mm) 1,686 fps (514 mps) 53.25
Note: The above information is from "Battleships: United States Battleships 1935-1992" for a muzzle velocity of 2,500 fps (762 mps) and is based upon the USN Empirical Formula for Armor Penetration. These values are in substantial agreement with armor penetration curves published in 1942.


This means that the Mk VII guns could start to engage the Scharnhorst from a massive 35000. Scharnhorst would need to close over 20000 yards relative range before she could effectively return fire. If the closure rate is say 2 knots (4000 yds per hour), it would take 5 hours of continous closure before the Scharnhorst could effectively return fire. Essentially, impossible I am afraid