# On the Stamina of Mikasa's Citadel. The Indestructible Slopes of the Japanese Battleship

Many thanks to everyone who took part in the discussion of my previous material, dedicated to the resistance of the Mikasa's defense against the domestic 12-inch armor-piercing shell. I express special gratitude for suggestions and constructive criticism to the respected Alexey Rytik and the commentator writing under the nickname Yura27.

The comments made forced me to reconsider the approaches to modeling the penetration of Japanese armor by Russian shells. I present to you the results of changing the methodology using the example of the destruction of the citadel of the battleship "Mikasa".

### Citadel - area of engine and boiler rooms

In this area, the citadel was protected by a 222 mm armor belt, coal pits, and a slope consisting of three steel sheets with a total thickness of 76,2 mm. Earlier, I calculated the resistance of the Mikasa's power plant protection based on the assumption that a shell should penetrate a 222 mm Krupp plate while maintaining a speed of 300 m/s, which it would need to overcome the coal in the coal pit and penetrate the slope. In this calculation, I assumed that the slope was located at the same angle as that of the Asahi, i.e. 30 degrees.

In fact, the bevel is not at 30, but at 35 degrees to the horizon line.

Accordingly, a projectile flying parallel to the deck, when hitting the slope, will have a deviation from the normal of not 60, but 55 degrees.

In addition, I mistakenly used the standard de Marr formula for the calculation, which is incorrect in this case, since it is intended for calculations on cemented armor thicker than 75 mm. For homogeneous armor, a slightly different formula should be used.

The Mikasa's bevel consisted of three steel sheets, each an inch thick. I calculated its resistance as the resistance of a "layered" barrier, in which the projectile penetrates each of the specified sheets successively, and this turned out to be correct. However, an error in the formula used and an incorrect bevel angle led to a large error in the calculations.

Earlier I determined that it was sufficient to break through the bevel of the flagship of the United fleet the speed of a 12-inch, 331,7-kg projectile is within 168 m/s, while it is no more than 116 m/s.

At the same time, the Berezan formula can be used to determine the loss of manpower of a projectile to overcome the contents of coal pits.

Unfortunately, like the de Marr formula, it is empirical, and the accuracy of its calculation directly depends on the correctly selected coefficient Kp, which characterizes the "projectile resistance" of a particular type of obstacle. At the same time, it was not possible to find the value of this coefficient for coal.

The whole point is that the Berezan formula is used to determine the parameters of land artillery, which is why its appendices include various types of soil, sand, limestone, brickwork and other materials that field artillery shells mostly encounter. Coal, however, is not listed among them for obvious reasons.

Nevertheless, Kp for coal can be, albeit very conditionally, determined somewhere at the level of 0,04, that is, it is slightly more resistant than compacted sand, and is twice as resistant as brickwork. This, of course, is a very rough estimate, which may be incorrect - however, it should be expected that such an approximation will still be more accurate than the "300 m/s behind an armor plate" I adopted earlier.

Of course, in addition to the "resistance coefficient" of coal, one should know the distance that a projectile will travel in a coal embankment. Considering that the main armor belt of the "Mikasa" only slightly rose above the water, one should consider hits on the upper part of the 222-mm armor plates - here the distance to the slope was about 2,5-3 m.

At the same time, after breaking through the slope, the shell did not fall into the next coal pit, but into the corridor along which ammunition was transported to the 6-inch and 75-mm artillery guns. The fragments of the Russian shell that exploded here, having broken through the relatively thin walls of the corridor, could easily disable steam engines or boilers, damage steam pipes and chimneys. If lucky, the shells moving along the corridor could detonate, which would increase the impact on the engine or boiler room opposite which the hit occurred.

In general, the calculation (K for Krupp armor - 2, steel - 275) gives the following figures. The speed required for a domestic 1-inch 000-kg projectile to overcome:

1) 222 mm armor plates of the main armor belt at a deviation from the normal of 0 degrees - 504 m/s;

2) 2,5 (3) m of coal – 175 (210) m/s;

3) a bevel of three steel plates, each 25,4 mm thick, with a deviation from the normal of 55 degrees – 116 m/s.

To overcome all three obstacles "at the limit", the projectile would have to have a speed of 545-558 m/s at the moment of impact with the 222 mm armor. Therefore, under ideal conditions, the projectile could reach the Mikasa's power plant from about 23-24 cables.

If the projectile hit a 222 mm armor plate with a deviation from the normal of 25 degrees, then the following would happen to it. When passing the armor belt, the projectile would normalize, turning approximately 19 degrees, which follows from the diagram given by Professor L. G. Goncharov in his book "Course of Naval Tactics. Artillery and Armor". The esteemed reader should pay attention to the leftmost curve: along the Y axis there is the deviation from the normal with which the projectile hits the armor, and along the X axis there are degrees of rotation of the projectile in the plate.

In the discussion of my previous material, the opinion was expressed that this diagram is not applicable to shells from the Russo-Japanese War era, since it was compiled for shells equipped with an armor-piercing tip, which was not present on the armor-piercing 12-inch shells of the Russian fleet during the Russo-Japanese War.

However, I am inclined to consider this opinion to be erroneous. L. G. Goncharov in his work gives an example of a solution to the problem of a projectile overcoming spaced armor, which takes into account the normalization of the projectile when passing both the 1st obstacle, which is helped by the armor-piercing tip, and the 2nd, which the projectile reaches without any tip.

Accordingly, the calculation is made based on the fact that the normalization of a 12-inch projectile when penetrating a 222-mm plate will be about 18,5–19 degrees, and a projectile that entered the plate with a deviation from the normal of 25 degrees will exit it with a deviation of 6,5–6 degrees. Nevertheless, this will slightly lengthen its path in the coal pit (by 1,3–1,6 cm) and slightly change the deviation from the normal when reaching the bevel (55,22 degrees instead of 55 degrees).

All of the above will lead to the fact that in order to overcome the protection of the engine and boiler rooms of the Mikasa, a 12-inch projectile will need a speed of 595–606 m/s, which approximately corresponds to a distance of 18–19 cables.

### Citadel - areas outside the power plant

The 222 mm thick section of Mikasa's main armour belt was longer than the boiler and engine rooms, and continued forward and aft of them. These sections lacked additional protection in the form of coal pits, but the slope was reinforced with an additional armour plate one and a half inches thick, or 38,1 mm.

Thus, in this section the bevel consisted of three sheets of steel and one sheet of armour with a total thickness of 4,5 dm or 114,3 mm.

Having made a calculation using a method similar to that used earlier, we find that such protection could be penetrated by a 12-inch domestic armor-piercing projectile at distances of 21–27 cables with a deviation from the normal of 25 and 0 degrees, respectively. Consequently, it can be stated that the appearance of a one-and-a-half-inch armor plate on the slope did not compensate for the absence of coal pits.

Further on, Mikasa had it even worse, since only 222 mm of armor extended from the 173 mm section to the bow and stern. Such protection could be penetrated at a distance of 31–37 cables with a deviation from the normal of 25 and 0 degrees, respectively.

The calculations performed show that the Mikasa's citadel outside the boiler and engine rooms was significantly less well protected than the central part. The reasons why British shipbuilders left such "windows" in the defense, especially opposite the ammunition magazines for the main caliber guns, are completely unknown to me, but such a practice was maintained even on the battlecruisers of the First World War.

I tried to guess and made the assumption that the British built their defense against shells flying perpendicular to the center plane of the ship. In this case, the shells will hit the armor plates in the center of the hull almost without deviation from the normal, but the armor plates located closer to the bow/stern will be located at an angle determined by the contours of the hull.

However, an attempt to measure these angles on Mikasa and the calculations performed on them show that even with this method, equal resistance of the various sections of the citadel is still not ensured.

But there are still some nuances.

### Nuance No. 1 – the distance an armor-piercing projectile travels before exploding

As was said earlier, to destroy the citadel in the area of the engine and boiler rooms, a 12-inch shell must penetrate the armor belt plate, pass through the coal pit and the slope. Having overcome all this, the shell will only have to overcome some very light structures (apparently, structural steel 8-12,7 mm thick, which I ignored in the calculation, due to their obvious insignificance), after which it will end up in the corridor for transporting ammunition to medium-caliber artillery.

If the projectile passes the slope without changing its direction, then its path will definitely be in the ammunition transportation corridor. But if the slope still manages to normalize the projectile (according to the diagram, it will change direction by only 13%), then in this case the projectile's passage into the coal pit is hardly possible.

Accordingly, the shell fragments will only have to overcome the thin bulkhead and then hit the contents of the engine or boiler room, which they will be quite capable of doing.

Consequently, the explosion of a Russian shell immediately after passing the slope within the power plant gives it a good chance of reaching its target (in this case, damaging the Mikasa's engines or boilers). But this cannot be said about shells penetrating the citadel outside the boiler and engine rooms. If such a shell explodes immediately behind the slope, the fragments will have to penetrate several bulkheads and then the feed pipe. High-explosive Russian shells were quite capable of this, but armor-piercing ones are questionable.

In view of the above, in my opinion, the scenario of successful target destruction for shells hitting the citadel outside the power plant should be changed - the shell explosion should occur at least 6 meters behind the armor plate. Accordingly, the shell after overcoming the slope should have enough energy to pass a couple of bulkheads, possibly - to penetrate some mechanisms and at the same time maintain sufficient speed to pass the above-mentioned 6 meters before the fuse is triggered.

However, calculations performed according to this scenario show that the required increase in the projectile velocity on the armor plate is no more than 10–15 m/sec, which results in a reduction in distance of at most 1,5–2,5 cables.

Therefore, even taking into account the above considerations, the penetration of the Mikasa citadel for 12-inch armor-piercing shells at a deviation from the normal of 25 and 0 degrees will be:

For a section of 222 mm + coal pit + 76,2 mm bevel – 18–23 cables (unchanged).

For a section of 222 mm + bevel of 114,3 mm – 19–25 cables.

For a section of 173 mm + bevel of 114,3 mm – 29–35 cables.

### Nuance No. 2 – rebound

Here we need to return to the diagram by L. G. Goncharov, which I have already cited above. However, now we should pay attention not to the left, but to the far right "squiggle". Its essence is very simple - along the Y axis we have the deviation from the normal of the projectile when it hits the armor, and along the X axis - the maximum armor thickness (in calibers) that the projectile can penetrate at all with such a deviation.

How does it work?

Let's look at this with an example.

Let's assume that our projectile hits the 173 mm armor plate of the Mikasa citadel from a distance of 20 cables, and the angle of deviation from the normal is equal to the angle of incidence of the projectile (something around 2,46 degrees). We look at the leftmost curve of the diagram and see that the plate completely normalizes such angles. Therefore, the projectile, having penetrated the 173 mm plate, will exit it with a deviation from the normal of 0. This means that it will reach the slope, moving parallel to the water surface, therefore, the deviation from the normal upon hitting the slope will be 55 degrees.

Now we look at the curve on the far right and see that with such a deviation from the normal, the projectile is capable of penetrating armor with a thickness of approximately 0,363 of its caliber.

Since we are considering a 12-inch shell, its caliber will be 304,8 mm, and the thickness of the armor penetrated will be 111 mm. But the slope of the Japanese battleship was 114,3 mm!

At the same time, L. G. Goncharov points out that:

**Thus, it turns out that the above calculations of the vulnerability of the citadel in areas protected by 114,3 mm thick slopes do not make sense, since shells hitting them should not penetrate such a slope, but ricochet off it.**

Of course, a weighty objection can be raised against this thesis.

The fact is that the Japanese bevel had a total thickness of 114,3 mm, but it was not monolithic, but consisted of 4 layers - three steel and one armor. Obviously, if a monolithic armor plate had been used instead of this pie, then its thickness, with equal resistance, would have been significantly less than both the 114,3 mm bevel and the 111 mm armor, which a 12-inch projectile could still penetrate at a deviation from the normal of 55 degrees. That is, if we count not by the actual, but by the reduced thickness of the armor, then the Russian projectile completely penetrates the specified bevel, and L. G. Goncharov's provisions on ricochet are not applicable to it.

But there is a counterargument to this objection. The fact is that the diagram of L. G. Goncharov is used for all types of armor, both cemented and homogeneous. It is quite obvious that homogeneous armor will be much inferior to cemented armor in terms of resistance with a relatively small deviation from the normal. However, this factor is ignored by Professor L. G. Goncharov - his curves are used for all types of armor.

This means that if the angle of the projectile's impact with the plate is close to the maximum at which it can be penetrated, then the armor's resistance does not affect the armor's thickness, but only the projectile's speed required to penetrate it. This thesis is difficult to understand, so I will explain it with an example.

In the diagram we see that at a deviation from the normal of approximately 26 degrees, the projectile is capable of penetrating armor with a thickness equal to its caliber.

That is, a 12-inch projectile is capable of penetrating (maximum) a 304,8-mm armor plate. Obviously, it will only penetrate it if it hits at a certain speed. For Krupp armor, with "K" = 2, this speed will be equal to 275 m/s. But even if we increased the projectile speed to 699,5, 750 or 800 m/s, this will not allow the projectile to penetrate armor more than 900 mm thick - this is the maximum thickness that can be penetrated at an angle of deviation from the normal of 304,8 degrees for a 26-inch projectile, and further increase in the projectile speed does not increase the thickness of the armor penetrated at this angle.

So, if we take ordinary homogeneous armor with "K" = 1 100 instead of Krupp cemented armor, then the 304,8 mm armor plate with the same deviation of the projectile trajectory from the normal of 26 degrees will be penetrated already at a projectile velocity of 338 m/s. But if we increase this velocity to 699,5 m/s, at which Krupp cemented armor is penetrated under these conditions, or even more, we still will not be able to penetrate homogeneous armor thicker than 304,8 mm.

This is the essence of L. G. Goncharov's diagram, it shows that there is a relationship between the angles of deviation from the normal and the thickness of the armor being penetrated, and it is not affected by the speed of the projectile on the armor (and therefore the durability of the armor). L. G. Goncharov himself speaks about this.

**Due to the above, the 114,3 mm bevel of the Mikasa cannot be penetrated at practically any reasonable combat distances for the Russo-Japanese War. Because a 12-inch shell, no matter what speed it has when it contacts the bevel, should not penetrate, but ricochet off it.**

Of course, when 331,7 kg pieces of steel start flying in the air, anything is possible. As I have shown many times before, armor penetration formulas are strictly probabilistic. It is quite possible that the 114,3 mm slope of a Japanese battleship will still be penetrated - even if according to the formulas and graphs it seems impossible. But the probability of such an outcome should be assessed as minimal - that is, with several hits on the slope, maybe one shell will not ricochet, but will penetrate it.

As always, I am ready to discuss the theses I have expressed above and would be very happy to hear constructive criticism from readers interested in the topic.

And – I’ll allow myself a little intrigue.

Regardless of whether my thesis about 114,3 mm bevels is correct or not, in the course of working on this article I came to very surprising and very different from generally accepted views conclusions about the armor systems of squadron battleships of the Russo-Japanese War. Which I will share in the next article, which I am currently working on.

*Продолжение следует ...*

## Information

sign in.