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“standard” US battleships, Germany and England. We believe armor

"Стандартные" US battleships, Germany and England. We believe armor

In this article we will try to deal with the armor penetration of the guns of battleship types Bayern, «Revenge», «Pennsylvania», as well as the comparative quality of German, American and British armor. This is extremely difficult to do., because the data on the American 356 mm, German 380-mm and British 381-mm guns are very sketchy and incomplete, and sometimes contradict each other, but we will try anyway.

In what, as a matter of fact, problem? Let's see, like most naval lovers (and not only) stories compare the armor penetration of certain guns. For example: in one edition, dedicated, for example, English dreadnoughts, contains information, that a British 381 mm shell of the First World War era pierced 381 mm armor plate at a distance of about 70 cable. In another edition, already dedicated to German "capital" ships - that a similar German 380-mm projectile "mastered" 350 mm armor only with 67,5 cable. It seems to follow from this, that the English cannon is more powerful - this is precisely the conclusion made.

However, in reality, comparing similar data in this way, very easy to get screwed.

Is the above data obtained from actual shooting, or they are calculated using armor penetration techniques? If these are the results of actual shootings, whether their conditions were identical for both guns? If the armor penetration is obtained by calculation, whether the same techniques were used? Are the obtained data the result of the work of specialists of the relevant ministries and departments?, or is it the result of calculations by historians, took up the calculator? understandably, that in the second case the accuracy will be much lower ...

You don't have to go far for examples: take the famous monograph C. Vinogradov, "Superdreadnoughts of the Second Reich" Bayern "and" Baden ". In Appendix No. 2, a respected historian together with V.L.. Kofman makes a large amount of calculations, so as, to compare the capabilities of the battleships "Rivenge" and "Bayern". but alas, just look at the table of parameters for 15-inch guns (pp. 124) and we will see, that according to the calculations of respected authors, British 381 mm cannon at elevation 20,25 hail has a range of only 105 cable, that is about 19,5 thousand. m. While foreign sources for the same initial speed (732 m / sec) and a slightly lower elevation angle (20 city.) give much larger distances - 21,3-21,7 thousand. m. Of course, such deviations from real values ​​have the most negative effect on the calculation results.

But even if the sources present the results of calculations by specialists, the accuracy of which there is no doubt, then another, complicating factor: it's the quality of the armor. clear, that the same Englishmen, calculating armor penetration when designing a particular dreadnought, enjoyed the corresponding rates of English armor, Germans - respectively German, etc. And the armor of different countries may differ in durability, but this is not so bad either: after all, in a single country, the same Krupp armor was constantly being improved.

In this way, obtained, that the calculations of artillery systems, completed, for example, in England, and seemingly for the same Krupp armor, but made at different times, may not be comparable. And if we add to this the almost complete absence of serious work on the evolution of the armor case in various countries of the world ...

Generally, more or less reliable comparison of armor penetration - the task is far from so simple, as it might seem at first glance. AND, in a good way, non-professional (what, without any doubt, is the author of this article) it's better not to take this business. But, alas - to our deep regret, the pros are somehow in no hurry to deal with these issues, so ... as they say, in the absence of stamp paper, we write in simple.

of course, it is no longer possible to carry out full-scale tests of the above-mentioned artillery systems, so our destiny is calculations. And if so, then it is necessary to say at least a few words about the formulas of armor penetration. Modern calculation methods even if published, then only in closed editions, and in popular literature the Jacob de Marr formula is usually given. Interesting, that Professor of the Naval Academy L.G.. Goncharov, in his artillery release textbook 1932 city, called it the Jacob de Marra formula. This formula, along with many others, was widespread at the beginning of the last century, and, need to say, it is quite accurate - perhaps, it is even the most accurate among similar formulas of those years.

Its peculiarity lies in the fact, that she is not physical, that is, it is not a mathematical description of physical processes. De Marra's formula empirical, it reflects the results of experimental shelling of iron and steel armor. Despite this "unscientific nature", de Marra's formula showed the best approximation to the actual results of firing and Krupp armor, than other common formulas, and therefore we will use it for calculations.

Those interested will find this formula in the appendix to this article., but there is no need to force everyone to understand it, reading this material - this is not necessary to understand the conclusions of the article. Note only, that the calculation uses very simple and familiar concepts to all those interested in the history of military fleets. This is the mass and caliber of the projectile, armor thickness, angle, under which the projectile hits the armor, as well as the speed of the projectile when it hits the armor plate.

However, de Marr, of course, could not limit myself only to the above parameters. After all, the penetration of a projectile depends not only on its caliber and mass., but also to a certain extent from its shape and quality of steel, from which it is made. And the thickness of the armor plate, which the projectile is able to overcome, depends, of course, not only on the performance of the projectile, but also on the quality of the armor. Therefore, de Marr introduced a special coefficient into the formula, which the, as a matter of fact, and is designed to take into account the specified qualities of armor and projectile. This coefficient increases with the quality of armor and decreases with the deterioration of the shape and quality of the projectile..

As a matter of fact, the main difficulty in comparing artillery systems of different countries is precisely "rests" on this very coefficient, which we, further, let's just call (TO). We will need to find it for each of the above weapons - if we, of course, we want to get a somewhat correct result.

so, let's first take fairly common data on the armor penetration of the German 380-mm / 45 gun "Bayern", according to which the weapon at a distance 12 500 m (the very 67,5 cable's length) could pierce 350 mm armor. We use a ballistic calculator to find the parameters of a 750-kg projectile, with initial speed 800 m / sec at the moment of impact on armor: obtained, that such a projectile hits a strictly vertically located armor plate at an angle 10,39 city., with speed 505,8 m / sec.

A small disclaimer - hereinafter, when we talk about the angle of the projectile, I mean the so-called "angle from the normal". "Normal" is when the projectile hits the bonneplite strictly perpendicular to its surface, that is, at an angle 90 city. Respectively, projectile hit at an angle 10 city. from the normal, means, that he hit the slab at an angle 80 city. to its surface, deviating from the "reference" 90 city. on 10 city.

But back to the armor penetration of the German gun.. Coefficient (TO) in this case it will be approximately (rounding to the nearest integer) is equal 2 083 - this value should be considered quite normal for armor of the era of the First World War. But here comes one problem.: the fact, that the source of data on armor penetration is the book «German Capital Ships of World War Two», where the 380-mm / 45 gun "Bayern" was compared with the main caliber of the battleship "Bismarck".

And couldn't it be, that the indicators of the Krupp armor were taken into account, created between the two world wars, which was much stronger, Cem th, what was installed on the "Bayenne", Rivenge and Pennsylvania? Especially, what the electronic encyclopedia navweaps reports, that there is evidence, what's in the distance 20 000 m German 380-mm shells were able to penetrate 336 mm armor plate, and we are talking specifically about the armor of the era of the First World War.

Well, We consider: on 20 km the angle of incidence will be 23,9 city., projectile speed on armor - 410,9 m / sec., and the coefficient (TO) - some unfortunate 1618, which does not fit at all with the armor resistance of the WWI era. A similar result generally brings German-made Krupp armor closer to homogeneous armor resistance ... Obviously, that the navweaps data contains some kind of error.

Then let's try to use another source of information.. Until now, we have used the calculated data, and now we will try to compare them with the results of actual tests of the German 380 mm / 45 cannon: those are given from. Vinogradov in the above-mentioned monograph, dedicated to German battleships.

It describes the consequences 3 rounds of armor-piercing shells, on armor plates thick 200, 290 and 450 mm, the latter being the most interesting for us: projectile weight 734 kg hit the armor plate at an angle 0 (that is, under 90 city. to the surface) and at speed 551 m / s struck 450 mm through the slab. A similar result corresponds to the coefficient (TO) 1 913, but, actually, it will be slightly lower, because the Germans found their shell already in 2 530 m behind the barrier he broke, and - as a whole. Alas, without any information about, how much of this distance the projectile flew through the air, how much - "rode" on the ground, it is completely impossible to determine the energy stored by it after armor penetration.

Now let's take the British 381 mm / 42 artillery system.. Alas, data on its armor penetration is rather vague: So, V.L.. Kofman there is a mention of, that these British guns pierced the armor, its own caliber at a distance of about 70 cable. But with what projectile and with what initial speed? Considering, that the reference is contained in the monograph, dedicated to the battle cruiser "Hood", and refers to the period of creation of this ship, It can be assumed, we are talking about 871 kg projectile.

However, another question arises here.: the official muzzle velocity of such a projectile was 752 m / sec, but some calculations by the British were carried out at a lower speed 732 m / sec., so what value should we take? However, whichever of the indicated speeds we take, coefficient (TO) will fluctuate within 1 983 – 2 048, and this is above, than we calculated for the value (TO) for the german gun. It can be assumed, what does this say about the superiority of the quality of English armor in comparison with German ... or is it, that the geometric shape of the German projectile was better suited for penetrating armor? Or maybe, the thing is, that V.L.. Kofman are calculated values, but in practice British shells would have achieved better results?

Well, we have data on the results of the shelling of the battleship "Baden".

"Стандартные" US battleships, Germany and England. We believe armor Photo «Badena» under fire

So, one of the English shells, hitting an angle 18 city. at a speed 472 m / sec., "Overpowered" 350 mm frontal armor of the German main battery turret. These data are all the more valuable., that in this case it was not the English, and german armor, that is, the tests of 381 mm / 42 and 380 mm / 45 guns are, thereby, in a single coordinate system.

Alas, and they don't help us too much. If we assume, that an English shell pierced the German tower, what is called, "With the last bit of strength", and be there armor 351 mm - he couldn't have done it anymore, then in this case his (TO) will be equal 2 021. Interesting, by the way, that C. Vinogradov is indicated, what a british shell, punched 350 mm frontal armor of the German turret, subsequently not found, but in fact the report states something else - it exploded, and there is a description of that, where the debris flew in the tower.

certainly, we have no absolute reason to believe, that this penetration was the limit for a 381 mm projectile, or at least close to that. But still, according to some indirect signs, one can assume, what exactly was it. This is hinted at by another hit: British 871 kg projectile, caught in 350 mm barbet at an angle 11 city., although he was able to make a hole in the armor with a diameter of 40 cm, but he himself did not go inside the barbet, bursting in the process of breaking through the armor. In this case, the hit occurred almost in the very center of the barbet, that is, the curvature of the armor plate if it had any effect, the very minimum.

From all of the above, you can try to draw some conclusions., but, due to the fragility of the evidence base, they, of course, will be highly speculative.

Conclusion 1: German armor during the First World War roughly matched the resistance of the English. This conclusion is true, if V.L.. Kofman about, that the British 381 mm / 42 gun was capable of penetrating armor, equal to its caliber 70 cbl, and if we were not mistaken in the assumption, what a break 350 mm of the frontal plate of the German tower at an angle 18 city ​​and speeds 472 m / sec. is the limit, or very close to, the penetration limit of the British 381 mm shell.

Conclusion 2. Apparently, the shape and quality of the German 380-mm projectile provided it with the best armor penetration, what did English have. Based on the above data, we can assume, what ratio (TO) the British 381-mm projectile when firing at German armor was about 2 000, while the German 380-mm projectile has about 1 900. If our first conclusion is correct, that the armor resistance of British and German armor is approximately equivalent, it is obvious, that the only reason for the lower coefficient (TO) maybe only the shell itself.

Why a German shell could have been better? Its caliber is slightly smaller, one millimeter, but, of course, this could hardly have any significant impact. Calculation shows, that with the same mass (750 kg) caliber change to 1 millimeter will lead to an increase in armor penetration by 1,03 millimeter. The German shell is also shorter - its length was 3,5 caliber, while the length of the British Greenboy is 4 caliber. maybe, there were other differences. of course, the quality of steel plays a significant role here, from which the projectile is made.

Now let's calculate the armor penetration of the German and British guns for a distance of 75 cables - the generally accepted distance for a decisive battle, on which one could expect enough hits to destroy the enemy ship of the line.

At the specified distance 871 kg British shell 381 mm / 42 cannon, launched at initial velocity 752 m / sec., hit a vertically positioned armor plate at an angle 13,05 city, and its speed "on the plate" was 479,6 m / sec. at (TO) equal 2000, according to Jacob de Marr's formula, the armor penetration of the British shell was 376,2 mm.

As for the German shell, then everything is a little more complicated. If our conclusion is, that he was superior to English in armor penetration, correct, then the capabilities of the German 380-mm / 45 guns on 75 the cable ones were approaching the English fifteen-inch.At this distance the German 750 kg projectile hit the target at an angle 12,42 city ​​of speeds 482,2 m / sec., and at (TO) equal 1 900 armor penetration was 368,9 mm. But if the author of this article is still mistaken, and for the German gun it is worth using the same coefficient, as for the English cannon, then the capabilities of the 380-mm projectile fall to 342,9 mm.

Nevertheless, According to the author, armor penetration of the German projectile is closest to 368,9 mm (after all, practical shooting gave a coefficient 1 913 despite the fact that the shell then flew to 2,5 km), but the armor penetration of the English projectile may be slightly lower than the calculated. In general, it can be considered, that at a distance 75 kabeltov British and German artillery systems are quite comparable in terms of armor penetration.

But with the American 356 mm / 45 gun, everything turned out much more interesting.. Canonical in the Russian-language literature should be considered the previously cited data for shells with a mass 680 kg.

"Стандартные" US battleships, Germany and England. We believe armor

As a matter of fact, the values ​​specified in it, kind of, lead to very obvious conclusions: even if 680-kg shells, appeared in the USA after 1923 city, in terms of armor penetration they are inferior to their 380-381-mm European "colleagues", what can we say about earlier 635 kg shells, which were equipped with 356-mm artillery of American dreadnoughts! They are lighter, which means they lose speed faster in flight, while their muzzle velocity did not exceed the heavier shells, and the shape and quality of ammunition 1923 g. should take precedence. Clear, like a day, that the American "Pennsylvania" at the time of entry into service were inferior in terms of armor penetration to the English and German dreadnoughts. Well, it's obvious, true?

This is exactly the conclusion the author made, considering the capabilities of American fourteen-inch aircraft in the article "Standard" US battleships, Germany and England. American "Pennsylvania" ". And then he picked up a calculator…

The thing is, that the calculation according to the de Marra formula showed that the American 356-mm / 45 guns had the armor penetration indicated in the table with the coefficient (TO), equal 2 317! In other words, the results shown in the table are American 680 kg shells demonstrated when exposed to armor, created not in the era of the First World War, and on much later and more durable samples.

Hard to tell, how much the strength of armor protection has increased in the interval between the first and second world wars. In Russian-language sources on this matter, there are only brief and, often, conflicting references, based on which we can assume, that the strength of Krupp's armor increased by about 20-25%. In this way, for large-caliber shells of the First World era, an increase in the coefficient (TO) will amount to 1 900 – 2 000 to 2 280 – 2 500, but here you need to remember, that with the growth of the quality of armor, of course, the quality of shells also grew, and therefore for heavy ammunition WWII (TO) maybe less. therefore (TO) at the rate of 2 317 for post-war shells, naturally, improved on the basis of previous experience, looks quite organic, but - for armor of the era of the second world war, not the first.

But by setting the coefficient (TO) for American 680-kg shells at the level 2 000, that is, bringing the quality of body armor to the era of the First World War, we are for distance 75 cables, we get armor penetration at the level 393,5 mm, that is above, than the British and German fifteen-inch guns!

"Стандартные" US battleships, Germany and England. We believe armor

Conversion to 635 kg projectile gives a very insignificant correction -ballistic calculator showed, that at a distance 75 cable, having an angle of incidence 10,82 city. and speed "on the armor" 533,2 m at (TO) equal 2 000, American projectile penetrates World War I era armor, thickness 380 mm, that is, significantly more than its own caliber!

On the other hand, quite possible, that such a calculation is still not entirely correct. The thing is, that according to some information, coefficient (TO) for the same armor decreases with increasing projectile caliber. So, eg, in our calculations the maximum value (TO) for the German 380-mm / 45 artillery system, obtained by calculation and published in sources, is 2 083. At the same time, the calculations for the same German 305-mm / 50 guns, which were installed on the ships of the Kaiserlichmarine since the "Helgolands", data from sources on armor penetration give (TO) at the level 2 145. Respectively, not excluded, what we took to calculate the armor penetration of American 356 mm / 45 guns (TO) = 2 000 still too small.

Besides, Unfortunately, the author does not have any "clues" for, to compare the armor resistance of American Krupp armor with its European counterparts. Nothing remains, how to consider it equivalent to German and English armor protection, although it, of course, maybe not so.

Let's summarize all these, rather messy data. Taking into account the errors of the "methods" used in the calculations, it is possible with a high degree of probability to assume, whatarmor penetration of vertical armor protection of the main caliber guns of the battleships "Rivenge", Bayern and Pennsylvania at a distance 75 cables were about the same, and was 365-380 mm roughly.

Despite a bunch of assumptions, the data at our disposal still allow us to draw some conclusions regarding vertical armor protection. But with the breakdown of horizontal barriers, what are the armored decks, everything is much more complicated. The thing is, that Jacob de Marr, Unfortunately, I didn't bother with creating a formula for determining the durability of horizontal protection. Its basic formula is, adapted to modern types of armor, only suitable for calculations of cemented armor, thick over 75 mm. This formula is given in Appendix No. 1 to this article, all the previous calculations in the article were made on it.

But the decks of the ships of those years were protected by non-cemented (heterogeneous) but with homogeneous armor, which lacked a surface hardened layer. For such armor (but - installed vertically!) a different formula is used, designed for evaluating non-cemented armor plates with a thickness of less 75 mm, it is given in Appendix No. 2.

Would like to note, that both of these formulas are taken from a more than serious source: "Course of naval tactics. Artillery and Armor " 1932 g. author - Professor of the Red Army Naval Academy L.G. Goncharov, one of the leading experts of the pre-war USSR in the field of naval artillery.

And alas, but none of them are suitable for assessing the durability of horizontal protection. Using the formula for cemented armor, then at a distance in 75 cable we get scanty armor penetration: 46,6 mm for 381 mm / British 42, 39,5 mm for 380 mm / 45 German, and 33,8 mm for 356 mm / 45 American guns. If we use the second formula for uncemented armor, we get, that when hit at an angle, characteristic of the distance in 75 cable, all three artillery systems penetrate easily 74 mm armor plates, after that, maintaining a huge supply of kinetic energy - so, eg, British 381-mm shell for penetrating armor of this thickness at a distance 75 cables will have enough speed 264,5 m / sec, while its speed will be 482,2 m / sec.

If you ignore the limitation on the thickness of the armor plate, it turns out, that the British 381 mm shell, according to the above formula, capable of penetrating deck armor over 180 mm! what, of course, absolutely impossible.

If we try to refer to the test results of the Bayerne-class battleship, we will see, that armor piercing 871 kg British shells twice hit the horizontal armor of the towers, having a thickness 100 mm angle 11 city., what is the distance 67,5 cables for a projectile with an initial velocity 752 m / s and 65 cables - for a projectile with an initial velocity 732 m / sec. Both times the armor was not pierced. But in one case, shell, ricocheted, made a groove in the armor 70 cm, that is, the slab is bent very strongly. And in the second, although the projectile, yet again, ricocheted, the armor was not only concave to 10 cm, but also torn.

"Стандартные" US battleships, Germany and England. We believe armor

This nature of damage suggests, what, although the German 100 mm armor and provided protection at specified distances, but - already if not at the limit of the possible, it is very close to that. But the calculation using the formula for cemented armor gives armor penetration for all 46,6 mm at greater distance, where the angle of incidence will be higher, and, respectively, it would be easier for a projectile to penetrate the deck armor. That is, according to the formula, it turns out, what 100 mm the deck had to jokingly and with a large margin of safety reflect British shells - however, practice does not confirm this.

In the same time, as calculated by the formula for non-cemented armor, obtained, that the roofs of the Baden's main caliber should have been easily pierced, and - with a large supply of energy of shells - that, yet again, absolutely not confirmed by practice.

Need to say, that such inaccuracies in calculations have a completely logical explanation. As we have said previously, de Marr's formulas are not a mathematical description of physical processes, but are just a fixation of patterns, obtained when testing armor. But vertical armor protection was tested, not horizontal, and not surprising at all, that the patterns in this case simply stop working: for horizontally placed armor, into which the shells fall at a very small angle to their surface, these patterns, naturally, completely different.

The author of this article came across opinions "on the Internet" that, that de Marr's formulas work effectively at angles of deviation from the normal not more than 60 city., that is, from 30 hail to the surface of the plate and more. One can assume, that this estimate is very close to the truth.

In this way, with regret we have to state, that the mathematical apparatus at the disposal of the author does not allow performing any reliable calculations of the horizontal protection resistance of the Rivenge battleships, Bayern and Pennsylvania. Due to the above, it will be difficult to use the data on the armor penetration of horizontal armor, given in various sources - as a rule, they are all based on the same calculations according to de Marr's formulas and are incorrect.

application 1

"Стандартные" US battleships, Germany and England. We believe armor

application 2

"Стандартные" US battleships, Germany and England. We believe armor

To be continued…

/Andrew from Chelyabinsk, topwar.ru/

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