Turret Armor

Discussions about the history of the ship, technical details, etc.

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George Elder
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From Pugliese...

Post by George Elder »

-c) The Admirals’ Committee discussed, in [light of] such circumstances, the advantages or not of equipping the bulwarks with a single thick [layer of] armor instead of a layered protection [scheme]. Following the presentation made by the representative of the Permanent Committee on the advantages of the latter, it was found worthwhile in every way to prefer over the first a protection divided between two plates, one decapping and one resisting.

More in a bit...
George Elder
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Testing program...

Post by George Elder »

This material was unknown to the British examiners in the above 320 mm trials, but it's part of Pugliese's record. It indicates quite an active R&D program:

++++++++++++++++++++++++++++++++++++++++++
Allegato No. 4: Translation

Addendum 4

Result of the firing trials on models of the armor belt at the waterline

The tests conducted at the Balipedio [ballistic firing range] in scale (1:1.88) were the following:

1° model = made out of two areas: one in which the decapping [plate] of 40 mm [thickness] was placed 20 cm from the resisting K.C. [plate] of 150 mm [thickness], and the other in which the said distance was 10 cm.

From the tests performed has emerged:

a) these two composite structures are equivalent, in regard to perforating shells, to similar structures with larger interspaces [that were] previously tested, meaning that the projectile is decapped by the external decapping plate and it breaks up upon hitting against the cemented plate, independently from what the distance is between the two plates, at least up to the minimum value tested

b) In regards to explosions [caused by high explosive shells hits] the structure with the smallest interspace behaves advantageously compared to similar structures with greater distance between the two plates.

2° model = with interspace of 6 cm, [we] fired sub-normal shots from which is was inferred that even with such a distance between the two plates, the projectile is decapped on the first [plate] and breaks up upon hitting the cemented plate, but the resultant plug [ejected] from the K.C. plate possessing a greater energy.

3rd model = with an interspace of 10 cm – the shot was fired at an inclination of 70 degrees from normal. The projectile perforates the decapping plate loosing its cap, scrapes against the resistant plate making room for itself in the interspace [region] between this one [the resisting plate] and the decapping [plate].

[end of document]
George Elder
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Lastly...

Post by George Elder »

There is a point in his text wherein Pugliese points toward yaw induction as being as important a factor in the array's ability to defeat shells as is decapping. This is probably why the space issue wasn't a critical factor in this particular system. I've written a 130 page paper on the subject, but it has languished. Alas, my job has taken me away from the hunt. But suffice it to say -- these are not odd ball results. They are the end result of an extensive R&D program -- the full depth of which the British and US did not know. Indeed, the Allegato refered to above is being seen for the very first time -- to my knowledge -- in its translated form. This is just a bit of what is contained in a remarkable document that Dr. Rastelli discovered. Will the project ever be completed? I don't know. But I do know this. A good degree of what is promoted as Hoyle within our community cannot withstand even a small bit of serious study.

George
Paul L
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Post by Paul L »

Quick response, Tanks never had spaced armor until they needed them. Since the vast majority of the projectiles produced could be countered by more of the existing armor it makes sence not to change that procedure. Some have doctrins that prohibit the use, like the Americans who despite wide spread use of side hull skirts on NATO tanks , rejected them cause they interfered with maintainance schedules...until 1973 then they were proposed for the M-60 tank. But that came to nothing since it was seen to compete with XM-1 programe.

UK had side screens since WW-II and I know an all round spaced armor was tested on the centurion. But against large shaped charges it didn't help enough to justifie the expense.
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Interesting

Post by George Elder »

You noted:

UK had side screens since WW-II and I know an all round spaced armor was tested on the centurion. But against large shaped charges it didn't help enough to justifie the expense.

How did this array work against AP and APC shells?

George
Paul L
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Re: Interesting

Post by Paul L »

George Elder wrote:You noted:

UK had side screens since WW-II and I know an all round spaced armor was tested on the centurion. But against large shaped charges it didn't help enough to justifie the expense.

How did this array work against AP and APC shells?

George
No they just were testing a SS-10 ATGM warhead against the tank target.
marty1
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Re: George Elder

Post by marty1 »

Paul L: Sounds good to me! I have toyed with a simplistic formula involving spaced armor effects. It seems to work well with smaller projectiles & APFSDS.

The net additional resistance of a given spaced plate is t/d [spaced plate] ^0.5 * Penetrator velocity at impact. Plate hardness would increase this spaced plate effectiveness by ~ 1.25 -1.34 for SHS or VHS/FH

So a 16 inch AP round striking a 3 inch FH spaced plate becomes
0.25^0.5*0.5 = 0.25 diameter additional resistance or LOS + 4 inches.

if the same array is struck by say a 14 inch shell, the resistance is
0.29^0.5*0.5 = 0.27 diameter additional resistance or LOS + 3.75 inches.

If the same array is struck by a 12 inch shell,the resistance is
0.33^0.5*0.5 = 0.29 diameter additional resistance or LOS + 3.46 inches.

This doesn't address the possibility of shell exploding inbetween spaced plate and main plate , or the amount that the ballistic cap affects the above figures?
This makes absolutely no sense.

In the first example 16" AP vs. 3" FH is an e/d ratio of 3/16 = 0.1875, not 0.25. So one would assume that the equation would be:

SQRT(3"/16") x Velocity = 0.433 x Velocity.

Are we to assume that the second example also applies to a 3" FH spaced plate??? If that is the case than e/d = 3/14 = 0.21, not 0.29. Or did you change the FH plate thickness and not tell the rest of mere mortal who maybe trying to follow along with this thread.

I assume the 0.5 is supposed to be 0.5km/s. This is a rather abnormal way of stating impact velocity for naval ordnance of this period. One would expect velocity units in either either feet/second or meters/second. If your using forms of velocity with non-standard units of measure, why not state what units you are employing so we can all follow along with your post?

Was this formula developed by you? Or is there is an applicable reference that disscusses development of this equation?
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Re: George Elder

Post by Paul L »

marty1 wrote:
Paul L: Sounds good to me! I have toyed with a simplistic formula involving spaced armor effects. It seems to work well with smaller projectiles & APFSDS.

The net additional resistance of a given spaced plate is t/d [spaced plate] ^0.5 * Penetrator velocity at impact. Plate hardness would increase this spaced plate effectiveness by ~ 1.25 -1.34 for SHS or VHS/FH

So a 16 inch AP round striking a 3 inch FH spaced plate becomes
0.25^0.5*0.5 = 0.25 diameter additional resistance or LOS + 4 inches.

if the same array is struck by say a 14 inch shell, the resistance is
0.29^0.5*0.5 = 0.27 diameter additional resistance or LOS + 3.75 inches.

If the same array is struck by a 12 inch shell,the resistance is
0.33^0.5*0.5 = 0.29 diameter additional resistance or LOS + 3.46 inches.

This doesn't address the possibility of shell exploding inbetween spaced plate and main plate , or the amount that the ballistic cap affects the above figures?
This makes absolutely no sense.

In the first example 16" AP vs. 3" FH is an e/d ratio of 3/16 = 0.1875, not 0.25. So one would assume that the equation would be:

SQRT(3"/16") x Velocity = 0.433 x Velocity.

Are we to assume that the second example also applies to a 3" FH spaced plate??? If that is the case than e/d = 3/14 = 0.21, not 0.29. Or did you change the FH plate thickness and not tell the rest of mere mortal who maybe trying to follow along with this thread.

I assume the 0.5 is supposed to be 0.5km/s. This is a rather abnormal way of stating impact velocity for naval ordnance of this period. One would expect velocity units in either either feet/second or meters/second. If your using forms of velocity with non-standard units of measure, why not state what units you are employing so we can all follow along with your post?

Was this formula developed by you? Or is there is an applicable reference that disscusses development of this equation?
The difference between 0.185 and 0.25 is the nominal difference between the resistance of FH armor vs the resistance of RHA reference , which was used as the basis of most ballistics. If the plate was mild steel, the adjustment would be 0.8 to 0.67 depending on the BHN of the mild steel plate.

The 0.5 figure refers to ~ 500 m/s velocity or 0.5km/s. Most modern ballistics forumlas work from such simple forms, inorder to simplify and stream line calculations...since the aim is to get a simplistic 'caliber' driven increase.

I choose 0.5km/s since it seems to roughly approximate the velocity of most long range naval AP shooting of the time period. That figure can be adjusted to fit what ever gun you need.
marty1
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Post by marty1 »

So than the initial format of your equation should be:

[(t/d x 1.34)^0.5] x V for FHA; or
[(t/d x 1.25)^0.5] x V for HHA

Going back to your first example of 16” AP striking 3” of FHA: Is the 3” FHA spaced plate supposed to be the total thickness of the spaced armor array? In other words is this two 1.5” FHA plates with some level of air gap between the two plates; Or is this a 1” plate followed by a 2” plate; Or..? Or is the 3” FHA an initial plate thickness with an arbitrary air-gap distance, followed by another plate of arbitrary thickness?

What exactly are you trying to say with the results of this equation? Does the 3” FHA plate (or is this an array with a total thickness of 3”) resist like 4” of FHA plate vs. 16” AP at an impact velocity of 500m/s?

Or does the 3” FHA plate resist like the line of sight thickness plus 4”? So at a 0-deg impact angle the 3” FHA plate resists like 7-inches?

In other words, does your complete equation for FHA look like:

{[(t/d x 1.34)^0.5] x V x d} + t/cos(Z)

Or does it look like:

[(t/d x 1.34)^0.5] x V x d

Projectile Diameter = d
Plate Thickness = t
Projectile Velocity = V (units: Km/s)
Angle of Impact = Z
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Dave Saxton
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Post by Dave Saxton »

Paul L wrote:[The difference between 0.185 and 0.25 is the nominal difference between the resistance of FH armor vs the resistance of RHA reference , which was used as the basis of most ballistics. If the plate was mild steel, the adjustment would be 0.8 to 0.67 depending on the BHN of the mild steel plate.
.....
So the resistance of the RHA is ~25% more than FH? This wouldn't happen to be British Type I RHA would it?
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Post by marty1 »

Dave:

I think what he is attempting to say is that RHA is the basis -- so you would multiple t/d x 1 for RHA. But it's tough to say for sure if this is what is meant. I'm not sure I even understand what his answer is supposed to imply. Is 3" of FHA supposed to be as good as 4"FHA or maybe 7" FHA??? :think:

marty
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Post by Dave Saxton »

Well, I think it must be 4". 7" seems a bit optimistic, but I could be wrong. I'm no math whiz, but hopefully getting better. Given a suitable arrangement of plates and given suitable materials, it possible for spaced armor to have an effective thickness equal to the sum thickness of the plates, or perhaps a bit more, as the Italian large caliber data and other data show. An effective thickness bump may be acrued simply to the resistance properties of the materials though. I thank Paul for sharing his ideas with us, and for tolorating the public scuitny.
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Post by Bgile »

It seems to be a commonly held belief here that multiple layers of armor represented a superior scheme to the "all or nothing" scheme employed on US, British, and Japanese new construction.

Why did those countries choose such an inferior means of protecting their battleships? I've read that the US plate of 8" and below was actually quite good, so why not use multiple layers of that? In view of what George has pointed out regarding poor armor quality control in the US, I question the value of having any at all, over the 8" thickness.

It also puts in serious doubt the ability of that armor to withstand heavy calibre shells at any reasonable range.

I do seem to recall that postwar testing against the thickest homogeneous armor showed that British projectiles broke up or deformed against armor as thick or thicker than the projectile diameter, such as that used on US turret faces. Maybe they made special high quality armor for that test.
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Post by Dave Saxton »

The Germans considered two types of limit velocities based on firing trials. One was the nessasary velocity to penetrate a given thickness of armour. The other was the limit that a typical shell could withstand, given it's condition and striking angle, and still penetrate, without breaking up. Were the nessasary velocity and the critical velocities are close to matching then the type of shell breakup you reffer to occurs. A spaced array can create a condition where the shell will breakup more easily. Were the effective thickness match up, the main advantage to using two plates instead of one, is that additional fail safes are introduced. If the system is designed well then a projectile carring enough energy to defeat the whole system will probably breakup. Slow heavy shells carring less energy than required for breakup, may be fuzed. However, not just any haphazard arrangement of plates will create these conditions, so great care must be taken; considering the critical velocities and the limit velocities, and the typical angles of attack. The material properties are also crucial. AoN is relatively simple but not nessasarly a bad approach either.

The Germans wanted to use scarps to backup their belts;extending the lower zone of protection into more likely battle ranges, so this requires the design of an effective spaced array system.
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Post by marty1 »

Dave:

I think in reality a spaced 3” armor array would resist like a single plate thickness of perhaps 2.8” or 2.7” -- particularly against 16” APC. You will find by studying the subject of spaced armor that it represents a mixed bag of potential results.

On the other hand spaced arrays ‘can’ be effective against small arms of dia=30mm or less, or against hard-core hyper-velocity shot like HVAP, APCR and APDS.

Neither you nor I seem to be able to figure out what the results of PaulL's equation are supposed to be. So while it may be interesting that PaulL presented his post on spaced armor, it really only succeeds in adding additional internet derived confusion to the subject.

Marty
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