IS-2 Tank Role and Gun Selection

[rexford]

If one goes to the following page on Russian Battlefield, and reads the three correspondence letter on IS tank use of 100mm gun, they will see interesting reasons why 100mm gun not chosen for IS tanks: http://www.battlefield.ru/library/archives/index.html

100mm gun outpenetrated 122mm at normal combat ranges but came with many drawbacks, including longer rounds to load into gun, increased fumes in turret, insignificant increase in ammo storage, etc. See letter 3.

More than penetration went into decision to stick with 122mm, which may have also considered HE capability. 122mm HE projectile is 15.2% HE filler by weight, 100mm HE is 9.3%, based on figures contained on Russian Battlefield.

Also interesting is discussion of ability of 122mm gun to fight German tanks, in particular (letter is written 1944):

"At the moment, "JS" tanks armed with the 122 mm main gun are successfully repelling all counterattacks by German tanks of all types at all ranges (i.e. up to 1500 metres);" IS tanks were expected to defeat German tanks, and were doing it.

U.S. firing tests with 122mm APBC predict that 122mm APBC will defeat excellent quality Panther glacis at 1450m against 85mm plate thickness, and 1850m against 80mm plate thickness.

[ February 09, 2002, 07:13 AM: Message edited by: rexford ]

[John Kettler]

rexford,

I've seen combat accounts by a surviving Panther commander where a Panther took a 122mm glacis hit

which penetrated the glacis, passed through the entire fighting compartment, smashed through the engine compartment, and drove the engine out through the rear plate and onto the ground. The writer reportedly saw this happen to a nearby tank and though his own tank was unhit, immediately ordered out the entire crew, being in shock, horror and fear of being next.

I first saw this account in a popular scale modeling magazine (SCALE MODELER, I think), have seen it elsewhere, but don't have it handy. Suffice it to say I found it stunning.

Regards,

John Kettler

[MajorBooBoo]

I find that hard to believe. The engine was driven out the back? The back armor being welded to the tank at the time? An armor piercing shell would just punch through an engine block, not get the whole thing moving.

More than likely (if it happened at all.. SCALE MODELOR?), woould be that an internal catastrophic explosion from the panthers own stored ammunition could detonate sympathetically and rip the tank apart.

I have read accounts of Panther AP (at point blank range) completely penetrating a Sherman's front/engine/rear and then bouncing off the front armor of the Sherman behind it. It didnt take any souveniers along for the ride!

[ February 09, 2002, 11:50 AM: Message edited by: MajorBooBoo ]

[PJungnitsch]

Originally posted by rexford:

U.S. firing tests with 122mm APBC predict that 122mm APBC will defeat excellent quality Panther glacis at 1450m against 85mm plate thickness, and 1850m against 80mm plate thickness.

The site also says here :

Further, after the first encounters between the JS-2 and German heavy tanks, it turned out that the sharp-nosed 122 mm APHE round - the BR-471 - could only penetrate the frontal armour of a Panther up to 600-700 metres. The less powerful frontal armour of a Tiger could be penetrated at distances up to 1200 metres. However, at such distances only very well trained and experienced gunners could score a hit. The vertical armour of a Tiger I, although thicker than that of a Panther, was more easily defeated by the sharp-nosed projectile of the JS-2 Main Gun, whilst it often ricocheted off the sloped armour of a Panther. Later, Soviet designers noticed the blunt-nosed projectiles worked fine against sloped armour....The first results of the IS-2 in combat (backed by the results of its tests at the Kubinka testing grounds in January of 1944) forced designers to look for new solutions to its problems....

...However, in the summer of 1944, the problem of the poor AP performance disappeared. The performance of the D-25T gun of the JS-2 against the German tanks improved dramatically. The reports from the front described cases where the BR-471 APHE round 122 mm projectile fired from 2500 metres ricocheted off the front armour of a Panther leaving huge holes and cracks in it.

This was explained by an interesting change of circumstances in the Summer of 1944. The Germans experienced a shortage of manganese and had to switch to using high-carbon steel alloyed with nickel, which made armour very brittle, especially at the seam welds.

Another report at the same website here gives the 122mm as penetrating Panther glacis as up to 2500, with the 100mm only up to 1500, and the L/71 88mm far behind at only 650 meters!

While tables here give different penetration values again! Here the 122 and 100 are very similar.

What gives? Are any of the tests accurate, or were they just skewed to fit the results wanted at the time by the particular government official?

[JasonC]

Major BB found it difficult to believe that a 122mm AP knocked the engine out the back of a Panther. If the armor is defeated, then the result is perfectly understandable. Large AP rounds have enourmous energies, and all of their energy and momentum has to go somewhere.

When the armor defeats the round, most of it is transfered to the tank in its entirety, or carried off by the round itself in ricochet. Notice, however, that a *direct* ricochet (straight back, as if that could happen) would impart *double* the shell momentum to the tank as a whole. Because the armor must accelerate the round off on its new vector, with a net new momentum vector of zero for tank and round taken together.

Imagine a tank on a sheet of ice or other frictionless surface, with a direct ricochet. In the case of a Panther and a 122mm AP, the tank would coast backward at about 2 mph after such a hit (yes Virginia, the shell may weigh a lot less than the tank but it is going quite fast). Friction prevents this in real ricochets; the momentum is imparted to the ground, though the tank rocks on its suspension.

But if the round enters and stops inside the tank, the momentum and energy are transfered to the pieces of the tank hit. A lot of it wherever the round stops, naturally. Imagine for a second that all the momentum of a Russian 122mm AP were imparted to the engine block, which weighs on the order of 4 tons. Then the engine block would exit the back of the tank at about 12 mph.

The energy of a 122mm AP is 7.6 million joules. The round is about 25 kg, and moving 780 m/sec when it leaves the muzzle. The energy falls off slightly with range due to air resistance. And of course, some of the momentum will be imparted to the tank as a whole, to the fragments of armor, to the shards of the rear plate when it breaks. But even 1/2 of the momentum of a round down 30% from its ME (down about 1/6th in MV) would move a 4-ton engine block to 2 m/sec, which is 4.3 mph. The engine would go for a brisk walk out the back.

Smaller tanks could be torned to pieces by high caliber AP even if the rounds did not penetrate at all. They just usually penetrate, too. But e.g. a turret can be lifted off by the energy of a hit from 120mm AP. Sherman turrets were often knocked askew by 88L71 rounds that went clear through them. A turret is heavier than an engine block. Those rounds have about 5 million joules energy at the muzzle, roughly 2/3rds that of the Russian 122mm.

[gunnergoz]

I'm sure that the IS-2 and it's powerful 122mm main gun will give rise to spirited debate about the merits of this tank vs it's German opponents. This will doubless include discussion of the equally awesome 100mm and 152mm weapons fielded on SP Gun chassis by the Russians.

One thing that should be kept in mind about the IS-2 in particular is that it was not primarily designed or fielded as an anti-tank weapon but rather saw it's most effective use in breakthrough and assault situations, moving up with or slightly in advance of the infantry. In those cases, the relatively low ammo loadout and slow firing rate were not so critical as is true in tank-vs-tank combat.

I, for one, am really jazzed about the prospect of fighting with these powerful beasts in CMBB.

[MajorBooBoo]

Originally posted by JasonC:

Major BB found it difficult to believe that a 122mm AP knocked the engine out the back of a Panther. If the armor is defeated, then the result is perfectly understandable. Large AP rounds have enourmous ENERGIES, and all of their ENERGY and MOMENTUM has to go somewhere.

When the armor defeats the round, most of IT is transfered to the tank in its entirety, or carried off by the round itself in ricochet. Notice, however, that a *direct* ricochet (straight back, as if that could happen) would impart *double* the shell momentum to the tank as a whole. Because the armor must accelerate the round off on its new vector, with a net new momentum vector of zero for tank and round taken together.

(To use a Jasonistic mannerism..um,no)

Most of IT? IT being what Jason? Energy or Momentum.

Momentum is conserved in this world.

The Principle of Conservation of Linear Momentum

During any interaction in which no external forces act, the total linear momentum is conserved.

Mass times Velocity. You described a perfectly elastic collision. That is, a ricochet that resulted in the round bouncing directly back from where it came. You never mentioned the tank moving in response , so its change in momentum is????

Before the collision:

(Mass of shell) times (velocity)= Value1

(Mass of Tank) times (Velocity, which is zero)= zero

After collision:

(Mass of Shell) times its (Velocity)= -Value1

(Mass of Tank) times its (Velocity)= zero

So we see that momentum is conserved in this case (as it should be).

Expressed in terms of masses and velocities this is equivalent to:

m1v'1 + m1v1 = m2v'2 + m2v2

where the primed (') symbols refer to the quantities after collision, and the unprimed symbols refer to the quantities before collision. Obviously one side of the equation is zero.

Momentum or Energy? Figure it out Jason. How did double the momentum enter into your post? Then I will respond to the rest of your post which also has holes in it. No pun intended.

http://www.usafa.af.mil/dfp/cockpit-phys/kw2th1a.htm

Heres a website that has an error(s) in it (I believe), anyone agree?

[ February 09, 2002, 05:49 PM: Message edited by: MajorBooBoo ]

[MajorBooBoo]

I'll bump this with a little more info:

The 122mm AP shell would have to:

1. Defeat the front hull armor

2. Move through all bulkheads/etc

3. Strike the engine block

4. Become a non-elastic collision at this point and start transferring all energy/momentum to said engine block

5. Break engine block off its mounts and disengage all exhaust fittings/DRIVESHAFT (!)hoses/lines connecting it to the tank

6. Move this mass towards the rear armor

7. This moving mass needs to strike the welded on rear armor and knock it off with its mass*velocity

The most ridiculous thing is the engine being accelerated in such as short distance to knock off the rear armor plate (basically one piece, I believe, that is 30-40mm thick?????) and jumping through!!!!

Almost as preposterous is the AP round "burying" itself in the engine block. Anyone that has worked on engines know that there is alot of hollow inside an engine. It is not made of armor and mostly cast iron/aluminum heads. I have fired rifles through engine blocks for christs sake!

Theres many stories of tanks being perforated all the way front to rear and this is the most unbelievable thing I have read. The superpershing firing on a jagdpanzer comes to mind in addition to the point blank panther vs shermans story I posted before.

[JonS]

Originally posted by MajorBooBoo:

(To use a Jasonistic mannerism..um,no)

...

Before the collision:

(Mass of shell) times (velocity)= Value1

(Mass of Tank) times (Velocity, which is zero)= zero

After collision:

(Mass of Shell) times its (Velocity)= -Value1

(Mass of Tank) times its (Velocity)= zero

So we see that momentum is conserved in this case (as it should be).

...

Um, no. In your example momentum is not conserved, it is reversed.

After the colision the vehicle must move away from the now reflected round with twice the initial momentum of the round in order for total momentum to be conserved. THe first lot equals the momentum of the round which is now heading back where it came from - this gives zero net momentum for the shell-tank system. However before the impact the shell-tank system had a positive total momentum, so the tank must start moving in the same dirrection as the round was initially travelling - this is where the second lot of momentum comes from.

Momentum_Shell_Before + Momentum_Tank_Before = Momentum_Shell_After + Momentum_Tank_After

If we take the initial momentum of the shell as being 1, we get

1 + 0 = -1 + x

I leave to you to work out the value for x ...

[Puff the Magic Dragon]

Mh, maybe the Panther simply did what I would do when a bullet walks throught my body: place a digested lunch into my panties...

Sorry, to much beer

[MajorBooBoo]

Originally posted by JonS:

...Um, no. In your example momentum is not conserved, it is reversed.

After the colision the vehicle must move away from the now reflected round with twice the initial momentum of the round in order for total momentum to be conserved.

If we take the initial momentum of the shell as being 1, we get

1 + 0 = -1 + x

I leave to you to work out the value for x ...

Everyone agree with this?

[JasonC]

Both momentum and energy are conserved in collisions. Either alone would not tell you the necessary shape of the collision, but both together do. The subtler aspect is that momentum is conserved as a *vector*, while energy is conserved absolutely, as a *number*.

Meaning, for the of momentum the net direction of all masses, in the sense of a sum of directed vectors weighted by the mass of each resulting object, is equal to the initial momentum, again weighted by the mass of all the different objects colliding. This is true in any reference frame, so in a two object collision it is easiest to use a frame with one of the objects stationary. That way you only have to worry about the momentum of the incoming object.

In addition, energy must also be conserved. Velocity enters the equation for energy twice rather than once, so this is a seperate and additional criterion, not reducable to the previous. There are an infinite number of ways of conserving the energy, since a slight addition to the velocity of object A can be compensated by a slight reduction in the velocity of object B, Among all the possible ways of conserving the energy, however, not all will conserve the vector momentum. And both must be conserved.

Notice, energy is a number, not a vector. It does not have a direction. Energy going west is not a conserved quantity. The energy going west can be reduced by a collision. But momentum is a vector, not a number. It does have a direction. Momentum going west is a conserved quantity, in the vector-addition sense anyway.

A perfectly inelastic collision is one in which the two colliding objects "stick together", becoming one object in effect, after the collision. It is particularly easy to figure out what form of a collision results from that, especially in the case of just two bodies, because the net momentum must all be in one mass, the new combined object.

In a frame of reference in which one object is stationary, the new velocity vector will just be in the direction of the initial vector of the other object, while the length or coefficient of the vector is just mass of the mover divided by sum of the masses times the initial velocity of the mover.

Applied to the example, the AP shell has an inelastic collision with the engine block. Meaning, it stops inside it and the two become one object. The stationary mass of the block beforehand was 4000 kg (say - that is probably a bit on the high side for the engine alone). The mass of the moving AP shell was 25 kg. The initial speed (that is, the absolute value or length of the velocity vector) was 780 m/s.

Then the new object (engine mit der shell inside) has a mass of 4025 kg. Its new velocity will be in the same direction as the initial path of the shell, and the new speed will be 780 * 25 / 4025, or 4.845 m/s, which is around 10 mph. Hence my first estimate.

But this assumes an inelastic collision between the AP round and the engine block, and ignores (1) the loss of speed to air resistance before hitting the tank and (2) the momentum carried off by flying bits and pieces of tank, or imparted to the tank as a whole and absorbed by the ground, etc. So I made some allowances for those - 5/6 speed at impact, only half the remaining impact momentum in the engine block "piece" - and you get around 2 m/s, a brisk walking pace. Hence my second estimate.

If the round bounced straight off, then yes the total momentum imparted to the tank as a whole would be twice the momentum of the shell. Because momentum needs to add as vectors, which include direction, thus the sign of the direction in a straight line case. Energy in effect squares the direction term, and makes the sign go away - just as the square of any integer is positive.

Consider a straight rebound of the 122 shell from a 45 ton tank, compared to an inelastic in which the shell remains inside. In the second case, the new velocity is 780 m/sec times 25 kg divided by 25 + 41000 kg, equals .475 m/s (~1 mph). If it bounces straight off, then you've got twice that.

But look at the energy in the two cases. In the first, the kinetic energy per collision is 7.6 MJ. Afterward, you've got 1/2 41025 .475 ^2 = 4.6 MJ. About 3 MJ would therefore have to be evolved in forms other than kinetic energy - heat, flash, bang. Not very likely. Parts would have to fly away considerably faster.

Suppose the shell rebounds with its full initial speed, and the tank thus jumps the other direction at 2 x .475 = .95 m/s. Then the tank's new kinetic energy is 20 MJ, and the shell by hypothesis is still booking enough to carry off 7.6 MJ. 27.6 MJ of kinetic afterward, and only 7.6 MJ coming in. The only way the shell could bounce literally straight back along its original flight path, is if it hit explosive material that blew up, imparting an extra 20 MJ (still conserved energy, but transformed from chemical energy in the material, into kinetic). Doesn't happen, obviously.

See the point? Both of the above conserve momentum, but neither conserves energy. There are, equally, lots of ways you can conserve the energy, that don't conserve the momentum. What actually happens is a particular arrangement of velocities and directions that satisfies both constraints.

After you've allowed for some portion of the energy to change form (out of kinetic, that is), you will have a number of constraints - total momentum of all the pieces, and total energy. You will have a number equation from one (energy), and a vector equation from the other (meaning, one with angles and cosines etc). There will then by a combination of velocities for the different bodies and angles of their scattering, that satisfies both equations. That will be the one you will see happen.

You do have to know how many pieces fly off, of what mass each, to solve it exactly. How many of those will form is a much more complicated bit of materials science, and unless it stays pretty simple, you are better off just testing to see.

[MajorBooBoo]

The shell has a scalar KE (kinetic Energy) of 1/2mv^2 coming in and 1/2mv^2 going back where it came from. Its energy is conserved (theoretically), since you are hypothesising no velocity loss. Wheres this extra energy coming from thats moving tanks and such?

Jason did a pretty good job of explaining himself, better than his first post here. He is hedging on the real issues about how the shell could possibly wedge in the engine (it might, if moving SLOWLY) and how its going to also then keep moving and pick up another mass (back armor) and all three are going out the back. No way! The engine/shell combo cant rebound back into the tank, they all gotta get out! If I was backed into a corner like he is, I would postulate that the engine/shell combo didnt go out the back but rather rode up the slope of the rear armor (the panther has that rakish look in the back) crashing through the thinner upper deck and landed on the ground.

JonS is correct because there isnt the case of conservation of momentum when one of the two elastic objects doesnt move. Its a trick question in physics. Ive seen it on tests and thought I would try my luck with it here. He should have just said so but insists on moving the tank!

For every complex problem there is a solution that is simple, neat, and wrong.

H. L. Mencken

[flamingknives]

I can't quite see something the size of a tank engine getting pushed out through 8-10 mm plate - more likely the welds fail and the plates fall apart.

[MajorBooBoo]

Originally posted by flamingknives:

I can't quite see something the size of a tank engine getting pushed out through 8-10 mm plate - more likely the welds fail and the plates fall apart.

I even think the scenario I proposed as a back peddle is also preposterous. Like someone telling me something they read in a Sgt Rock comic book.

I can explain the turret stuff he posted earlier because it has to do with rotational dynamics/internal explosions and not a strait linear brute force tale like the engine thing.

[gunnergoz]

Please excuse me if at first this seems off-topic:

I have an MPG of a Javelin missisle test fired against a fairly modern Russian tank, T-72 or T-80 era. The Javelin is a lightweight HEAT missle that uses a pop-up atttack to hit top-down on the target where the armor is thinnest.

Anyway, in the film, the tank is hit and literally flies to pieces. The turret lands some distance away, perhaps 10 meters, and buries itself several feet into the earth. The chassis, sides and track of the tank fly in many large and small pieces up to 50 meters away. Finally, the engine block lands about 70 meters from the tank.

Admiteddly, this is a modern weapon hitting a modern tank. Also, there is no way to know if the tank had ammunition within it for the test.

What I'm getting at is the stark reality of catastrophic failure of a tanks' armor is awesome indeed to behold. I've seen photos of hundreds of destroyed tanks, old and modern, and I've never seen this scale of destruction.

So...is pushing an engine block out of a tank possible? I don't know for a fact that it is in the scenario of the IS-2 hit on a German tank, but somehow, having seen that film, I'm a bit less inclined to say anything is impossible. The level of violence inherent in these weapons' effects is a little hard to comprehend from the comfort of my computer hutch. Let me just say, I wouldn't want to be on the receiving end of one.

[JasonC]

Sorry you guys don't like physics problems, lol. The brute force is enough to do it. Since people seem to be having trouble seeing how the end result is basically like an inelastic collision between engine and round, perhaps it will make more sense to them if they track the momentum through the tank.

When the shell punches through the front plate, yes it is slowed down. But only because the part of the front plate it went through is speeded up. Instead of a 25 kg shell doing 780 m/s, then you've got 25 kg of shell and ~25 kg of broken armor doing an average of 390 m/s. Actually, the armor half is probably going somewhat faster, and the shell somewhat slower, making the collision somewhat elastic. Think of the armor "plug" in front of the shell as billard ball #1, and the shell as billard ball #2.

Notice that the momentum can't go away. And keeps its direction.

Some portion of the armor "spall" will fan out to the sides. But the cloud of it will be going much faster along the axis of the tank, than in the sideways direction. Each bit drifting right matchs a bit drifting left, so that the net vector sum is the original flight path. Some of the pieces flying off at angles will hit the sides of the tank, but they will then bounce. Converging back toward the original flight path.

The same process recurs with every obstruction hit. Instead of a small amount of matter going fast, you get a larger amount going somewhat slower. It is a cascade. By the time you've burst the forward wall of the engine compartment, perhaps you've got 2-3 times the material going 1/2 to 1/3 as fast. The momentum is nearly the same (only a bit transfered to rocking the suspension has been transfered away from hitting engine).

The shell is probably going faster than the individual pieces by the time you get deep into the engine, since it is probably the largest and longest single projectile involved. Of course parts of the engine are being broken off too, and joining the "matter flow". The shell has perhaps lost 3/4 of its speed by then. It or something it hits going at least as fast hits the back plate, with a remaining energy on the order of hundreds of thousands of joules. That plate doesn't withstand that from the outside when a shell does it, and won't from the inside either.

The overall process remains - from all the momentum being in a small object going very fast, it is transfered to a much larger body going proportionally slower. There is no need to track every sub-projectile in the cascade. The momentum and its direction are conserved, so it is going to get to the back of the tank. If the tank maintains integrity completely, then this may just rock it, the dirt under the tracks taking up the momentum. But if it doesn't hang in one piece, sizable pieces of the tank are going to be left with the momentum - which means they are going out the back.

[ February 10, 2002, 01:16 PM: Message edited by: JasonC ]

[offtaskagain]

That was a T-72 I believe. It was fully fueled and loaded with ammo which led to the awesome secondary effects.

[rlh1138]

Not to prolong this, as I think everyone has expressed their view on the likihood/plausibility/math of this event, but... I don't see anything in the discussion about the HE inside the shell and how it's detonation would affect the formulas. Is this a 'solid shot' shell and I'm the only one that doesn't know that?

[The Commissar]

Originally posted by gunnergoz:

I have an MPG of a Javelin missisle test fired against a fairly modern Russian tank, T-72 or T-80 era. The Javelin is a lightweight HEAT missle that uses a pop-up atttack to hit top-down on the target where the armor is thinnest.

Anyway, in the film, the tank is hit and literally flies to pieces. The turret lands some distance away, perhaps 10 meters, and buries itself several feet into the earth. The chassis, sides and track of the tank fly in many large and small pieces up to 50 meters away. Finally, the engine block lands about 70 meters from the tank.

An off topic reply to a slightly off topic post:

That test of the Javelin against the T-72 was supposed to replicate battle field conditions (the tank was stocked full of ammo and fuel) but was inherently flawed.

The major flaw being that the tank in question was not using any of the modern Russian anti-missile systems like Shtora for example. It would be interesting to see how a modern missile like the Javelin would cope with modern anti-missile systems the Russians are mounting on their tanks to give them much higher chances of survival in a TOW-heavy enviorment of the modern battlefield.

Other flaws include the tank standing nice and still in the open with no cover for miles.

Still, the fireworks display was pretty damn cool!

[ February 10, 2002, 02:17 PM: Message edited by: The Commissar ]

[Iron Chef Sakai]

Will the Germans be getting anything equiped with their 128mm gun? I know the 128mm FlaK was not produced enought to be in the game, but did the Germans use the 128mm in anything besides the Mause?

[109 Gustav]

Originally posted by gunnergoz:

I have an MPG of a Javelin missisle test fired against a fairly modern Russian tank, T-72 or T-80 era. The Javelin is a lightweight HEAT missle that uses a pop-up atttack to hit top-down on the target where the armor is thinnest.

Got a link to it? I'd love to see the mpeg.

[gunnergoz]

1. The 128mm was only used in the Jagtiger AFAIK.

2. The MPG is pretty large at several megs and I downloaded it from Usenet (alt.binaries.military so you may get someone to post it again for you.

[The Commissar]

Originally posted by 109 Gustav:

Got a link to it? I'd love to see the mpeg.

Its not the real movie, but a frame-by-frame annalesis. Shows the highlights of the movie pretty well though.

http://www.strategypage.com/gallery/default.asp?target=jslide1.htm

Modern war is scarry. Thank goodness for anti-missile systems.

[ February 10, 2002, 06:06 PM: Message edited by: The Commissar ]

[Thomm]

Originally posted by JasonC:

When the shell punches through the front plate, yes it is slowed down. But only because the part of the front plate it went through is speeded up. Instead of a 25 kg shell doing 780 m/s, then you've got 25 kg of shell and ~25 kg of broken armor doing an average of 390 m/s. Actually, the armor half is probably going somewhat faster, and the shell somewhat slower, making the collision somewhat elastic. Think of the armor "plug" in front of the shell as billard ball #1, and the shell as billard ball #2

This neglects the shear forces between "plug" and tank body.