T-80 Tanks


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fishhead

Banned Idiot
After thinking about it, if it is impact, then momentum can be used and actually in most physics book, it is the only correct one to calculate the one deminstional inelastic collision. In that scenario. we have to consider impulse, J=F*t, the bullet travelling at a much faster speed would have a much less impact time. Thus the force acting on the tank armor would be:
F=mv/t
One reasonable assumption would be since the bullet travelling 1000 times faster, the impact time would be a thousand times less than the coconut. Actually the bullet would probably have a much less impact time, because of its shape.

Then the large amount of F acting on rather small area, would have a even bigger pressure, and that is how do you have deformation.
In theory I woundn't dispute you too much. But it's noted that the impact is affected not only by some physics unit, also time and space interval, that's where your t comes from.

The problem of momentum as the measurement is that, it may have something to do with the time(speed), but it's hard to express it in space.

But energy can, you can condense the energy both in time and space interval, exactly why people choose it.
 

Skorzeny

Junior Member
After thinking about it, if it is impact, then momentum can be used and actually in most physics book, it is the only correct one to calculate the one deminstional inelastic collision. In that scenario. we have to consider impulse, J=F*t, the bullet travelling at a much faster speed would have a much less impact time. Thus the force acting on the tank armor would be:
F=mv/t
One reasonable assumption would be since the bullet travelling 1000 times faster, the impact time would be a thousand times less than the coconut. Actually the bullet would probably have a much less impact time, because of its shape.

Then the large amount of F acting on rather small area, would have a even bigger pressure, and that is how do you have deformation.

If there is no deformation, the impact time would be the same. hypothetical example
 

zraver

Junior Member
VIP Professional
Optionsss,

lets compare gnerally accpeted penetration values of given rounds, hier leangth and thier weight. We will assume that all rounds have basically the same design and level of machining.

We will look at the 3BM-42, DM-53,M829E3 and then try to extrapolate to the PLA round.

3-BM-42

penetrator weight 4.85kg (7.05w/sabot* we will use 2.2kg for all rounds sabot as size is so close and all use the spool style sabot) @ 1760m/s fired from a 2A46M5 L/52-56 Round L/D 16:1

leangth 560mm 560/16 D= R^2=35mm

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Note thier energy estimate is wrong (see below)

4.85/2 (1/2M)= 2.425* 1760^2 (V^2)= 3,097,600J

est performance 560mm 3,097,600/560= 5531.43J per mm of RHAe

M829E3

projectile weight 7.8kg (3.9=1/2M)* 1555^2 (v^2)= 9,430,297.5J

L/D=? 830/? if we assume 30:1/830/ R^2= 27.66mm

est perfromacne 850mm vs RHAe 9,430,2957.5/900= 11094.5J per mm of RHAe

DM53
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Length of dense core: 646 mm
Equivalent cylinder diameter: 20.8 mm
L/D: 31:1
Penetrator mass: stated to be "approx" 5 kg.

2.5(1/2M) * 1750^2* (V^2)= 7,656,250j

* fired from 120L/55

est performance 650mm per Andrew Jaremkow on tanknet
11778.8j per mm of RHAe


Note I high lighted the performance of the longer western styleguns beucase Fishhead claimed the L/D=30/1 long and thin vs short and thick like the Russian round.

PLA 125mm APFSDS

est mass 6.5-2.2= 4.3kg (2.15kg= 1/2M)

leangth 650mm L/D30/1 650/30 R^2=21.66

Now we have to workbackwards we have mass and estiamted performance but not V^2

using the 11k-11.9Kfigure of the western guns and picking 11.45 gives us a joule vlaue of 10,305,000j ^2root(10305000/2.15)= 2189.3 V=2189.3m/s

the chinese gun is traveling way to fast to be feasable for such a light round. This brings up 3 very serious questions

1- how can the Chinese gun contian the pressures with out bursting? (metallurgy miracle #1) The thickness of the barrel does not appear to be much thicker than a 2a46 or Rh120

2- How can the round with stand the pressures (metallurgy miracle #2)

3- Where did this super propellant come from (Chemistry miracle number 1) The Chinese breach is not signifigantly larger and might be smallker than the breach of the 2A46m5 and Rh120L/44 and is almost certainly smaller than the Rh120L/55 yet it produces 22% more energy than the western guns.

I can maybe buy 1 miracle, possibly 2, but not three from a nation that thad to import tank cannon well into the 1980's. Russia and Rheinmetal have been building tank cannons in thier favored caliber for 40 years. and have either or both a better brain trust and industrial resources. Like wise America has been refining DU rounds for 30 years now. Yet out of nowhere China passes them all up. Something stinks!

We don't see these metallurgical and chemistry miracles reflected elsewhere in the Chinese industry or military.
 

fishhead

Banned Idiot
PLA 125mm APFSDS

est mass 6.5-2.2= 4.3kg (2.15kg= 1/2M)

leangth 650mm L/D30/1 650/30 R^2=21.66

Now we have to workbackwards we have mass and estiamted performance but not V^2

using the 11k-11.9Kfigure of the western guns and picking 11.45 gives us a joule vlaue of 10,305,000j ^2root(10305000/2.15)= 2189.3 V=2189.3m/s

the chinese gun is traveling way to fast to be feasable for such a light round. This brings up 3 very serious questions
LOL, where do you get the "est mass"?

Show some math skill here please. You know the length and radius you should know how to get there.

And I will tell you why this kind of estimate is WAY WRONG.
 

optionsss

Junior Member
zraver, I really have some trouble believe the mass is only 4.3, but assuming what you said was right. Then let look at the BIG miracle here.

在发射第三代 钨合金尾翼稳定脱壳穿甲弹时初速为1780/秒,可在2000米距离上击穿850毫米的均质装甲,而最新的特种合金穿甲弹在该距离上的穿甲厚度可达97 0毫米
From this section they said the Chinese round capability is 850 Sobot and the newer Special round will go as far as 970 with an speed of 1780m/s.

Since we are assumpting that you are right and the round mass is indead 4.3kg . Then lets look into how much pressure each gun have to sustain for the DM-53,M829E3 and the PLA round.

First lets look at the speed and total KE of each:
PLA::
11.45kj*850=1/2mv^2
v= 2127.6m/s
KE=9732 kj

How much force required to accelerate an object from rest to 2127.6m/s over a distance of 6.412 meters (barrel length):
We know F*D=W=KE, thus:
F=KE/D
F=9732kj/6.412
F=1517000n
Then what about the pressure: P=F/A
The internal surface area is here: 2rL*3.14
125*3.14*6412= 2516710mm^2
Then pressure
p=1517000/2516710mm^2
p=0.60277

M829E3:
v=1555m/s
KE=9,430 kj
If you use the L44 with 5280 barrel length you get
F=1780000n
P=0.897


DM53:
v=1750m/s
KE=7656 kj
Do the same with barrel length 6600, you get
F=1160000
P=0.582


The pressure measure as newton per mm^2. PLA tank barrel has to sustain just a slightly higher pressure than the DM52 with a L55, but they are very close. This would explain the shorter life span for the PLA barrel, but clearly it is not nearly the pressure the L44 DU rounds have to take. So, really not a BIG miracle, looks more like a reasonable break through to me, nothing really special. Like I said before the 850mm seems possible.
 
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fishhead

Banned Idiot
This kind of estimate make very little sense, nobody really know how much exactly it is.

Look at its shape, you can't calculate it as well

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zraver

Junior Member
VIP Professional
Optionsss, I am not even going to claim to be able to follow your math, my skills are very limited in math.

That being said, can you please square Fisheads article claim=

From this section they said the Chinese round capability is 850 Sobot and the newer Special round will go as far as 970 with an speed of 1780m/s
.

Also please explain how the Chinese round which weighs less and according to the above has more KE than either the DM-53 or the M829E3 and how the it achieves this with such low pressures?

Fishhead I esitmated the weight based on the Russian round (bore size) and the German round (leangth) thats why I said estimated.

Using your figures of 960mm of RHAe penetration and 1780 m/s and the 11.4K est round performance. I am subtracting 2.2kg from each rounds projectile weight to account for the spool.

est 9kg total weight-2.2kg 6.8kg/2= 3.4kg= 10,772,560/960=11221 which in the proper range.

Optionsss, what the pressure with a 9kg round accelerated to 1780m/s from a 6.25m barrel?
 

LIGO

New Member
How much force required to accelerate an object from rest to 2127.6m/s over a distance of 6.412 meters (barrel length):
We know F*D=W=KE, thus:
F=KE/D
F=9732kj/6.412
F=1517000n
Then what about the pressure: P=F/A
The internal surface area is here: 2rL*3.14
125*3.14*6412= 2516710mm^2
Then pressure
p=1517000/2516710mm^2
QUOTE]

Well, I think you can just use force divided by the cross-section of the barrel to calculate the pressure. (This is by assuming that the pressure is the same in all direction everywhere in the barrel.)

I think a better way to calculate the pressure is to use the adiabatic expansion (no heat exchange between the gas in the barrel and the surrounding) of the ideal gas. Then we know that Pressure*(Volumn^gamma)=constant, which gives us a relation between pressure and volume. Then integrating Pressure*dVolume from the start to the end of the barrel will give us the kinetic energy obtained by the round. This will help fixing the initial pressure in the barrel. (Roughly speaking, as the volume of the gas increases, the pressure decreases in the barrel as the round moves forward, the temperature of the gas decreases, and the gas looses internal energy, which is converted to the kinetic energy obtained by the round.)

This is as far as I can go. If someone here knows the correct initial condition (pressure, volume or temperature) then he/she can continue further…
 
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optionsss

Junior Member
Also please explain how the Chinese round which weighs less and according to the above has more KE than either the DM-53 or the M829E3 and how the it achieves this with such low pressures?
The Chinese round achieved high KE with a much higher velocity than the other two round, i used your data.
11.45kj/mm * 850 mm= 9732 kj. and that means the velocity have to be about 2127.6m/s to achieve. I think you know how to do the math here.

However the pressure is a different story. I first get the force required to move the projectile through the barrel to their required velocity. To get that, the Work done on the projectile by the round is same as the KE the round process after it had left the barrel. With work equals to the KE and the equation to get work is force times distance. The Pressue if just the amount of force acting on a unit area.
The Chinese gun had a much lower pressure than the M829E3 L44 with 5280 barrel, because the gun had a much bigger internal surface area. The equation to get internal surface area of a cylinder is length*diameter*3.14.
For PLA: it is about 6412mm length and a diameter 125mm,
The L44 have a barrel length of 5280mm and diameter of 120mm.
The L55 barrel length is about 6600mm and diameter of 120mm.

The main adavantage for the Chinese gun is the bigger barrel and if you look, the KE is about the same, this makes sence if you think about it.

The PLA gun is pushing a smaller object over 6.5 meters and the L44 is pushing a bigger object over 5.28 meters. So the Chinese gun have a longer period of time to transfer the energy, so it does not have to use as much force than the L44 and the force acting on the gun is distributed over a much bigger area than the L44.

Optionsss, what the pressure with a 9kg round accelerated to 1780m/s from a 6.25m barrel?
From this, I got 0.929N/mm^2 with a 125mm diameter.
 

fishhead

Banned Idiot
Also please explain how the Chinese round which weighs less and according to the above has more KE than either the DM-53 or the M829E3 and how the it achieves this with such low pressures?

Fishhead I esitmated the weight based on the Russian round (bore size) and the German round (leangth) thats why I said estimated.

Using your figures of 960mm of RHAe penetration and 1780 m/s and the 11.4K est round performance. I am subtracting 2.2kg from each rounds projectile weight to account for the spool.
1.
You estimate never comes with any mathematic prove, so I can't take your words as its face value. It has no credit for me.

2.
The penetrating power not only depends on the mass, also the L/D ratio impacts a lot on it, as well as speed. Although I don't know exactly the relationship, I don't think using only mass is right.

3.
All M829 family have very low speed, so don't use them to compare with Chinese shell.

Don't believe too much of your propaganda, here is the official DOD document: All M829 muzzle speeds are 15xx m/s something, way lower than Chinese 1780m/s.
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You M829E3 is still in test stage, so don't use it to compare. Even that we know it has a low speed, 15xx m/s something.
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Stop using Russian or American shells to compare with Chinese round.
 
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