Physics of shooting a rifle - bolt thrust

[tangolima]

On the head of a bottle necked brass, there are 3 force components.

Fh = head bearing area * p, backward
Fbt, bolt thrust, forward
Ft, tension in the brass body, forward

Ft+Fbt-Fh=0

On the front end of the brass (body taper, shoulder, bullet), it has total projected area same as the head bearing area. There 4 force components

Fh, forward
Fs, force exerted on chamber shoulder, backward
Fb, reaction force caused by bullet acceleration, backward
Ft, brass body tension, backward

Fh-Fs-Fb-Ft=0

Combining the 2 equations

Fbt=Fs+Fb

Before the chamber pressure builds up to expand the brass, there is always clearance between the brass shoulder and chamber shoulder.

Fs=0
Fbt=Fb<Fh

The tension force in the brass body

Ft=Fh-Fb != 0

Once the brass shoulder contacts chamber shoulder it is fully supported. The tension in the brass disappears

Ft=0
Fs=Fh-Fb
Fbt=Fh

If the brass is so strong against the chamber that it does expand enough, the bolt thrust is lower. The tension in the brass body balances out part of the force.

To push it to the extreme, the bullet is welded to a super strong casing (probably not brass), it is effectively a self contained bomb inside the rifle. The tension force balances everything out, and the bolt thrust is zero.



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[JohnKSa]

The brass is part of the rifle...

The net force on the brass is much lower because of its light weight.

???

If it's part of the rifle, then the force on it is calculated as if it's part of the rifle, not as if it's an independent entity.

[tangolima]

Originally Posted by JohnKSa

???

F=ma

F_rifle = m_rifle * a
F_brass = m_brass * a

F_rifle / F_brass = m_rifle / m_brass



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[JohnKSa]

F_rifle = m_rifle * a
F_brass = m_brass * a

The brass is part of the rifle...

If it's part of the rifle, how does it make sense to calculate the forces on it as if it is not part of the rifle?

The force on the brass is what accelerates the brass and the rifle, which, as you say, are essentially all one entity for practical purposes.

[tangolima]

Originally Posted by JohnKSa

If it's part of the rifle, how does it make sense to calculate the forces on it as if it is not part of the rifle.

What I meant is they have identical speed and acceleration.

-TL

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[JohnKSa]

In order for the brass to gain that speed and acceleration, the force on it has to accelerate the entire rifle with it. So your comment that the brass is part of the rifle is accurate.

To push it to the extreme, the bullet is welded to a super strong casing (probably not brass), it is effectively a self contained bomb inside the rifle. The tension force balances everything out, and the bolt thrust is zero.

So the bullet doesn't move, and that somehow negates the force to the rear on the brass? How does that work? <<<Edit. Assuming that this theoretical cartridge can't stretch or fail, this assumption is correct.>>>

[tangolima]

Originally Posted by JohnKSa

If it's part of the rifle, how does it make sense to calculate the forces on it as if it is not part of the rifle.

What I meant is they have identical speed and acceleration.

More rigorously,

F_rifle = (m_rifle+m_brass) * a

F_brass = m_brass * a

F_rifle / F_brass =( m_rifle+m_brass) / m_brass -&gt; m_rifle / m_brass as m_rifle &gt;&gt; m_brass


-TL

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[JohnKSa]

Right, they have identical speed and acceleration because they are being accelerated to the rear as a unit by the force on the brass.

All of the force from firing applied is applied to the brass. All of the acceleration from firing is the result of force on the brass.

Since the brass and the rifle are being accelerated as a unit and all of the force applied by firing the rifle is applied to the brass, how does it make sense to divide up their masses and consider them separately?

[JohnKSa]

To push it to the extreme, the bullet is welded to a super strong casing (probably not brass), it is effectively a self contained bomb inside the rifle. The tension force balances everything out, and the bolt thrust is zero.

So a sensor placed between the brass and the bolt would register zero, regardless of the chamber pressure, as long as the bullet doesn't move? That means that we could drive the chamber pressure as high as we want to and the bolt will never come out the back of the rifle as long as we insure that the bullet can't move forward.

That is obviously incorrect.

<<<Edit. Ok, I get it, you are assuming that the cartridge doesn't stretch or fail. Yes, with those (unrealistic) assumptions your conclusion is correct.>>>

[tangolima]

What made me think about this is my dubious AR-10 in 7mm SAUM as quoted in the example in post #4.

I realized I need to stay below .308's chamber pressure of 62kpsi that AR-10 is designed for. But its larger head bearing area is much bigger than .308. The bolt thrust is magnified. It could become a problem considering the same number of lock lugs, and that they are somewhat thinned out to make room for the larger bolt face.

Sounds like I should lower the chamber pressure further. How much lower?

-TL

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[tangolima]

Originally Posted by JohnKSa

Right, they have identical speed and acceleration because they are being accelerated to the rear as a unit by the force on the brass.



All of the force from firing applied is applied to the brass. All of the acceleration from firing is the result of force on the brass.



Since the brass and the rifle are being accelerated as a unit and all of the force applied by firing the rifle is applied to the brass, how does it make sense to divide up their masses and consider them separately?

The force is indeed going to the rifle through the brass. But that also means the rifle is also applying reaction force to the brass. The difference of this two forces is the net force that accelerates the brass. That net force is close to zero because of the brass' light weight.

-TL

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[tangolima]

Originally Posted by JohnKSa

So a sensor placed between the brass and the bolt would register zero, regardless of the chamber pressure, as long as the bullet doesn't move? That means that we could drive the chamber pressure as high as we want to and the bolt will never come out the back of the rifle as long as we insure that the bullet can't move forward.



That is obviously incorrect.



&lt;&lt;&lt;Edit. Ok, I get it, you are assuming that the cartridge doesn't stretch or fail. Yes, with those (unrealistic) assumptions your conclusion is correct.&gt;&gt;&gt;

The precondition is that the brass is strong enough to hold the pressure BY ITSELF, i.e without chamber support.

Don't try it. It is a theoretical scenario for discussion only.

-TL

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[JohnKSa]

For practical purposes, the rearward force on the brass is the force on the bolt face. With the material properties of the bolt head, and the dimensions and shape of the locking lugs, the yield strength can be calculated. That will let you know if the rifle is going to come apart from a single firing incident.

The fatigue strength should also be calculated to ensure that it doesn't fail after repeated firings.

[JohnKSa]

But that also means the rifle is also applying reaction force to the brass. The difference of this two forces is the net force that accelerates the brass. That net force is close to zero because of the brass' light weight.

Ok. From a theoretical standpoint, I see where you are coming from. From a practical standpoint, there's enough force on the brass to stretch and deform it, and potentially enough to even turn it to liquid or gas if the pressure gets high enough.

And the latter can happen even without the rifle recoiling at all, so it's important to pay attention to the actual force involved.

[tangolima]

Originally Posted by JohnKSa

For practical purposes, the rearward force on the brass is the force on the bolt face. With the material properties of the bolt head, and the dimensions and shape of the locking lugs, the yield strength can be calculated. That will let you know if the rifle is going to come apart from a single firing incident.



The fatigue strength should also be calculated to ensure that it doesn't fail after repeated firings.

I went through that with the assistance of chatgpt. Looks like I will need to drop the load quite a bit. Still unknown is whether special steel is being used in the kak magnum bolt. For now I just assume not.

Actually I'm still working on it. Before I am absolutely sure, I just reduce the load. MV drops from 3000fps to 2850fps for 140gr bullet. It is still respectable.

-TL

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