Physics of shooting a rifle - bolt thrust

[tangolima]

Saw this debate on the other forum. Assuming no friction on chamber wall and the brass is intact. What's the bolt thrust, or the force exerted on the bolt face by the brass when the gun is fired? There are 2 schools of thought.

Head area x chamber pressure. Reason: gun literatures say so. Principle of hydraulics (pneumatics).

Bullet cross section area x chamber pressure. Reason: Sir Issac Newton's said so. Action and reaction.

What say you?

-TL

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[JohnKSa]

If pressure is a thing and it is really measured in pounds per square inch, then you should be able to use it to calculate force.

But it's not really the case head area that should be used for the calculation, it's the sum of the force inside the shell that is directed into the bolt. The area that the pressure acts on is inside the casing, pressure doesn't act on directly on the area of the case head. The area inside the case exposed to pressure where there will be resultant forces in the direction of bolt thrust will be smaller than the actual area of the case head except in cases with rebated heads.

I haven't messed around to see if the difference is significant. Or maybe the convention is done that way because it will underestimate bolt thrust and give an additional safety margin in conventional cartridges. It's worth pointing out that in cases with rebated heads, the bolt thrust could actually be more than the calculation of case head area times chamber pressure suggests.

Same thing with using the bullet cross sectional area with the pressure measurement. It's going to be less than the case head area except in cases with rebated heads.

[tangolima]

Originally Posted by JohnKSa

If pressure is a thing and it is really measured in pounds per square inch, then you should be able to use it to calculate force.

But it's not really the case head area that should be used for the calculation, it's the sum of the force inside the shell that is directed into the bolt. The area that the pressure acts on is inside the casing, pressure doesn't act on directly on the area of the case head. The area inside the case exposed to pressure where there will be resultant forces in the direction of bolt thrust will be smaller than the actual area of the case head except in cases with rebated heads.

I haven't messed around to see if the difference is significant. Or maybe the convention is done that way because it will underestimate bolt thrust and give an additional safety margin in conventional cartridges. It's worth pointing out that in cases with rebated heads, the bolt thrust could actually be more than the calculation of case head area times chamber pressure suggests.

Same thing with using the bullet cross sectional area with the pressure measurement. It's going to be less than the case head area except in cases with rebated heads.

Correct. It should be the inside diameter of the head that bears the chamber pressure. Unfortunately such dimension varies among manufacturers. For now we just call it "head area" with understanding that it refers to the inside diameter.

You are correct that the bolt thrust is sum of all forces on that plane. Now you know which way I am leaning towards.

-TL

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[tangolima]

I think we can use some real numbers as illustration.

Say my 7mm SAUM. Bullet diameter 0.284" and cross section area 0.063 sq-inch. Head inside diameter 0.5", and bearing area 0.196 sq-inch. Peak chamber pressure 50,000 psi.

Should the bolt thrust be

50,000 * 0.063 = 3150 lbf or
50,000 * 0.196 = 9817 lbf?

-TL

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[mehavey]

Head bearing area (0.196 in-sq) is the proper multiplier (ignoring brass stretch force reduction)

[tangolima]

Originally Posted by mehavey

Head bearing area (0.196 in-sq) is the proper multiplier (ignoring brass stretch force reduction)

That's the majority opinion.

But isn't the bolt thrust the same as the force on the bullet to accelerate it down the bore? Action and reaction, right?

-TL

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[JohnKSa]

Momentum is conserved, but momentum is based on the mass velocity product. Force is based on the mass acceleration product. If the mass velocity products are equal, it's pretty simple to see that the mass acceleration products can't also be equal except in the trivial cases.

Besides, given that motion results from the forces involved, it's safe to assume that the forces are not balanced.

Before anything moves, the force on the case head from the inside is the same as the force on the base of the bullet, assuming they have the same area.

[P Flados]

The pressure pushes the base of the case rearward to generate "bolt thrust". The inside diameter near the rear of the case is an approximation. Case friction and case stretch do matter, but can be neglected to bound the "well lubricated & very smooth chamber" condition. A slightly higher thrust would occur with a case separation near the rear of the case. This increases effective diameter and eliminates and reduction due to friction for the forward half of the case.

Now lets look at the other end of the cartridge.

For a straight wall case, the pressure is acting on the base of the bullet that has nearly the same area what we considered for bolt thrust.

For a bottleneck, there is a projected area of the shoulder. This area is equal to the max case ID Area minus the case neck ID area. So the pressure pushes forward against the bullet base and forward against the chamber at the neck. The sum of these two forces will be very close to the bolt thrust force. The force required to stretch the case is a small reduction in bolt thrust and the push forward against the shoulder when looking at bottleneck applications.

Now the above also assumes that the gun remains near stationary during firing. Thrust against the breech measured in pounds is reduced by how much the breech face accelerates rearward. The amount of breech face force reduction is a function of the F = M*A equation.

[mehavey]

But isn't the bolt thrust the same as the force on the bullet to accelerate it down the bore?

Yes (sorta, but....)

Pressure (e.g., 50,000 psi) is equal all through the chamber.
(However) Pressure = Force per square inch

So the absolute Force on any segment of that chamber (e.g., the bolt) = Pressure x the Area of that segment (the bolt face)








(Think of it as a classic hydraulics force multiplier... in reverse)

.

[tangolima]

Originally Posted by JohnKSa

Momentum is conserved, but momentum is based on the mass velocity product. Force is based on the mass acceleration product. If the mass velocity products are equal, it's pretty simple to see that the mass acceleration products can't also be equal except in the trivial cases.



Besides, given that motion results from the forces involved, it's safe to assume that the forces are not balanced.



Before anything moves, the force on the case head from the inside is the same as the force on the base of the bullet, assuming they have the same area.

Interesting point. I need to think about it more. At first glance, that's what comes to mind.

Conservation of momentum is derived from Newton's 2nd (F=ma) and 3rd (action and reaction) laws. It can't be turned around to invalidate Newton's 2nd law.

preriquisite of conservation of momentum is no addition of external force or energy. Not sure burning of propellent is considered external force or energy. The principle is valid at the moment when bullet exits the muzzle, when pressure drops to zero. We use that all the time. But not sure it is applicable while the projectile is accelerating in the bore.

-TL

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[tangolima]

Originally Posted by P Flados

The pressure pushes the base of the case rearward to generate "bolt thrust". The inside diameter near the rear of the case is an approximation. Case friction and case stretch do matter, but can be neglected to bound the "well lubricated & very smooth chamber" condition. A slightly higher thrust would occur with a case separation near the rear of the case. This increases effective diameter and eliminates and reduction due to friction for the forward half of the case.



Now lets look at the other end of the cartridge.



For a straight wall case, the pressure is acting on the base of the bullet that has nearly the same area what we considered for bolt thrust.



For a bottleneck, there is a projected area of the shoulder. This area is equal to the max case ID Area minus the case neck ID area. So the pressure pushes forward against the bullet base and forward against the chamber at the neck. The sum of these two forces will be very close to the bolt thrust force. The force required to stretch the case is a small reduction in bolt thrust and the push forward against the shoulder when looking at bottleneck applications.



Now the above also assumes that the gun remains near stationary during firing. Thrust against the breech measured in pounds is reduced by how much the breech face accelerates rearward. The amount of breech face force reduction is a function of the F = M*A equation.

Good point. I'm thinking alone the similar direction, only that the brass shoulder is not in contact with chamber brass. The gas also pushes the brass shoulder forward.

-TL

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[tangolima]

Originally Posted by mehavey

Yes (sorta, but....)



Pressure (e.g., 50,000 psi) is equal all through the chamber.

(However) Pressure = Force per square inch



So the absolute Force on any segment of that chamber (e.g., the bolt) = Pressure x the Area of that segment (the bolt face)

















(Think of it as a classic hydraulics force multiplier... in reverse)



.

The hydraulic principle is not wrong, neither is the action and reaction . But yet the conclusions are different. They cannot possibly be both correct, or sorta correct. There is something missing to reconcile these two theories.

-TL

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[JohnKSa]

But not sure it is applicable while the projectile is accelerating in the bore.

Yes, it is. Recoil begins and follows the law of conservation of momentum as soon as there's motion.

But yet the conclusions are different. They cannot possibly be both correct, or sorta correct. There is something missing to reconcile these two theories.

The forces are balanced (equal in all directions) as long as there's no movement. Once there's movement, the forces are unbalanced. They have to be or there would be no movement.

[tangolima]

Originally Posted by JohnKSa

Yes, it is. Recoil begins and follows the law of conservation of momentum as soon as there's motion.The forces are balanced (equal in all directions) as long as there's no movement. Once there's movement, the forces are unbalanced. They have to be or there would be no movement.

Recoil begins as dictated by action and reaction. We only apply conservation of momentum to calculate the rearward speed of the rifle at the moment the bullet exits muzzle. You may be right. I just need to think it through.

Use the brass as reference. It has no motion, so net force on it is zero.

-TL

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[JohnKSa]

We only apply conservation of momentum to calculate the rearward speed of the rifle at the moment the bullet exits muzzle.

That's only because it's hard to measure the speed of the bullet while it's in the muzzle. If we could measure the speed of the bullet while it's in the bore, we could apply conservation of momentum to determine the recoil velocity of the rifle the instant that the bullet starts moving.

We know that recoil begins the moment that the bullet moves because:

1. Acceleration can't be instantaneous except with zero mass or infinite force. The gun can't, therefore abruptly gain all its recoil velocity at the moment that the bullet exits the bore. It has to have started moving before that and gradually accelerated to full recoil velocity by the time the bullet exits the muzzle.
2. High speed video (if it is high enough speed and good enough resolution) will show that the gun is already recoiling before the bullet exits.
3. Conservation of momentum tells us it must.

Use the brass as reference. It has no motion, so net force on it is zero.

It moves with the rifle in recoil.

Once the bullet exits the case, the net force on the brass is no longer zero.

[tangolima]

Originally Posted by JohnKSa

That's only because it's hard to measure the speed of the bullet while it's in the muzzle. If we could measure the speed of the bullet while it's in the bore, we could apply conservation of momentum to determine the recoil velocity of the rifle the instant that the bullet starts moving.



We know that recoil begins the moment that the bullet moves because:



1. Acceleration can't be instantaneous except with zero mass or infinite force. The gun can't, therefore abruptly gain all its recoil velocity at the moment that the bullet exits the bore. It has to have started moving before that and gradually accelerated to full recoil velocity by the time the bullet exits the muzzle.

2. High speed video (if it is high enough speed and good enough resolution) will show that the gun is already recoiling before the bullet exits.

3. Conservation of momentum tells us it must.It moves with the rifle in recoil.



Once the bullet exits the case, the net force on the brass is no longer zero.

You are right. The brass does move with the rifle. But the its mass is so little that the net force needed to accelerate it is insignificant compared to all the force components involved, so practically we can assume it zero without incurring much error.

-TL

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[JohnKSa]

The force that accelerates the brass accelerates the brass AND the entire rifle rearward. In this situation, that's the physical mechanism that creates recoil.

Where else is there any other unbalanced force acting on the rifle in the direction of recoil?

[tangolima]

Originally Posted by JohnKSa

The force that accelerates the brass accelerates the brass AND the entire rifle rearward. In this situation, that's the physical mechanism that creates recoil.



Where else is there any other unbalanced force acting on the rifle in the direction of recoil?

The brass moves the rifle. That's true. But the net force on the brass alone is a tiny portion of the force to accelerate the rifle backwards.

-TL

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[mehavey]

.

[tangolima]

Originally Posted by mehavey

.

Imagine the bullet is welded to the casing, and the casing is strong enough to hold the pressure alone. There is no bullet acceleration. The bolt thrust is zero.

I'm not saying you are wrong necessarily. The discussion is how to reconcile those 2 theories.

-TL

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[mehavey]

There is no bullet acceleration. The bolt thrust is zero.

Bullet (or any any other movement) isn't part of the equation.
The bolt is part of the chamber. And the chamber experiences Force/Thrust over its entire inner surface area

Now "Work" (force over distance)... that does involve movement.
But Force (e.g., bolt thrust) exists even in the total absence of movement.

[mehavey]

Y'all may be trying conflate conservation of momentum with static thrust force.

Yes... mv = MV between the bullet/gases going one direction, and the whole rifle
(of which the bolt is only a small part) going the other direction.

But again... actual movement of any part of the System isn't necessary for thrust force to exist at any instant in time.
AND... note that the thrust force rises -- and falls -- over all those instants of time

[JohnKSa]

But the net force on the brass alone is a tiny portion of the force to accelerate the rifle backwards.

What is the source of the rest of the force that accelerates the rifle backwards, and what is it acting upon?

But again... actual movement of any part of the System isn't necessary for thrust force to exist at any instant in time.

Correct. If the bolt couldn't withstand the thrust force, the brass would fail.

[FrankenMauser]

And what about a tapered case acting as a wedge?...

[tangolima]

The bullet and rifle are subject to the same force but in opposite directions. That force is bullet cross section area x chamber pressure =3150 lbf. The brass is part of the rifle, so it has the same speed and acceleration as the rifle. The net force on the brass is much lower because of its light weight.

Say the rifle weighs 8 lb, and the brass 100gr. The net force on brass is

3150 * 100/(8*7000)=5.6 lbf

which is insignificant compared to the magnitude of bolt thrust force we are trying to determine. For practical intent and purpose, we can assume zero net force on the brass without incurring too much error.

Bearing that in mind, we can try calculating the bolt thrust. Straight wall is no dispute. Bolt thrust force

Fbt=bullet cross section area* chamber pressure p = brass head bearing area * p

Focus will be on bottle necked brass.

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