Range Report Physics of recoil

[Chanonry]

OK apologies in advance. I can't get this straight in my head. I have been thinking about the bullet energy and rifle recoil energy.

When the primer is struck the powder produces a gas which applies a pressure in all directions accelerating the bullet and the rifle in opposite directions. Fine. The pressure x the cross-sectional area of the bullet equates to the force applied which when multiplied by the distance through which this is applied (roughly the bll) gives the energy imparted or work done. This force accelerates the bullet depending on its mass. OK

The opposite reaction causes the rifle to recoil but with a lower accelration due to its larger mass. It therefore achieves a lower velocity. I guess calculating kinetic energy from this say that the energy is low (relative to the bullet) as the velocity is considerably lower.

BUT why does the powder apply more energy to the bullet than the rifle? Shouldn't it be the same?

Nerdy but I can't get it out of my head

Andy

 

[mattj]

Re: Physics of recoil

Instead of trying to think about conservation of energy (not really helpful for simply calculating recoil), you want to apply the law of conservation of momentum:

(M1*U1) + (M2*U2) = (M1*V1) + (M2*V2)

M1 is mass of first object (rifle).
M2 is mass of second object (bullet).
U1 is velocity of M1 before.
U2 is velocity of M2 before.
V1 is velocity of V1 after.
V2 is velocity of V2 after.

...for convenience, lets just overlook the "hot escaping gas" factor (which is actually significant, but I think ignoring it will still provide sufficient insight):

Before the rifle fires, U1 and U2 are zero, so:

0 = (M1*V1) + (M2*V2)

...or shuffled around:

M1*V1 = -M2*V2

The negative is basically just telling us that the velocity of the rifle and the bullet are in different directions.

Lets just say, for example, that the rifle and everything attached to it weights 10 pounds. There are 7000 grains in a point, so the rifle weighs 70,000gr. For our example, lets use a 175gr bullet. Plugs those in to the last equation, and we get:

70,000gr*V1 = -175gr*V2

Divide both sides by 70,000gr, and we get:

V1 = -V2 / 400

In other words, the bullet will accelerate to 400 times the speed it its direction than our rifle will in the other direction (it was just floating and allowed to accelerate freely).

So, if our bullet accelerates to 2750fps, our rifle (if allowed to accelerate without hindrance from anything) would accelerate to just under 7fps (2750fps/400). Coincidentally, our convenient ignoring of the "escaping hot gasses" adds up to another couple fps for the rifle.


I think the simple concept you are looking for in putting things right in your head is: "a gun is a LOT heavier than a bullet"

Hope that helps,

-Matt

 

[Chanonry]

Re: Physics of recoil

Thanks for doing the math.

So if we look at the relative kinetic energy then the bullet will have KE of say 175.(3000)^2 while the rifle has 70,000. (3000/400)^2. So the rifle's kinetic energy is only 0.25% of the bullets? Is that right, 3000fps is a bit zippy but the logic is right?

I was getting confused by some bloke who wrote a loooong article on terminal ballistics. He was poking holes in every theory under the sun but I think he had a few holes in his own.

His assertion was that the kinetic energy of a bullet at the target isn't actually that great (whatever that means he was comparing it with 3 lb gel bags at 50fps etc, I calculate the bullet as having 3 times the energy) and arguing the bullet doesn't have enough energy to knock a deer off its feet despite what we have seen in the field. At one point he made an argument that the recoil impact was analogous to the bullet impact (ignoring penetration - small point!!). That just didn't compute for me but explains my energy focused thinking. I guess the charge imparts more energy to the bullet than the rifle because the bullet has much lower inertia.

Anyway the physics says there is a LOAD more energy at the deer's end than mine...

Andy

 

[Greg Langelius *]

Re: Physics of recoil

The physics are more than this.

Once the bullet has exited, there is still all the mass of the propellant gas at the muzzle pressure, which exerts additional force on the rifle in the form of gaseous thrust Ala a rocket.

There is less energy it the deer's end than at the muzzle, because much of that energy is consumed by air friction.

As for why bullets do so much damage, imagine a .30 caliber pointed rod being placed against one's chest while lying back on the ground, and then having a 1000lb weigh dropped on its end from a height of one foot above the rod. It'll go pretty deep, won't it? That's the effect of a bullet that carries 1000ft/lb of terminal energy.

Greg

 

[Chanonry]

Re: Physics of recoil

Greg

Hadn't thought about the gas. So does that mean we can consider the mass of gas to be equivalent to the powder mass, and it's velocity is what - the same as the muzzle velocity of the bullet?

So ejecta momentum is much higher as the ejected mass is circa 175gr + say 42gr load x exit velocity from the barrel? Therefore doing the calc properly implies 20% increase in rearward velocity of the gun (in this case)?

Interesting

Andy

 

[Greg Langelius *]

Re: Physics of recoil

Andy;

The whole question is a matter of academics, and my interests are practical ones. My point is simple, once you have precise numbers what value do they have? As a matter of comparison, or for the purposes of discussion, they might be germaine.

But we don't need numbers to know that two rifles kick about the same, or one more than another, and that a bigger bullet will mean more recoil, that a muzzle brake will reduce some, but not all recoil, and that a recoil pad and/or ballast will reduce felt recoil.

The question has about the same degree of meaning for me as asking how many angels can dance on the head of a pin. Perhaps it's interesting for some, but not for me.

Greg

 

[kmussack]

Re: Physics of recoil

Mass of propellent (not weight)

Gas Velocity =4X muzzle velocity of bullet (approximate)

Thrust: F = mVe + (pe – po) Ae

The first half of this equation will get you close.
The second half is the temperature/aperture contribution.

 

[Phil1]

Re: Physics of recoil

Newtons third law of motion(grade 10 physics);
Third law
Whenever a particle A exerts a force on another particle B, B simultaneously exerts a force on A with the same magnitude in the opposite direction. The strong form of the law further postulates that these two forces act along the same line. This law is often simplified into the sentence, "To every action there is an equal and opposite reaction."

Don't overthink the obvious.

 

[Killswitch Engage]

Re: Physics of recoil

holy shit!!!!! i am backing out now. sorry

 

[jwp6114]

Re: Physics of recoil

My head hurts...

i'z just's piont it an pull da triga boss.

 

[The Mechanic]

Re: Physics of recoil

I think some energy is imparted in the form of resistance to the bullets forward movement by friction in the bore. That should pull the rifle forward as friction attempts to slow the forward motion

 

[The Mechanic]

Re: Physics of recoil

forgot about muzzle breaks..........

 

[noname]

Re: Physics of recoil

total recoil is sum of ejecta, powder charge in mass + plus mass of bullet, or in other words, ejecta from gas discharge can be substantial on a proportional basis, eg, .223 say 25 gr of powder + 55 gr bullet == 25/80 or 31% tot. as opposed to say a 45 acp 5 gr. powder+230 gr bullet or 5/235 or .02%, why the higher the mass ratio is-- the better the muzzle brake works. (escaping gases accelerate to above 4000 fps because of lower mass when the exit barrel into the atmosphere. acceration vector of bullet stays same, while heavier gun with break shows a smaller (in comparison), vector.

 

[Rovieairto]

Re: Physics of recoil

This is REALLY simple.

Work = force over a distance (measured in Ft. x Lbs.)

There are two projectiles in this equation (bolt gun for simplicty)

One is the bullet and powder gasses and the other projectile is the rifle.

Although they both have equal forces acting on them the bullet travels a far greater distance than the rifle does in the same time period.
Therefore much more work is done to the bullet than the rifle.

Actually the rifle may move say: 0.05" while the bullet travels 24.0".

If the bullet energy were to be 2000 ft. x lbs. and we were to employ a 100% effective muzzle brake (you can exceed 100% BTW)
then in this sample problem the rifle's recoil energy would've been 4.17 ft. x lbs.

Bullet) 1000 lbs. force (ave) acting over 2.0 ft. = 2000 ft. x lbs.
Rifle) 1000 lbs. force (same) acting over 0.00417 ft. = 4.17 ft. x lbs.

Knowing bullet energy loss at x distance you can solve for average bullet drag over that distance and many other fun things that you might want to know.

 

[noname]

Re: Physics of recoil

Hamilton please explain how the brake can be 100% effective let alone above that figure?

 

[wadebrown]

Re: Physics of recoil

I have done a little physics in the past, but do not consider myself a Physicist.

Stick with the momentum arguments and stay away from trying to calculate the recoil by energy/work techniques. Unless you are ready to make complicated measurements of the temperature increase of the firearm and brass your results using energy /work are not going to be accurate. I fully agree with cons of energy and the second law of thermo but a lot pf energy goes in heating the barrel and if you do not account for this your results will not be very good.

Using momentum as long as you include the mass of the powder, most people just use the bullets mass, is much more accurate and easier.

 

[Rovieairto]

Re: Physics of recoil

In this case the energy method requires the fewest operations so it should be the simplest method for folks to understand.
If you can't accept that work = force x distance then there will be no progressing further.
Using momentum (excluding propellant gasses) both rifle and bullet have equal quantities. That doesn't help him understand how they are not equal in energy.
So I showed how they can have the same momentum, but with vastly different energies.

About the brake being 100% effective at reducing propellant effects:
Suppose that the brake were to vector the gasses rearward (as many do) so they are, in effect, ejecting a mass with a counter velocity.
If no projectile was involved (but the exhaust gasses remained the same) the rifle would actually recoil forward because of this "thrust" effect. Newton's 3rd law of motion.

A 100% effective brake would eject matter to the sides with some leaking forward and the same amount rearward so that their collective momentum about a point were zero.
I did this because propellant gas momentum is irrelevant to the core of the solution and for simplicities sake it is best left out.
I really think that this is a simple calculation and so I wanted to show just how simple it really is.
I've studied physics so long that I may see the world a little different and it might actually be complex but if it is then I apologize.

 

[Downzero]

Re: Physics of recoil

I have never taken a physics class for credit, so bear with me here.

Why do people think that heavier bullets make for more recoil? If f = mv^2, it would seem that a lighter bullet would have more recoil.

My experience with pistols is that heavier bullets produce slower, and what feels like lighter, recoil events. And perhaps my perception sucks, but I can't tell the difference between a 155, 168, and 175 grain .308 bullet, and all .223 feels the same to me (light).

Explanation?

 

[Rovieairto]

Re: Physics of recoil

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: wwbrown</div><div class="ubbcode-body">
I fully agree with cons of energy and the second law of thermo but a lot pf energy goes in heating the barrel and if you do not account for this your results will not be very good.
</div></div>

The results will remain unchanged if we account for it or not.

It is a simple matter to measure the bullet's velocity.
Any energy that went into the barrel through friction or heating was not available to do work on the bullet and so it has no effect on your velocity measurement.
Meaning that you don't need to know one to know the other.
It doesn't matter if the velocity would've been higher, so then would have the been the recoil.
We are only concerned with what is and not what could have been in this sample.
It is propellant energy lost and only relevant if you wanted to know the efficiency of energy transfer.

 

[Rovieairto]

Re: Physics of recoil

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Downzero</div><div class="ubbcode-body">I have never taken a physics class for credit, so bear with me here.

Why do people think that heavier bullets make for more recoil? If f = mv^2, it would seem that a lighter bullet would have more recoil.

My experience with pistols is that heavier bullets produce slower, and what feels like lighter, recoil events. And perhaps my perception sucks, but I can't tell the difference between a 155, 168, and 175 grain .308 bullet, and all .223 feels the same to me (light).

Explanation? </div></div>

F= MA
E= 1/2 M V^2

M= mass in slugs/ft.^3

A heavier bullet that gets the same amount of energy from the propellant as a lighter bullet will have more momentum.
Usually it is a wash because you end up using a little less propellant with heavier bullets, but the propellant is more efficient at transferring energy because it does not have to waste as much of it's energy getting itself up to the slower speed of the heavier bullet.

BTW:
(There is a maximum speed that the propellant can get itself going with no bullet mass)

Please understand that the effects of propellant gasses only add to the difficulty in understanding something that folks are already having trouble with. It really is just more of the same and so we should not try to include it just yet.

So now: If the bullet now has more momentum then so do the recoiling parts.
Since the mass of the recoiling parts is constant and it's momentum was increased, then so is the velocity and energy of the recoiling parts.

Therefore the recoil will be increased if bullet weight increases but bullet kinetic energy remains the same.

 

[noname]

Re: Physics of recoil

switching from bookeeping for a moment, why not just solve the problem with thermodynamics? the work done is equal to the integral of the force applied. since pressure can be measured anywhere in the length of the barrel (complicated or not), the amount of work at that point can be evaluated. Also heat differential can be easily calculated as power lost in the barrier transfer. If you plot the results of the two curves, velocity in the barrel, against kinetic energy they will cross about 1/3 the distance from the chamber at which velocity increase contributions by the propellant become almost linear.
No brake directs all the gases back and few even get to 45 degrees. Reduction of recoil by brakes is reduction by subtraction not decceleration and the derivative is a constant or an impulse not thrust which is per sec/ per sec. (at least first order).

 

[Rovieairto]

Re: Physics of recoil

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: shooter2</div><div class="ubbcode-body">switching from bookeeping for a moment, why not just solve the problem with thermodynamics? the work done is equal to the integral of the force applied. since pressure can be measured anywhere in the length of the barrel (complicated or not), the amount of work at that point can be evaluated. Also heat differential can be easily calculated as power lost in the barrier transfer. If you plot the results of the two curves, velocity in the barrel, against kinetic energy they will cross about 1/3 the distance from the chamber at which velocity increase contributions by the propellant become almost linear.
No brake directs all the gases back and few even get to 45 degrees. Reduction of recoil by brakes is reduction by subtraction not decceleration and the derivative is a constant or an impulse not thrust which is per sec/ per sec. (at least first order). </div></div>



If you can't dazzle them with brilliance..........................

 

[noname]

Re: Physics of recoil

you didn't answer the question

 

[Rovieairto]

Re: Physics of recoil

I don't see the question and I apologize for the wisecrack.
I'm trying to help the OP and those who want to learn.

This problem can be solved many different ways.
This is a physics problem IMO.

 

[Downzero]

Re: Physics of recoil

You still haven't satisfied my question either.

If energy = mass times the square of velocity, there is no justification for heavier bullets having a higher felt recoil. If F=ma is the right formula, then it'd depend on how acceleration is properly calculated, which I have on idea (but I recall it involving a derivative).

 

[Rovieairto]

Re: Physics of recoil

For gravity use 32.2 ft./sec.^2
(The metric system is king for a reason)

The momentum an object has tells you nothing about how much kinetic energy it has, you need to know more to find that.

This is how to calculate momentum (neglecting powder mass effects for simplicity, and we need it)

Vg x Mg = Vb x Mb

If you increase Mb (mass of the bullet) see what happens to the Vg.
Vg would = mass of recoiling gun and for a 1911 it would include the slide, barrel, and the empty brass case.

Here is a funny story that can help the visual crowd:
I watched this video of the fellow who decided that it would be a good idea to jump off the top of an aluminum step ladder into a pool.
So he sets the ladder next to the pool and he climbs to the top. With a stupid look on his face he smiles, waves, and leaps with all he's got!
Of course he never made it.

In this analogy:
The man is the gun (because he's heaviest), the ladder is the bullet, and his legs are the propellant.
His legs supplied the force over a distance (energy) that could have gotten him there and his brain knew that, it's just that the ladder took most all of the energy he supplied.
If he had used a heavier ladder (bullet), he would've attained more velocity (gun) and thus possibly made the pool and not the internet.
Recoil is simply you arresting the guns rearward velocity.

If you're really interested in the physics of projectiles and firearms then get this book:
Modern Practical Ballistics
It's a great start.

I also have many E-book versions of studies conducted by the war dept that are full of empirical data.
For those who REALLY want info I can help out.

 

[Kinsman]

Re: Physics of recoil

I'm too simple minded for all this.
For a given energy produced by the propellant, the gun recoils at a slower velocity because it is heavier than the bullet.
The bullet is traveling at many times the velocity of the gun.
As speed doubles, the energy quadruples.
That's the simple reason that the gun recoils with significantly less energy than the boolit.

 

[Rovieairto]

Re: Physics of recoil

exactly.

Any time you have a force over a distance, work has been done and you could express this with ft. x lbs. (energy).

While the propellant supplies the energy the bullet and recoiling parts share it.
(still shouldn't throw in gas mass just yet)

The bullet experiences the same forces as the recoiling parts except that during that time frame it travels considerably farther.
Read: same force over much greater distance.

Actually the bullet is slowed somewhat because the platform it is launching from recoils to the rear.
Just like the idiot on the ladder.
This loss in velocity of the bullet, is exactly equal to the velocity of the recoiling parts.
The energy it lost due to this velocity loss is also equal to the energy in the recoiling parts.

What is interesting is about this is say the recoiling rifle attained a rearward velocity of 10 ft/sec.
The bullet then also lost 10 ft/sec.

That last 10 ft/sec of the bullet contained as much kinetic energy as the (much heavier) recoiling rifle.

Energy increases by the square of the velocity so it can compound quickly.

 

[BLAWTON]

Re: Physics of recoil

Quickload has this handy little feature that figures recoil for the load you are working on.

 

[BugSniper]

Re: Physics of recoil

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Greg Langelius *</div><div class="ubbcode-body">The physics are more than this.

Once the bullet has exited, there is still all the mass of the propellant gas at the muzzle pressure, which exerts additional force on the rifle in the form of gaseous thrust Ala a rocket.
</div></div>

Don't forget that gaseous thrust actually starts with the displacing of at least 16 inches of air before the loss of the bullet's mass creates thrust at T-1 before the pressurized gas gets to enjoy the depressurizing freedom of the open air. It may not be as spectacular as the pressurized gases, but the same thrust principal is already at work once you have ignition.

 

[Rovieairto]

Re: Physics of recoil

How about a 9x19mm AK vrs. a 45ACP AR15 debate?

Anyone?

 

[GuinnessNM]

Re: Physics of recoil

Both energy and momentum are conserved. The math goes like this -

Energy conservation can be written as:
E(powder)=K(bullet)+K(gas)+K(recoil)
This says that the chemical energy released burning the propellant is shared as kinetic energy among the bullet, gas created by the burning, and the recoil of the rifle (some also goes into heating up the barrel and the world). The kinetic energy is K=1/2*M*v^2, where M and v are the mass and velocity of the component (bullet, gas, or recoiling rifle).


Momentum conservation can be written:
p(total)= 0 = p(bullet)+p(gas)-p(recoil)
where the minus sign indicates that the rifle recoils in the opposite direction to the motion of the bullet and gas. Before the powder touches off nothing is moving and the total momentum is zero. After the explosion the momentum of the recoiling rifle balances that of the bullet and gas. p=Mv for each component, thus K=p^2/2M for each.

It is simplest to ignore the gas for the moment because its energy and momentum are small compared to the bullet. Then the momenta are equal for the bullet and rifle, call this p, and the energies are:
K(bullet)=p^2/2M(bullet) and K(rifle)=p^2/2M(rifle).
This means that the energies divide in the ratio K(rifle)/K(bullet)=M(bullet)/M(rifle) <<1. The mass of the bullet is only about .2% of the rifle mass (for a 150gr bullet and 10 lb rifle), so the recoil energy of the rifle is .2% that of the bullet. Energy is what kills, not momentum (Jeff Cooper got it wrong), so the bullet can kill but not the recoiling rifle. (Cooper's conclusions might be right, but for the wrong reason.)

In terms of physical understanding the relevant quantity is the impulse of the explosion, not the work done.

The momenta of the bullet and the recoiling rifle are equal because they experience equal forces (F) from the explosion acting over the same time (t = the time it takes the bullet to leave the barrel). p = F*t = impulse. The rifle and bullet experience the same impulse.

The energies are not the same because the work done on an object is given by the force times the distance the object moves. The energy gained by the object equals the work done on it by a force, K = W = F*d. The bullet gets more energy than the rifle because the bullet moves much further under action of the propellant force than does the rifle during the same time period.

Hope this helps.

 

[badbot]

Re: Physics of recoil



I bow to the math wizards but my simpleton vision is to break the question into smaller questions.

1 everything before the bullet exits the barrel.
2 everything before the bullet reaches the target after exiting the barrel.
3 everything before the bullet stops after reaching the target.

These need to be separate issues, as they really do not interact.
I.e. internal ballistics, external ballistics, terminal ballistics.

Trying to put all of these into your head as a single thought will make you nuts and confuse others.

--------------------------------

The original poster asked “BUT why does the powder apply more energy to the bullet than the rifle? Shouldn't it be the same?”
No.
Energy is expended in dozens of places and ways. The bullet and gun are only 2 of these.

The powder expends X energy! Most of the things that use X have been tossed in somewhere.
Expanding the brass, friction, heating many things brass-bolt-barrel-bullet-air, overcoming inertia of the bullet-gun-shooter-bipod-bench-earth. The contained pressure acts on many surfaces, the base of the bullet is the smallest, but it is the one we want. After the bullet leaves the barrel the powder is still expending energy but not getting much for us beyond thrust. And some powder goes unburned.

---------------------------------

Another poster mentions lighter bullets having higher power should have higher recoil.
Terminal force for the lighter bullet enjoys the square of velocity so the light bullet looks like it has more force.

Recoil is a completely separate thing!.
A 100-grain bullet has much less moment of inertia than a 200-grain one. About half as much I would think.
Overcoming that is what affects recoil more with heavy bullets.

Putting the weight of the bullet over the weight of the gun makes nice math but the real world is much harder to accurately measure.
OK the gun is say 10 pounds, but I weigh 200 I am pulling the gun into my shoulder with 3 pounds of force starting the compression of flesh and bone. The resistance of my body on the butt of the gun starts there and is a rising curve until I am moving at the top speed that the gun reaches.
That curve depends on many things am I leaning over a bench? Standing on the balls of my feet? Or on my heels heading for my ass hitting the ground?

---------------------

One other poster thought the friction of the bullet in the bore could pull the barrel forward.
I don’t think so the force of friction in the bore merely reduces total acceleration. Remember the gas is pushing the gun to the rear with more force than the bullet while it also pushes the bullet forward deforming the bullet into the rifling and overcoming the friction and air in the barrel.
I hear you asking more force??? Yep, well I think so, the pressure is close to equal while the bullet is in the barrel but the surface area of the bullet base is smaller than the base of the cartridge. Hydraulics would tend to have more force at a slower speed. Inertia should be equalized between the opposing forces.

-----

too many questions huh? simple is better, lets go blow something up.

 

[Chanonry]

Re: Physics of recoil

Guys

i am glad i started this thread despite arsy comments from earlier participants!

my initial problem was that i couldn't see how energy was conserved despite the fact that my education says it must be. having filled in all the elements of the energy equation i am getting there.

Ta

Andy

 

[skinnypitt]

Re: Physics of recoil

It sounds like the paper you are talking about is one called "Shooting holes in wound theories, The mechanics of terminal ballistics" It is a very good paper and has a broad range covered by something like 20-30 pages. I think a few calculations on my part could not do as good a job as he did in his paper. http://www.rathcoombe.net/sci-tech/ballistics/myths.html

But basically I think he was pointing to the fact that you not only have to look at the force, or mass x accelleration, but you also have to look at the area over which that force is applied (P) and additionally the time over which that force is applied.