Interesting: Recoil due to powder weight.


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Macchina
March 23, 2013, 03:13 PM
I'm sure many (most) of you know this, but I was reading about internal ballistics and a light bulb turned on. I often wondered why efficient cartridges, especially pistol caliber carbines, have such low recoil for the energy they deliver. One of the many reasons is a light-weight powder charge. I never thought about, but the mass of that powder is exerting a force on the breachface just like the mass of a bullet and the 24g of H110 in my .44 Mag is a heck of a lot lighter than the 70+ grains I used to less into my 300 Win Mag.

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danthearmyman
March 23, 2013, 03:44 PM
It wouldn't be the mass of the powder. The powder burns to create a expanding gas. This expanding gas pushes on the bullet and the breach face equally. The bullet accelerates like hell because of its low mass as compared to the firearm. More powder=more gas produced.

Newtons Law: Every action has an equal and opposite.

If you push the bullet forward, the gas pushing has to push back against something.

Macchina
March 23, 2013, 04:54 PM
That is originally what I thought, but all that gas weighs something (exactly as much as the powder out came from) and is traveling at near the same velocity as the bullet. Think about why a compensator works ..

Haxby
March 23, 2013, 05:04 PM
Yep, all that powder goes out the end of the barrel. And since the powder is behind the bullet, pushing, as soon as the bullet gets out of the way, the gas exits at a faster speed than the bullet did.

ljnowell
March 23, 2013, 05:32 PM
Recoil difference could also be because the 300 win mag runs at 64,000 psi whereas the 44 Rem Mag runs at 36,000 psi. Big case, large volume of high pressure gas = more recoil. Note where I said large volume of higher pressure gas. That volume plays a role in it.

Macchina
March 23, 2013, 05:49 PM
I Understand there are substantial differences between the .300 and .44. I'm only taking about powder weight.

Take for example a rocket in outer space, very simplified: the exploding fuel exits the rocket at a velocity (a gun with no bullet), it is only the mass of that fuel being propelled away that that pushes the rocket.

rhadamanthos12
March 23, 2013, 06:09 PM
What happens with combustion is you end up with by products, if the combustion isn't complete which usually it isn't you don't use all the available energy, the unburned components appear as smoke,soot, etc. You can use the by products to determine how much air was used or wasted in your combustion.

Energy in has to equal energy out, that is why muzzle energy would work better for recoil calculation, you just need to use the proper equation for it. Also it is going to be a multi dimensional calculation, granted we most likely only care about the horizontal component.

Lost Sheep
March 23, 2013, 07:09 PM
Try this: Using a long barreled gun, load up with a stout charge of an energy-dense powder. Measure the velocity.

Get a slower, heavier powder and load whatever it takes to match the Bullseye velocity with the same bullet and same gun. (The powder will certainly be 10-20 grains more)

Compare the felt recoil. Compare the actual recoil.

You would expect the heavier charge of powder to give more recoil. But I am given to understand that the lighter powder charge will FEEL heavier in recoil.

Some say it has to do with the fact that the peak pressure of the faster, lighter powder charge is greater. Some say it has to do with the fact that the slower powder delivers its acceleration to the bullet over a longer period of time.

I will leave it to each of you to judge for yourselves.

Lost Sheep

Haxby
March 23, 2013, 07:24 PM
The online recoil calculators work well to calculate recoil.

kingmt
March 23, 2013, 07:28 PM
Lost Sheep
Problem with that theory is the slower powder reduced to the same speed probably won't burn fully. Unless you just mean the difference between Red Dot & Blue Dot. There isn't really much difference in the recoil of those two in a rifle.

Lost Sheep
March 23, 2013, 08:11 PM
Lost Sheep
Problem with that theory is the slower powder reduced to the same speed probably won't burn fully. Unless you just mean the difference between Red Dot & Blue Dot. There isn't really much difference in the recoil of those two in a rifle.
I don't understand the problem.

The bullet winds up downrange at the same velocity.

The powder, whether burnt, unburnt or in-process ejects from the muzzle. We are not talking about cartridge efficiency, but only about recoil.

I don't understand your question. PM if the point is off-topic and the answer is too long to post.

Lost Sheep

Macchina
March 23, 2013, 09:29 PM
100% of the mass of the powder (in the form of gas) exits the muzzle at the velocity of the bullet. How could the mass of the powder contribute anything less to recoil than the bullet itself does.

My original point was:
.44 Mag with a 200 grain bullet + 24g of powder = 224g exiting the muzzle.
.300 Win Mag with a 200 grain bullet + 74g of powder = 274g exiting the muzzle.

The 300 has a 22% higher load mass, all of which is seen completely in recoil. I understand that the mass CONTRIBUTED to a much higher velocity, but in the end two 200 grain bullets are still exiting the muzzle.

Perhaps a better example is a 357 Mag with 2 different powder charges of 6 and 16 grains, both of which deliver the same velocity with the same 158gr bullet. The 16 grain load will have a higher felt recoil with no terminal advantage.

bds
March 23, 2013, 09:47 PM
I often wondered why efficient cartridges, especially pistol caliber carbines, have such low recoil for the energy they deliver. One of the many reasons is a light-weight powder charge. I never thought about, but the mass of that powder is exerting a force on the breachface just like the mass of a bullet and the 24g of H110 in my .44 Mag is a heck of a lot lighter than the 70+ grains I used to less into my 300 Win Mag.
It's simple application of physics (Newton's second law of motion (http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion)) where force (F) is expressed as:
F = ma (Force = mass x acceleration)
It doesn't matter whether it's pistol/carbine/rifle cartridge, greater the force the bullet exerts on the breechface from expanding high pressure gas, more the recoil.

rhadamanthos12
March 23, 2013, 09:52 PM
I don't think you can count it that way, the powder is a potential energy. Upon ignition it become kinetic energy, if there was another force acting on the powder and the powder was not acting as a source of work on the bullet then you might be able to include it into your recoil calculation

ljnowell
March 23, 2013, 09:52 PM
100% of the mass of the powder (in the form of gas) exits the muzzle at the velocity of the bullet. How could the mass of the powder contribute anything less to recoil than the bullet itself does.

My original point was:
.44 Mag with a 200 grain bullet + 24g of powder = 224g exiting the muzzle.
.300 Win Mag with a 200 grain bullet + 74g of powder = 274g exiting the muzzle.

The 300 has a 22% higher load mass, all of which is seen completely in recoil. I understand that the mass CONTRIBUTED to a much higher velocity, but in the end two 200 grain bullets are still exiting the muzzle.

Perhaps a better example is a 357 Mag with 2 different powder charges of 6 and 16 grains, both of which deliver the same velocity with the same 158gr bullet. The 16 grain load will have a higher felt recoil with no terminal advantage.
__________________

100% of the powder will never make it out of the barrel, a large majority yes, but not 100%. Further, to illustrate the point that pressure is the commanding factor, lets take a 44 mag load with a 240g bullet and 22g of powder. Leave the same amount of powder in it and put a 300g bullet on top of it, now what happens to recoil? Or a 340g bullet?

rcmodel
March 23, 2013, 09:56 PM
Or use this recoil calculator.

http://www.handloads.com/calc/recoil.asp

Notice the powder weight is part of the 'ejecta' used to figure recoil.

All the powder gas mass isn't used as 'Ejecta" in the recoial formula because some of it stays in the barrel pushing the other way and not all of it is ejecta.

rc

Walkalong
March 23, 2013, 10:02 PM
Yep, powder weight figures into recoil. The weight gets pushed forward just like the bullet. It isn't a big part of the equation, but it is a part.

jstein650
March 23, 2013, 10:08 PM
The mass of powder in used in any cartridge most certainly contributes to real free recoil energy, whether it's completely consumed in combustion or not. In fact, the factor used in most calculations, the velocity of the powder gases (part of the ejecta) is multiplied from 1.25 for long shotguns to 1.75 for high powered rifles which definitely contributes to recoil, since a lot of that mass is moving faster than the bullet or shot.

.22-5-40
March 23, 2013, 10:08 PM
Most 12GA. shell boxes are printed..3 Dram Equlivant..this means the smokeless powder charge, even though of much lighter weight gives the equal velocity to shot charge as the old 3 Drams of black powder...And from what I have heard..the old timers agree the black powder loaded shells kicked harder than smokeless.

kingmt
March 24, 2013, 12:38 AM
Actually I think I'm pulling out of this one. I didn't really understand the OP anyhow.

As for unburnt powder tho it is wasted because it never burnt to release it's energy. I always welcome PMs.

Jim Watson
March 24, 2013, 12:41 AM
In a .223 M193 the powder charge is about half the weight of the bullet, therefore 1/3 of the ejecta and contributes 1/3 of the recoil without even considering flow faster than bullet velocity after exit.
With a muzzle pressure of several thousand psi, I doubt much is staying in the barrel at 15 psi.

Lost Sheep
March 24, 2013, 01:54 AM
It's simple application of physics (Newton's second law of motion (http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion)) where force (F) is expressed as:

It doesn't matter whether it's pistol/carbine/rifle cartridge, greater the force the bullet exerts on the breechface from expanding high pressure gas, more the recoil.
The bullet exerts no force on the breechface. (edited to add: Because the bullet does not actually touch the breechface, how can it apply force?) Indeed, friction with the barrel actually acts to pull the gun forward, thus reducing felt recoil.

The force applied to the breechface is by the base of the cartridge. The cartridge is being acted upon by the pressure of the powder and that pressure only.

Thus a 25,000 psi application for so many microseconds can deliver a given bullet where 12,500 psi applied for twice that length of time will deliver the same bullet at the same velocity. It is not as simple as that, but you get the picture.

Which do you think has the greater recoil (if either)? The 12,500 psi or the 25,000 psi applied for half the time?

Newton's Laws (principles) are simple, as most physics is. Nature is simple and direct. How these simple principles combine to effect can be very complex.

I am still trying to figure it out. But making progress little by little.

Lost Sheep

bds
March 24, 2013, 03:32 AM
The bullet exerts no force on the breechface.
I tried to keep the concept simple ...

If you don't believe me, remove the bullet from the cartridge and cover the opening with some tissue paper and fire to see how much recoil you get. ;):D

What I said was:
greater the force the bullet exerts on the breechface from expanding high pressure gas, more the recoil.
It is ultimately the mass of the bullet and the acceleration caused by the expanding high pressure gas that exerts force on the breechface ... which we feel as recoil.

I understand there are other forces at play such as escaping high pressure gas through the muzzle pushing the gun back, friction between bullet and barrel, rotation of the earth etc. but they are minor in comparison.

hentown
March 24, 2013, 08:42 AM
I'm sure many (most) of you know this, but I was reading about internal ballistics and a light bulb turned on. I often wondered why efficient cartridges, especially pistol caliber carbines, have such low recoil for the energy they deliver. One of the many reasons is a light-weight powder charge. I never thought about, but the mass of that powder is exerting a force on the breachface just like the mass of a bullet and the 24g of H110 in my .44 Mag is a heck of a lot lighter than the 70+ grains I used to less into my 300 Win Mag.

Not meaning to flame, but that has to be the silliest, most uninformed post I've ever read on a forum like this. JHC!!!! Makes Gump look like Einstein! :evil:

Macchina
March 24, 2013, 01:24 PM
Not meaning to flame, but that has to be the silliest, most uninformed post I've ever read on a forum like this. JHC!!!! Makes Gump look like Einstein! :evil:
And in what way is it silly? The way I stated that the lower the mass of powder, the lower the recoil, or in the way you probably didn't read any of the discussion below that goes more in-depth into the topic?

jstein650
March 24, 2013, 01:38 PM
Don't want to beat this to death, but I think it's important, and interesting to a lot of folks. I posted above (#18), but was a little vague. Here is the formula for calculating free recoil energy. The k here is a constant applied to the velocity of the powder gasses. Different constants are used because of the different dynamics of firearms. Shotguns with a long barrel use a low of 1.25 due to the relatively large bore, small powder charge, and low velocity. HP rifles use a high of 1.75, again relatively small bore, large charge and high vel. The reason it's not an 'exact' is that the powder gas column is expanding and the front of that expanding column will be different than the rear.

R.E. = [0.5 / M(gun)] * [M(bullet)*V(bullet) + M(powder)*k*V(bullet)]^2

Of course, you need to keep all the units in check; lbs, ft., sec.

Clark
March 24, 2013, 02:28 PM
The book "Elements of Ordinance" by Hayes 1938 to get the gas portion of the recoil. Chapter VII page 241 to 278 "The theory of recoil and recoil systems". On page 241 it says that:
Mvf = mv + uVc
That is the [mass of the gun] times [the free recoil gun maximum velocity] is equal to the [mass of the bullet]times [ the velocity of the bullet] plus the [mass of the powder] time the [velocity of the powder].
On page 242:
"experiments with the Sebert Velocimeter indicate..... the center of mass of the powder leaves the muzzle at 4700 fps"

My father used that book in the design of the M55, M107, and M110 for Detroit Arsenal, and the XM-70 for the marines.

jstein650
March 24, 2013, 04:57 PM
I'm not very familiar with those weapon systems, but I believe they are pretty high velocity numbers, which puts my figure in the ball park with regard to powder gas speed. I doubt though, that any part of the ejecta from a shotgun gets anywhere close to 4700 fps, hence the lower constants for those guns.

1911Tuner
March 24, 2013, 05:57 PM
The bullet exerts no force on the breechface.

In any action/reaction system, there must be a reactive mass or resistance. The bullet is the reactive mass. It provides a resistance for the force to push off of.

Recoil is more about the rate of acceleration and the force requirement to achieve that rate. The higher the rate of acceleration, the higher the force requirement. Force forward is force backward.

Let's do a hypothetical.

Two .44 Magnum revolvers, identical in every way...including weight...except for the barrel. One has a 10-inch barrel, and the other, a 2-incher.

Let's use a 240-grain bullet for both, and let's assume equal muzzle velocities of 1200 fps.

For 240/1200 in the 10-inch barrel, we'll load up with 2400

For 240/1200 in the two inch barrel...we'll use Bullseye.

Let's assume that the Bullseye charge doesn't blow up the gun, and...for the sake of simplicity...let's assume equal pressures.

Let's also assume that acceleration is linear, for the same reasons.

Let's ignore the powder/gas/ejecta exit mass and its effect on total recoil, for simplicity's sake, and just focus on a 240-grain projectile.

In the 10-inch barrel, the rate of acceleration is 120 fps per inch.

In the 2-inch barrel, the rate of acceleration is 600 fps per inch.

Which revolver will kick harder?

jstein650
March 24, 2013, 06:08 PM
OK. Same gun weight, even though barrel lengths are different, fine. You say ignore powder weight/ejecta. (I understand the 2400 & B'Eye hypothetical, but again we're ignoring those weights)
"Kick" is a subjective thing, that's why real recoil energy is used for a consistent definable quantity. Time is a factor in acceleration. Long barrel less force over more time, short barrel, more force over less time. The free recoil energy will be the same. One probably will perceive the short barrel to be 'sharper', and the long barrel to be 'more as a shove' as they say, but the energy will be the same.

bds
March 24, 2013, 06:22 PM
1911Tuner did a good job of illustrating.

Hence they call Newton's laws of motion (http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion) "laws" and not "theories" (like the "theory" of evolution which has yet to be proven ;)).

The law of physics is something that's been proven and although we could talk all we want and "theorize" but in the end, it will be the law of F=ma. :D

Yay, I remembered something from my physics class!

Lost Sheep
March 24, 2013, 06:40 PM
When I stated the bullet exerts no force on the breechface it was to emphasize the fact that


The bullet does not actually touch the breechface, thus cannot apply force to it.

I regret any misunderstanding

Lost Sheep

jstein650
March 24, 2013, 06:44 PM
bds: F=ma is true. Again, as I've stated above, recoil energy involves the factor of time, which is being ignored in 1911's example. Energy is not the same as force.

bds
March 24, 2013, 07:07 PM
bds: F=ma is true. Again, as I've stated above, recoil energy involves the factor of time, which is being ignored in 1911's example. Energy is not the same as force.
The factor of time is part of acceleration as it is defined as distance/time squared - http://en.wikipedia.org/wiki/Acceleration
In physics, acceleration is the rate at which the velocity of a body changes with time


The bullet does not actually touch the breechface, thus cannot apply force to it.
A fan may not actually "touch" the curtain but will "exert" force on the curtain to move.

In the same manner, a bullet may not "touch" the breechwall but will "exert" force on the breechwall and push it back which we feel as recoil.

jstein650
March 24, 2013, 07:35 PM
I think where we're getting our signals crossed with regard to how one looks at recoil. Yes, we're talking about a recoil FORCE, but how that force is applied over time is, what I would argue, is more relevant. Recoil ENERGY accounts for force over time. This is why in 1911's example, the recoil energy is identical; although the force on the bullet, and its opposite counterpart recoil force, will differ, if the same velocity is reached, the total recoil energy will be identical. More specifically, in the short barrel the force would be 5 times higher, but applied for a fifth of the time. I the long barrel would require a fifth of the force applied for 5 times longer.

bds
March 24, 2013, 08:28 PM
OK. Same gun weight, even though barrel lengths are different, fine ... The free recoil energy will be the same. One probably will perceive the short barrel to be 'sharper', and the long barrel to be 'more as a shove' as they say, but the energy will be the same.
I think I am beginning to understand at least one aspect of your posts.

Let's say two different bullet weights/powder charge loads produce the same amount of muzzle energy but different muzzle velocities. This would mean they will produce the same amount of "total" recoil but not the same "felt" recoil.

The lighter bullet with more powder charge will need greater acceleration or higher chamber pressure which will produce more "felt" recoil than heavier bullet with less powder charge that needs less acceleration or lower chamber pressure which will produce less "felt" recoil. Even though muzzle velocities will vary for the two bullet weights, acceleration/chamber pressures will differ inversely to produce the same amount of muzzle energy or total recoil.


I am not sure if this will help but here's one example using 230 gr 45ACP bullet vs 155 gr 40S&W bullet to produce the same 165 power factor (PF) USPSA major loads:

PF = (bullet weight in grains x velocity in fps) / 1000 or Velocity = (PF x 1000) / bullet weight

So for the heavier 230 gr bullet, only 717 fps and about 10,000 PSI is needed but for the lighter 155 gr bullet, over 1060 fps and about 28,000 PSI to make 165 PF.

jstein650
March 24, 2013, 08:59 PM
Kind of... and this sort of brings us back to the beginning. Muzzle energy normally only refers to the energy of the moving bullet after it's been propelled out of the muzzle. If recoil, however it is perceived, only depended on its velocity and mass, the charge required to get it there wouldn't matter. Since the the cartridge does require the expulsion of nearly all the powder in the case at higher than the bullet velocity, it does matter. In the same rifle, if I can get a 180 gr bullet to 3000 fps with a 45 gr charge it will have less recoil than a load that required 60 gr, by a calculable amount.
1911's example sort of took that out of the equation, and brought up the possible dynamics of how a short quick acceleration would 'feel' compared to a necessarily longer gradual acceleration. The free recoil concept is an attempt to quantify ejecta mass, gun mass, and the energy that imparts to the shooters shoulder or hand as 'recoil'. And felt recoil is a rather nebulous thing to really measure. The effect of chamber pressure (and we usually only see 'peak' pressure), combustion time, barrel friction, firearm design and its center of gravity, among other factors have all been debated and certainly contribute to 'felt' recoil though.

Clark
March 24, 2013, 09:01 PM
jstein650
I'm not very familiar with those weapon systems, but I believe they are pretty high velocity numbers, which puts my figure in the ball park with regard to powder gas speed. I doubt though, that any part of the ejecta from a shotgun gets anywhere close to 4700 fps, hence the lower constants for those guns.

I have been building some rifles and one shotgun that have gas escapement below 1 atmosphere above ambient. That is the threshold of super sonic gas.

http://i757.photobucket.com/albums/xx220/ClarkM/775pounds50CBshortintheshoulder1grRedDotthen3grRedDotdownthethroat10-4-2012.jpg
50CB short rifle

http://i757.photobucket.com/albums/xx220/ClarkM/ShotgunThreMetroChokesAddedDSCF0103-1.jpg
12 ga shotgun with long choke tubes that have choke tubes

But when Quickload calculated the muzzle pressure of my loud guns, 257 Roberts Ackley and 7mmRemMag, they are often 10kpsi at the muzzle.

And gas is going to come out of a big bore faster than a small bore, given the same pressure.

jstein650
March 24, 2013, 09:13 PM
bds: PF = (bullet weight in grains x velocity in fps) / 1000 or Velocity = (PF x 1000) / bullet weight

So for the heavier 230 gr bullet, only 717 fps and about 10,000 PSI is needed but for the lighter 155 gr bullet, over 1060 fps and about 28,000 PSI to make 165 PF.

One thing though, Power Factor is really momentum - mass x vel., while energy figures, whether muzzle or recoil use a square of the velocity x mass.

bds
March 25, 2013, 01:11 AM
Well, I am still trying to grasp the OP questions and thought PF example might shed some light on the discussion. :D
I am not sure if this will help but here's one example using ... 165 power factor (PF) USPSA major loads:

Clark
March 25, 2013, 09:16 AM
We have a book that says for gun design, the center of mass of the gas leaves at 4,700 fps.
And I have guns that no gas, not even the leading edge, goes faster than 1,125 fps.

The variables to deal with are multiple;
1) the gas pressure at the muzzle [not max chamber pressure]
2) the impedance of the bore [small hole, slow flow]
3) the total volume of the bore and chamber

jstein650
March 25, 2013, 01:58 PM
Clark:
Exactly. By the way, those are some pretty wild looking guns! I imagine that shotgun is pretty quiet too, no?
I think the 4700 fps figure could only apply to a relatively short barrel, high muzzle velocity gun, on the order of 3000+ fps. Which is why as I was trying to explain different constants are used for different types of guns when figuring free recoil energy. The factors you mention, bore diameter, chamber and bore volume, etc. do affect this then as your experiments have shown.

1911Tuner
March 26, 2013, 12:48 PM
More specifically, in the short barrel the force would be 5 times higher, but applied for a fifth of the time.

Let's revisit this for a minute, and define what recoil actually is.

What we perceive as recoil is mostly momentum after recoil has ended. Recoil itself is only present while the bullet is being accelerated...because recoil is nothing more than backward acceleration.

In my hypothetical, time is much more of a factor than it is in an actual recoil event.

With even slow burning pistol powders, pressure and force rise rapidly, often reaching peak within a half-inch of bullet travel. With really quick numbers, it can happen before the bullet clears the case mouth. Bullseye is such a powder.

Since peak pressure equates to peak force on the system...maximum recoil acceleration occurs before the bullet even gets close to the muzzle, and any acceleration that occurs after the peak only adds to the total. We perceive this as a longer
recoil event...a "push" after the initial impulse...for lack of a better description, and by the time our brains can even process the fact that something has happened...it's over, and momentum is all that's left.

Free recoil energy is one factor and momentum is another...but the sudden acceleration...the "punch"provided by acceleration... is what it's about. The boxer's hard right cross stays in contact with his opponent's jaw for a brief instant after the initial shock and pushes him backward...but it was the violent acceleration...the punch...that put his lights out. The shove after the punch is just a bonus.

In the hypothetical, the 10-inch revolver would recoil pretty hard, but it's nothing that we can't handle. The 2-inch gun would probably fracture a few bones.

jstein650
March 26, 2013, 01:19 PM
How about your example like this: Same gun in every way. Same bullet and MV. Same 10" barrel. One uses a heavy charge of 4227 or H110, the other a light charge of, say B'Eye. How do you think the recoil would compare?

ATLDave
March 26, 2013, 01:24 PM
I don't think you can count it that way, the powder is a potential energy. Upon ignition it become kinetic energy, if there was another force acting on the powder and the powder was not acting as a source of work on the bullet then you might be able to include it into your recoil calculation

We're not doing nuclear fission in the chamber. The powder is not "converted to" energy. Chemical bonds are broken and created during combustion, and that releases energy, but the mass remains the same. That mass is ejected from the gun. It contributes to recoil. Whether the difference in the weight of one powder charge versus another is detectable by a human shooter is another matter.

1911Tuner
March 27, 2013, 09:41 AM
How about your example like this: Same gun in every way. Same bullet and MV. Same 10" barrel. One uses a heavy charge of 4227 or H110, the other a light charge of, say B'Eye. How do you think the recoil would compare?

The recoil impulse would naturally be less due to ifferent force requirements to achieve the rate of acceleration.

You can lift a 50-pound weight over your head at a rate of one foot per second pretty easily. If you try to lift the same weight...accelerate it off the floor...at a rate of 100 feet per second, it becomes a much different problem due to the force requirement needed to achieve that rate.

Another look at it, just to clarify.

An action/reaction event requires an accelerating force on to interacting objects. Once that accelerating force has been removed...as in at the point of bullet and residual ejecta exiting the muzzle...the system is no longer closed, and action/reaction ends.

Another way:

The only time that recoil is occurring is while the reaction side of the system is being accelerated by a force. No force, no acceleration. No acceleration...no recoil.

If you stand close to a wall and shove yourself away from the wall, you are being accelerated by the force provided by your arms. Once you've pushed yourself far enough for your hands to lose contact with the wall, you are no longer being accelerated, and an action/reaction system no longer exists. You have momentum and energy...but there is no recoil.

Back to the gun.

When the bullet/ejecta leave the system, the force that drove the system leaves behind it. The action side of the equation no longer exists, and without action, there is no reaction. Neither the bullet nor the slide/breechbolt can accelerate to a higher velocity. All they can do is decelerate from outside forces.

Macchina
March 27, 2013, 09:25 PM
Well, I am still trying to grasp the OP questions and thought PF example might shed some light on the discussion. :D
It was not a question, it was a statement. Rephrased: we often forget to add the weight of the powder when THINKING (not calculating) recoil as we plan new loads or decide on new calibers.

The point of this thread is that when we think of recoil, we tend to think of bullet weight but forget the total load weight which includes powder. My favorite .38 Special load (158g LSWC over 4g W231) has only 2.5% of the load made up by powder. Let's compare the powder/load ratios of a 300 RUM to a 300 WSM:

300 RUM: 130g bullet over 107g of H1000 gives us 45% of the load as powder. Velocity: 3615 fps.

300 WSM: 130g bullet over 77g of Superformance gives us 37% of the load as powder. Velocity: 3645 FPS

That's an 8% decrease in total load weight with a slight increase in velocity! You may not think 8% is that big a deal, but think about the recoil difference between a 130g bullet and a 160g bullet traveling at the same speed: it's very noticeable.
P.S. I used the first offered load data for 130g bullets from Hodgdon's Website (http://data.hodgdon.com/cartridge_load.asp) for the above figures.

jstein650
March 27, 2013, 09:36 PM
I'm afraid, that after all of this discussion, and it has been enlightening, that a lot of casual readers do not equate, that, differences in the 'LOAD' (powder mass in your example) really contribute to recoil energy (however that may 'perceived'). And, I agree, these differences are realized in real life much more in the realm of higher powered rifles, than in, say a .45 Colt Cowboy load or a 12 ga. skeet load.

jstein650
March 27, 2013, 10:06 PM
"The recoil impulse would naturally be less due to different force requirements to achieve the rate of acceleration."

"With even slow burning pistol powders, pressure and force rise rapidly, often reaching peak within a half-inch of bullet travel. With really quick numbers, it can happen before the bullet clears the case mouth. "

I'm not sure in the 10" vs. 10" barrel example, which you think would have been different due to 'different force requirements to achieve the rate of acceleration'. Or which you think would have been greater.

"recoil is mostly momentum after recoil has ended".

The momentum imparted to the firearm; M=vm , which is why my (Lyman's) formula neatly takes into account the momentum of the mass of the firearm, as well the ejecta, based upon the VELOCITY of the ejecta, in the real world, and the effect the atmosphere has upon that moving mass. 'Clark' above, has experience with this, since he's been kind of basing his subsonic loadings based on atmospheric pressures.

Your information thus far, seem to concur with mine that the nature of the INTERNAL dynamics that occur during the cycle from ignition to muzzle exit; the DIFFERENCES in peak chamber pressure, the linearity of the acceleration, etc., all occur so very quickly, that the effect on felt recoil is negligible or intangible at best. The OP's question was always whether or not, and by how much, the powder mass contributed to recoil.

Captcurt
March 27, 2013, 10:29 PM
The online recoil calculators work well to calculate recoil.
And if you check you will see that powder weight is in the equation used to calculate recoil.

Macchina
March 27, 2013, 10:32 PM
Don't want to beat this to death, but I think it's important, and interesting to a lot of folks. I posted above (#18), but was a little vague. Here is the formula for calculating free recoil energy. The k here is a constant applied to the velocity of the powder gasses. Different constants are used because of the different dynamics of firearms. Shotguns with a long barrel use a low of 1.25 due to the relatively large bore, small powder charge, and low velocity. HP rifles use a high of 1.75, again relatively small bore, large charge and high vel. The reason it's not an 'exact' is that the powder gas column is expanding and the front of that expanding column will be different than the rear.

R.E. = [0.5 / M(gun)] * [M(bullet)*V(bullet) + M(powder)*k*V(bullet)]^2

Of course, you need to keep all the units in check; lbs, ft., sec.
Also don't forget to take the acceleration due to gravity out of the weight (pounds) of an object to get its mass (slugs). Acceleration due to gravity is obviously 32.2 ft/s^2.

I.E. A 180 grain bullet weights 0.0257 pounds (weight). The mass of the object is 0.0257/32.2 = 0.000798 slugs (mass) for use in the equations above. By squaring the velocities in the equations above, you are converting the slugs back into pounds.

Note: Pounds Mass (Lbm) are not to be confused with Pounds Force (Lbf)... The beauty of the English System...

jstein650
March 27, 2013, 10:39 PM
Quote:
Originally Posted by Haxby
The online recoil calculators work well to calculate recoil.
And if you check you will see that powder weight is in the equation used to calculate recoil.
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Of course, and little else! No mention of chamber pressure, time in barrel, length of barrel, powder burn rate... or anything else.

The reason I wanted to avoid any automatic on-line calculators, was to get to the heart of the matter as to what we're really looking at in the world of real world physics.

jstein650
March 27, 2013, 10:42 PM
"Also don't forget to take the acceleration due to gravity out of the weight (pounds) of an object to get its mass (slugs). Acceleration due to gravity is obviously 32.2 ft/s^2.

I.E. A 180 grain bullet weights 0.0257 pounds (weight). The mass of the object is 0.0257/32.2 = 0.000798 slugs (mass) for use in the equations above. By squaring the velocities in the equations above, you are converting the slugs back into pounds."



michaelmcgo:

That kind of pertains to relative ballistics in 1g, doesn't it? More of a trajectory thing... Now there I will use a computer. I'll pass on an enormous multi point by hand calculus, thank you.

1911Tuner
March 28, 2013, 04:45 AM
The momentum imparted to the firearm; M=vm , which is why my (Lyman's) formula neatly takes into account the momentum of the mass of the firearm, as well the ejecta, based upon the VELOCITY of the ejecta

Yes. Momentum is a factor of mass and velocity, but momentum isn't recoil by the Newtonian definition. What we perceive as recoil is mostly momentum. The actual recoil...by Newton's definition...is over so quickly that our brains don't have time to process it until after it's over.

Again:

Recoil is the reaction side of an action/reaction system and it's only in play while the system is being accelerated. Once the accelerating force has been removed from the system...action AND reaction end, and any motion of the components are due to conserved momentum.

Analogy:

The boxer's sharp left jab snaps his opponent's head back, but the actual punch only exists while his glove is in contact with his opponent's face. When the glove loses contact, the punch ends.

In an alternate universe, where the known laws of physics don't exist and momentum isn't conserved...motion would stop the instant that an accelerating force is removed.

brickeyee
March 28, 2013, 01:32 PM
You use the momentum of the ejects to compute the momentum imparted to the gun.

Momentum MUST be conserved.

You then use the momentum of the gun to determine a velocity of recoil.
(dived by the mass of the gun, watch out for units).

The velocity of recoil is used with E=(mv^2)/2 to compute a free recoil energy of the gun.

There are more complicated ways of doing all this based on impulse that take into account how fast the acceleration occurs.

They are rarely needed since FELT recoil is dominated by other things anyway (stock shape, etc.).

Get a rifle stock cut at the wrong angle and get smacked in the face with every shot and the felt recoil is worse.

Njal Thorgeirsson
March 28, 2013, 02:41 PM
The recoil impulse would naturally be less due to ifferent force requirements to achieve the rate of acceleration.

I disagree with this statement (I'm 99% sure this is false).

Here is why:

Things we can establish:
-The force felt by the bullet is at any given time the same as the force felt by the back of the case and subsequenty, the bolt face and gun.
-The Force vs. Time (Pressure vs. Time) profiles (of forces felt by the bolt gun and bullet) WILL differ between the two loads. HOWEVER, the integral of these graphs (which is momentum) will be identical, which should make sense, as both bullets will have the same momentum upon exiting the gun.
-Now, as there will be a different F vs. T profile for each gun, the recoil impulse will differ for any given increment of time during which and only during which the bullet is in the barrel. Once the bullet leaves the barrel, the total change in momentum (net impulse) will be the same for both guns. Both guns will have identical velocities into your hand as a result.

Yes, the impulse will differ during times in which the bullet is actually in the barrel, but the net change in gun velocity will be the same.

Unless you can feel slight variations in the gun's acceleration during the tiny fraction of a second that the bullet is accelerating down the barrel (which you can't), the recoil will feel identical, because a) both guns will accelerate to the same velocity in the same amount of time, b) suring that amount of time, average acceleration of both guns will be the same, and c) the increment of time during which the acceleration profiles of both guns differ is imperceptibly small.

But of course the one with the heavier powder load will be slightly heavier in recoil due to more mass being pushed out of the gun, but that has nothing to do with the above explanation.

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