Should it really be all about Momentum?

[lbmii]

When we talk of ballistics we talk of bullet weight, bullet velocity and bullet energy. We focus a lot on the energy when comparing different rounds. Energy is mass times velocity times velocity. Energy favors lighter bullets going fast over heavy bullets going slow.

But what of momentum? Is not Momentum the measurement of an object's ability to resist a change in direction or velocity? Is this not what it is all about when talking about ballistics?

Momentum is mass times velocity. Both the mass and the velocity have equal roles when calculating momentum. Could it be that the ballistic tables are showing energy when they should be showing momentum? Have the ballistic tables in only showing energy skewed our perceptions to place greater emphasis on velocity over bullet mass? Should ballistic tables be rewritten to show momentum instead of energy?

Does it make any sense that when one wants to hunt big game one chooses a heavy bullet over a light bullet for a given type of round even though in doing so one is often choosing to go down in energy? In reality with a given round, in choosing to go up in bullet mass one is going up in momentum or staying equal in momentum even though one is going down in energy.

Momentum in pounds feet equals bullet weight in grains times velocity in FPS divided by 225120.

357 Mag 125 grain @ 1450 fps = 0.81 momentum but 584 in energy.
357 Mag 158 grain @ 1240 fps = 0.87 momentum but 539 in energy.

So the 125-grain bullet has about a 7 % lower momentum in relative percent difference to the 158-grain bullet. However the 125-grain bullet is about 8 percent greater in energy over the 158-grain bullet.

So in the case of the 357, the energy oriented ballistic tables show favor to the lighter 125-grain bullet even though experienced hunters will choose the heaver 357 Mag bullets for deer.

It seems to me that momentum should be included in ballistic tables.

Look at the most common and highly reputable handgun calibers for personal defense and you will find the momentum figures remarkably close. Suggesting some degree of equivalency of these rounds that tend to not show that well in the energy tables. The 230 grain 45 auto appears much weaker to the 40 S&W and the 357 Mag when you focus on energy but look how the 45 compares to the others when you focus on momentum.

45 auto….230 @ 830 = 0.85
45 auto…185 @ 1000 = 0.82
40 S&W 155 @ 1205 = 0.83
40 S&W 165 @ 1190 = 0.87
40 S&W 180 @ 1000 = 0.80
357 Mag 125 @ 1450 fps = 0.81
357 Mag 158 @ 1240 fps = 0.87

I personally calculate the momentum of a round when I do comparisons.

 

[lbmii]

Just to add some more. The 44 Special appears anemic when its’ energy is compared to more modern rounds. But its’ momentum is right in there with the 45 auto, 40 S&W and the 357 MAG.

44 Spc 246 @ 755 = 0.83 Compare this with the numbers in my first post.

If your Grandpa used to proclaim the power of the old 45 Colt, he was right:

45 Colt 250 @ 860 = 0.96

For the never-ending debate between the 38 special crowd and the 9 mm crowd; let’s look at the momentum:

38 Spc +P 157 @ 890= 0.62
38 Spc +P 125 @ 975= 0.54
9 mm…...147 @ 990 = 0.65
9 mm…...124 @ 1110= 0.61
9 mm…...115 @ 1135= 0.58

 

[Yooper]

I've accepted momentum as a better determinant of ballistic potential than energy and I think you're on the right track. Now look at what happens to the velocity of the projectile over time/distance to find the actual momentum at the time of bullet strike.

 

[BigG]

It looks like you are calculating it the way old Elmer Keith did. He always went with a heavy bullet for the caliber which makes sense to me.

The ammo companies have marketing departments like everything else these days. By using the energy figures they get higher i.e., more impressive figures.

 

[Hatton7]

First I'm a 9mm fan but that was a great post. What about rifle loads like 5.56,7.62 and 12 guage shotgun?

 

[lbmii]

Oh yes the 12 Gauge!

12 pellet OO buck = a total of 645.6 grains @ 1275 fps = 3.66

9 pellet OO buck = a total of 484.2 grains @ 1330 fps = 2.86

12 pellet O buck = a total of 579.6 grains @ 1240 fps = 3.18

16 pellet 1 buck = a total of 640 grains @ 1236 fps = 3.51

27 pellet 4 buck = a total of 556.2 grains @ 1330 fps = 3.29

8 pellet OOO buck = a total of 544 grains @ 1330 fps = 3.21

I have a spread sheet that I created that does a great many calculations and rankings.

 

[Jim K]

Not only exterior ballistics. Momentum is what recoil and recoil operation are all about. A Model 1911 functions because the momentum of the bullet moving forward causes an equal and opposite momentum in the barrel/slide unit. Good old Law of Conservation of Momentum, the Borax freshened, lemon scented version of old Newton's "For every action..."

Jim

 

[McCall911]

I almost hate to bring this up, but here's the big monkey wrench in the momentum theory, and the energy theory to some degree:

5 ounce baseball thrown at 100 mph:
Momentum--1.44 ft-lbs/sec
Energy--106 ft-lbs

As we know, a hard-thrown baseball is painful or could even break small bones, but this projectile couldn't normally be called lethal on a human being.
Sorry...

 

[JohnBT]

Baseballs are like bullets, they're only lethal when they hit the right spot.

Doesn't anybody else pick ammo according to how it shoots, and how well they can shoot it, in their gun? If you can't hit what you're aiming at the numbers are just about meaningless.

John

 

[0luke1]

Regarding the baseball analogy - wouldn't you have to factor in reasonable comparisons - such as size and density of projectile?

 

[Deavis]

You are simply playing with equations to get an answer you favor. You are presenting momentum because you seem to favor heavy bullets and momentum makes them look good on paper. This is just like comparing torque and horsepower! Here is an example..

If you want to go fast you need horsepower. You can get it by
1) increasing torque (bullet weight)
2) increasing RPM (velocity).

This is the Energy analogy
If you go the RPM route you need an engine capable of revving very high. If you rev an engine very high, the torque must be very low because you cannot physically get high RPMs with long stroke engines. Just like you can't get a .45 moving at the speed of a .22LR out of a canonical pistol due to physical constraints. Now, this method gives really huge horsepower, like in F1 racing, but while you may go really fast, your acceleration sucks. Just like a smaller projectile trades total expansion size for velocity and thus very high energy.

Here is the momentum analogy
If you go the torque route you naturally lose RPM because you have to lengthen the stroke of your engine. It becomes impossible to turn that crank very fast due to the stresses that build incredibly quickly on the rods. You don't make as much HP as the RPM route (Momentum looks worse than energy) but since you torque is insanely high, you end up with killer acceleration! The huge bullet really knocks you over, but it doesn't always penetrate or expand as much (normalized to starting size) as the smaller bullet with high velocity.

In each case, the end result is a high HP figure, but what you choose depends on what you want. Are you an F1 driver going top speed all the time (RPM) or are you drag racing and need that acceleration out of the hole (TORQUE)? Are you hunting squirrels or elephant? In a race, if we have equal horsepower and I make more torque, I will win as long as we don't go over my gearing limited speed (Shooting conditions).

Remember, I can make a round, say a 22, that looks incredible on paper, but sucks in reality if I'm hunting elephant. Likewise, I can take a medicore round and turn it into an amazing round by simply modifying the dynamics of the system. I can add barrel length to get the heavy projectile moving faster. I can use a bottleneck cartridge to get velocity up. I can make an incredibly strong gun and pack it with a keg 'o' powder to get that .45ACP moving supersonic.

Don't let the numbers obscure your thought process. The only thing that is important is the end result. For the ammunition that I carry, tests on ammolab show that a 165gr bullet out performs a 180gr bullet (in ballistic gelatin with a few layers of denim over the front). Since that most resembles the situation I might face (unarmored threat, no cover, close range) that is what I choose to carry. If my situation were different, I would choose the round best suited to it.

They are only equations! You can manipulate them in all sorts of ways to get the answer you want. You can invent relationships based on equalities that make your idea look awesome, but only one answer matters. Does it work for its intended job?

 

[Molon Labe]

As stated by Jim Keenan, momentum is most useful when looking at the recoil of the gun. The momentum you receive when firing the gun is the same momentum experienced by the target.

Contrary to popular opinion, the kinetic energy of the bullet is not equal to the amount of energy received by the gun & shooter during recoil. The kinetic energy of the bullet is always much, much greater than the amount of energy received by the gun & shooter. I once did a calculation using the FN/FAL battle rifle, and concluded the 30-cal bullet has 400 times the energy as that received by the gun & shooter.

There’s also no such thing as a bullet’s “knock down†power. A bullet cannot knockdown a person strictly via energy transfer. Not even a .50 BMG.

 

[mfree]

You could probably fix the baseball problem by mixing in a hefty dose of point pressure... 100 ft-lbs spread over the surface area of a baseball isn't highly lethal, where 100 ft-lbs spread over the area of the nose of a .22 bullet would probably penetrate, leading to a much higher lethality level.

Call it "contact PSI" or along those lines. Could even help you predict if a round would make it through a vest, since kevlar's yield strength is a known figure.

 

[Hypnogator]

357 Mag 125 grain @ 1450 fps = 0.81 momentum but 584 in energy.
357 Mag 158 grain @ 1240 fps = 0.87 momentum but 539 in energy.

And to further throw a monkey wrench into the equation, the 125-gr .357 has the absolute best percentage of one-shot stops in actual gunfights, if one is to believe that particular set of statistics.

I, too, tend to favor momentum over energy as a measure of effectiveness, not the measure. FWIW, penetration beyond the target = wasted momentum. Thus the 125-gr .357 may have greater real-world stopping ability than the more powerful (from a momentum standpoint) 158-gr load, because the 125-gr dumps all of its energy (hence, momentum) into the target, while the 158-gr bullet penetrates completely and wastes, say, 1/4 of it's energy/momentum on the landscape behind the target.

Ain't armchair ballistics fun?

 

[Deavis]

As stated by Jim Keenan, momentum is most useful when looking at the recoil of the gun.

I hate to beat a dead horse but the bullet's momentum is not the only factor dealing with the recoil of the gun in a strict sense. The momentum imparted to the bullet just happens to be a pleasant side effect (depending on which side of the gun you are on) of the system, it does not independently drive the system. You could get a better measure of the potential recoil of a load (in a recoil operated pistol) by blocking the chamber and not having a bullet at all.

The felt recoil is dependent on many factors, not just the momentum of the bullet. You do not need a moving bullet to get recoil in a recoil operated gun since there is more than one degree of freedom in the system. You can get the same effect by blocking the barrel and making a very small hole in the end of it with a bolt action rifle. It won't be the normal "recoil" force as you think of it but it will be recoil.

Looking at some bullet momentum numbers and trying to judge recoil is very short-sighted.

 

[Molon Labe]

The felt recoil is dependent on many factors, not just the momentum of the bullet. You do not need a moving bullet to get recoil in a recoil operated gun since there is more than one degree of freedom in the system. You can get the same effect by blocking the barrel and making a very small hole in the end of it with a bolt action rifle. It won't be the normal "recoil" force as you think of it but it will be recoil.

I’m not a physicist, so correct me if I’m wrong, but it is my understanding that the primary reason you feel recoil is because you’re throwing an object that has mass. (Conservation of momentum, theory of elastic collisions, etc.) In the case of shooting a gun, you feel recoil because the bullet has velocity and mass. The momentum of the bullet (mv) must equal the momentum imparted upon the gun+shooter+Earth.

 

[lbmii]

I have little disagreement with the majority of you. My point is momentum is never even brought up. It is always energy and I believe that because it is always energy that is given in published data, our perspectives have been heavily skewed. There might even have been some degree of manipulation on the part of the manufactures by focusing us on energy over momentum due to the fact that they benefited from us replacing our old calibers with new calibers.

Concerning self-defense handgun rounds. What is striking is that the major common handgun rounds that have long established good street reputations are all remarkably close in momentum. In fact the momentum spread of the different common loads of these major common calibers all overlap the other major common rounds.

The momentum spread for the different common bullet weights for the 45 auto, the 40 S&W, the 44 special, the 38-40, and the 357 mag all are remarkably very close and all overlap each other. However the above rounds all differ significantly from each other in energy. My point is that both the street and the momentum figures place these round as being closely equivalent given equal bullet development and optimization for each particular round. I should also point out that remarkably all of the above-mentioned rounds have momentums that fit within the momentum spread of the different common bullet weights of the 5.56/223 rifle. I cannot help but conclude that the engineers of the more modern above mentioned rounds were matching up the momentum of the newer rounds with the momentum of the established rounds before it. I suspect that these rounds were designed purposely with a particular momentum range in mind.

So there you go, these round are way too close in momentum for it to be not designed in, but the manufactures never ever mention momentum.

The modern +P 38 special and the 9 mm and are very close in momentum. Even the old 38-200 is close in with them.

For all you 10 mm folks; you like to carry a little bit of a bigger stick; the 10mm compares in momentum to the 45 Colt that was carried by those of the old days that liked to carry a little bit of a bigger stick as well.

I think it comes down to picking a momentum level that you can handle and then deciding if you want that momentum level in the form of mostly velocity or mostly bullet weight, as well as wide diameter or smaller diameter. You can also calculate the PSI of a bullet by using the momentum and diameter of the bullet, for a given momentum the smaller the diameter on the bullet the higher the PSI.

As for the momentum of a baseball, imagine that baseball with a long blunt spike stuck to the front of it.

 

[RyanM]

How about kinetic pulse? Energy * momentum.

.45 LC 250 gr @ 850 fps = 378
.45 ACP 230 gr @ 850 fps = 320
.45 ACP 185 gr @ 1050 fps = 391
.357 mag 158 gr @ 1250 fps = 481
.357 mag 125 gr @ 1450 fps = 470
.40 S&W 165 gr @ 1150 fps = 408
.40 S&W 180 gr @ 1000 fps = 319
9mm para 124 gr @ 1150 fps = 231
9mm para 147 gr @ 1000 fps = 213
3.5" Baseball 2187.5 gr @ 68 fps = 15

Of course, any number, energy, momentum, or KP, is only going to tell you the potential damage. It's bullet design that turns that potential damage into real damage.

I think the best "formula" for bullet effectiveness would go something like Kinetic Pulse * Caliber * Sectional Density, since lots of sectional density can translate to either deep penetration, big expansion, or a combination of the two, depending on bullet design.

.30-06 180 gr @ 2700 fps = 525
12 ga slug 1 oz @ 1400 fps = 443
.454 casull 300 gr @ 1600 fps = 326
.44 mag 240 gr @ 1200 fps = 78
.44 mag 180 gr @ 1600 fps = 78

.45 LC 250 gr @ 850 fps = 30
.45 ACP 230 gr @ 850 fps = 23
.45 ACP 185 gr @ 1050 fps = 23
10mm 180 gr @ 1150 fps = 31
.357 mag 158 gr @ 1250 fps = 30
.357 mag 125 gr @ 1450 fps = 23
.40 S&W 180 gr @ 1000 fps = 21
.40 S&W 165 gr @ 1150 fps = 24
.357 Sig 125 gr @ 1350 fps = 19
9mm para 147 gr @ 1000 fps = 13
9mm para 124 gr @ 1150 fps = 12
.38 spl +P 158 gr @ 900 fps= 11
.38 spl +P 125 gr @ 1000 fps= 8

.380 ACP 95 gr @ 850 fps = 2.1
.32 ACP 71 gr @ 800 fps = 0.8
.25 ACP 50 gr @ 750 fps = 0.3
3.5" Baseball 2187.5 gr @ 68 fps = 1.3

Not very surprising results, to me.

Edited to add baseball.

 

[Deavis]

I’m not a physicist, so correct me if I’m wrong, but it is my understanding that the primary reason you feel recoil is because you’re throwing an object that has mass. (Conservation of momentum, theory of elastic collisions, etc.) In the case of shooting a gun, you feel recoil because the bullet has velocity and mass. The momentum of the bullet (mv) must equal the momentum imparted upon the gun+shooter+Earth.

The felt recoil, or kick, is different from the actual physics definition of recoil in a gun. The kick you feel depends on many things, not just the bullet weight. If you had a 1000 lb gun and fired a .22 bullet out of it, there would be no kick even though a recoil does exist. Let's take the other extreme, if the same exact gun wieghed less than the .22 bullet you are trying to push out the kick will be tremendous even the the recoil would be the same as with the 1000lb gun.

Your last sentence is almost correct. The energy of the whole system must remain constant. That means that the energy you get from igniting the primer and powder must go somewhere. The momentum of the bullet will be a function of many variables since its velocity is based on more than one factor (length of the barrel, friction from rifling, etc...) Not to mention, you can impart momentum on the gun after the bullet has left the barrel if the powder is still burning (just like a rocket). Will it add to the kick? No, it is too small in normal circumstances, but it does exist if you break the system down carefully.

 

[Molon Labe]

The felt recoil, or kick, is different from the actual physics definition of recoil in a gun. The kick you feel depends on many things, not just the bullet weight. If you had a 1000 lb gun and fired a .22 bullet out of it, there would be no kick even though a recoil does exist. Let's take the other extreme, if the same exact gun wieghed less than the .22 bullet you are trying to push out the kick will be tremendous even the the recoil would be the same as with the 1000lb gun.

I concur, as can be demonstrated when I said “momentum is conserved.†The mv of the bullet must equal the mv of the gun:

m_gun * v_gun = m_bullet * v_bullet

Now this begs the question: What is recoil? Is it proportional to v_gun? Is it proportional to m_gun * v_gun? Is it proportional to the time integral of the force? I don’t know. Perhaps someone a lot smarter than me could chime in here.

The energy of the whole system must remain constant.

True, but trying to keep track of energy is a pain in the butt. This is because energy can take many different forms (e.g. kinetic, heat, chemical, potential, electrostatic). In other words, energy likes to "hide." Suffice to say, it would be difficult to come up with a model based on conservation of energy. Unlike energy, momentum can't "hide," and thus a model based on momentum would be simpler and more accurate.

The momentum of the bullet will be a function of many variables since its velocity is based on more than one factor (length of the barrel, friction from rifling, etc...)

It is true there are many factors that influence bullet velocity, and thus influence momentum. But I think you're overcomplicating the situation. If you want to know the momentum of the bullet, simply measure the velocity using a chronometer, and multiply the reading by the mass of the bullet.

 

[Deavis]

The mv of the bullet must equal the mv of the gun

NO! You are assuming that 100% of the energy transfers to the bullet and the gun from the chemical reaction driving the system. That is simply not true. As I stated before, if you used an slow burning powder or a really short barrel you could still be adding momentum to the gun after the bullet has left the barrel if the powder continues to burn.

True, but trying to keep track of energy is a pain in the butt. This is because energy can take many different forms (e.g. kinetic, heat, chemical, potential, electrostatic). In other words, energy likes to "hide." Suffice to say, it would be difficult to come up with a model based on conservation of energy. Unlike energy, momentum can't "hide," and thus a model based on momentum would be simpler and more accurate.

Any model you make neglecting a portion of the energy will be incorrect, thus inaccurate. There is a reason that physicists say, "assume x..." and that is why models have a fudge factor in them. While you may consider it a pain, your gun is not that complex of a system (compared to say surface physics problems) and there are only a few ways for energy to get out of it.

But I think you're overcomplicating the situation. If you want to know the momentum of the bullet, simply measure the velocity using a chronometer, and multiply the reading by the mass of the bullet.

I'm not complicating anything, I'm simply asking you to look at the totality of the system and realize what you are saying. This is a simple system, but even simple systems are complex when you want to look at the real physics of it.

Yes, for all intents and purposes you can do what you say and get great results. Your shoulder isn't sensitive enough to detect the differences we are discussing. Just remember, the reading of velocity that you get will only infer the momentum of the gun since you still don't know what the momentum of the bullet was when it left the gun, do you? Sure, you can work backwards but now let's toss in the chrony error and rounding the uC does. What is the delay in that analog circuitry due to mismatched components? Don't forget about the exterior ballistics the bullet is now subject to!

Have you read Rinker's book on ballistics? It is a good read and goes into many of the details for you. Plus it does provide some simplified formulas for doing exactly what you propose. Complete with fudge factors!

 

[lbmii]

RyanM,

I have myself developed and I have read where others have developed various formulas that place different factors together to come up with numbers to compare two different rounds.

In your first one you are giving equal weight to momentum and energy, in your second one you are evenly dividing the equation between momentum, energy and bullet sectional density.

After trying various equations similar to what you have done I have concluded that really the first big step is to select what level of momentum and then go from there.

I really think that this is what the manufactures have been doing the whole time. Is it just by dumb luck that the original loading for the 30.06 has a momentum very similar to the 45-70? I think it was thought out. The military was disappointed with the 30-40 and wanted to go back to the stopping power (whatever that means) of the 45-70 so the engineers created an updated modern flat shooting round with the same momentum as the 45-70.

My question is why is momentum never discussed?

 

[Molon Labe]

NO! You are assuming that 100% of the energy transfers to the bullet and the gun from the chemical reaction driving the system. That is simply not true. As I stated before, if you used an slow burning powder or a really short barrel you could still be adding momentum to the gun after the bullet has left the barrel if the powder continues to burn.

If you’re saying that gas and un-burnt powder will continue to shoot out the barrel after the bullet has left the barrel, and that the gas & powder contribute to recoil, then you are correct. But is this a relatively small contributor to total recoil? I would assume the total momentum of the ejected gas & powder would be a couple orders of magnitude smaller than the momentum of the bullet, but I could be wrong. If we assume momentum of gas + momentum of powder << momentum of bullet, then my previous posts are 100% correct, i.e. m_gun * v_gun = m_bullet * v_bullet.

Yes, for all intents and purposes you can do what you say and get great results. Your shoulder isn't sensitive enough to detect the differences we are discussing. Just remember, the reading of velocity that you get will only infer the momentum of the gun since you still don't know what the momentum of the bullet was when it left the gun, do you? Sure, you can work backwards but now let's toss in the chrony error and rounding the uC does. What is the delay in that analog circuitry due to mismatched components? Don't forget about the exterior ballistics the bullet is now subject to!

Surely there would be measurement error. Heck, I could probably think of 20 sources of measurement error off the top of my head. (I’m an EE with a background in metrology.) But my point was not to perform an uncertainly analysis on this subject; my point was simply to say that momentum is conserved, and that the momentum of the bullet when it leaves the barrel is the same (of nearly the same!) momentum imparted onto the gun. And if someone were to try and determine if a gun had a lot of recoil, it would make more sense to look at the bullet’s momentum vs. its kinetic energy.

 

[RyanM]

My question is why is momentum never discussed?

I think it might be because momentum has more to do with terminal ballistics, while energy is more internal ballistics. It seems like V grains of W powder in X case through Y gun impart Z ft-lbs of KE on almost any bullet (provided the gun doesn't blow up).

So, for instance, if you used a load of whatever that's going to propell a 180 grain bullet with 400 ft-lbs of energy, then use the exact same load, gun, etc. on a 150 grain bullet, you'll still get roughly 400 ft-lbs, with lower pressure. You could also use the same load on a 200 grain bullet, and still have around 400 ft-lbs, assuming it's within pressure limits.

I guess energy ratings were originally meant to tell you about how much performance was being squeezed out of the cartridge, in terms of space being used up by powder (more energy = more powder), and not what the comparative effect would be downrange.

 

[lbmii]

I have seen some huge recoil equations! If you really want to get accurate about recoil you need to factor in the weight of the powder because it to is mass that leaves the barrel. But understand; as one of my physics professors often said: “You only need 3 significant digits of accuracy to land on the moon.†A quick bullet momentum calculation is a very good way to determine relative recoil between two rounds.

But really what I have been concerned with here is terminal ballistics.