Ballistics by the inch

Velocity is great, but mass penetrates.

OK, kiddies, it’s time for SCIENCE!

Ballistic science, specifically. I promise to keep the math to a minimum, because I don’t like it much, either. Jim Kasper is the one who thinks in terms of equations, not me.

If you look at any of the various pages for test results on BBTI you will see that each caliber/cartridge also has a link for a Muzzle Energy (the kinetic energy of a bullet as it leaves the muzzle of a gun) graph for that set of results. That’s because Muzzle Energy can also give an idea of the effectiveness of a given ammo, since it is a calculation of both the weight of a bullet as well as the velocity it is traveling. This calculation, specifically:

E_\text{k} =\tfrac{1}{2} mv^2

Here’s what that says in English, taken from the explanation that goes with that image on Wikipedia:

The kinetic energy is equal to 1/2 the product of the mass and the square of the speed.

In other words, you multiply the weight of the bullet times the square of the velocity, then take half of whatever number you get. And that gives you the Muzzle Energy, usually (as on our site) expressed in foot-pounds of energy.

So there are two ways you can change the result: change the amount of weight, or change the amount of velocity.

But since it is the square of the velocity (the velocity times itself), changes to the velocity have a larger impact on the final amount of Muzzle Energy. That’s the reason why how the velocity changes due to barrel length is such a big deal, and why we’ve done all the research that we’ve done over the last seven years.

But while Muzzle Energy gives you a good way to compare the power and potential effectiveness of a given cartridge as a self-defense round, there are a couple of other factors to consider. A couple of VERY important factors.

One is the shape and composition of the bullet itself. There’s a very good (surprisingly good, in fact — I heartily recommend you read the whole thing) discussion of the basic shapes and how they interact with the human body in this online teaching tool intended for medical students. The relevant excerpt:

Designing a bullet for efficient transfer of energy to a particular target is not straightforward, for targets differ. To penetrate the thick hide and tough bone of an elephant, the bullet must be pointed, of small diameter, and durable enough to resist disintegration. However, such a bullet would penetrate most human tissues like a spear, doing little more damage than a knife wound. A bullet designed to damage human tissues would need some sort of “brakes” so that all the KE was transmitted to the target.

It is easier to design features that aid deceleration of a larger, slower moving bullet in tissues than a small, high velocity bullet. Such measures include shape modifications like round (round nose), flattened (wadcutter), or cupped (hollowpoint) bullet nose. Round nose bullets provide the least braking, are usually jacketed, and are useful mostly in low velocity handguns. The wadcutter design provides the most braking from shape alone, is not jacketed, and is used in low velocity handguns (often for target practice). A semi-wadcutter design is intermediate between the round nose and wadcutter and is useful at medium velocity. Hollowpoint bullet design facilitates turning the bullet “inside out” and flattening the front, referred to as “expansion.” Expansion reliably occurs only at velocities exceeding 1200 fps, so is suited only to the highest velocity handguns.

Now, while that last bit about needing to exceed 1200 fps may have been true, or a ‘good enough’ approximation a few years ago, it isn’t entirely true today. There has been a significant improvement in bullet design in the last two decades (and these innovations continue at a rapid pace), so that there are now plenty of handgun loads available which will reliably expand as intended in the velocity range expected from the round.

The other REALLY important consideration in bullet effectiveness is penetration. This is so important, in fact, that it is the major criteria used by the FBI and others in assessing performance. From Wikipedia:

According to Dr. Martin Fackler and the International Wound Ballistics Association (IWBA), between 12.5 and 14 inches (318 and 356 mm) of penetration in calibrated tissue simulant is optimal performance for a bullet which is meant to be used defensively, against a human adversary. They also believe that penetration is one of the most important factors when choosing a bullet (and that the number one factor is shot placement). If the bullet penetrates less than their guidelines, it is inadequate, and if it penetrates more, it is still satisfactory though not optimal. The FBI’s penetration requirement is very similar at 12 to 18 inches (305 to 457 mm).

A penetration depth of 12.5 to 14 inches (318 and 356 mm) may seem excessive, but a bullet sheds velocity—and crushes a narrower hole—as it penetrates deeper, while losing velocity, so the bullet might be crushing a very small amount of tissue (simulating an “ice pick” injury) during its last two or three inches of travel, giving only between 9.5 and 12 inches of effective wide-area penetration.

As noted above, the design of the bullet can have a substantial effect on how well it penetrates. But another big factor is the weight, or mass, of the bullet relative to its cross-section — this is called ‘sectional density‘. Simply put, a bullet with a large cross-section and high mass will penetrate more than a bullet with the same cross-section but low mass moving at the same speed. It isn’t penetration, but think of how hard a baseball hits versus a whiffleball moving at the same speed. They’re basically the same size, but the mass is what makes a big difference. (See also ‘ballistic coefficient‘).

With me so far?

OK, let’s go all the way back up to the top where I discussed Muzzle Energy. See the equation? Right. Let’s use the baseball/whiffleball analogy again. Let’s say that the baseball weighs 5.0 ounces, which is 2,187.5 grains. And the whiffleball weighs 2/3 of an ounce, or 291.8 grains. A pitcher can throw either ball at say 60 mph, which is 88 fps. That means (using this calculator) that the Kinetic Energy of a baseball when it leaves the pitcher’s hand is  37 foot-pounds, and the whiffleball is just 5 foot-pounds. Got that?

But let’s say that because it is so light, the pitcher can throw the wiffleball twice as fast as he can throw a baseball. That now boosts the Kinetic Energy of the whiffleball to 20 foot-pounds.

And if you triple the velocity of the whiffleball? That gives it a Kinetic Energy of 45 foot-pounds. Yeah, more than the baseball traveling at 88 fps.

OK then.

Now let’s go look at our most recent .45 ACP tests. And in particular, the Muzzle Energy graph for those tests:

What is the top line on that graph? Yeah, Liberty Civil Defense +P 78 gr JHP.  It has almost 861 foot-pounds of energy, which is more than any other round included in those tests. By the Muzzle Energy measure, this is clearly the superior round.

But would it penetrate enough?

Maybe. Brass Fetcher doesn’t list the Liberty Civil Defense +P 78 gr JHP. But they did test a 90 gr RBCD round, which penetrated to 12.0″ and only expanded by 0.269 square inch. Compare that to the other bullets listed on his page, and you’ll see that while the depth of penetration isn’t too bad when compared to other, heavier, bullets, that round is tied with one other for the least amount of expansion.

Driving a lightweight bullet much, much faster makes the Muzzle Energy look very impressive. Just the velocity of the Liberty Civil Defense +P 78 gr JHP is impressive — 1865 fps out of a 5″ barrel is at least 50% faster than any other round on our test results page, and almost 400 fps faster than even the hottest of the .45 Super loads tested.

But how well would it actually penetrate? Without formally testing it, we can’t say for sure. But I am skeptical. I’m not going to volunteer to getting shot with one of the things (or even hit with a whiffleball traveling 180 mph), but I’m also not going to rely on it to work as it has to in the real world, where deep penetration is critical. I want a bullet with enough punch to get through a light barrier, if necessary. Like this video from Hickok45, via The Firearm Blog:

Personally, I prefer a heavier bullet. Ideally, I want one which is also going to have a fair amount of velocity behind it (which is why I have adapted my .45s to handle the .45 Super). All things being equal (sectional density, bullet configuration and composition), velocity is great, but mass is what penetrates.

Jim Downey


November 8, 2015 - Posted by | .45 ACP, Data, Discussion. | , , , , , , , , , , , , , , , , , , , , , , , , , , ,


  1. Thanks for the write up. I enjoy reading this kind of content.

    I’ve leaned toward heavier bullet weights due to the reasons you’ve outlined above. I was a little surprised, though, when I compared the results Lucky Gunner recently posted in their “Handgun Self-Defense Ammunition Ballistics Test” article (link at the end of this comment). I compared the 9mm Speer 147 grain load to the 9mm Speer 124 grain +P load and found that, in this case, the 124gr had a slightly better average penetration depth. As well, they seemed to expand more reliably/consistently.

    The test barrel is shorter and perhaps the results would be closer to what we expect (heavier = deeper penetration) if a standard length barrel were used. As well, it is just one example of this comparison. As well, either way, the data set is too small to draw conclusions.

    I’m on my phone, otherwise I’d go look at other 124 gr +P vs 147 gr example as well as look for similar relationships in the other calibers to see how often this pattern is found.

    I suppose the gold nugget to take from this is that guidelines are guidelines and testing your own loads for your own circumstances is likely to offer a clearer picture.

    Comment by Stephen | November 9, 2015 | Reply

    • Yeah, there are enough variables to make trying to document a overall conclusion really difficult. Even with all the testing we’ve done (25,000+ rounds fired) I still have to remind people that our results are only indicative, not definitive — the only way you’ll really know how your ammo in your gun will perform is to actually test it yourself.

      So it’s almost impossible to *prove* a lot of this stuff. What you have to do is look at the data, think through the implications carefully, and then come to a tentative conclusion which is always subject to change with new data.

      Comment by James Downey | November 9, 2015 | Reply

  2. […] Source: Velocity is great, but mass penetrates. […]

    Pingback by Velocity is great, but mass penetrates. | Rifleman III Journal | November 29, 2015 | Reply

  3. […] discussed this previously: Velocity is great, but mass penetrates. In that post, I used the example of a whiffle-ball versus a baseball, where they both had the same […]

    Pingback by Shooting big stuff. « Ballistics by the inch | May 19, 2021 | Reply

  4. If keeping MASS CONSTANT, lets say an arbitrary m = 2, making (m * 1/2) = 1, leaving E = (1)*(v^2), we notice that a velocity of 1 (v=1) results in E equaling 1 (E=1^2 ;E=1), we notice that doubling the velocity (from 1 to 2; v=2) results in 4x the energy (E=2^2; E=4), and tripling the velocity (from 1 to 3; v=3) results in 9x the energy (E=3^2; E=9).
    When MASS is constant: E = v^2
    v=1; E=1
    v=2; E=4
    v=3; E=9
    v=4; E=16

    If keeping VELOCITY CONSTANT, lets say an arbitrary 1, making the (v^2) = 1, leaving E = 1/2m, we notice that a mass of 2 (m=2) results in E equaling 1 (E=1/2m; E=1), we notice that doubling the mass (from 2 to 4; m=4) results in 2x the energy (E=1/2m; E=2), and tripling the mass (from 2 to 6; m=6) results in 3x the energy (E=1/2m; E=3), and quadrupling the mass (from 2 to 8; m=8) results in 4x the energy (E=1/2m: E=4).
    When VELOCITY is constant: E = 1/2 * m
    m=2; E = 1
    m=4; E = 2
    m=6; E = 3
    m=8; E = 4

    P = E * A
    Penetration (P) is equals (=) the total energy (E) * the surface area of impact (A) (all * the force pushing back by target per unit A). If the bullet doesnt deform then ALL the energy is put into penetration. This is why you want a solid. Anyways, penetration comes down to total energy (E) vs surface area (A) of which this energy is being applied. We can see that when we increase the velocity we observe greater resultant energies vs the scenarios in which we increased the mass by the same proportion as we did velocity (doubling mass = 2x the E, but doubling velocity = 4x the E). So what? Of course the faster projectile will penetrate deeper if everything else is the same (mass and surface area), and of course the heavier projectile will penetrate deeper if everything else is the same (velocity and surface area), but we notice that there is a greater increase in total energy when increasing velocity by 40% compared to increasing mass by 40% (assuming same deformation pattern and surface area of projectile). This means that velocity will have a larger influence on penetration than mass bc velocity influences energy exponentially (2 to 4, 3 to 9, ect) while mass influences energy linearly (2 to 2, 3 to 3, ect).

    Comment by SirPunchalot | November 29, 2022 | Reply

    • **Correction, it should be delta E or the change in E in relation to delta m or v or the change in mass or velocity

      All are the same:
      E = 1/2 * m * v²
      E = (1/2 * m) * v²
      E = (1/2 * v²) * m

      When MASS is constant (set m = 2 so that the term (1/2 * m) = 1, leaving ΔE = 1 * Δv²), and comparing the difference in energy under different velocities, then the change in energy is equal to the change in velocity, squared:

      When MASS is constant: ΔE = 1 * (Δv)²
      Δv=2; ΔE = 4
      Δv=3; ΔE = 9
      Δv=4; ΔE = 16

      When VELOCITY is constant (set v = √2 so that the term (1/2 * v²) = 1, leaving ΔE = 1 * Δm), and comparing the difference in energy under different masses, then the change in energy is equal to the change in mass:

      When VELOCITY is constant: ΔE = 1 * (Δm)
      Δm=2; ΔE = 2
      Δm=3; ΔE = 3
      Δm=4; ΔE = 4

      When deformation = 0: Penetration = E * A * f
      E = total energy of bullet
      A = impact area of bullet
      f = force of target pushing back

      All things being constant (A and f), greater E means greater Penetration. Velocity effects E exponentially while mass effects E linearly. This means changes in velocity effect penetration more so than equal changes in mass.

      Comment by SirPunchalot | November 29, 2022 | Reply

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