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Some thoughts on bullet performance.
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Old 11-09-2004, 02:15 PM
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Default Some thoughts on bullet performance.

Many, many times, I have observed bullet performance that was either not what the “experts” swore it was, or even what I thought it might be. In most cases, these surprises were with hollow point bullets. As I have recently begun to use slow, “fat” bullets, I am seeing more surprises in performance. In the end, I have come to the personal conclusion that there’s a lot more to “knockdown power”, or “hammer effect” than can be easily quantified. Furthermore, I am unimpressed with kinetic energy as THE SOLE measure of terminal performance.



I have read several authors that have developed “equations” or formulas to try to quantify the contribution of bullet diameter on terminal performance. Most were either poor results due to the author’s lack of math or physics training, or they were simply self-aggrandizement. The only one I have ever cared much for was Water’s formula, and truthfully, even that one is of marginal value. At least as importantly to me, is the fact that there is no way of quantitatively evaluating the effects of high speed and/or hollow point bullets.



Lest someone jump to a conclusion, I am not about to propose any such equation in this post. I have no great new idea about how to easily quantify the effects of bullet diameter on terminal performance. Nor do I have any such idea about how to easily quantify the effects of fast hollow points. But I do have some opinions on WHY those bullet characteristics are so effective, and I do have an idea about how to test these ideas, albeit neither simply nor inexpensively.



First I want to introduce a new measurement characteristic to the discussion of terminal performance and "lethality"– work: more specifically, work per unit time, AKA power. In classical mechanics, work is defined as force applied over a distance. English system units of work are foot-pounds (don’t confuse with energy – very easily done) metric system is joules. How much work is accomplished per unit time, power, is watts in the metric system. The math and terminology is very confusing in the English system of units, so I am going to stick to doing calculations in the metric system. The ratios would be exactly the same in the English system of units. Also, before you take me to task on terms and units, 1) hear me out, and 2) please only correct my arithmetic errors, or erroneous use of a term.)



Before I get to the description of how to measure the work done and power exerted by a bullet in an animal, let me talk about my ideas regarding what’s going on in the animal as the bullet transfers energy and disrupts tissue. I think most of us will agree that tissue disruption, in all its many forms, will kill the (big game) animal. If we poke a .50 caliber hole through the heart of an elephant, sooner or later, it will bleed to death. What many do not agree on, is the effect of a phenomenon generally referred to as “shock”. “Shock” is a nebulous “force” that is not connected to the direct physical damage that the physical bullet does. It is instead often referred to as hydro-static shock, implying (and often stating outright) that a shock wave created by the bullet compressing body fluids (interstitial fluid and blood), create a physically destructive wave that disrupts tissue or “shocks” (whatever that is) the animal. Others argue that there is a physiological shock to the central nervous system (hereafter CNS) created by the shock wave. The effects of the physical shock wave are easily demonstrated – those of the physiological shock wave to the CNS, are not. Personally I have faith (since I cannot see it, or otherwise empirically demonstrate it), in both phenomena contributing to the “lethality” of a bullet. So, let me put succinct terms to these three components of “lethality”. The hole I will refer to as direct tissue disruption (hereafter DTD), the tissue damage caused by the hydrostatic shock wave I will refer to as cavitation damage (hereafter CD), and the physiological shock to the CNS simply as “shock”. Let me define another term – dead. Doesn’t seem to be too ambiguous, but I want to define it in terms of hunting, not physiology. Therefore, for this discussion, I consider an animal “dead” if it is on the ground and cannot gather its feet underneath it to escape me. If I can walk up on an animal to apply a coup de grace it was “dead” when it fell to the ground.



OK, if I don’t think the bullet’s kinetic energy is the sole explanation of why “fast” bullets, especially HPs seem to “hammer” most critters to the ground, than what is it? In a word – power. The question of lethality is answered in my opinion by how FAST a bullet drops its energy in the animal (power), AND over how much AREA is this force applied (pressure).



Let’s take a big, fat 300-grain, ‘solid’ .458 caliber bullet doing 2000 feet per second (~610 m/s) at IMPACT. (Muzzle velocities are irrelevant.) Its impact energy is 2664 ft-lbs (~1964 J). The frontal area over which that bullet transfers energy to the animal is 0.165 square inches (~106 square mm).



Now let’s take a 225 grain, round-nosed, ‘solid’ .308 caliber bullet doing 2309 f/s (~704 m/s) at impact. Impact energy is the same 2663 ft-lbs (~1964 J). Frontal area over which that bullet transfers energy to the animal is 0.075 square inches (~73 square mm).



Now let’s take a 115-grain hollow point .284 caliber bullet with an impact velocity of 3230 f/s (~985 m/s). Again, the impact energy is the same 2664 ft-lbs (~1964 J). Frontal area is 0.063 square inches (~41 square mm).



Now let’s “shoot” a couple of animals behind the shoulder and between the shoulder blade and the heart, hitting neither the shoulder blade or the heart, and let me suggest what I think happens up to the point that the bullet stops or exits.



The animal will be a 220-pound (100 kg) whitetail. It is my belief that the .458 will exit the offside (hereafter ‘perforate’) of the deer. Therefore, some unquantifiable amount of its 2663 ft-lbs of energy will be “wasted” on travel past the deer. However, it will make a .458” hole creating a DTD volume of at least ~1.5 cubic inches, (the volume of the tissue disturbed by the bullet assuming it travels through 9” of deer), plus there will be some CD, and probably considerable shock. Unfortunately, we can’t really calculate any numbers other than a conservative wound volume, because the bullet perforates the deer. However, it is well recognized by those that have shot deer with this caliber bullet, that they die (remember the hunting definition) quickly. Whatever proportion of the 2664 ft-lbs of energy was dropped on the deer, was dropped ahead of a half-inch frontal area. We cannot calculate the power (work per unit time) accomplished by the bullet because it perforated the deer. However, if we assume that the bullet traveled trough the deer at impact velocity, it then spent 0.000375 seconds in the deer. If we assume it dropped of its energy in the 9” of deer, then the power exerted was 1964 j divided by 2 (the change in KE is the work) or 982 j. 982 j divided by 0.000375 seconds (the time it takes the bullet to travel the 9 inches), yields 2.619 megawatts – the power needed to cause the bullet to drop of its energy. Or, put another way, the 982 j was the work needed to convert half of the KE to other forms of energy and rend flesh.



Next is the .308 bullet. Same impact energy, but about 300 fps more speed at impact. Being a round-nosed solid, I will say this bullet also perforates the deer. It makes a wound channel with a volume of ~1.0 cubic inches. As a RN solid, the frontal area it has with which to create a shock wave is less than half of what the .458 has. And while we cannot calculate exactly what that value is, we can know that the energy transferred is proportional to the resistance, and the resistance of the .308 is probably about 45% of the resistance of the .458. Furthermore, the .308 will actually spend less time in the deer by virtue of increased impact velocity, and decreased resistance. Therefore, it is a certainty that it will convert less of its 2664 ft-lbs of energy to tissue damage, than the .458 will. Let’s assume some numbers: since the frontal area ratio is 75/165, let’s say that the dropped energy is 45% of the .458’s dropped energy, or 444 j. The bullet stayed in the deer only 0.000322 seconds, therefore, the deer only generated 1.389 megawatts of power in slowing down the bullet as it did.



Now to the .284 HP. It is unlikely that the HP will perforate the deer, and for this discussion we’ll say it doesn’t. At impact, the HP starts to open up, and is fully “mushroomed” within 6 inches of the entrance hole. It stops inside the chest cavity without penetrating the far side of the rib cage. The whole 2664 ft-lbs of kinetic energy are transferred to the deer in the form of DTD, CD and shock. The DTD will be significant, but indeterminate prior to a necropsy. Whatever physical wave was created by the bullet, the frontal diameter would likely have been on the order of 2x the original diameter, or .568”. But to my way of thinking, the most significant point is the TIME over which the energy is dropped. The power required to stop the bullet will be considerably more than the power dissipated in the above bullets. 1964 J/0.000155 seconds equals 12.687 megawatts, or almost 5 times the power needed to slow the .458 down, and 9 times the .30 caliber bullet. I want to stress that these calculations are strictly “seat of the pants” stuff based on reasonable guesses and proportions. While the exact values are most likely wrong, the proportions are probably not far off.



Let’s take those values on their face for the moment, and talk about “toughness” of critters. Personally, I DO think some critters are tougher than others. Also, I think there are two parts to “toughness”. The first is purely physical. Elephant bones are denser (tougher) than most smaller animals because they have to hold up such enormous weight. Whether Cape buffalo bones are tougher than other large ungulates is debatable, but measurable. (However I do not have access to such measurements.) The second part of “toughness” is what I will call “attitude”. Now before someone takes me to task for anthropomorphizing, (ascribing human traits to animals), let me clarify that opinion. Certain animals, especially after they reach sexual maturity, “realize” or have an understanding of their “vulnerability” to predation. Here’s where I would acknowledge the “special” toughness often ascribed to most of Africa’s Big Five – elephant, rhino, Cape buffalo, lion, and instead of leopard, I would substitute hippo. I would add to the African list, eland and giraffe. These animals, especially mature bulls, “know” (have a conscious awareness) that they are basically “beyond” danger from predators. (Size does matter.) As such it doesn’t enter their ‘psyche’ that they are ‘mortal’. Hence when they get shot, they’ve got an “attitude” that may very well sustain them long enough to stomp mud holes in you before the effects of a non-shocking bullet take effect. Hence the need for brain shots in elephants and bone-breaking shots in the others.



Smaller herding herbivores, otherwise known as “food for everybody” never develop this “attitude”, and when shot are just as likely as not to lay down and die or run off, instead of “defending” themselves. While I have not hunted Africa, I have seen this very “attitude” many times in brown bears (”immortal”) vs. caribou (“I’m food”).



Now, “attitude” and actual physical “toughness” aside, in my opinion, there is another factor that plays a very important role in an animal’s ability to withstand “punishment” as gun writers are wont to say. That is sheer mass. Consider the above numbers. Which animal would be more capable of withstanding a 12-megawatt power surge, a 100 kg (220 lb) whitetail, or a 1000 kg (2200 lb buffalo)? We can actually calculate that to a certain degree in the form of watts per kilogram. In the whitetail, the figure is 12.687MW/220kg, which equals 57.668 kW/kg. In the buffalo, the figures are 12.687MW/2200kg, which equals 5.767 kW/kg. What that means is, each kilogram of the buffalo only has to dissipate about 6 kW of power, whereas every kilogram of the whitetail’s flesh has to dissipate 10 times as much, or 60 kW.



The larger the animal, the more flesh it has available to dissipate power. If it has more flesh to dissipate power, the less any one kilogram of “flesh” will have to take. That means that a specific organ, or the CNS will dissipate less power per unit mass in a large animal than it will in a small animal. The less power dissipated per unit mass, the less the damage incurred.



So… not only do very fast bullets cause massive DTD, and CD, they also cause the critter’s body to dissipate a greater quantity of power even if the two bullets are delivering exactly the same kinetic energy. In the case of “big” critters, where their mass is sufficiently large so as to be able to "manage" the power dissipation, only big, fat, deep –penetrating bullets can provide the necessary DTD (especially to skeletal structures) to disable the critter until it “realizes” it is “dead”.


Here’s how these “theories” can be quantitatively tested. A 12”x12”x24” (or any appropriately sized) block of ballistic gelatin is set up at the appropriate distance to permit the desired impact velocity. In the gelatin at specific and uniform distances in 3 dimensions, 3-axis strain gauges are imbedded. Upon bullet impact and transit through the block, the strain gauges will measure not only the pressure, but more importantly, the pressure per unit time. The pressure-time curves, will illustrate clearly the effect of each type of bullet tested. Of course, this would be very labor and instrument intensive. Which translates to expensive. Why would premium bullet manufacturers want to conduct such a test if the results might demonstrate that their “premium” bullets aren’t so “premium” after all?


Paul

Last edited by gitano; 03-24-2018 at 01:23 PM..
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cns, dead right there, drt, energy, hydrostatic, kinetic energy, lethality, mass, physics, power, shock, work


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