Range Report -Equation for Transfered Energy...?-

[NYSharpShooter]

Recently I was having an argument as to the appropriate uses of cartridges considered "too much gun" with proper shots and bullets (i.e. .375 H&H on Deer) and wondered if there might be an actual equation that would help me prove my point. This may get a little messy, but follow me here...

What I am trying to determine is if actual transferred energy can be measured as a correlation between a bullet's weight, velocity prior to entering a target, and resting velocity within, or resulting velocity after exiting a target. I would also think that the diameter of the projectile at full expansion would come into play.

To use my previous example to convey what I'm trying to determine here, I would think that in technical terms some accepted whitetail cartridge like a .30-06 with a bullet that expands reliably and retains most of its weight would really be more gun than a .375 H&H with a solid that zips through. Furthermore, the latter remains at or very near it's nominal diameter whereas the former's expanding .308" bullet, say one of Barnes' offerings, will end up much wider. This all results in more transfered energy and damage than a solid punching cleanly through, no matter it's ballistic data on paper.

So, weary of hearing arguments about what's too much gun, and "bear rifles" that will vaporize medium game, is there a way to actually measure the transferred energy from a projectile, or is it really down to field results and word of mouth?

Thank you fellas, shoot true.
Alex

 

[THUNDERBOLT68]

Re: -Equation for Transfered Energy...?-

yes shoot them into ballistic gel look at and measure wound cavity but my thoughts are dead is dead. i love the 300wsm i use it to hunt all ove world. yes it is over kill for texas hill country whitetail at 75 yards under a feeder. but some day i might want to shoot the deer under my buddies feeder at 700 yards away therefore use one rifle know how it shoots and shoot it well

 

[The Mechanic]

Re: -Equation for Transfered Energy...?-

Try the Taylor knock down calculator. Some like it some don't.
http://www.flyinglead.info/Ballistics/ballistics.html#taylor

How it is derived

The Taylor K O Formula is an empirical measurement and is calculated as follows...

Bullet Weight (in grains) multiplied by Caliber (in inches) multiplied by Velocity (in fps) divided by 7000 = TKO rating.

This figure is basically the momentum of the projectile modified by a factor proportionate to the caliber.

 

[Scuzy]

Re: -Equation for Transfered Energy...?-

You could use the kinetic energy formula k=(1/2)mv2
If the round dose not exit then all the energy was transferred.
If it did exit then you would have to use the formula a second time on the round after it exits and find the difference. As for the expansion, it has an indirect effect, but it dose take some energy to deform the round. But this wont tell you how much is overkill.

 

[noname]

Re: -Equation for Transfered Energy...?-

one difficulty is chronographing the bullet as it comes out the other side, the amount of energy transfer is the 3rd derivative of the basic formula and is called the jerk. while the momentum can be completely transferred the results of the kinetic energy dump can be many faceted, as when it strikes a steel target the ke splays the bullet, but in live tissue perhaps only expansion. The system is non-deterministic and sensitive to initial conditions as all 3rd order expressions.

 

[noname]

Re: -Equation for Transfered Energy...?-

heres a link for what was going on last week, it seems this thing ain't gonna die yet.
http://www.snipershide.com/forum/ubbthreads.php?ubb=showflat&Number=1428400#Post1428400

 

[BryanLitz]

Re: -Equation for Transfered Energy...?-

I agree with Scuzy's approach for calculating transferred energy. It's the 'right' way to calculate it, but how do you recover the bullet when it comes out, and how do you know how fast it exited?

Hmmm, well...

Putting that question aside for a minute;

IF you could somehow know the mass and velocity of the bullet that penetrates, you would calculate it's KE as such:

KE = BW*V^2/450800 (it's Scuzy's 1/2mv2 formula simplified for units)

So if a 180 grain bullet goes in at 2000 fps, it's going in with: 180*2000^2/450800 = 1597 ft-lb.

If 120 grains of bullet exits at 700 fps, it only has: 120*700^2/450800 = 130 ft-lb

So that means the bullet lost (transferred) 1597-130 = 1467 ft-lb in the animal.

So that's how the math works, but the important question is, how much energy do you <span style="font-style: italic">want</span> to transfer?

I won't pretend to dictate the answer here, just illustrate some opposing points of view.

One point of view is the penetrating/weight retaining approach which states that you want the bullet to zip thru the animal with minimal expansion/fragmentation so that it penetrates as deeply as possible and exits, creating a blood trail. This approach has the appeal of minimizing meat damage. Maximum energy transfer is not guaranteed in this case because the bullet is likely to exit with a good deal of it's energy.

The opposing point of view favors an expanding/fragmenting bullet that basically 'blows-up', and transfers all the energy inside the animals vitals. Exit wounds and blood trails are not critical if the animal falls in it's tracks. The bullet may damage more meat by physical trama, but some say that an animal that runs after being shot fill the meat with adrenaline, affecting the taste.

Proponents of both views, and those in the middle have good points, it's difficult to say one is more 'right' than the other. The conclusion I've come to regarding lethality is that shot placement and the animals biology are a much bigger consideration than the math/ballistics involved.

You were asking specifically about the size/energy of the gun used on certain game. There are many theories and formulas out there that attempt to 'match' a rifle/bullet to a particular sized game. They're all a little different, and you'll find passionate advocates and detractors of each. Mantunas Optimal Game Weight formula appeals to me because it identifies the size of the game that can be taken based on the kinetic energy AND the momentum of the bullet. Some may advocate other approaches like the Taylor KO formula, and I wouldn't argue. The only absolute truth about lethality that I will maintain is that shot placement trumps all. How much KE do you think a typical arrow has? The key to killing with a bow and arrow is also shot placement.

-Bryan

 

[BallCoeff_606]

Re: -Equation for Transfered Energy...?-

Simply measuring energy transferred into a target is pretty simple, but you have to look at what that energy is actually doing, to see what minimal amount you'd need.

If you're impacting and destroying the brainstem, then you need to punch through the hide, either find a soft channel to the brain base or punch through skull, then destroy spinal cord material. The biggest factor here is how much bone you have to go through; the rest could be done with .22LR at shorter ranges.

It you're after a heart and lung shot on a bigger animal, then you have to punch through the hide, maybe go through a major legbone and rib, possibly frontal cartilage, then shoulder muscle and tendon, then chest membrane, then lots of lung material, then heart muscle. Each one of those components will consume a measurable amount of energy, after the ballistics are taken care of, assuming the bullet holds together long enough to get there.

So, it then becomes a statistical 'worst case' of what you have to get through and destroy, that determines minimum bullet mass, terminal ballistic velocity, and bullet construction. Once you have that, then you have to figure out how to package that mass and minimum terminal velocity in order to best get it downrange to the target. I can only assume that rule-of-thumb formulas like Taylor's etc are a generalization on those statistical worst case requirements made from observation over a long time.

So, transferred overall energy and/or momentum are probably just another kind of generalization for the statistical worst case of what you have to destroy. It's too rough and too simple to be an "accurate equation" that will indicate how much KE is necessary to kill a target.

I generally agree with Bryan that you have to take into consideration your target's anatomy and physiology much more than simple energy or momentum transfer formulas could.

The way I'd approach it is to generalize on the minimum KE necessary at point of impact for a particular shot placement on a particular kind of animal and particular weight of animal. Then determine the appropriate bullet construction from frangible to solid to deliver that KE into the target. Then package all that into the most appropriate bullet type/caliber and velocity to get that KE downrange to the target. Then find the best firearm to launch that particular bullet at said velocity.

It would end up being a set of relations best laid out in tables of observations, like steam tables, rather than trying to calculate everything from first principles.

 

[MitchAlsup]

Re: -Equation for Transfered Energy...?-

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: shooter2</div><div class="ubbcode-body">the amount of energy transfer is the 3rd derivative of the basic formula and is called the jerk. </div></div>

Third Integral not third derivative.

 

[noname]

Re: -Equation for Transfered Energy...?-

its the 3rd derivative of position, and in this discussion is negative (deceleration) and advantage of doing the calculas method rather than the algebraic subsitution presented above is the ability to integrate the expressions and derive values in energy,force and work either through transforms or thermodynamics. Using the algebraic methods would require iteration then conversion.

 

[MitchAlsup]

Re: -Equation for Transfered Energy...?-

OK, I misread you original post; sorry:

for completeness:
The first derivative of position wrt time is velocity
The second derivative of position wrt time is acceleration
The third derivative of position wrt time is jerk
The third integral of jerk wrt time is position

The first derivative of energy is power
The second derivative of energy is force

I used to teach calculus (at community college level). I recommend calculus for solutioins of all kinds of physics problems over straight algebraic methods (expecially itteration.)

 

[toddk]

Re: -Equation for Transfered Energy...?-

The unfortunate reality for this conversation is that terminal performance is about destroying vital tissue, not the right formula.

The "right formula" to calculate "transferred energy" is pointless as it does not tell us how much tissue is/can be destroyed.

 

[noname]

Re: -Equation for Transfered Energy...?-

that is correct the -- either of the algebraic methods or the calculas method, (or if you want to do numerical integrators and differential equations) only define the relationship between the variables. the amount of entropy lost into the target resulting in different observations is a reference frame based upon "time sensitive intial conditions- 3rd order- and deterministic chaotic.

 

[steelcomp]

Re: -Equation for Transfered Energy...?-

What about the energy rquired to decelerate a given mass traveling at a given rate? Is there anything to that?
Reason I ask is that many seem to elude to the fact that you can't impart any more load on the receiving end of a bullet than was imparted on the delivering end. On the delivering end, a bullet is accelerating from a dead stop to lets say 2500fps in 24". That requires X amount of energy. In reverse, it should also require X amount of energy to <span style="font-style: italic">de</span>celerate that same bullet from say 1500fps at the time of impact, to zero fps in say 12", or less.
Just thinking out loud.

 

[noname]

Re: -Equation for Transfered Energy...?-

the energy required is physics just like the acceleration process -- assmuming no change in the physical properties, on impact the process may take on the attributes of a non-elastic impact (tissue) and/or elastic collision (billard balls). the problem with the inelastic collisions are that there are many paths "to putting the bullet" back together, thus the probability of the reverse path becames a creature of inital conditions. the integral of the process can be used to calculate the work,energy,power etc. in elastic collisions, but not inelastic because you can't run the movie backwards.

 

[noname]

Re: -Equation for Transfered Energy...?-

think of target impact as a transition to the lowest possible ballistic coefficient possible-- or the ultimate "drag" coefficient, how well the bullet makes this transition, in terms of hanging together, staying in a straight line, etc, gives an insight into its use, ie, barrier penetration, varmit vaporization etc.

 

[skinnypitt]

Re: -Equation for Transfered Energy...?-

This is a good paper that I think will give you more info than you could ask for.

http://www.rathcoombe.net/sci-tech/ballistics/myths.html