The question of "How much energy is needed to kill an animal is older than I am. Energy is the wrong value to use because it favors the very fast, very light bullets. When shooting at long ranges a light bullet will be blown around by the heat coming off the ground and the gentlest winds while a heavier bullet will be affected less. There is a value that is used to determine the impact effects of a moving object on a larger (relatively) motionless object that I have found works well to determin the effect of a bullet with a given weight will have on a target with a given weight.
The formula is here:

( V^3 / 450380 X Wb^2 / 7000 ) / (2GxPi + (G / Pi))

to make it easier you can do it this way:

Energy X Momentum (in pound/foot seconds [Wb x V /7000]) / 212.37

Calculate the values at the impact range and then you have the weight of the animal that will die with a shot placed to the vitals. This is supposed to work regardless of the makeup of the projectile.

Yah, I think I got it now. I think.--tr

OK, I am sure that in my rush to get that posted before I had to run out to the three appointments I had this afternoon I did not give a decent explaination of the math - I appologize.
Here is a better example:

3006 165 gr bullet with a muzzle velocity of 2650.
I am shooting a coyote at 400 yards
velocity at 400 yards is 1981
the energy at 400 yards is 1438
the bullet weight (Wb) is 165 so 165 x 1981 =326865
divide that by 7000 and you have 46.695 which is the momentum

next multiply the energy X the momentum
now 1438 x 46.695 = 6714741

divide that by 212.37 (2G x Pi) + (G / Pi) = [64.34 x 3.14] + [32.17 / 3.14]
divide 6714741 by 212.37 = 316.1812403

At 400 yards I should be able to kill an animal that weighs 316 pounds - if I hit it in the vitals - that is one big coyote.

Use it and see if you agree with the results. Test it and see if the results agree with the math.

Welcome to the HuntingNut PaulS. That sounds interesting I'll have to try it.
Joe

shrpshtrjoe,

Thanks for the welcome. I worked the formula out so it didn't depend on velocity too much or on bullet weight. It is a mix of energy and momentum so that each plays a role in the results. It also is based on a hit to the vitals - if you are used to shoulder shots, the formula might not work as well - especially with lighter bullets.
Like all formulas you have to use common sense with it and it won't make up for the important stuff like shot placement and careful thought on the conditions.

PaulS

Paul,

Forgive the pun but what is the circular reference from? What part of Sir Issac's laws are you combining to get this number? Using G and Pi give me a gravitional frame of reference. Can you give me a more complete "proof" for the theorem? It looks very interesting and I have just enough physics back ground to be dangerous! Physics is phun after all.

I'm with you Dawgdad

Caveat Calculatis

Dawgdad wrote:

Paul, Forgive the pun but what is the circular reference from? What part of Sir Issac's laws are you combining to get this number? Using G and Pi give me a gravitional frame of reference. Can you give me a more complete "proof" for the theorem? It looks very interesting and I have just enough physics back ground to be dangerous! Physics is phun after all.

Dawgdad,

G (the acceleration of gravity) shows up as a constant in any equation that involves decceleration or acceleration as a result of a collision.
To calculate the energy you use G (actually its 2G) Take the velocity squared times the weight (in pounds) and divide by 2G = Kinetic energy. The 2G is used to convert weight to mass. because the formula to find the kinetic energy is actually Velocity squared times Mass (V^2M). In finding energy in our bullets flying we use the Velocity squared times the weight (of the bullet in grains) divided by 2G (64.34) times 7000 (the number of grains in a pound) which equals 450380 (64.34 x 7000)
Pi occurs in hundreds of equations in many sciences including those describing the DNA double helix, a rainbow, ripples spreading from where a raindrop fell into water, chemistry, superstrings, general relativity, normal distribution, distribution of primes, geometry problems, waves, navigation and many more.

In the process of looking for a formula that would convert known quantities of the bullet in flight to the actual killing effect I stumbled (spelled worked for years trying to narrow it down) upon this formula and every test I have conducted shows that it works (providing the limitations of the formula are adhered to). It won't work if you shoot the animal in the toe or if you use head and neck shots you can kill with a lot less bullet than the formula recommends. If you shoot into the vitals the formula works. See the next post.

lets say I have a weapon that will fire a 5 grain sewing needle at
13422 FPS, which gives it 2000 ft.lbs. of kinetic energy
and one that launches a 12 lb. spear at 103.5 FPS with the
same 2000 ft. lbs. of energy. Which of these projectiles
will make a 300 lb. inpenatrable object move more? (the
target will catch and hold either projectile) Which of
these "bullets" will kill the larger animal?

The answer is:

The spear has 1242 ft. lb. seconds of momentum and will
accellerate the 300 pound target to 4.14 FPS.

The sewing needle has 9.6 ft. lb. seconds of momentum and
will accellerate the 300 pound target to .032 FPS.

Even though both projectiles have 2000 Ft. Lbs. of energy
and will both go completely through most animals the
difference in killing power becomes very clear when using
the "hunting index" to calculate their potentials.

Energy x Momentum / 212.37**

The spear would most likely kill a five ton elephant.
The "hunting index" for the spear is 11697 lbs

The needle would likely not kill any but the smallest of animals.
The "hunting index" for the needle is 90 lbs.

**the constant derived from (2G x Pi + (G / Pi))

Hi PaulS...What is being questioned is the fact that the term for energy is already corrected to mass units when first used.

Shootist

I appreciate the work you have put into you theorem and I am sure the field tests were the best part, but adding Pi to the linear equation "just because" lends a bit of hocus pocus to it even though you cancel it out by dividing by Pi too. (Pi/Pi=1) Also KE = 1/2 mv^2, the use of G in collison equations is generally used to convert weight to mass units. Force is measured in joules or slugs.

I fully agree with you analogy of the spear and the needle. Energy and momentum are not equivalent. You need the bullet to give up its energy and transfer it to the target. The balance of penetration and disruptive energy transfer is what keeps the bullet makers in business.

Dawgdad,

the equation doesn't cancel PI out of itself it is not Pi/Pi its Pi +G/Pi
Here it is with the numbers replacing the Constants:
2*32.17*3.14159 + (32.17 / 3.14159)
64.34*3.14159 + (10.24) (note that nothing on the left side is canselled)
202.13 + 10.24 = 212.37
The equation never started out as linear - it was a differential equation when it began. I just couldn't correlate the data through the relationship between the mass-energy of the bullet and the mass-energy of the animal. I finally realized that the animal is more a liquid than a solid and then I was able to get some information while destroying my Fackler box instead of actual animals. Some of the bullets energy produces a series of waves (circular) that propagte outward from the path of the bullet. At least with the Fackler box I was able to make measurment of deceleration, expansion and penetration with a valuable record of approximate wound channel size. Discounting the "temporary wound channel" and comparing it to actual results in the field. This was a ten year project (off and on) and I made some wrong assumptions along the way.
As I said earlier (I think) the formula isn't perfect but it is more reliable than the TKO formula or using energy alone or even using just the momentum alone.


Shootist,

The weight of the bullet has been converted to mass but the weight of the animal has to be dealt with as well as the wave front and decelleration. That is why there is G and G-force components.

When all is said and done it is just a tool to check the potential of any given round to cleanly kill an animal with a single shot to the vitals before you go out and shoot the animal. Its just one tool amoung many - it is not an icon that I hold as absolute or that I would expect anyone else to use blindly without some critical thought processes going on. If you don't need it then it is as worthless as the shoes you threw away last year but if you have some uncertanty about the use of a cartridge on a particular animal you can use it to add to the evidence for or against. Although I am the one who developed it there have been times when I have disagreed with what it said - usually because it said a smaller round could do a job that I would use a larger round on. It has proven me wrong in the past but I still question it from time to time.

0,1,1,2,3,5,8,13........

Thanks for the expanded explanation. At this time I will do as you suggest and keep my grain of salt and defer to your 10 years of development and testing. It has been too many years since I have done Diff EQ's and the calculus makes my head hurt just thinking about it!!! (I am still confounded with the Pi though?)

In my opinion use a 100 grain bullet in your .243, and you will have a good hitter. True you may have to think just a little bit more about trajectory, but you will have a better BC. and at range the litle extra may make the difference between dead dogs or smarter ones.

I use 100 grain bullets in my 25-06. At long range, it make the difference for me.

longwalker

I found the math to be very interesting. When I started reloading it was all about "working up" a load for a "particular gun". In getting the job done at distance I've always believed in a heavier bullet pushed to a higher velocity is all that's needed in my "working up" theory. I like to experiment in that regard, but, I keep it in the ranks, meaning I wouldn't step up to large game calibers for taking varmints. Should I rethink my thinking?