500 AHR
570 Grain Woodleigh soft muzzle velocity 2400fps
Pentration in game approximately 6 feet.

Let us calculate the force exerted upon the animal by the stopping of the bullet in the animal's tissue. We will make an assumption that the decelleration is linear (which is valid) and we will not take into consideration the mass of the disrupted animal tissue. By not considering the tissue mass we will obtain a calculated force which is less than the actual force.

Bullet mass = 2.532E-3 slugs
Decelleration = 897,067 feet per sec per sec
Force = mass X decelleration = 2271 pounds

This calculation is lower than actual becuase we did not take into consideration the additional mass of the animal's disrupted tissues moving with the bullet path. Also, the deceleration will not be linear and will in reality be greater than the linear approximation suggests.

This buffalo was knocked on it's ass. Since the animal weighed approximately 1300 pounds live weight and the bullets force was 1.75 times the animals weight I have to agree that the bullet knocked the animal down. The force was realized by animal in the incredibly lengthly amount of time of
2.586E-3 seconds (0.002586 seconds). There is not way that an animals brain can comprehend and respond to an inputted force of that magnitude in that amount of time.

Now have fun arguing with me.

Todd E

But how long was the force applied?

Mike

What about this.

Your load has about 7000 ft/lbs of kinetic energy, so that is enough to lift the 1300 pound buffalo about 5 feet.

The problem though is a "gearing" problem. Like putting an 800 HP Forumla 1 engine in the big Kenworth. But if we reduced the gear ratios by maybe 15 times, then we would have one Kenwort with big pulling power.

Now if you could somehow attach a block and tackle system to your bullet and we ignore friction and other problems, and if the block and tackle greared down 100s or maybe 1000s of times, then by the time your bullet (assume we fire straight up)came to a stop, the buffalo would have been raised about 5 feet.

The next thing to happen would be for the buffalo to start falling back to the ground. By the time the buffalo hit the ground, your bullet would be wound back and be back at 2400 f/s.

Perhaps this is an experiment Saeed can do but with a camel

Mike

I have know that the bullet can knock over a animal.

If the buffalo went down, and died without hurting anyone, the numbers don't matter to anyone but the guy reading them! Certainly not to the Buffalo!

Everybody wants to use numbers to define killing power on game, but some cartridges just seem to defy all accepted theories, and just do it right, all out of proportion to the numbers generated by the cartridge. This can be done with a small fast bullet, or a large slow bullet,or a dozen other combinations, but the end result is the same, a dead Buffalo!

It has been shown that, for some reason, a 500 gr bullet pushed at 2150 FPS, is more effective, than the same bullet pushed much faster. One would normally think the higher , weight of the bullet/FPE/FPS the More knock down power, but experiments prove otherwise. The bullet actually penetrated less with the higher velocity. Addtionally,one knows, as the FPS goes up, and the weight of the bullet goes down, the FPE is increased. This tells me PFE is not a good indicator of killing power! There are simply too many verribles, to use simple math to define killing power. A bullet placed 1/2" one way or another, or a degree or two of different angle, and the numbers would change drasticly.

IMO, the best indicator of killing power on game is "results",nothing more.

The numbers, are fine fodder for discussion, however!
------------------
..Mac >>>===(x)===>
also DUGABOY1
DUGABOY DESIGNS
Collector/trader of fine double rifles, and African wildlife art

[This message has been edited by MacD37 (edited 01-29-2002).]

[This message has been edited by MacD37 (edited 01-29-2002).]

Also, if you do not understand physics (and from your rantings you do not) you should just sit back and be quiet. Power is work done over time. A formula one engine does not produce the same amount of work at 1500 rpm as a Cat diesel does. That is why the Cat diesel rated at 400 Hp performs better than a formula 1 in a Kenworth. I warn you against arguing with me in the realms of engines and gearing. I have been a project leader on many motorsports powertrains. I have some small understanding of how to accelerate a 2300 pound car from a standing start to over 200MPH in a 1/4 of a mile.

Overkill,

I figured that you did and have been having a great deal of fun with the ignorant.

Todd E

[This message has been edited by Todd E (edited 01-29-2002).]

[This message has been edited by Todd E (edited 01-29-2002).]

quote:

---

Originally posted by Todd E:
I have some small understanding of how to accelerate a 2300 pound car from a standing start to over 200MPH in a 1/4 of a mile.

Todd E

[This message has been edited by Todd E (edited 01-29-2002).]

---

how bad was the buffalo hurt by that 2300 lb bullet @ 200 MPH? Did it pennetrate, or was the bullet destroyed?

------------------
..Mac >>>===(x)===>
also DUGABOY1
DUGABOY DESIGNS
Collector/trader of fine double rifles, and African wildlife art

460 WM; MV 2600, Impact Vel 2500 Bullet: 500
Force acting on animal = 2313 lbs

500 AHR MV 2500, Impact Vel 2440 Bullet: 570
Force acting on animal = 2512 lbs

577 NE MV 2050, Impact Vel 1960 Bullet 750
Force acting on animal = 2133 lbs

585 Nyati MV 2470, Impact Vel 2400 Bullet 750
Force acting on animal = 3198 lbs

300 Win MV 3050, Impact Vel 3000 Bullet180
Force acting on animal = 1199 lbs

Now all of the above numbers are low. In reality the larger the bullet the more tissue will be disrupted; therefore, the greater the mass being decellerated. Furthermore, field experience indicates that the smaller bores typically completely penetrate and therefore do not obtain the decelleration assumed in the above model.

To clarify the tissue issue. If a .510" upsets to 1.2" and we assume that the wound channel is 2.5 times the projectile's upset diameter we would then have a column of flesh with a diameter of 3" and a length of 6 feet in this example.

Making the same assumptions for a .308" bullet we would have an upset diameter of .72". The column of flesh would then be 1.8" in diameter and 6 feet long.

The respective volumes would be:
.510" 509 cubic inches
.308" 183 cubic inches

The masses of these two volumes would be additive to the mass of the bullet and would greatly influence the force calculation. The modelling of the flesh mass addition is much too complex for for to try and explain on the web, but it is there. I believe that it becomes rather obivous why the big bore has more "stopping power" than the smaller bores.

MacD,

I guarantee you that the 2300 pound bullet going 200 MPH would kill and knock down any buffalo (except the old class 9 Chevy truck). But let me tell you that is one EXPENSIVE buffalo bullet!
Todd E

[This message has been edited by Todd E (edited 01-29-2002).]

Tod, this is the first time I have seen this engineering term. Out of curiosity is this your training? I'm curious 'cause U. of AK at Fairbanks is an engineering school and I know many who are aspiring engineers.

I liked what you did. Thanks for putting the time and effort into it for our benefit.

------------------

I tried a little once before to explain this phenomena by stating that the knock down is a froce moment relationship. Those statements are on Overkills thread about shooting a moose in the shoulder with a 585 Nyati. The energy and momentum are not really applicable to our discussions.

Slugs in this case does not refer to bullets. The slug is the name for the english systems unit of mass. There are 32.2 pound masss in 1 slug. To my knowledge all english based calculations were mass is considered are done with the mass in slugs. There may be some equations now that have been adjusted were one uses pounds multiplied by the local acceleration of gravity with this product divided by the gravitational constant. In this case you are taking a weight (which is by definition a force) and converting it to mass with the unit being slugs.

I hope that answered your question.

Todd E

Buff weighs 2000x7000 = 14,000,000 grains. Divide 2400 by 14,000,000 and get 0.00017 feet per second. That's the final velocity of the buffalo. He fell down. Period.

Shotguns won't knock down people, Hollywood aside. And big guns won't knock down buff, either. I'm not denying that the animal appeared to be knocked down. But his center of mass did not move appreciably due to the bullet, in the short time it took him to fall.

Pertinax
(a physicist)

And I suppose I should point out that a bullet with sufficient momentum to knock a 2000 lb. buff off its feet would do the same to you-- times 10 (assuming 200 lb. man/gun combo). You'd not be delivering a follow up, though none would be needed.

The bullet energy is what does the damage to the animal, but that's not the property of the bullet that moves the buffalo. What will move it is momentum, pure and simple. Sure, the bigger, faster bullet will do far more tissue damage, and possibly cause enough shock to the nervous system to cause its legs to collapse beneath it. But it won't "knock it down", in the sense that bowling pins get knocked down or off a table in pin matches. If you've ever tried that, you won't hold out much hope of your handgun knocking a human being down.

Pertinax

A formula 1 engine that can develop 800HP can do twice as much work in the same time as 400HP Cat.

Note that my post did not state 1500 RPM but rather to lower the gear ratios by perhaps 15 times or so.

Next, I did not link kintetic energy to F=ma

Since you are a physics expert, you might like to tell us how far the force which you calulated would move the buffalo in the time the force was applied.

In addition you might like to tell us all how the ballistic pendulum works.

Mike

[This message has been edited by Mike375 (edited 01-29-2002).]

My knowledge of physical sciences is very rusty, but I don't think that itty bitty bullet knocked anything over. As far as all your pounds applied, where were they applied? Over what area? Got a lot of pressure in that area. That's why it made a deep hole.

See also Pertinax, above.

And i agree whit Todd E. I have shoot big mooses whit small calibers like the 6,5x55 and the moose have been knocked down by shoulder shoots.

And my friend shoot a moose at 30 yards in the front off the moose whit a 30-06 180 grains soft nose and the moose were knocked on it�s ass.

In the real world a big bore can knock over a animal

Now lets think about this. I punch you in the jaw with a force of about 150 pounds. This is a hard hard hit. I'll bet I knock you flat on you arse. Did the momentum of my fist knock you over? No idiot your brain was not capable of comprehending the force suddenly induced against you jaw.

Mike,

Horsepower is torque X rpm / 5250. The 5250 number is a conversion constant. Ok brainiac. The 800 horsepower formula 1 engine is making 800 horsepower at about 10000 rpm. It is not making much horsepower at 1500 rpm which is a "normal" driving speed for most over the road vehicles. That is why it is not uncommon to see these cars pushed out of the pits (I may be thinking of CART there).

Mike why don't you explain to me how a ballastic pendulum works. Here is the example problem. You have a projectile which weights 437 grains. It strikes the pendulum bob weigh and causes the pendulum to move 3 feet in an arc about its fulcrum point. The length of the pendulum arm is 8 feet and the bob has a mass of 25 slugs. The arm itself has a mass of 1.5 slugs and is a retangle in section.

Todd E

I will try again on the 800 HP Formula 1 engine.

If the truck is geared about 15 or so times lower etc. etc. etc.

In other words the F1 engine won't be doing 1500 RPM in the truck.

Likewise, if we have the 400HP Cat engine in the formula 1 car, the the car will be lucky to reach 20mph in top gear. So we would have to gear up 10 or 15 times which would mean the performance would be terrible.

If we used your buffalo theory for the pendulum, the pendulum would fly through the the air with the greatest of ease

Mike

I assume you want to know the velocity of the bullet. No problem to calculate that, but you've left out one needed detail: the size of the rectangle, so I can figure out how far the center of mass is from the fulcrum.

Or just stipulate something; I suspect I'm the only one interested in playing.

To every one else: Newton was (and is) correct. Large animals cannot be knocked off of their feet by a bullet that you can fire.

Pertinax

... I mean the size of the "bob", or should I just assume the bob is a point mass on an eight foot arm?

Methinks the non-physicists here would expect a bob massing 600 slugs to spin around and around with their mighty 500s. You and I know better.

Pertinax

quote:

---

Originally posted by ramrod340:
I have to agree with everyone that says a rifle isn't going to "move" a buffalo.
. . . Apply the movement of the 39# box to a 2000# buffalo at distance and the buff isn't going to be moved enough to knock him down.

---

Unless he's standing on his tiptoes to get a beter look at what's going on, and you catch him off-balance.

Rule of thumb I read somewhere: model the powder as exiting at 5000 fps. Add that component to that of the bullet to get a rough estimation of the momentum going into your shoulder.

BTW, it's a good thing for us all that ENERGY isn't conserved in "equal but opposite" fashion. Thank goodness your shoulder only gets a few dozen foot-pounds of energy, rather than the thousands the buff gets.

Folks, the math on all this works out quite well. While intesting to discuss, it does NOT tell us anything about the effect of a bullet on a game animal. The biology is far more important than the physics. If you really want an argument, start talking about stretch and crush cavities.

Pertinax

What kind of force is needed to knock over a 150 man? If he stands still you can push him over easily...if he braces himself with a good base it takes quite a wollop. animals are not pins knocked over with a little push, they are pretty stable standing on four legs.

In my experience with big game in North America the only time you knock an animal over is with 1)breaking major bones(shoulders, hips, legs) 2)spine/head shot 3)shooting a jackrabbit with a 30'06.

Is a 577 t. rex on an elk like a 3006 to a rabbit? 10,000ft/lbs of enerygy to a 1000lb elk and 2500ft/lbs of enery to a 8 lb hare..... :0

I know when I shoot 10 lb foxes with a 17 rem and 900 ft/lbs of energy through the chest they die....but they are not flipped over or pushed back...some even take a few more steps. All that energy is left in the fox..no exit hole.

About engines the key is torque curve, if one engine has flat curve 100Nm up to 2000rpm and other 200Nm up to 1000rpm, second one could be transmited to 100Nm and 2000rpm or back - first to second. But torque curve are not flat, Cat engine will has much flatter curve than F-1 engine, that's all . . .

[This message has been edited by Jiri (edited 01-29-2002).]

25 slugs at sea level is equivalent to 805 pounds force. No I do not expect it to fly around and around even with the 585 Nyati. You will find if you do the problem that the this is a much bigger gun than you would shoot from your shoulder I think.

You may consider the bob to be a point mass, and that the bullet strikes at the center of it mass i.e. the point.

Mike,

In your analogy I can take a small electric motor out of a laundry drier and if I have enough gear mechanical advantage I can drive that Kensworth down the road. I will go very slowly.

Now you stated that we could go 15 to 1 gearing with the formula 1 engine. That means that the kenworth while go 15 times slower than the formula 1 car. If we assume that the formual 1 can make it to 230 MPH then the Kenworth would make it to a top speed of 15.333MPH. Now that is cooking!

Now the Cat diesel with all of it low end torque does not require that steep gearing to move that little formula 1 car chassis. So if we assume that the formula 1 car has a final drive of 3:1 we could and reduce that for use with the Cat diesel to an overdrive of lets say 0.6:1. Then the formula 1 car with the diesel will be capable of a new top speed of 1150 MPH.

Now noone is going to use a 2000 pound Cat diesel in a formula 1 car because the engine is so heavy that the car's chassis cannot support the engines mass. Furthermore, the engine's torque (Cat 3406B) is rated at
1375 lb-ft at 1200 rpm. The horsepower rating for the Cat 3406B is 400 Hp at 2100 rpm.

You need to boost your formula 1 engine's work potential by gearing and that is correct Mike. The problem is that is is reduction gearing which means that the little engine is still operating within it's power range, but the vehicle speed is greatly reduced. Actually the vehicle speed is reduced by the reduction gear ratio.

Rune,

Did your 416 bullet stay within the animal or did it pass completely through? The asumption in my calculations is that the animal's tissue decellerates the bullets so that it (the bullet) is stopped and remains within the animal. I have witnessed similar results with a 416 Rigby, but the bullet passed through with little expansion. I make this statement because of the very small (nearly cliber diameter) wound channel produced.

You guys should just shut up and accept the truth. Every one of the negative agruments are either not representative of the mathematical model i.e. a pass through, or just misunderstanding of physical law.

To explain a little further. I have made no attempt to determine at what speed (velocity) the animal is moved by the projectile. I have made a simplistic representation of the decelleration of projectile as it travels through an animal and comes to rest within said animal. The decelleration curve is conservative I assure you. It is not my fault that some individuals are so weak in physics and mathematical modelling that you cannot see how simple this is. I am sorry if you cannot accept the facts. Do not try and twist physics around to suit your opinions. Some on this sight have and they come across as morons.

Todd E

[This message has been edited by Todd E (edited 01-29-2002).]

Numbers are meaningless, unless they are shown to correlate to the real world. Since we know from real world experiences that bullet construction and placement are very important factors, and that they are ignored in your calculations, to say that your numbers correlate is wishful thinking at best.

quote:

---

Originally posted by Todd E:
(sic)
You guys should just shut up and accept the truth.
(sic)

---

Very convincing argument.

56.2 fps/s times .002586 seconds gives the buff a final velocity of 0.145 fps. That's more than 30 times slower than a normal human's walking speed. That's if his leg muscles didn't push back at all.

Make him fall down? At that velocity it would take 1.72 seconds to move him sideways 3 inches. That's plenty of time to compensate and keep his balance, unless, of course, he was drunk!

I forgot to add, from conservation of momentum you would have 570*2400/9100570 = 0.150 fps with fewer rounding errors.

[This message has been edited by Jon A (edited 01-30-2002).]

You seem to be having more difficulty with "power" than you do with momentum and energy etc.

We are not talking about just moving a chassis.

I will try and make and simpler for you with the engine.

Let us say we have a crane.

Now the crane has a 400 HP Cat engine.

Said crane with 400 HP Cat engine can lift a certain weight and at a certain speed.

If we fit an 800 HP Formula 1 engine and alter the gearing to be suitable for that engine, then the crane will be able to lift a weight twice as heavy at the same speed or a weight of the same at twice the speed etc.etc. In fact we could use the crance to calculate the "power".

Remember 1 horsepower will lift 550 pounds 1 foot in 1 second or 33000 pound 1 foot in 1 minute.

By the way, you have not worked out for me how far your force acting ofr .00256 seconds will shift the 1300 pound buffalo.

Mike

Mike375, You are wrong if I understand you correctly. A diesel makes gobs of torque down low. You can gear it down and go alot faster. Hell my diesel pickup with 3.73 gears kicks the crap out of a friends with 4.56 gears and a chevy 454! He has more horsepower than me, but I have more torque.

If you are reducing that race engine it will run slower than the diesel becuuase for each turn of the race engine the axle shaft will only make 1/15 of a turn. The diesel is just the opposite.

I have seen deer fall over with 12 gauge slugs every fall. The ones I shoot.

Kent

Dude, relax-- I'm agreeing with you. And while I haven't checked my work all the way through, looks like 45,750 fps. I must be converting the slugs raised up (.1418 ft for the bob and half that for the arm) to energy wrong... You can't get that velocity without a rail gun, I think-- that's faster than the detonation rate of most high explosives.

Pertinax
(Who, while a physicist by schooling, doesn't do much of this anymore)

P.S. And yes, I did not follow the significant figures through the equation. Take off a point.

I assume you are referring to diff ratios only.

Lest take an example.

Say you had a deisel that produced 1000 ft/lbs of torque at 1500 RPM. That would mean if you had a pulley 2 feet in diamter and a cable around the pulley, then the diesle could lift 1000 pounds.

Now let us say that you hada Formula 1 engine that could 300 ft/lbs of torque at 15000 RPM. Witht e same pulley set up it could only lift 300 pounds but lift it 10 times quicker.

So, we gear the Formula 1 engine down 10 times so its pullet is also doing 1500 RPM. But now, this pulley will have 3000 ft/llbs of torque and will therefore lift a weight 3 times heavier than the deisel.

On the other hand, if we gear the deisel up to 15000 RPM, then that pulley will only deliver 100 ft/llbs of torque.

Mike

Nice example problem. Unrealistic as hell, but nice. Show me a transmission which will allow you to start off with a 15000 rpm speed differential for very long. The best part of this entire discussion is that you formula car has a many ratioed transmission to allow you to get the car moving and get up to those horsepower ratings you love to talk about so much. The diesel in my example makes the 1000 lb ft of torque at 2100 rpm not 1500.

The most interesting part of this entire discussion is that it has obsolutely nothing to do with stopping rifles. I apologize if I upset you.

Paul H,

You did not read my original post very well. The physics was correlated to a real experience that happened last fall shooting buffalo. JonA is right though the animal was not knocked off of his feet. It did fall right down on it's ass and then got back up to be finished immediately there after with a head shot. Again the animal went right down as struck by lightning. This phenomena has been experienced in previous hunts on bison and moose also.

Rune,

You example does not fit the models assumptions. The reason is that your bullet failed. The fragmentation was spread out over a greater area which allowed for a less efficient tranformation of energy. JonA understood that the conservation of momentum was in fact applicable because the momentum of the bullet was conserved since the bullet was stopped by the animal. The force calculation confirms the law of conservation of momentum. The animal was dropped though.

If the bullet did not do this then why did the animal fall?

Would the animal have fallen if the bullet had not struck?

How fast does the nervous system comprehend and react to a force acting upon it if that force is not expected?

What effect will the force have since if it is not applied directly to the center of mass of the animal?

Could a yaw force cause the animal to fall?

Answer these questions instead of crying about physics problems which are fundamentally correct. Also, can we stop arguing about race cars vs. semitractors. I mean it can get interestin if we get realistic with the power transmission instead of the what ifs, but it is not related.

Todd E

What about an animal hit in mid shoulder, the slug doesn't exit but comes to rest under the skin on the off side. The animal has no clue you are there are about to shoot. The bullets hits as described above and the animal flips onto its side hard enough that all four legs are in the air kicking. The bullet went through the lungs below the heavy back muscle. I have witnessed this kind of response to many shots.

What about the buffalo shot last fall. Hit in the front chest. Bullet travelled through the chest and abdomin ending in the right hind quarter under the about an inch from the skin. The animal was unaware of the impending shot and fell onto it's haunches when hit. The animal then got back onto it's feet and stood there in what appeared to be a daze until I shot it again. This time a head shot. Speaking of hydrodynamic force, both of the buffalos eyes were blown out of their sockets.

Explain these to me. Did the bullet cause this or was it something else.

Todd E

Dead tissue laying on the ground will not behave like a living creature standing on its feet. Shoot an engine block does it move? The shooting of a downed animal is totally different. The muscles are relaxed and the mass is not supported solely by the legs. The equilibrium of balance is also no longer in existence. That means that a downed animal cannot lose it's balance and fall over. Can it? Shooting dead animals proves for some insight into bullet performance and that is about it!

I have shot over 100 deer and I cannot remember one that did not drop on the spot and stay down with the shot placement that I described in the previous post. The weapon was a 12 gauge slug at between 20 and 50 yards.

Todd E