zrexpilot

gorse you make a lot of sense and it's all true what you say except the part about energy, energy does nothing, it doesnt kill, it doesnt create wound channels, it doesnt expand bullets, its a theory, it's mathamtical equasion and depending on who's energy theory you use you come up with two different numbers. Einstiens theory or newtons, two different formulas that come up with two different numbers, which means nothing really. Wound channels are what kills and a pass through is a bigger wound channel. bigger bullets also leave bigger wound channels, but so do small bullets at high velocity , they just dont penetrate as well as bigger heavier bullets. however were talking a whiteatail deer here which is rarely wider than say 18 inches and does not offer much resistance in stopping a bullet when shot in the vitals. There is no way a little 'ol deer rib is gonna stop a .22 bullet traveling at or over 3500 fps. The .22 centerfires are just as poisonous to deer as any other caliber. how can 130 gr bullet be adequate for a 1000 + lb animal but a 50 gr not be for a 150 lb animal.

green01

i see alot of this guys read like they shoot. i knew they would not catch the two differant works. read error being rimfire vs centerfire. .223 will do the job anyday. man you guys are easy to stir up.[:'(]

gorse

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zrexpilot

Yes gorse energy is a theory. Here is some copy and pasting. You'll see that einstein and newton did in fact have two different theory's for energy.




The Kinetic Energy Theory:

Energy is defined as the ability to do work. The energy of an object in motion is called kinetic energy (KE) and is expressed in foot-pounds. It is not measured directly, but is calculated from the bullet's mass and velocity. The basic formula for bullet energy is:

KE = (m * v²) * ½: Where; m is the mass of the bullet, v is the velocity.

Mass and weight are not the same. To get the results to come out correctly in the English unit of energy, the foot-pound, the following formula is used. It contains the necessary conversions into English units:

KE (in foot-pound) = bullet weight (in grains) * velocity² ÷ 450400.

(If you recall from "Math Definitions" to get 450400 we multiply in the denominator the acceleration of gravity, which is 32.1734ft/sec², by the conversion from grains to pounds, which is 7000, and by the 2 in the above formula, this comes out to be 450427.6 but the scientific community uses 450400 and so will we).

Now, let's work an example, say were shooting a 185-grain bullet at a muzzle velocity of 2530 fps.

KE = (v² ÷ 450400) * m
KE = (2530² ÷ 450400) * 185
KE = (6400900 ÷ 450400) * 185
KE = 14.2115896980462 * 185
KE = 2629.14409413854
KE ~ 2629 foot-pounds

Now, let us see what we get if we use the same example with the chart above: look for 2500, then read across to 30, you should get 14.20. Now, multiply that number by 185 and we get 2627 foot-pounds, pretty close, just 2 foot-pounds off I think we can live with that. The above-described method is the one used by the late Charles Newton.

If we know the kinetic energy and the velocity we can find the weight of the bullet by changing the formula to look like:

m = KE * 450400 ÷ v²
m = 2629 * 450400 ÷ 2530²
m = 1184101600 ÷ 6400900
m = 184.989860800825 or ~ 185 grains.

On the same thought, if we know the kinetic energy and the weight we can find the velocity of the bullet by changing the formula to look like:

v² = KE * 450400 ÷ m
(OR)
v = the square root of (KE * 450400 ÷ m)
v = the square root of (2629 * 450400 ÷ 185)
v = the square root of (1184101600 ÷ 185)
v = the square root of (6400549.18918919)
v = 2529.93066885027 or ~ 2530 fps.

For relatively slow moving objects, objects well below the speed of light, Newton’s third law of motion, that deals with kinetic energy, is normally used in calculations whereas Einstein’s equation of energy is used for velocities close to the speed of light. Not because Newton’s equation is correct for low velocities but because it is far easier to use than Einstein’s equation. But as we can see from the example above a 185-grain bullet moving at 2530 fps is suppose to generate approximately 2629 foot-pounds of kinetic energy, energy of an object in motion. For curiosity let’s see what Einstein’s equation of energy has to say to this:

KE = (m * c2) - (mo * c2)

where m is the mass of the object in motion, mo is the mass of the object at rest, and c is the speed of light. We would need to find the mass of our 185-grain bullet traveling at 2530 fps and that number is 185.00000000061202800418747854534-grains, and the speed of light in feet is 983571056.688 fps.

KE = [(m * c2) - (mo * c2)] ÷ 450400
KE = [(185.00000000061202800418747854534 * 983571056.6882) - (185 * 983571056.6882)] ÷ 450400
KE = [(185.00000000061202800418747854534 * 967412023554348909.529344) - (185 * 967412023554348909.529344)] ÷ 450400
KE = [178971224358146631512.9315781475 - 178971224357554548262.92864] ÷ 450400
KE = 592083250.00293814750051965679658 ÷ 450400
KE = 1314.572047075795 ~ 1315 ft-lbs
KE ~ 1315 ft-lbs


Einstein’s equation shows more than a 50% reduction over Newton’s third law of motion in the bullet’s true kinetic energy. The kinetic energy of 1315 ft-lbs is easier for the layperson to believe but still hard to accept.

To a lot of lay people the term foot-pound is misleading. The layperson thinks the term, for example, 2174 foot-pounds of energy of a bullet means that the bullet is able to lift or move a 2174-pound object a distance of one foot or a one-pound object a distance of 2174 feet. It also seems that a time interval is missing in the results of both equation. If a time interval is part of the foot-pounds then as the time interval is allowed to get shorter that the energy that is imparted is also smaller. But neither of these is correct. In the first case, the wording of foot-pounds is more kin to the application of leverage. In this case the energy of 2174 pounds is applied to a leverage that is one foot long. And in the second case the time variable in both equation has, mathematically, been canceled out. The total amount of energy is not dependant upon the amount of time the force is applied.

The energy required to move an object at rest to a specific velocity or change the object’s direction to a specific velocity can be calculated. But the energy cannot be calculated of the object at rest if the velocity or the resistance is unknown. But the energy required to move, say a metal silhouette, can be calculated by its mass and the maximum velocity it achieves. Thus, the amount of energy that is loss between the total kinetic energy of the bullet minus the required energy to move the metal silhouette. Likewise, the energy of the ground resistance, at the point of the bullet’s impact, can be calculated by how deep the bullet penetrates the ground. The amount of kinetic energy of a little bullet may sound large but likewise the total energy of an object to overcome its inertia and the resistance to surface interface is also large.

The energy of objects in motion, kinetic energy, makes sense to me but I also know that it does not make sense to a lot of other people. I may understand the subject of kinetic energy and its significance but there are many answers to this complex topic of “Bullet Energy” that has a lot of people wondering what is the answer? So I do not present this material as an answer but merely as food for thought.


Now what ya got ta say mister smarty pants ?

zrexpilot

and heres the link

http://www.aeroballisticsonline.com/ballistics/bulletenergy.html

nh bowman

just say no

gorse

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clearanceman

Wow, you guys are really into this. Where .22 centerfire is legal, it is an ethical decision. No doubt .223 can kill deer. Are you good enough to do this or are you having some get away. I know I used to have some deer get away with 30-06, now I don't but I don't shoot at as many deer. I use a single shot 25-06 now mostly. If you are willing to wait for proper shot placement, probably anything will work ok. Personally I would rather use at least a .243. A bigger caliber gives you more certainty but is no substitute for proper shot placement. Bullet construction is also a factor. I hunted for awhile with a .44 mag revolver and the results were disapointing I gave it up. But that was before there were these new copper bullets with mechanical hollowpoints. I would probably have better results now.

I think if hornady releases a .223 or 22-250 in SST interlock it would kill pretty well. There are probably decent bullets already, especially when you get over 50 grains. Part of the poor performance has to be related to people using varmint bullets in .223/22-250 on deer. They are not designed to kill these animals although it is possible to kill a deer with them. Still, there are fewer bullets designed to kill deer in .22 cal than in .243 and up.

Interestingly, here in WV ANY centerfire rifle is legal for deer. I guess a .17 centerfire would be legal too. But I would not go under .243 despite how the law reads. It is just a personal choice, I hate seeing a deer get away and I want to have a good advantage along with waiting for a good shot. Good bullets designed to kill deer and at least a .243.

zrexpilot

Duhh . I said I did. To bad you didnt read it. It clearly talks about the different THEORY'S of energy. You said newton and einstein's energy formula were the same and I just proved to you they are not , I think your the one that slept in physics class. cant admit your wrong, I'll say it for you "your wrong".
Ever here of einsteins THEORYof relativity ? Duh !
Energy is a theory.

gorse

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