Is a 30 cal big game rifle needed anymore?

[ARK_Ginger]

Not true at all. Energy = mass x velocity. Any bullet you can design for lighter projectiles you can also do with a larger projectile. No advancement in bullet design is going to make a .223 hit as hard as a .50bmg

The key word is transfer here. A big caliber has more energy than a small caliber at similar velocity. The design of the bullet determines how this energy is transferred in both.

The velocity is squared by the way.

 

[RobHazmat89]

Bullet design can only close the gap so much for a lighter projectile to a heavier one moving at the same speed. And again, any design you come up with for a smaller caliber can be used on a larger caliber to get the same effect. Again, you can have the most perfected .223 bullet in the world, and it will still never be as devastating as a .50bmg that's only using simple ball ammo.

Thanks Einstein, we all know the equation is e=mc². But in simple man's terms, mass×speed=energy

 

[omicron1792]

Bullet design can only close the gap so much for a lighter projectile to a heavier one moving at the same speed. And again, any design you come up with for a smaller caliber can be used on a larger caliber to get the same effect. Again, you can have the most perfected .223 bullet in the world, and it will still never be as devastating as a .50bmg that's only using simple ball ammo.

Thanks Einstein, we all know the equation is e=mc². But in simple man's terms, mass×speed=energy

Yeah but you’ve missed the point again. Velocity has more of an impact because it is squared.

Also a bullet that pencils through puts most of its energy in the dirt on the backside of the animal.

 

[ARK_Ginger]

Bullet design can only close the gap so much for a lighter projectile to a heavier one moving at the same speed. And again, any design you come up with for a smaller caliber can be used on a larger caliber to get the same effect. Again, you can have the most perfected .223 bullet in the world, and it will still never be as devastating as a .50bmg that's only using simple ball ammo.

Thanks Einstein, we all know the equation is e=mc². But in simple man's terms, mass×speed=energy

It closes the gap more than enough for killing game animals which has been demonstrated ad nauseam on several threads here ranging from 224 to 6.5 bullets. It’s also been demonstrated that similarly designed bullets in larger calibers cause more damage that most people want.

Advocating big calibers= more energy = more better killing Is just tiresome. Hunt with your 50bmg will ball ammo if you’d like.

 

[RobHazmat89]

Yeah but you’ve missed the point again. Velocity has more of an impact because it is squared.

Also a bullet that pencils through puts most of its energy in the dirt on the backside of the animal.

How did I miss the point lol? The velocity doesn't not have more of an impact when the velocities of the two different weight projectiles are so similar. Feel free to do the math and tell me what you come up with lol.

Again, shoot an deer/elk with a super advanced .223 that doesn't pencil through, and shoot another one with a .50 bmg using normal ball ammo and tell me what you get. Now imagine using that .50 bmg in the same super advanced projectile as the .233. Not matter what the .50 bmg is gonna be more devastating.

Again, bullet design only does so much to close the gap, and anything you can do to improve the lighter/smaller projectile, you can also do to improve the heavier/larger projectile.

You can kick an scream as much as you want, but the math doesn't lie lol.

 

[RobHazmat89]

It closes the gap more than enough for killing game animals which has been demonstrated ad nauseam on several threads here ranging from 224 to 6.5 bullets. It’s also been demonstrated that similarly designed bullets in larger calibers cause more damage that most people want.

Advocating big calibers= more energy = more better killing Is just tiresome. Hunt with your 50bmg will ball ammo if you’d like.

You could probably kill an elk with a .22lr if you're lucky enough as well. Just stating that bigger heavier projectiles makes it easier.

 

[Bluefish]

Bullet design can only close the gap so much for a lighter projectile to a heavier one moving at the same speed. And again, any design you come up with for a smaller caliber can be used on a larger caliber to get the same effect. Again, you can have the most perfected .223 bullet in the world, and it will still never be as devastating as a .50bmg that's only using simple ball ammo.

Thanks Einstein, we all know the equation is e=mc². But in simple man's terms, mass×speed=energy

No one here is arguing that a larger bullet going faster doesn’t have more energy. The argument is once you have enough energy, why suffer more recoil to have more when it doesn’t result in a more dead animal. That extra recoil effects accuracy and how much you practice. No one can say they shoot a 8lb win mag better than a 8 lb 223.

Mass x speed = momentum. Energy is 1/2 mass x velocity ^2. Thus velocity is the major component of energy.

 

[RobHazmat89]

It closes the gap more than enough for killing game animals which has been demonstrated ad nauseam on several threads here ranging from 224 to 6.5 bullets. It’s also been demonstrated that similarly designed bullets in larger calibers cause more damage that most people want.

Advocating big calibers= more energy = more better killing Is just tiresome. Hunt with your 50bmg will ball ammo if you’d like.

No one here is arguing that a larger bullet going faster doesn’t have more energy. The argument is once you have enough energy, why suffer more recoil to have more when it doesn’t result in a more dead animal. That extra recoil effects accuracy and how much you practice. No one can say they shoot a 8lb win mag better than a 8 lb 223.

Mass x speed = momentum. Energy is 1/2 mass x velocity ^2. Thus velocity is the major component of energy.

Arguing semantics isn't going to make you right

 

[yeti12]

Bullet design can only close the gap so much for a lighter projectile to a heavier one moving at the same speed. And again, any design you come up with for a smaller caliber can be used on a larger caliber to get the same effect. Again, you can have the most perfected .223 bullet in the world, and it will still never be as devastating as a .50bmg that's only using simple ball ammo.

Thanks Einstein, we all know the equation is e=mc². But in simple man's terms, mass×speed=energy

From the deer I seen that was hit with a 416 Barrett with a 395gr fmj the wound channel was less impressive than a 147gr from a 6.5 creed. Still lethal, but a 40lb rifle isn't the most comfortable to carry.

 

[ARK_Ginger]

Arguing semantics isn't going to make you right

Arguing FMJ bullets do more damage won’t make you look smart either but here we are.

 

[RobHazmat89]

From the deer I seen that was hit with a 416 Barrett with a 395gr fmj the wound channel was less impressive than a 147gr from a 6.5 creed. Still lethal, but a 40lb rifle isn't the most comfortable to carry.

Sure it was

 

[RobHazmat89]

Arguing FMJ bullets do more damage won’t make you look smart either but here we are.

If that's what you got from what I said you're extra special needs

 

[yeti12]

Sure it was

I don't think you understand how bullets perform. If hornady made a varmint bullet or very thin jacket match bullet for that 416 it probably would have made a few separate 40lb pieces of the deer.

Fmj's do very very little damage.

 

[Article 4]

This paper seeks to support the notion of ‘hydrostatic shock’, but it fails to prove how meaningful it is to incapacitate as it relates to time since it is just simply a review of the existing literature on ballistic pressure waves. It even fails to prove that hydrostatic shock would be the correct terminology for the observed effect, since ballistic pressure wave is a much more accurate description.

If someone wanted to use this paper to ‘prove’ hydrostatic shock exists I guess you could. I think it would be more accurate to say that it shows there is research measuring tissue damage from ballistic pressure waves/the temporary stretch cavity and cause damage at energy transfer levels above 300ft/lb and is more evident above 600ft/lb, but how this aids in time to incapacitation is undetermined currently.

May also want to read this.

https://medcoeckapwstorprd01.blob.core.usgovcloudapi.net/pfw-images/borden/conventional-warfare/Chap4Pages13-25.pdf

Then this

Just re-reading this, and thought I would pull out a bit more, as I think its really relevant to the topic, even though its very general info:

View attachment 724948

So here’s the recommendations from this paper based on the finding that a “ballistic pressure wave” exists, and likely contributes to incapacitation:
“Minimum penetration requirement of 12” in ballistic gelation” : so hunters need a minimum penetration requirement as well. Is more better? Maybe, it does not say so.

“Maximizing pressure wave effects requires transferring maximim energy in a penetration distance that meets this requirement”. Key word being “TRANSFERRING”. In other words, a bullet that passes through on most shots in order to be able to take wacky off-angle shots, is going to carry a lot of its energy out the other side…ie that energy is NOT TRANSFERRED, ie it has NO BENEFIT until it is transferred. Also, it it’s “enough” energy for the “normal” shots, then we should quantify that and discuss the “enough” concept relative to smaller cartridges.

“In addition, bullets that fragment and meet minimum penetration requirements generate higher pressure waves than bullets which do not fragment”. Self explanatory, ie take two bullets that are identical in caliber, weight, shape, etc except that one fragments and one doesnt…the one that fragments generates more ballistic pressure wave. OR, the same thing means that a smaller bullet that fragments can make an equal ballistic pressure wave as a larger bullet, by virtue of its fragmentation.

The rest is speaking specifically to defensive handguns, but seems very transferrable to hunting.
“No wounding mechanism can be relied on to produce incapacitation 100% of the time in the short span of most gunfights”. Ie no bullet will get a DRT 100% of the time, which is relevant when basing decisions on small sample sets such as “I used x and it worked great” or “I used x and it ran off, never again”.

“Selecting a good self-defense load is only a small part of surviving a gunfight. You have to hit an attacker to hurt him, and you need a good plan for surviving until your hits take effect”. Sounds to me like they are advocating for shootability, and this is a good argument for being able to make fast, accurate follow-up shots.

“Get good training, practice regularly, learn to use cover, and pray that you will never have a lethal force encounter armed inly with a handgun”. Hmmm, training. Maybe something inexpensive, easy to shoot in some
volume would be a good choice for a friend with a “one-gun” situation to amke practicing easy and likely??

This paper also doesnt say if ballistic pressure wave is a SIGNIFICANT part of incapacitation, it only says it exists and does contribute. Is it 90% of incapacitation? Doesnt sound like it. Is it 10%? No idea. But depending on how large of a role it plays it should dictate how large of a priority we place on creating a ballistic pressure wave. If it isnt a MAJOR factor, it might make sense to prioritize other factors in picking a cartridge??

This is not directed entirely at you however given your thorough response, which I appreciate, this is directed here as well to everyone else that has provided their opinion.

Saying ballistic pressure is a part of incapacitation is confirming the existence of energy and its affect on the target and tissue. "It exists and does contribute" is all we need to read from that portion of the paper to know. Furthermore the paper continued to reference ballistic pressure waves, calculations of energy, and transmission to surrounding tissue including the brain at impact. Think about when you spine an animal and it drops dead on the spot. The energy transfers through the spine and into the brain, causing a rupture at the medula and instant death.

So whether we discuss higher pressure waves, frangibles, or blunt force trauma - physics has proven through mathematical calculations that have been peer reviewed and verified through multiple outlets too numerous to quote - the existence of it energy in projectiles is proved. The amount is also proved depending on the size and speed of the projectile.


In simple terms, if you shoot an elk in the back hip with a 223 will it kill? Probably not. If you shoot an elk in an a vital area with a bullet that perhaps deviates from its original path, do you have a better chance of a kill by creating a hole PLUS a significant pressure wave - absolutely and without argument YES

Here it is in its simplest form

1. How does energy transfer occur in ballistics? In ballistics, energy transfer occurs through the transfer of kinetic energy from the bullet to the target. When the bullet strikes the target, the kinetic energy is transferred to the target, causing damage or penetration. e.g. "ringing steel"
2. What factors affect the transfer of energy in ballistics? The transfer of energy in ballistics is affected by several factors, including the mass and velocity of the bullet, the density and composition of the target, and the distance between the bullet and the target.
3. How does momentum play a role in ballistics? In ballistics, momentum is the product of an object's mass and velocity. It plays a crucial role in determining the trajectory and impact of a bullet. The greater the momentum of a bullet, the more force it will exert on the target upon impact.
4. What happens to energy and momentum after a bullet impacts a target? After a bullet impacts a target, the energy and momentum are transferred to the target, causing damage or penetration. Some of the energy and momentum may also be transferred to the surrounding environment, such as the adjoining tissue, in the air or in some cases pass to the ground.
5. How is the transfer of energy and momentum in ballistics calculated? The transfer of energy and momentum in ballistics can be calculated using the laws of conservation of energy and momentum. This involves measuring the mass and velocity of the bullet and the target, as well as the distance and angle of impact.

Reference: https://www.physicsforums.com/threa...of-energy-and-momentum-in-ballistics.1014644/

HITS - HORNADY INDEX OF TERMINAL STANDARDS - Yes HORNADY​

You can effectively calculate the impact of a bullet via the following series of steps.

Collect the required information for the ammunition under consideration. You will need to know the bullet weight, measured in grains, which is listed on the ammunition box. You must also know the velocity the bullet is moving at impact. Each ammunition manufacturer publishes ballistic tables for their ammunition. These tables will normally list the bullet velocity at the muzzle of the gun and at intervals of 100 yards out to the useful range of the ammunition (usually 300 yards or so). For distances between the listed increments, you must estimate.

Ammunition manufacturers typically measure rifle bullet velocities using a 24-inch test barrel. As a rule of thumb, for every additional inch of barrel length beyond 24 inches, the velocity increases by 20 feet per second. Likewise, for every inch of barrel length below 24 inches, the velocity decreases by 20 feet per second. For example, if your rifle barrel is 20 inches long, then subtract 80 feet per second from the manufacturer's stated velocity. There is no such rule of thumb for pistol ammunition.

You also will need to know the bullet diameter. Precise diameters can be found in tables published by ammunition manufacturers. Otherwise you can use the bullet caliber as an approximation. A 30-06 bullet has a diameter of 0.308 inches, but you could use 0.300 inches in your calculations without significantly compromising accuracy.

Calculate the energy the bullet will deliver to the target on impact using the formula

𝐾𝐸=𝑤𝑏𝑣2450,437KE=450,437wbv2
In words, the bullet energy KE (in foot-pounds) is equal to the bullet weight (in grains) wb times the square of the bullet velocity v (in feet per minute) divided by 450,437.

Calculate the Hornady Index of Terminal Standards (HITS) number using the formula

𝐻𝐼𝑇𝑆=𝑤𝑏2𝑣700,000×𝐷2HITS=700,000×D2wb2v
In words, the HITS number is equal to the square of the bullet weight (in grains) times the velocity (in feet per second) divided by the square of the bullet diameter (in inches) divided by 700,000.

This is why energy is also listed on nearly every single box of ammunition you can buy, to guide the hunter is picking a bullet that has the HIGHEST probability of exerting the amount of energy to kill, in addition to making the hole.

Once again, the transfer of energy by projectiles is a fact. The amount is variable based on many factors. To say it does not have an affect is - to use a term folks use here all the time - "fuddlore"

 

[JBradley500]

If that's what you got from what I said you're extra special needs

I think everyone agrees that larger caliber bullets that have the same design and are going the same velocity will cause more damage than the smaller caliber.

The argument here is current bullets in small calibers cause all of the tissue damage you ever need to quickly and ethically kill most North America big game animals, and do it with less recoil (allowing the shooter to make better shots) and can even be in more compact and lighter rifles. Practice will also be cheaper and easier on the shooter.

Moving up in caliber will not make a hunter more effective, and more than likely will cause the average hunter to make worse hits on target and lose more animals.

 

[omicron1792]

How did I miss the point lol? The velocity doesn't not have more of an impact when the velocities of the two different weight projectiles are so similar. Feel free to do the math and tell me what you come up with lol.

Again, shoot an deer/elk with a super advanced .223 that doesn't pencil through, and shoot another one with a .50 bmg using normal ball ammo and tell me what you get. Now imagine using that .50 bmg in the same super advanced projectile as the .233. Not matter what the .50 bmg is gonna be more devastating.

Again, bullet design only does so much to close the gap, and anything you can do to improve the lighter/smaller projectile, you can also do to improve the heavier/larger projectile.

You can kick a scream as much as you want, but the math doesn't lie lol.

You missed yet again. The square means small changes in velocity have huge impacts. Much much more than similar changes in grain size.

ie. A 10% change in velocity has orders of magnitude greater impact on energy than 10% change in mass.

This doesn’t even begin to take into account that 10% increase in size of bullet causes negligible increase in tissue damage or energy transfer to animal. It doesn’t take into account the 25-50% increase in felt recoil that causes less practice, less accurate shooting.

That’s as simple of math as you get. You are dead wrong in this path. Pick another

Swing and a miss

 

[roadrunner]

The correct terminology is the conservation of energy.

Velocity is a dependent variable, mass being the independent variable.

Also something to note; faster moving objects will decelerate more rapidly than slower moving objects. Usually, the objects of less mass that decelerate more rapidly for context are the smaller calibers. Their bullet design will require a higher degree of fragmentation at lower velocites for downrange terminal performance.

 

[WeiserBucks]

@RobHazmat89 have you ever seen a deer shot with a 50 BMG FMJ? There's more than a few videos out there and about 1/2 the time the deer turns and runs as if it had been shot with archery equipment.

 

[PBBananaHammock]

I’m getting the feeling the initial post was not a question of relevance for 30s still being a do-all for new hunters, but is actually a phishing mission to argue the relevance of energy making up for poor shot placement…

That’s one way to inject its relevance, because it certainly hasn’t been a relevant factor or consideration on any animal I’ve shot.


EDIT: sorry that’s not true, I had one that got away 5 years ago. It was a high shot that struck between the vitals and spine. 40 yards offhand, from a substantial elevation. Was it an elevation miscalculation, scope/rings, or did I flinch? Idk, maybe all 3. However, I calculate that interlock hit with more than 2100ft/lbs of energy, at over 2500fps. It didn’t make a difference to that 100lb deer. I messed up and missed. There was no margin of error granted by the 308 and its “sufficient” energy. No hydrostatic shock came to save the day. It wasn’t a reliable wounding mechanism.

However if the bullet had been frangible, it may have had a better chance of hitting the spine with shrapnel. If my rifle had less recoil, I may have gotten a follow up shot off before it could get away. Hitting a desired figure of energy is a waste of mental bandwidth. It does not make up for a bad shot and shouldn’t give any false sense of security that it will.

 

[ARK_Ginger]

May also want to read this.

https://medcoeckapwstorprd01.blob.core.usgovcloudapi.net/pfw-images/borden/conventional-warfare/Chap4Pages13-25.pdf

Then this

This is not directed entirely at you however given your thorough response, which I appreciate, this is directed here as well to everyone else that has provided their opinion.

Saying ballistic pressure is a part of incapacitation is confirming the existence of energy and its affect on the target and tissue. "It exists and does contribute" is all we need to read from that portion of the paper to know. Furthermore the paper continued to reference ballistic pressure waves, calculations of energy, and transmission to surrounding tissue including the brain at impact. Think about when you spine an animal and it drops dead on the spot. The energy transfers through the spine and into the brain, causing a rupture at the medula and instant death.

So whether we discuss higher pressure waves, frangibles, or blunt force trauma - physics has proven through mathematical calculations that have been peer reviewed and verified through multiple outlets too numerous to quote - the existence of it energy in projectiles is proved. The amount is also proved depending on the size and speed of the projectile.


In simple terms, if you shoot an elk in the back hip with a 223 will it kill? Probably not. If you shoot an elk in an a vital area with a bullet that perhaps deviates from its original path, do you have a better chance of a kill by creating a hole PLUS a significant pressure wave - absolutely and without argument YES

Here it is in its simplest form

1. How does energy transfer occur in ballistics? In ballistics, energy transfer occurs through the transfer of kinetic energy from the bullet to the target. When the bullet strikes the target, the kinetic energy is transferred to the target, causing damage or penetration. e.g. "ringing steel"
2. What factors affect the transfer of energy in ballistics? The transfer of energy in ballistics is affected by several factors, including the mass and velocity of the bullet, the density and composition of the target, and the distance between the bullet and the target.
3. How does momentum play a role in ballistics? In ballistics, momentum is the product of an object's mass and velocity. It plays a crucial role in determining the trajectory and impact of a bullet. The greater the momentum of a bullet, the more force it will exert on the target upon impact.
4. What happens to energy and momentum after a bullet impacts a target? After a bullet impacts a target, the energy and momentum are transferred to the target, causing damage or penetration. Some of the energy and momentum may also be transferred to the surrounding environment, such as the adjoining tissue, in the air or in some cases pass to the ground.
5. How is the transfer of energy and momentum in ballistics calculated? The transfer of energy and momentum in ballistics can be calculated using the laws of conservation of energy and momentum. This involves measuring the mass and velocity of the bullet and the target, as well as the distance and angle of impact.

Reference: https://www.physicsforums.com/threa...of-energy-and-momentum-in-ballistics.1014644/

HITS - HORNADY INDEX OF TERMINAL STANDARDS - Yes HORNADY​

You can effectively calculate the impact of a bullet via the following series of steps.

Collect the required information for the ammunition under consideration. You will need to know the bullet weight, measured in grains, which is listed on the ammunition box. You must also know the velocity the bullet is moving at impact. Each ammunition manufacturer publishes ballistic tables for their ammunition. These tables will normally list the bullet velocity at the muzzle of the gun and at intervals of 100 yards out to the useful range of the ammunition (usually 300 yards or so). For distances between the listed increments, you must estimate.

Ammunition manufacturers typically measure rifle bullet velocities using a 24-inch test barrel. As a rule of thumb, for every additional inch of barrel length beyond 24 inches, the velocity increases by 20 feet per second. Likewise, for every inch of barrel length below 24 inches, the velocity decreases by 20 feet per second. For example, if your rifle barrel is 20 inches long, then subtract 80 feet per second from the manufacturer's stated velocity. There is no such rule of thumb for pistol ammunition.

You also will need to know the bullet diameter. Precise diameters can be found in tables published by ammunition manufacturers. Otherwise you can use the bullet caliber as an approximation. A 30-06 bullet has a diameter of 0.308 inches, but you could use 0.300 inches in your calculations without significantly compromising accuracy.

Calculate the energy the bullet will deliver to the target on impact using the formula

𝐾𝐸=𝑤𝑏𝑣2450,437KE=450,437wbv2
In words, the bullet energy KE (in foot-pounds) is equal to the bullet weight (in grains) wb times the square of the bullet velocity v (in feet per minute) divided by 450,437.

Calculate the Hornady Index of Terminal Standards (HITS) number using the formula

𝐻𝐼𝑇𝑆=𝑤𝑏2𝑣700,000×𝐷2HITS=700,000×D2wb2v
In words, the HITS number is equal to the square of the bullet weight (in grains) times the velocity (in feet per second) divided by the square of the bullet diameter (in inches) divided by 700,000.

This is why energy is also listed on nearly every single box of ammunition you can buy, to guide the hunter is picking a bullet that has the HIGHEST probability of exerting the amount of energy to kill, in addition to making the hole.

Once again, the transfer of energy by projectiles is a fact. The amount is variable based on many factors. To say it does not have an affect is - to use a term folks use here all the time - "fuddlore"

I do appreciate you providing papers, it definitely adds more depth to the discussion.

I should probably clarify my position though, I’m not saying energy in a projectile and It’s transfer doesn’t exist. That is obviously false. My argument would be that the raw KE value isn’t super useful for determining wound channels, and that there isn’t much evidence that the hydrostatic shock/pressure waves/temporary cavities cause enough tissue damage to aid in incapacitation times.

I say not useful for determining wound channels because the design of the bullet will determine how much energy is transferred to the target. We see from both of these studies you provided fragmenting bullets transfer more energy than a tougher projectile. Personally, I think the industry would be better off providing ballistics data (like Hornady tap and Federal LE websites) instead of KE numbers on their ammunition. Combine that with measurements on the permanent wound cavity in inch diameter and volume and we’d really have the complete picture.

I haven’t seen a study yet that shows how ballistic pressure waves help in time to incapacitation. I bring that up because it doesn’t seem to me that ‘extra’ energy not transferred is useful, so to me select a bullet that transfers as much energy as you need to kill your selected target. How much energy that is is up to debate since it can be so variable, but I think the 1000ft/lb or 1500ft/lb often quoted isn’t accurate. I’ve attached this clip from that paper to back my point here.