quote:

If the technical stuff bores you or you do not understand it

Gerard,

I suggest you do the same.

I have no argument with what Norbert has to say.

It just bothers me when people start making up shit, for whatever reason.

OK, so you do not understand the technical stuff. Now I know not to respond when you ask a question.

.

Not to steal Geralds thunder but to see if he concurs. I would postulate the following.

The smaller diameter bullet being driven by the exact same force as the larger one creates a higher pressure upon striking the object. This pressure is more destructive to the target object allowing the smaller bullet to penetrate further.

For the same reason a sharper knife cuts easier than a dull one.

Now penetration is a different matter entirely than wound channel diameter. Killing effectiveness is Wound channel Diameter and Penetration together.

Big Guy,

It sounds good at first blush, but I'd disagree a bit. FIrst, the smaller bullet has that same pressure acting on it, so the softs would deform more, hindering penetration. With the "more destructive to the target object" portion, when that target object is an animal there's a lot of elasticity to overcome, so the destruction might not be as much as with something with a larger surface area.

My own backyard experiments show me it's all about sectional density and bullet construction.

The higher the sectional density - the deeper the penetration - all things being equal. The only time that formula doesn't work is with "conventional" lead core bullets. They penetrate far deeper when they are going so slow that they don't expand (around 1900) or so, for most of them.

quote:

Originally posted by ALF:
Gerard:

Can I ask you a question:

If we shoot a 400 gr 45 cal non frangible, non deforming bullet at 2000 fps into a uniform test medium and we shoot another bullet 400 gr of 40 cal of similar construction at 2000 fps into the same test medium why does the 40 cal bullet penetrate deeper than the 45 caliber bullet ?

Both have equal momentum as momentum is defined as mass x velocity.

What is the physical difference between these projectiles that cause the one to outpenetrate the other under identical physical conditions?

Alf,

I gave you the formula the other day. Because the 416 bullet has a higher sectional density than the 45 cal. bullet.

But it only takes a lathe and some round stock to make you a scientist!

.

Alf,

quote:

If we shoot a 400 gr 45 cal non frangible, non deforming bullet at 2000 fps into a uniform test medium and we shoot another bullet 400 gr of 40 cal of similar construction at 2000 fps into the same test medium why does the 40 cal bullet penetrate deeper than the 45 caliber bullet ?

Both have equal momentum as momentum is defined as mass x velocity.

What is the physical difference between these projectiles that cause the one to outpenetrate the other under identical physical conditions?

I am not Gerard, but to answer your question. The .40 caliber out pentrates the 45 caliber even though the have the same momentum, because the 40 caliber has a high momentum density.

Now I have never really understood this fanscination with "momentum". What really describes penetration is "FORCE density". "Force density" has a more common name, STRESS. Specifically with regard to penetration it is "shear stress". Force is the change in momentum with respect to time (see Norbert's post were he posts this differential equation). Shear stress is force divided by cross sectional area. I figure the light bulbs are lighting up nice and bright for you by now (if they weren't already lit).

ASS_CLOWN

Ah, so, some red herrings were inserted here and there in the foregoing discussions. I have a bit better grasp of stagnation pressure significance related to soft and solid bullets. Thanks to Gerard and Norbert. It has been more than 25 years since I studied fluid mechanics in my former chemical engineering curiculum, so please be patient. I haven't worked with these concepts for a quarter century.

Norbert, you should point out that V x SD is proportional to (not equal to) momentum density due to the arbitrary/artificial definition of SD related to square bullet cross section.

And It is said that some bullet yaw does not settle down until 80 yards? Not just a few inches? Maybe on average about 20 yards? That was the former gossip.

Norbert,
What was the G11 and how did they verify that it went to sleep at 80 yards?
Good to hear that increased twist decreases yaw, or speeds the damping of it.

Gerard, give Will a break. He does not mean to be a grouch.

Alf,
Two non-deforming solids of same weight and different caliber, and same form factors otherwise: say a .416/400gr FN/FP versus a .458/.400gr FN/FP ....

The .416 will have a higher momentum density AND less resistance to penetration due to smaller frontal area, so obviously the drag from the sides of the bullet is negligible, and is reduced by cavitation.

Only if stresses are such that the longer bullet bends will the .416 fail to out penetrate the .458 when both weigh 400 grains.

Is there a trick answer or trick question here? Seems obvious.

.

Alf,
Capstick said it best, about Professional Small Boys. Hope I can always be one.

The synthesis of all this:

Use a monometal copper FN/FP, get as close to 0.3 SD as possible, not much more, and go as fast as tolerable to both the shooter and bullet, but use at least a 300 grain .375 caliber bullet to turn out the lights.

(The smiley below applies to the following posts.)

Alf,
Asked and answered on that other thread. Remember I added a .50 cal 400 gr bullet at 2000fps to the sample. But to repeat the reply, TheBigGuy is on the right track and in practise it is called momentum density or momentum/XSA.. When same caliber examples are compared, the XSA can be dropped and momentum be used on it's own for the comparison. When dissimilar calibers are compared, it has to be with momentum/XSA..

Now that I have answered yours, tell me this. Which of two identical solids would go deeper if one was 100fps faster than the other and why?

RupertBear,
Alf's example was for non-deforming bullets but in essence you are not wrong.

BCBrian,
Can you give us some examples?

(Cancel smiley.)

Will,
You are wrong.

(re-apply smiley)

Alf again,
Have another look at Poncelet where you posted it, he uses mass, cross sectional area and velocity. That is not Sd.

AC and RIP,
Thank you for posting it was getting lonely here.

Alf,
Just when I thought I should tell you to listen to RIP, he goes and insists on that infernal Sd thing again. This is like herding a bunch of cats.

.

.

Geez, Bekker boo boo:

V x SD = Mo/XSA

Wrong!!!
Does anyone else understand this besides me???

Alf, I will discuss your post above fully, but only after you have given me this from my post above.

"Now that I have answered yours, tell me this. Which of two identical solids would go deeper if one was 100fps faster than the other and why?"

Gerard,
Of course the 100 fps faster one would penetrate a little deeper.

Use sectional density as a crutch, an easy shorthand. It has no real units or meaning. Too much SD is bad.

Gerard,
why do you make stupid my comments?

quote>> Let us see you do some expanding bullets and then tell me again that stagnation pressure has nothing to do with penetration.<<
I excluded expanding bullets to give an easy understandable comment!

quote>> It is of course true that two bullets with exactly the same momentum density number will penetrate equally. This means that a 300gr .375 bullet at 2500fps will penetrate exactly to the same depth as a 270gr .375 bullet at 2780fps if niether deform. Of course the Sd of the 300gr bullet is .305 and the 270gr bullet is .274 but we see here that Sd is no longer relevant.<<
??What is that for an argument? If you accept momentum density, it must also SD. Or is the velocity also irrelevant? This comment is rabulistic BS.

quote>> Norbert, you must also be careful about giving out technical information about your experience and the bullets you have designed and sell. Will does not like it and believes it is a ploy of yours to sell more bullets.<<
This is an imperdinent comment. I am not acting like you, giving always reference to your FN and HV bullets which you sell.
I do not manufacture bullets.
I do not sell bullets.
The results of my research are open to the public and everybody can use it and implement it in his manufacturing of bullets.

Cutting through the BS and therory it boils down to this:

You can drive a monolithic as fast as you want and the faster the better for penitration in most cases..The same appears to apply to a cross bred bullet like the Northfork, a half monolithic, half bonded core, it seems to penitrate better at high velocity and certainly will not come apart...

A conventional bullet will many time penitrate better at slower speeds, but not always...

Many varibles exist but basically the above applies...Nothing on this subject is written in stone, and a few kills proves zilch, time will tell the real story.

Which of these bullets would penetrate best, and which is most suitable for a thick skinned animal:

500 grains at 2000 fps?

400 grains at 2500 fps?

350 grains at 2857 fps?

Assume these are 458 caliber non-deforming bullets. They all have the same momentum. The only difference is how much they weigh.

I am not trying to get anyone's goat. I think everyone who cares about his goat has left the room.

At some point, as this hypothetical bullet gets shorter and shorter, I am starting to worry not so much about the absolute distance it penetrates, but also about the straightness of its path within heterogeneous tissue.

At some point, I expect a shorter, faster bullet to start flying like a frisbee and missing the vital organs I aimed at, even if the bullet does penetrate several feet of meat and bone before stopping or exiting.

H. C.

RIP,

you wrote:
quote>>Norbert, you should point out that V x SD is proportional to (not equal to) momentum density due to the arbitrary/artificial definition of SD related to square bullet cross section.<<
I pointed out this in different AR threads and on my website. Alf elaborates on this above. It is somewhat arbitrary with respect to physical units, but: Hunters buy either low-SD bullets or high-SD bullets over the counter and not velocity or momentum.

It is boring to bring same facts in AR forums again and again!!

quote>>And It is said that some bullet yaw does not settle down until 80 yards? Not just a few inches? Maybe on average about 20 yards? That was the former gossip.<<
Hunting bullets will come to rest below 20 yards. But their angle of yaw can amount to several degrees. Mean value 2° to 6°. Extreme: Tungsten core .308 Nato: 10°.

quote>>What was the G11 and how did they verify that it went to sleep at 80 yards?>>
It was a Nato development, replacement of .308 G3 and the american .223 M16. A small calibre combat rifle with caseless ammunition. They tested it at the proving grounds in germany. They stopped it, too many problems. Now H&K supplies a .223 more conventional rifle, G 36 ? or so.

quote:

Originally posted by Atkinson:
Cutting through the BS and therory it boils down to this:

You can drive a monolithic as fast as you want and the faster the better for penitration in most cases..The same appears to apply to a cross bred bullet like the Northfork, a half monolithic, half bonded core, it seems to penitrate better at high velocity and certainly will not come apart...

A conventional bullet will many time penitrate better at slower speeds, but not always...

Many varibles exist but basically the above applies...Nothing on this subject is written in stone, and a few kills proves zilch, time will tell the real story.

Ray,

You say it. I always wonder, why so much posts are generated to this simple question. And so much BS included.

H.C. :

quote:

Originally posted by HenryC470:
Which of these bullets would penetrate best, and which is most suitable for a thick skinned animal:

500 grains at 2000 fps?

400 grains at 2500 fps?

350 grains at 2857 fps?

Assume these are 458 caliber non-deforming bullets. H. C.

This is the dilemma I described in a post above.
Under ideal condition there would be no difference in penetration, all parameters else are equal.

But: With our hunting cartridges it is impossible to compensate mass with velocity in safe pressure limits. Only the 500 grs at 2000 f/s is a realistic choice.
To achieve the other velocities the pressure would be as follows:

500 grains at 2000 fps? 3600 bar

400 grains at 2500 fps? 4600 bar

350 grains at 2857 fps? 5800 bar

The maximum pressure (piezo) allowed is 4300 bar, copper crusher 3800 bar.

quote:

Originally posted by Norbert:
RIP,
It is boring to bring same facts in AR forums again and again!!

Jawohl,
but this is no excuse for a false equation:

velocity x Sd = Momentum density

is meaningless in terms of real units, and you agree. You should simply say it is proportional to or directly proportional to, not egual to.

Henry,
You are undoubtedly right.
It is easier to get high momentum out of high Sd bullets, than to load the low Sd bullets to equal momentum at excessive pressures/velocities.

Even the "useless" Sd term is good guidance. Around 0.3 is all that is needed, to maximize momentum and minimize bending of overly long bullets.

Get used to this compromise use of Sectional density, Gerard. As "meaningless" as the term is, it is still a "useful approximation" for some purposes.

Norbert,

Thanks for concisely clearing than one up for me. I sometimes get a headache trying to follow the mixture of logic and other substances in these "technical" threads.

H. C.

RIP,

Norbert's use of the "equals" sign in his definition of momentum density is absolutely correct, and there is no need to use qualifiers or the term "proportional". Momentum density is mass times velocity divided by cross sectional area. The units your scale and chronograph and dial caliper are calibrated in do not change the momentum density of your bullet. They only change the number you wrote down after you calculated it.

There are no "real units". A person may prefer one set over another, but it doesn't change the quantity being measured.

You can't make yourself more manly by measuring in centimeters, and if a girl lists her weight on a dating service as "100", you wouldn't know whether or not to actually show up for a blind date.

H. C.

quote:

Originally posted by HenryC470:
RIP,

Norbert's use of the "equals" sign in his definition of momentum density is absolutely correct, and there is no need to use qualifiers or the term "proportional". Momentum density is mass times velocity divided by cross sectional area. The units your scale and chronograph and dial caliper are calibrated in do not change the momentum density of your bullet. They only change the number you wrote down after you calculated it.

There are no "real units". A person may prefer one set over another, but it doesn't change the quantity being measured.

You can't make yourself more manly by measuring in centimeters, and if a girl lists her weight on a dating service as "100", you wouldn't know whether or not to actually show up for a blind date.

H. C.

Henry: Hooey. You are wrong.

Momentum density is momentum divided by true cross sectional area of the bullet.

Sectional density uses an imaginary square bullet. Artificial. Contrived. You do not understand.

quote:

Originally posted by RIP:
Sectional density uses an imaginary square bullet.

Nope. Sectional density is calculated using cross sectional area. Convenient units are square inches or square centimeters. We could use hectares instead. That doesn't make the bullets square.

H. C.

quote:

Originally posted by HenryC470:

quote:

Originally posted by RIP:
Sectional density uses an imaginary square bullet.

Nope. Sectional density is calculated using cross sectional area. Convenient units are square inches or square centimeters. We could use hectares instead. That doesn't make the bullets square.

H. C.

Henry! You are showing a fundamental misunderstanding of the definition of sectional density!

Sd = (Bullet weight in pounds)/ (bullet diameter in inches)^2
Sd = (bullet weight in grains/7000 grains per pound)/caliber squared

It is a quick and dirty convention that avoids the use of A = (pi)r^2

Remember "pie are squared" ???

Sectional density's true definition is as if the bullet had a square cross section, and the square is a caliber diameter per side.

Units of sectional density have no physical reality, but they are directly proportional to the bullet weight divided by the true bullet cross-sectional area.

Calculate some Sectional densities for yourself. Try a 300 grain .375 caliber bullet.

You will have to use a square bullet to get the proper answer of Sd = 0.305

End of lesson.

Norbert,
Some still do not understand the definition of sectional density. Henry will listen to you, not me. Straighten him out please.

Yes Norbert, please straighten Henry out. "sectional density" aint cross sectional area. It ain't density either since the "mass" term is actually a weight.

grains divided by 7000 = weight NOT mass.

Ron, "sectional density" does in fact have units. They are pound-force / inches^2.

I do not like the term "sectional density" becuase it is mathematically INCORRECT. This FACT does not mean that "Sectional Density" does not follow the same trend line, with regard to penetration, as the mathematically correct momentum density

Penetration of the so-called "non-deforming" bullet (doesn't exist a more accurate description would be NON-PLASTICALLY DEFORMING) is easy!

Therefore, I recommend a new more interesting discussion. The penetration characteristics of the jacketed soft pointed expanding bullet. How is everyone at solving higher order differential equations? I guarantee you all a discussion on expanding soft point's penetration will separate the "men from the boys"!

ASS_CLOWN

.

quote:

Originally posted by RIP:
Norbert,
Some still do not understand the definition of sectional density. Henry will listen to you, not me. Straighten him out please.

I understand that the mass of a bullet (in whatever units) divided by its cross sectional area (in whatever units) is the bullet's sectional density. I understand that for convenience, the sectional density is commonly computed using diameter rather than radius and with the omission of "pi".

I (and not you) understand that this amounts to reporting sectional density in units of pounds per square inch that are off by a factor of pi/4 from doing it with pi in there and everything. I (and not you) understand that this amounts to a change in unit of of measure and does not require the use of real or imaginary square bullets.

The square bullets are all in your mind. If they help you do the math, that's fine. They are not making you get the wrong numerical answer. The fact that you think about square bullets when you do math problems does not render the concept of sectional density any less useful in guessing how well those bullets will penetrate.

H. C.

Alf,

I have seen and used a number of military derived penetration (armor) equations. They tend to be ponderous.

Your comments in the last thread intrigue me, if it isn't too much trouble would you mind posting the equations used to determine penetration using sectional density.

This is a fundamental fact of penetration. In order to penetrate the target the shear stress imparted to the target by the projectile MUST exceed the shear strength of the target. Penetration ceases when the projectile's force/cross sectional area (imparted shear stress) can no longer exceed the shear strength of the target.

I could care less about momentum. What I care about is force. Force divided by area is stress, stress penetrates when it exceeds target's strength. That is fundamental, and fact.

Like I said before I have used armor penetration equations in the past and they tend to be ponderous. Many date back to a time when it was practically impossible to measure the forces, time, etc occuring during the penetration event. As such they contain lots of, for lack of a more politically correct term, fudge factors to get the mathematical model's results to match up with the REAL world results of experiments.

I know with regard to armor penetration, most (all the equations I can remember right now but memory cannot be trusted anymore) rely upon momentum density. Penetration in soft targets may be "fudge factored" to SD.

Anyway, if you could post the equation(s) I would greatly appreciate it.

Henry,

RIP is right.

Area of a square = side X side or side^2

Section density uses bullet diameter^2 . In effect providing the area of a square with sides the length of the bullet's diameter.

ASS_CLOWN

AC,
Thanks for the help with Henry. Maybe he will believe you?
However, you must admit that the sectional density units (of pounds force or pounds weight per inch squared) have no physical reality, by definition, because THE BULLET IS ROUND NOT SQUARE IN CROSS-SECTION!!!

Sectonal density = W/d^2

This is from Hatcher's Notebook, where W = weight of bullet in pounds and d = diameter of bullet in inches.

Alf,
I have taken "momentum density" bandied about here for granted as being = MV/(pi)r^2
where r = radius of bullet shank or (1/2)d.

I will pay closer attention. Where is Harold when you need him? I'm looking into this too.

quote:

Originally posted by HenryC470:

quote:

Originally posted by RIP:
Norbert,
Some still do not understand the definition of sectional density. Henry will listen to you, not me. Straighten him out please.

I understand that the mass of a bullet (in whatever units) divided by its cross sectional area (in whatever units) is the bullet's sectional density. I understand that for convenience, the sectional density is commonly computed using diameter rather than radius and with the omission of "pi".

I (and not you) understand that this amounts to reporting sectional density in units of pounds per square inch that are off by a factor of pi/4 from doing it with pi in there and everything. I (and not you) understand that this amounts to a change in unit of of measure and does not require the use of real or imaginary square bullets.

The square bullets are all in your mind. If they help you do the math, that's fine. They are not making you get the wrong numerical answer. The fact that you think about square bullets when you do math problems does not render the concept of sectional density any less useful in guessing how well those bullets will penetrate.

H. C.

Henry,
Do the math and prove it to yourself.
All the quoted sectional densities quoted everywhere are as if the bullets are squares of one caliber on a side. You are being childish about this, or just plain ignoring facts. That is ignorant.

AC,
Yes a fudge factor applied to a sectional density dependent term would make things square with reality. 4/pi or pi/4, depending on the direction of correction needed, would figure into this fudge factor, as Henry has pointed out.

Where is the definition of Momentum Density?