Y'all do realize we're crapping all over the momentum vs. energy thread when we probably should be discussing this in the sectional density thread, don't you? It's like a barroom brawl spilling out into the street and into the building next door.

quote:

We're barely scratching the surface of the subject as far as I can see. If you don't need to learn anything more, by all means ignore the thread.

I'm having fun learning.

More information please...!

Nah, I can never say I don't need to learn anything more, RC.

Where in the equations for downrange momentum or energy, do you put the SD of bullets in a box?

"Momentum Density = Mo/XSA"

You are making science up as you go. That is no good.

quote:

Originally posted by Rat Motor:
"Momentum Density = Mo/XSA"

You are making science up as you go. That is no good.

It's momentum divided by cross sectional area (which is NOT merely the square of the caliber!) Momentum density does look simplistic and maybe like junk science, I agree. But look a little closer.
You have to remember that momentum can also be expressed by a unit of force times a unit of time.
So momentum density would be:

Force / area x time

But wait. What is force / area? It's pressure.

So the momentum density can be further reduced to a unit of pressure times a unit of time.

The next problem then would be determining what the stagnation pressure might be. That probably isn't so easy.

Finding the area of an expanding bullet is what troubles me.

quote:

Originally posted by 900 SS:
Finding the area of an expanding bullet is what troubles me.

That really is a big problem! About the best you can ever hope to do is to estimate it, if you're interested in making a mathematical model. That's why I think that it's never going to be possible to make an entirely accurate mathematical model for penetration. There are just far too many variables involved. All you can ever hope for is an approximation.

quote:

Originally posted by someoldguy:

quote:

Originally posted by Rat Motor:
"Momentum Density = Mo/XSA"

You are making science up as you go. That is no good.

It's momentum divided by cross sectional area (which is NOT merely the square of the caliber!) Momentum density does look simplistic and maybe like junk science, I agree. But look a little closer.
You have to remember that momentum can also be expressed by a unit of force times a unit of time.
So momentum density would be:

Force / area x time

But wait. What is force / area? It's pressure.

So the momentum density can be further reduced to a unit of pressure times a unit of time.

The next problem then would be determining what the stagnation pressure might be. That probably isn't so easy.

It might be well to remember that all of these formulas are mathematical models and as such are simply estimates. In that light how many of us pick a load off the shelf or for reloading and use energy, momentum, shoulder stabilization or stagnation pressure (what ever that is)to select the best load for our intended use. What we do select is bullet weight, velocity, bullet shape and construction and how accurate that load is in our rifles. The others are just an esoteric discussion of academic importance only.

Almost all of the formulas such as energy, momentum etc. are made up of such important things such as bullet weight, velocity and some include caliber. All things that we have direct control over at the loading bench. Non of these formulas would work with out some of these basic ballistic values. Seems to me that makes those values more important than the formulas in a practical sense.

465H&H

It might be well to remember that all of these formulas are mathematical models and as such are simply estimates. In that light how many of us pick a load off the shelf or for reloading and use energy, momentum, shoulder stabilization or stagnation pressure (what ever that is)to select the best load for our intended use. What we do select is bullet weight, velocity, bullet shape and construction and how accurate that load is in our rifles. The others are just an esoteric discussion of academic importance only.

Almost all of the formulas such as energy, momentum etc. are made up of such important things such as bullet weight, velocity and some include caliber. All things that we have direct control over at the loading bench. Non of these formulas would work with out some of these basic ballistic values. Seems to me that makes those values more important than the formulas in a practical sense.

465H&H
------------------------------------------------

Amen

SSR

You'll notice that all of these complicated factors that we are discussing are pretty much taken into consideration in The Optimal Game Weight Formula. Basically take your hunting rifle and load the heaviest weight bullet of good construction that it can stabilize, then go hunting, obviously matching cartridge to game.

For what most of us hunt in North America, you'll be good to go. For dangerous, massive African game that can kill you, I think it wise to keep learning and researching as to what is best.

We've talked about momentum and sectional density as being the major factors in bullet penetration. Of course we haven't scratched the surface of every variable that will have bearing on penetration/killing ability.

Bullet construction, nose profile, the way the bullet acts under drag in the target, just to name a couple that I'm aware of; I'm sure there's more and they've been discussed all over these forums.

The links Gerard posted to the 5 year old thread were interesting, but I'd have to take two weeks off from work to read them and digest their contents

I use these penetration models, etc. not just to give specific results but as a kind of tool to give me some intuition and insight into how a bullet will potentially perform. I think it's good to have an idea, even if it's not precise.

RC, you may have a point about the OGW formula. Some mathematically-inclined hunters seem to like it. I just haven't looked at it that closely, but apparently you have and it matches your own observations.

Dear Gerard,

I find it perplexing that you have not responded to my question below. It deserves some attention.

Your friend, rcamuglia

quote:

Originally posted by rcamuglia:

quote:

Specifically the problem in my understanding of your statement:......


quote:
The penetration depths will differ because of different momentum levels, not becuse of SD or weight.



....is that since velocity is constant, the difference in momentum levels between the two projectiles is directly attributable to the Group 1 factors of weight and Sectional Density.

Are you saying that penetration is ONLY DEPENDENT on momentum and if the lighter aluminum ball is fired at an increased velocity to give it identical momentum to the lower velocity lead ball in the given problem, that penetration will be identical?

rcamuglia,

11 posts up from this one is my reply. I was wondering how you could have missed it.

quote:

Originally posted by Gerard:
rcamuglia,

11 posts up from this one is my reply. I was wondering how you could have missed it.

Trying to find a single verse in the entire Bible can be difficult eh?

quote:

Originally posted by Gerard:
rcamuglia,

11 posts up from this one is my reply. I was wondering how you could have missed it.

I understand that there are many factors that affect penetration and they work in concert or in opposition with each other.

quote:

Theoretically, by increasing the speed (2) of the aluminum (1) ball (1) to 4.5 times that of the lead (1) ball (1), momentum (2) can be equaled. All we can say for sure is that, although momentum (2) will be equal, the other factors (2) will be so far from reality that any attempt at guestimating an outcome would be impossible. Also take note of the 'and so on' above, because the knock on effect of changing a parameter (1 or 2), extends beyond what I have mentioned there by way of example.

I was hoping for a simple "Yes" or "No" answer because your statement.....

quote:

The penetration depths will differ because of different momentum levels, not becuse of SD or weight.

....was very "matter of fact", straightforward, and simple. If the penetration depths differ simply because of the different momentum levels, and not weight and SD as you state, the answer would be "Yes"


I guess I have to answer the question myself.


No.


It's NO because in reality, weight and sectional density play a far more important role in penetration than velocity all else being equal.

It would be fantasy to believe you could first of all, fire the light projectile at 4.5 times the velocity of the heavier one. I'm not sure of the actual weight difference between the AL and PB balls of identical size, but I can resort to reality. Take an actual, real chambering, like the .243 WSSM. The lightest available bullets, 55 grain, can be fired at over 4000 fps. The heaviest, 107 grain?, can be fired at around 3100.

The lightest bullet to be fired at 4.5 times the velocity of the heaviest is impossible in the same rifle because of constraints that are in reality. The gun would blow up well before 13,950 fps could be obtained.

I would have to venture a guess as well that in reality, any bullet fired at that velocity would vaporize.

Remember, we are assuming that neither projectile can deform. We assume the construction of each is perfect so construction is not a factor in the problem.

Chapter 5:



Warrior

So changing the nose profile of two projectiles identical in weight changes the shape of them (length).

M/A is equal to sectional density

The sleeker nose profile has increased penetration due to its better ability to fight drag, but also because changing the nose profile increased SD?

These drawings/charts are obviously theory, not actual tests.

quote:

Originally posted by rcamuglia:
So changing the nose profile of two projectiles identical in weight changes the shape of them (length).

M/A is equal to sectional density

The sleeker nose profile has increased penetration due to its better ability to fight drag, but also because changing the nose profile increased SD?

The use of m and A using Warriors definition of A (wetted surface) is not the definition of SD. SD uses mass and cross sectional area in the formula. I don't know what Warrior would call his m/A but it is not SD. I wonder about his formula though as it would show a straight line relationship between m/A and penetration. The smaller A the greater the penetration. michael458's testing has shown that the realtionship of A to penetration is a bell shaped curve, with the best penetration seen when meplat size is in the high 60's to low 70's of percentage of bullet diameter. Above or below these levels penetration drops off.

465H&H

Last week I went for a walk in the last remaining snows of winter.

As I walked I stepped on the snow, broke through the crust and sank to my knees.

I then remembered put on snowshoes and behold I was walking on the snow instead of in it !

It got me wondering, my mass did not change, nor the velocity with witch my feet impacted the snow, what did change when I put on the snowshoes was the surface area with witch I was making contact tthe snow!

That reminded me of an article I read in the Air Canada Book they have in the pocket in front of your seat.

Something to do with the amazing caribou herds that habit the tundra North of Yellowknife.

Their body mass to footprint area ratio is the second lowest of all the mammals that habit the frozen North, an adaptation so as not to fall through the snow and they believe, in the short summer an adaptation that does the least damage to the delicate tundra Lichens.

Now that was profoundl Perhaps theory but my walking in the snow that was real and not theory! if it were I would still be out there trying to get around

Alf!!!

Welcome back. I have missed your participation although there is much of them I don't understand.

465H&H

Vapodog,

quote:

Trying to find a single verse in the entire Bible can be difficult eh?

This is true. Of course, you can always Google it. It may tell you where to find it but not what the meaning is. These drawings are getting further removed from reality as we go along.

If the shape of the bullets tip had anything to do with SD then a (perfect) roundnosebullet would have indefinite SD.

All this is much easier to understand if we forget the SD value and consentrate on the bullets penetration-resistance in whatever media. That would make expansion characteristics and shape more important. SD dont tell me anything important without knowing a lot of other factors to consider how the bullet will perform.

If I already know bullet construction, velocity, mass and caliber, calculating SD doesnt give more meaningful information.

As an example: adding a knife to the buttstock doesnt alter SD of the rifle significantly. But the shape of the knife vs. a rubberpad will affect felt recoil.

quote:

Originally posted by 465H&H:
Alf!!!

Welcome back. I have missed your participation although there is much of them I don't understand.

465H&H

Yeah!!!, warrior is getting behind in transcribing things he does not understand(I wonder "who" he learned that from )

Hopefully you can, "Dumb it down!" for all of us.

quote:

Originally posted by ALF:

quote:

Originally posted by rcamuglia:

quote:

Originally posted by ALF:

Thank you, ALF!

quote:

As I walked I stepped on the snow, broke through the crust and sank to my knees.

I then remembered put on snowshoes and behold I was walking on the snow instead of in it !

It got me wondering, my mass did not change, nor the velocity with witch my feet impacted the snow, what did change when I put on the snowshoes was the surface area with witch I was making contact tthe snow!

I have been saying this since I was 7.
This is better than any bullet experiment.



Warrior

quote:

Originally posted by Warrior:
This is better than any bullet experiment.

Warrior

If you want to avoid bullets falling through the snow.

quote:

Originally posted by Warrior:

quote:

As I walked I stepped on the snow, broke through the crust and sank to my knees.

I then remembered put on snowshoes and behold I was walking on the snow instead of in it !

It got me wondering, my mass did not change, nor the velocity with witch my feet impacted the snow, what did change when I put on the snowshoes was the surface area with witch I was making contact tthe snow!

I have been saying this since I was 7.
This is better than any bullet experiment.



Warrior

Warrior,

How often do you walk on snow in South Africa?



465H&H

quote:

Originally posted by 465H&H:


Warrior,

How often do you walk on snow in South Africa?



465H&H

Now that Alf has had his fun with his examples and strung some of you along, we can examine the facts within the examples.

Snowshoes:

1. Without snowshoes, Alf falls through the surface of the snow.
2. With snowshoes, which Warrior has been using since age 7, Alf does not fall through the snow surface.
3. Alf has not increased his weight/mass. Alf has not increased his waistline/diameter.
4. SD is weight divided by diameter squared. Therefore, Alf has not increased his SD.
5. If Alf's SD has not changed, why does he not suffer from snowballs when he wears snowshoes?
6. Alf has not changed the speed with which his feet impact the surface, so the momentum and energy levels are the same. He said so himself.
7. What has changed is the shape with which the snow surface is impacted. Instead of Alf's widdle pointy feet, the shape is now big and flat.
8. As a result, the cross section over which Alf's momentum is applied to the snow has changed.
9. So, the fact that Alf no longer falls through the snow-surface, is because momentum/cross section has changed as a result of the change in footsole shape.
10. Nothing to do with SD, everything to do with the change in surface area to resist the momentum applied to the snow surface.

Knife:

1. To be more accurate in this analogy, in picture 1 we ask Alf to cut off the tip of the blade and tape it to the other side of the stock. This is to keep SD of 1 and 2 identical.



2. The weight of the two kniferifles are therefore the same and the diameters are the same.
3. Therefore, the SD of the two kniferifles are the same.
4. It is clear that kniferifle 2 will penetrate deeper than kniferifle 1, if both are put into backwards motion at the same speed.
5. If the SD, momentum and energy of the two kniferifles are the same, why will 2 penetrate better than 1?
6. The shape is different. Could it be that the shape change causes the difference in the momentum applied to the cross section, thereby increasing penetration?
7. Once again it is proved that SD remains unchanged and, that the important change that increases penetration, is the change in shape.

Alf, are you baiting Warrior again? You know he agrees with everything you post. Like Pavlovs dogs, he cannot help himself. Shame on you.

It's a strange, strange world we live in, Master Jack.



Warrior

No technical input, no opinion, just GoogleFu. What is the problem? Has Alf stopped feeding you charts?