1. Objects penetrating a surface

Here is another question I've actually had for a while now. It came up when I was trying to make a pen and paper role playing game, but it's come up outside of that a few times. It relates to a moving object penetrating a stationary one. I know that a portion of that energy will be turned to heat as one or both deform. I also know that the math on that part is probably way over my head (two years of algebra I in high school). Also, the tiny bit of physics I know I learned on my own and none of it was in metric units.

So, here's the situation. I was trying to make a spreadsheet to compare all different types of projectiles, from bullets, to arrows, to baseballs in a consistent manner. I could figure the energy using mv^2. From there, I got the momentum (mv) and divided it by the surface area (pi()r^2). My thinking was that the momentum/surface area would be a good indicator of how well something would penetrate (ignoring deformation). The analogy was that a 10 pound pick falling on the dirt would go in much further than a 10 pound hammer, due to the smaller contact area.

First question, is this a valid method to get a ball park comparison?
Second question, would it be better to use the energy or the momentum if this is a valid method.
Third question. If I use ballistic units, say
Energy= 400 ft-lb
Mom= 21 Pounds/Feet per second (?) 0.016 pounds/ 1280 fps
Area= .100 in^2
Would the units for either Energy/Area or Momentum/Area be in pounds per square inch?

Some explanation about some of my recent oddball questions. I signed up over at the Mythbusters site a while back and they have a section on weapons. Most of the stuff people post are pretty easy to confirm or refute, but now and then I get something that has me wondering, whereas other I kow the answer but can't really explain it with out saying "trust me". That's what brought this one up. Some people are insisting that an Airsoft gun can be lethal. It fires a 0.2 gram plastic pellet at a speed of about 100 meters per second tops.

BTW, their forum software sucks so hard it bends light.

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My guess is the errors are large, if we we assume penetrating styrofoam, diamond and rubber can be calculated the same. Some of the kinetic energy changes the molecular arrangement. Some of it heats the target, projectile and nearby air, and some of it is converted to sound waves. At high enough speed some of the energy is released as photons as in a flash of light. Neil

3. But if we make the surface a single material, let's say the end of a pine log, when the point of entry is into the end, using the rings as a target; and the composition of the bullets or pellets is the same for all sizes and speeds, would the method described be correct?

This also assumes that all bullets are full metal jacketed round nose, so no variation for a hollow point deforming more than an armor piercing would apply.

4. I believe one standard is a set thickness of "homogenous steel", and you can then pretty much guess the rounds characteristics by how much it will penetrate at X distance. Like a round that goes through a quarter inch at 300 yards is different from one thats goes through an inch at 300 yards.

(if I understand the question correctly.) I believe the power needed to penetrate different thickness' of "homogenous steel" is available somewhere.

5. Thanks BigDon. Sort of what I was after, but not quite.

Let me try to say it another way. Let's say we have a sheet of magic plastic. This plastic is 100% rigid and cannot be bent by anything, regardless of temperature changes. It has a "puncture rating" of 100kg/square cm. this means that if any single square cm has more than 100 kg resting on it, it will poke a hole in the exact shape of whatever is resting on it.

For the first test, we make a little metal table that has 4 legs, each with a foot that is exactly .25 square cm, for a 1 square cm contact area on the surface. This table has a mass of 1 kg and is made if some sort of impervium-handwavium alloy.

If we set an additional 98.9 kg on the table, it will hold, but of the combined weight of the table and whatever it's supporting goes over 100 kg, the pressure from the feet will push through the magic plastic. That's the set up.

Now, we drop a 1 kg weight from a whatever height it needs to reach 10 m/s. The kinetic energy from this would be 10*10*1/2 or 50 joules. As I understand it, this means that the weight will have a momentum of 10 kg m/s. If those two values are right, and the able still rests on the sheet with a 1 square cm area of contact, is it right to say that the pressure felt by the sheet is 10 kg/cm^2? If not, what would the pressure felt by the sheet be, and how could I determine it?

6. Originally Posted by Tog_
Thanks BigDon. Sort of what I was after, but not quite.

Let me try to say it another way. Let's say we have a sheet of magic plastic. This plastic is 100% rigid and cannot be bent by anything, regardless of temperature changes. It has a "puncture rating" of 100kg/square cm. this means that if any single square cm has more than 100 kg resting on it, it will poke a hole in the exact shape of whatever is resting on it.

For the first test, we make a little metal table that has 4 legs, each with a foot that is exactly .25 square cm, for a 1 square cm contact area on the surface. This table has a mass of 1 kg and is made if some sort of impervium-handwavium alloy.

If we set an additional 98.9 kg on the table, it will hold, but of the combined weight of the table and whatever it's supporting goes over 100 kg, the pressure from the feet will push through the magic plastic. That's the set up.

Now, we drop a 1 kg weight from a whatever height it needs to reach 10 m/s. The kinetic energy from this would be 10*10*1/2 or 50 joules. As I understand it, this means that the weight will have a momentum of 10 kg m/s. If those two values are right, and the able still rests on the sheet with a 1 square cm area of contact, is it right to say that the pressure felt by the sheet is 10 kg/cm^2? If not, what would the pressure felt by the sheet be, and how could I determine it?
We cannot calculate the force exerted on a target by a moving projectile without knowing the distance the projectile moves after the moment of contact and before coming to a stop.

If your magic plastic is perfectly rigid, meaning it either holds or breaks but does not bend, the thought experiment blows up if the projectile also is rigid. Stopping the projectile instantly would require a momentarily infinite force, and the plastic would be punctured no matter how slowly the projectile is moving. At least one object has to be deformable.

Your plastic can support a gravitational static load of up to 100kg per square cm. That is a force of about 1000 newtons, to use standard metric units of force and rounding Earths gravity to g = 10m/sec^2.

Suppose your falling kilogram is going 10m/sec. Its kinetic energy is K = 50kg m^2/sec^2
Maximum resistive force F = 1000kg m/sec^2
Force*distance = change in energy

Thus the force has to be applied over a distance of 1/20 meter to stop the projectile.

If the falling mass is soft gob of putty that is over 10cm long, and it flattens out into a pancake with its center of mass stopping in 5cm, and it does so smoothly enough, your plastic would just be able to stop it without being punctured.

More thoughts on this topic later.

7. Let's extend our thought experiment. Suppose the target is now a block of magic styrofoam-like stuff that has the same load bearing strength of 1000nt/cm^2. If a projectile has enough punch to penetrate it, the block exerts an opposing force of that same amount until the projectile is stopped. Now a rigid 1kg projectile with a frontal area of 1cm^2 will penetrate 5cm if it hits at 10m/sec.

A real target probably would not have the same resistance at different velocities of penetration. We would have to evaluate it empirically by shooting real bullets into it.

8. Originally Posted by Hornblower
Let's extend our thought experiment. Suppose the target is now a block of magic styrofoam-like stuff that has the same load bearing strength of 1000nt/cm^2. If a projectile has enough punch to penetrate it, the block exerts an opposing force of that same amount until the projectile is stopped. Now a rigid 1kg projectile with a frontal area of 1cm^2 will penetrate 5cm if it hits at 10m/sec.

A real target probably would not have the same resistance at different velocities of penetration. We would have to evaluate it empirically by shooting real bullets into it.
Okay this more like what I was looking for, I just had the set up completely backwards. Go me.

Let me see if I understand the basic bits of this. Say we have a 10 gram bullet moving at 1000 m/s, and has a frontal area of 0.5 cm^2.

1 Newton = the force needed to accelerate 1 kg at a rate of 1 m/s. The force of the powder in the cartridge exerted a force of (0.01*1000) 10nt on the bullet. Is that right?

1 Joule= the energy carried by a moving object (mv2)/2, or 5000j for our bullet.

The magic foam can resist a force of 10nt/cm^2, and the bullet carries a force of 20nt/cm^2, so the bullet will penetrate 2 cm of the foam. Is any of that right? Also, the energy in Joules doesn't matter a bit?

9. The trick to not mess up your powers of 10 when doing metric is to start in kg/m/s, once you remember that, you can stop worrying.

Incidentally, the symbol for Newton is N, not nt.

Bullet 1 has a mass of 4*10-3 kg and a speed of 1000 m/s with a surface area of 0.25*10-4m2. (M-16)
Bullet 2 has a mass of 25*10-3 kg and a speed of 400 m/s with a surface area of 1*10-4m2 (.45-70)

For deceleration, we've been using an ideal material which regardless of deformation and velocity applies a constant pressure on the surface of the projectile of 10N/cm2=1*105Nm-2 (very low incidentally, it's equal to the atmospheric pressure)

This gives deceleration forces of
F1=1*105Nm-2*0.25*10-4m2=2.5N
F2=1*105Nm-2*1*10-4m2=10N

This gives deceleration of
a1=F1/m1=2.5N/4*10-3kg=625ms-2
a2=F2/m2=10N/25*10-3kg=400ms-2

This means, following d=½v2/a, that penetration in this hypothetical material will be:
d1=½v12/a1=½*10002m2s-2/625ms-2=800m
d2=½v22/a2=½*4002m2s-2/400ms-2=200m

10. Originally Posted by HenrikOlsen
This means, following d=½v2/a, that penetration in this hypothetical material will be:
d1=½v12/a1=½*10002m2s-2/625ms-2=800m
d2=½v22/a2=½*4002m2s-2/400ms-2=200m
This is the part I'm having a hard time understanding. Real world testing (in the form of hunting bullets) has shown that small, light bullets are effectively useless against large animals because they don't penetrate deeply enough. the 5.56mm bullet from an M-16 is considered too ineffective for deer where I live, but the .45-70 was used for decades to hunt buffalo. A lot of that may have to do with the energy of the bullet being used to deform it, which is a lot more than I would be able to get into, but if the resistance of air is around the 1*105Nm-2 used here, and the penetration of the second bullet is only 200m, then wouldn't that mean that the bullet would have come to a stop long before say 500 m?

11. First of all, remember that in reality nothing acts like the hypothetical material used in the calculations, at the very least there's a velocity dependent part, easily seen since the force on the bullet when stopped is 0, this would probably mean the faster bullet has a higher pressure stopping it initially, making the penetration relatively shorter.

From what I understand, there's two ways a bullet can kill, one is from the hydrodynamic shock-wave triggering cardiac arrest, the other is when the passage through the body sever blood-vessels or nerves.
For the latter case, a wider hole will give a higher chance of intersecting interesting things, which is the point in expanding or disintegrating bullets.

I would guess the problem with the 5.56mm for hunting is that it makes a too narrow wound so don't intersect enough bloodvessels.

12. Originally Posted by HenrikOlsen
I would guess the problem with the 5.56mm for hunting is that it makes a too narrow wound so don't intersect enough bloodvessels.
Yes, those are the two main reasons bullets and bodies don't mix well, but the problem with a small bullet used for hunting is simply that the bullets won't go in deep enough to do any of that stuff on a larger animal.

All of the hunting books say basically the same thing. Energy is not the important factor to consider. Momentum is. While a small, fast bullet might make a something like a soda bottle explode from the shockwave moving through the liquid, it won't make it through many of those bottles before stopping, whereas a larger bullet will, even if it's moving more slowly.

The definition used by the books in this case is the bullet mass in pounds/the speed in fps, so kg*(m/s). For the 5.56 round, the momentum would be 28, vs. the 65 of the .45-70. This made sense to me, since it seemed that a bigger bullet would be more resistant to slowing down.

Then I tried a baseball. 6 ounces at 150 fps which is 56 which really doesn't seem right. As I understand it, that means a person in a bulletproof vest would feel twice the impact force from the baseball than from a 5.56 bullet.

So I guess the logical question first off would be, am I doing Momentum right? Mass times Speed?

Second, does momentum matter at all, and if not, why not?

13. Originally Posted by Tog_
Yes, those are the two main reasons bullets and bodies don't mix well, but the problem with a small bullet used for hunting is simply that the bullets won't go in deep enough to do any of that stuff on a larger animal.

All of the hunting books say basically the same thing. Energy is not the important factor to consider. Momentum is. While a small, fast bullet might make a something like a soda bottle explode from the shockwave moving through the liquid, it won't make it through many of those bottles before stopping, whereas a larger bullet will, even if it's moving more slowly.

The definition used by the books in this case is the bullet mass in pounds/the speed in fps, so kg*(m/s). For the 5.56 round, the momentum would be 28, vs. the 65 of the .45-70. This made sense to me, since it seemed that a bigger bullet would be more resistant to slowing down.

Then I tried a baseball. 6 ounces at 150 fps which is 56 which really doesn't seem right. As I understand it, that means a person in a bulletproof vest would feel twice the impact force from the baseball than from a 5.56 bullet.

So I guess the logical question first off would be, am I doing Momentum right? Mass times Speed?

Second, does momentum matter at all, and if not, why not?
It is vitally important to remember that our recent thought-experiment exercises were done with simplistically idealized targets whose assumed penetration-resistance characteristic probably does not resemble that of the torso of a bear or a lion. The latter may well have more drag at higher velocities, which I would expect to adversely affect the faster but lighter bullet. In addition, real gunpowder pushing a real bullet through a real barrel that has friction might not get a lighter bullet going enough faster to make up for its lesser mass.

Hunters and others who make informed choices of ammunition might not know diddly about textbook physics, even at the entry level, and they might use terms such as force and momentum incorrectly. Nevertheless they may know ten times as much as do most physicists about practical real-world firearms ballistics.

I would take their word for it and use a heavier bullet for a large animal. The fact that they are still alive after hunting big game suggests that they know the optimum combination of muzzle velocity and bullet mass.

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