Regarding Force, physicists please regard

Factual Questions
1

Ok, this is a pretty complex question and I am not sure how to properly ask it.

I am running a campaign in a role-playing game in which there are no Dice. Some dopers out there might know what it might be, but really the game is irrelevant. In this game there is an Attribute named Strength. Which in game terms is pretty simplistic. In the game it covers physical strength and resiliance. After so many points of strength are reached, the person becomes ultimately bulletproof. (depends on the type of gun really)

I like to make my games pretty realistic in terms of this attribute and the others (psyche, strength, endurance and warfare), but run into some problems when their strength comes up against high caliper rifles. There strength protects them against the bullet punching through them, but in many places force still gets carried through. (i.e some universes the rules of the universe may not be the same as here)

My question is how much force does differing caliper guns and rifles produce and how much strength/mass would it take to not get thrown on your ass from the concussion. Obviously if your muscle acts as armor a .45 magnum might not penetrate your skin, but are you strong enough and have the leverage necessary to not get thrown on your ass from the force that still gets transfered to your body? Is there a possible formula for figuring this out?

For example, my formula for strength and armor capabilities are as follows. Strength- Every point above Amber basic (name of the game), is about 100 lbs of mass you can lift at earth level gravity. Amber basic is the equivlant of the world’s strongest powerlifter in terms of absolute strength. Lifting about 1/2 ton from a deadlift position. Every 10 points is another 1000 lbs. At 10 points of strength above amber basic you can basically lift a short ton. Pressing it over your head is a different matter entirely.

Ok, resistance to damge. The book has values for armor qualities already established. They have I believe, 1 pt armor is resistance vs conventional weapons, (swords, knifes, etc), 2 pts is resistance to firearms, and 4 pts is Impervious to all conventional weapons including firearms. I don’t like to use these because not all firearms are created equal. Is a high powered plasma gun a firearm? How about a sidewinder missle?

My thing I generally do to make it simple is every 15 pts of strength gets you the next armor quality. 15 pts you get the 1pt armor, 30pts you get the 2pt armor, 45pts you get the 3 pt version (which there isn’t one of, but is better than the 2 pt and less than the 4pt) and finally at 60pts of strength you are Immune to most firearms. (within GM’s discresion of course, thats me) Possibly upper limit resistance will be the 50 caliper rifles and such.

Now, saying that the muscle tension awards such immunity (and even organs and bones somehow magically get this bonus), Would a guy with 60 points of strength *Raw physical power of lifting 7000lbs off the ground, be able to get hit with a high power rifle and remain standing. Say it caught him off guard and he couldn’t brace himself in time. Does such a high power rifle exceede 7000lbs of mass/velocity? (not that he could just shrug off what he can maximally lift mind you)

What are some of the high velocity/mass/acceleration of such high moving objects?

Also, how much pressure per square inch can some of the hardest and/or most resiliant materials take?

2

Here is one way to get an intuitive feel for it.

Find out the muzzle velocity and weight of the actual slug that is fired. Multiplying these two values together gives the momentum of the projectile.

Now with regard to knocking someone down (as opposed to puncturing the skin, breaking bones, other injuries) it is only the change in momentum that matters. So you can compare this projectile to something more familiar with the same momentum, a baseball for instance.

Find out the weight of a baseball, and use

mass of bullet X muzzle velocity / mass of baseball

to give the speed of a baseball with the same momentum as the bullet. A baseball with this speed would have the same knocking-down power as a bullet with the muzzle velocity of the gun. So then just ask yourself, would a baseball traveling this fast knock down this character?

3

For example (knowing nothing about guns and using some random numbers I pulled off of the Internet), something called a “470 Nitro Express” (the most powerful gun on the list) delivers a .0714 pound projectile at 2150 feet per second. This would give it a momentum of 153.5 “foot-pounds/second” (I’m too lazy to convert to metric). Dividing this by the mass of a baseball (5 ounces (OK, that’s weight. Sue me.), that gives us…the momentum of a baseball thrown at 491 feet per second, or 335 miles per hour. If that number is still incomprehensible, let’s do it with a 12-pound bowling ball…I get that the momentum transferred is the same as that for a 12-pound bowling ball thrown at nine miles per hour. If your character can withstand that, he can withstand a pretty powerful rifle round.

4

What you are looking for is the “Law of Conservation of Momentum”.

p=mv

-FK

5

Another way to look at it is that momentum (which is the bit of the equation that would knock you flat on your ass) is conserved in the absence of external work. When you fire a gun, there is no external work done. So just before you pull the trigger, the gun-bullet-gunpowder combination isn’t moving, so its momentum is zero. After you pull the trigger, the total momentum is also zero. But now the bullet has a whole whack of momentum going forwards, and the gun has a bunch going backwards. There’s also a bunch of gases moving forwards at more-or-less variable speeds. So in fact, the gun has more momentum than the bullet. Now, granted, dude shooting the gun is probably braced in some way against the kick-back, but if you could fire the gun off-hand and not be bowled over by the kickback, then the bullet hitting you (assuming you’re wearing a Kevlar vest or whatever, so the bullet doesn’t penetrate) would also not knock you over.

6

Okay, when does the momentum mv become kinetic energy, 1/2 mv^2?

Conservation of Energy vs. Conservation of Momentum used to get me confused in HS physics.

7

One doesn’t become the other in any meaningful sense. They are two different properties.

Total energy is always conserved, but can be converted into all sorts of wild and whacky forms (e.g. in this example, chemical energy in the powder, heat, and kinetic energy), so it can be hard to do useful calculations with except in certain specific cases where it’s safe to assume no conversions happen. So for instance in the classic billiard-ball example, we say the collision is elastic and kinetic energy is conserved, because neither billiard ball permanently deforms, converting kinetic energy into heat. On the other hand, in this bullet-on-bullet-proof-material example (in the absence of the bullet proof material, the target would also, um, deform, turning even more kinetic energy into heat, which is not very pleasant for the target) , the bullet would deform, turning energy into heat. So while total energy would be conserved, kinetic energy would not.

On the other hand, momentum is conserved unless there’s a force external to the system that does work. In the current example, air friction would be considered external to the system, so the bullet would lose momentum between the muzzle and the target (as friction generates heat, kinetic energy would also be lost). However, at the instant (OK, few milli-seconds to be precise) that the gun fires, however, we can say that momentum is conserved. The same goes for the instant of target impact, although there are some variations. If the bullet either penetrates into or remains stuck to the target, the bullet-target combination has the same momentum as the bullet alone before collision (assuming the target was stationary). If the bullet bounces back, it ends up with negative momentum (in the other direction), so the target gets more momentum than the bullet had. And if the bullet goes right through the target, it may keep some momentum, so the target would have less than the initial bullet. Clear as mud?

If we assume the former, we can get another handle on the effect of said momentum. Let’s assume a relatively massive 500 grain bullet (500/7000lb) travelling at 2000fps and striking a 200lb target. Then
m1v1=m2v2
and
v2=[(500/7000)x2000] / [200+(500/7000)] ~ [(1000/7)] / [200] =5/7
v2=0.71 fps or 0.8 km/h
Now, a leisurely stroll for me is around 4 or 5 km in an hour, so suddenly finding myself moving at 0.8 will probably not knock me over too badly…

8

Conservation of momentum is key here. Is the momentum of the bullet leaving the gun high enough to knock an unbraced shooter over? I don’t believe so. If it cannot knock the shooter over, it cannot knock the shootee over.

Shotguns, rather powerful weapons, are sold with pistol grips, the shooter cannot even brace those against his shoulder, how could it generate enough force to knock him over? All pistols are operated at arms length. Even the most powerful rifles are designed to be shot by a person, braced only by a shoulder. If they generated enough momentum to knock you over, they would surely injure the shooter.

At best, I would think the person would be knocked off balance, and would have to do nothing more than shift his feet to keep upright.