Space Lobsters and Gravity....

[Bruce70]

I didn't say KE was conserved, I just said energy. But in any case, if there is no deformation, KE is pretty much conserved (not much heat generated in the collision AFAIK), whereas if there is deformation (and of course there is) there is not much you can say except that energy is conserved.

As for angular momentum, I was just being facetious.

Now back to momentum being the only thing that matters... bollocks. If you are catching a baseball without a glove, there is a huge difference between catching it with "soft" hands and catching it with firm hands. Momentum is conserved in both cases, but one hurts like hell - even though it didn't hurt the pitcher. Same with the tack turret, momentum is conserved, but the forces involved are enough to knock the turret of it's rim and gravity does the rest. The gun barrel is like the pitcher slowly accelerating the ball, and the target is like a frightened child trying to catch it with stiff hands.

[ScouseJedi]

It would be interesting to see what happens when the pig is rotated 180 degrees and the Barrett is fired into the 2" steel sheet. Would there be a bigger deflection?

Surely a bullet passing into soft tissue cannot rapidly transfer its energy into the spongy body whereas the same bullet striking a solid object transfers all of its energy at once.

Its been a long time since my Physics A-Level but I can still remember the Knight Rider episode where KITT was going to ram KARR and I believe the principles in there somewhere.

btw, Have we finalised the Space Lobster Weapons as being projectile based rather than energy weapons?

[Drusus]

From Young & Freedman, University Physics, 9th edition, page 238:

In any collision, momentum is conserved and the total momentum before equals the total momentum after; in elastic collision only, the total kinetic energy before equals the total kinetic energy after.

What this means is that in calculating the knock down effect of the bullet the energy doesn't matter. The bullet stays in the body (inelastic) or even worse, goes through. What happens to the energy? It does a lot of internal damage to the target, thus there is a lot of deformation. Interestingly enough if you fire at the 2" steel plate, then the energy does matter, as the collision isn't inelastic. It isn't compeletely elastic either (nothing in reality is) so calculating the end result without knowing the speed of the bullet after the collision isn't possible.

As to the catching ball effect, I think there is two things that matter here. First is human anatomy which I don't know much about and the second is the thing that the thrower is applying the force to the ball a lot longer than the catcher, thus it doesn't hurt the thrower. The thing is that the energy does matter, but it doesn't matter in the end speed of the catcher + ball in any way, if the collision is inelastic.

Actually I think what happens is that with firm hands, the energy goes to a lot bigger area and your muscles are absorbing it. In the soft hands case the energy is able to do damage which hurts. Like punching somebody in the stomach when he is ready for it vs when suprised. The latter knocks you down as in bullet knocking somebody down.

[SSgt Viljuri]

Kurki-Suonio does the trick better...

[c3k]

Impulse?

[acrashb]

Originally posted by Bruce70:

I didn't say KE was conserved, I just said energy.

[...]

As for angular momentum, I was just being facetious.

[...]

If you are catching a baseball without a glove, there is a huge difference between catching it with "soft" hands and catching it with firm hands.

1) KE is the only kind that matters for 'knock down', as it is the only kind with a vector (direction coupled with magnitude), so there was a) a reasonable assumption that you were referring to it and the necessity that you were if it was to matter.

2) Oh... I should have caught that.

3) Drusus has taken care of the last statement.

Originally posted by Bruce70:

btw, Have we finalised the Space Lobster Weapons as being projectile based rather than energy weapons?

No, and I would imagine that they would use meson guns, bypassing armour altogether and mostly eliminating recoil in the first place.

[Bruce70]

Drusus:

"The bullet stays in the body (inelastic) or even worse, goes through. What happens to the energy? It does a lot of internal damage to the target, thus there is a lot of deformation."

Ah yes, fair point. For some reason I was only thinking about deformation of the bullet - pretty stupid huh? However, it's not an all-or-nothing equation. The less deformation, the more KE is important. You can't just write off KE, that energy has to go somewhere. However, I guess that your statement: "in calculating the knock down effect of the bullet the energy doesn't matter" is fair enough. But don't forget that classic physics toy with the five metal balls. Yes I know this is an elastic collision, but it just illustrates that it is dangerous to assume that KE doesn't matter.

c3k:

Impulse = Force * Time.

To stop an object, or change it's momentum in any way, you need to apply an impulse. If the amount of time that you apply the impulse is short, the force required is much greater. e.g. car breaks to a stop vs car crashes into a brick wall. The impulse is the same in each case, but the amount of time and hence the forces involved are much different...

...and this is also what explains the baseball example. Not human anatomy. That just explains how the body deals with these forces.

BTW Drusus, you got this the wrong way around, you should catch with soft hands because it hurts less. Maybe it's different when you have a glove (I don't play baseball), I was thinking of cricket but trying to phrase it for a larger audience.

[Neutrino 123]

Impulse is the integral of force over a time period, not just F*t. That will only work if F is constant during your interval, and it usually isn't. You could take the average F and multiply by the time period, though to determine impulse.

Originally posted by acrashb:

</font><blockquote>quote:</font><hr />Originally posted by Bruce70:

I didn't say KE was conserved, I just said energy.

[...]

As for angular momentum, I was just being facetious.

[...]

If you are catching a baseball without a glove, there is a huge difference between catching it with "soft" hands and catching it with firm hands.

1) KE is the only kind that matters for 'knock down', as it is the only kind with a vector (direction coupled with magnitude), so there was a) a reasonable assumption that you were referring to it and the necessity that you were if it was to matter.

2) Oh... I should have caught that.

3) Drusus has taken care of the last statement.

Originally posted by Bruce70:

btw, Have we finalised the Space Lobster Weapons as being projectile based rather than energy weapons?

No, and I would imagine that they would use meson guns, bypassing armour altogether and mostly eliminating recoil in the first place. </font>

[acrashb]

Originally posted by Neutrino 123:

KE does not matter for knockdowns and is not a vector. Momentum does, and it is a vector. Someone needs to pay more attention in physics class...

[...]Also, mesons do have mass, so the meson gun would have signifigant recoil. Assuming the mesons are pi-mesons, the most common, they would probably decay before they reached their target into a bunch of photons, muons, and......

NEUTRINOS!!!!!!!!!!!

That's what I get for rushing. KE is a scalar - I lost sight of the difference between speed and velocity.

Now, regarding meson guns, the damage mechanism is the decay, not the impact (I think that ordinary matter is largely transparent to mesons), so they might or might not have a lot of recoil depending on the mass of mesons you send downrange. It would be interesting to do the math, since the mechanism that gets the mesons on the target before decay is relativistic velocities (to extend the apparent lifespan), and 'relativistic' implies increased mass. But I don't know how many mesons you'd need for a given useful (not counting, eg, the neutrinos) energy release or what they'd mass. For that matter (pun intended), I don't know how fast you'd have to get them going to get any significant range, not knowing the decay time (or the distribution of same) and being rusty on observer time effects near the speed of light - I think if the stationary decay time is 'sdt', the total distance travelled ('d') at a given speed 'v', with the speed of light being 'c', would be

d=v*sdt*(1/(1-(v^2/c^2))^(.5))

Feel free to correct me

Meson guns showed up in 'Traveller', and are now in various offshoots.

As flexible as CMx2 is touted to be, we will need at least a patch to do meson-gun-armed space lobster grav tanks.

[Peter Cairns]

But what would the space lobsters be.

Given that we are talking about near enugh in the future to be within 100 years, anything that could coss stellar distances would probably wipe the floor with us.

What about the SL's being "genetically" produced say as life forms that can survive in adverse enviroments such as Mars, as an alternative to robots or humans, but they sort of develope a mean streak....

Peter.

[Bruce70]

"Impulse is the integral of force over a time period, not just F*t"

That's like saying that distance isn't v*t. You are 100% correct, but it is also just being pedantic.

"KE does not matter for knockdowns and is not a vector. Momentum does, and it is a vector. Someone needs to pay more attention in physics class..."

Yes, someone does. You cannot calculate the outcome of a collision by considering momentum only, except in special circumstances. Now I conceed that if you know that the bullet is lodged in the body, that is one of those special cases, but if the bullet bounces off, then there are many possibilities. The problem can only be solved by considering KE (assuming you know the coefficient of elasticity), and the fact that it is a scalar is completely irrelevant.

Perhaps you have not seen the toy I was talking about (although I find that hard to believe). If a steel ball (10grams) travelling at 1m/s hits a group of four steel balls (same weight and size) lined up in a row and touching. And if the collision is perfectly square:

o -> oooo

What will happen?

There are many possibilities if you consider momentum only, but only one if you know that KE is conserved.

Now as I said, in the bullet lodging in the body case, you are quite right - you can work it out by considering momentum only, just as you can if you know that the 5 balls all stick together. But for the turret example momentum will not give you a conclusive answer. However, in that case it makes more sense to consider the forces involved and hence the impulses as has been pointed out already.

And in fact it is probably easier to consider impulses in both cases. If the bullet accelerates over a distance of 1m and decelerates over a distance of 0.1m (lodging somewhere inside the body) then the firer will experience a force 10 times less than the target, but for 10 times longer... simple. I think that could easily be enough to make you lose balance if you weren't expecting it (in fact my grandfather was often knocked over by the recoil in his old age, let alone the impact), but not enough to send you flying through the air. No idea about a moose, but I would guess that it would barely notice it, if it wasn't for the damage caused.

[Neutrino 123]

Originally posted by acrashb:

Now, regarding meson guns, the damage mechanism is the decay, not the impact (I think that ordinary matter is largely transparent to mesons), so they might or might not have a lot of recoil depending on the mass of mesons you send downrange. It would be interesting to do the math, since the mechanism that gets the mesons on the target before decay is relativistic velocities (to extend the apparent lifespan), and 'relativistic' implies increased mass. But I don't know how many mesons you'd need for a given useful (not counting, eg, the neutrinos) energy release or what they'd mass. For that matter (pun intended), I don't know how fast you'd have to get them going to get any significant range, not knowing the decay time (or the distribution of same) and being rusty on observer time effects near the speed of light - I think if the stationary decay time is 'sdt', the total distance travelled ('d') at a given speed 'v', with the speed of light being 'c', would be

d=v*sdt*(1/(1-(v^2/c^2))^(.5))

Feel free to correct me

Meson guns showed up in 'Traveller', and are now in various offshoots.

As flexible as CMx2 is touted to be, we will need at least a patch to do meson-gun-armed space lobster grav tanks.

You've got the decay time right (from time dilation), and the decay time for a charged pi-meson at rest according to my particle data book, is 2.6*10^(-8) seconds and for a neutral pi-meson 8.4*10^(-17) seconds. However, a meson gun would really be an extremely inefficient weapon if the decay was what caused the damage. This is because these deay times are AVERAGE times. For a group of mesons, this means that half of the starting mesons would decay after this time (if they were at rest). The decay is continuous, so mesons from a gun would give a continuous stream of decay, with the decay the most intense at the gun barrel. Only the part of the beam overlapping with the target would do damage. I have no clue how the meson gun works in any games, though...

Originally posted by Peter Cairns:

Given that we are talking about near enugh in the future to be within 100 years, anything that could coss stellar distances would probably wipe the floor with us.

Judgeing by the current rate of human technology developement, that is undoubtedly true, but there is no reason to assume that technology advances the same between species . Besides [ FRIENDLY QUIP ALERT ], it couldn't be much worse then U.S. vs. Syria...

Originally posted by Bruce70:

That's like saying that distance isn't v*t. You are 100% correct, but it is also just being pedantic.

I guess it's sort of pedantic, though 't' is commonly associated with the variable time and not a time interval. Also, in the problems classes I've taught and taken, impulse is normally used in integral form, while acceleration or velocity are normally constant.

Originally posted by Bruce70:

[QB] Now I conceed that if you know that the bullet is lodged in the body, that is one of those special cases, but if the bullet bounces off, then there are many possibilities. The problem can only be solved by considering KE (assuming you know the coefficient of elasticity), and the fact that it is a scalar is completely irrelevant. [QB]

Bullet bouncing off of a person? In the previous examples, the bullet entered the target, and came to rest in the body, making it essentially a perfectly inelastic collision (if the bullet went through, the momentum transfer would be even less). In the general case, there are many possible things to consider in collsions that are not elastic or perfectly inelastic.

Originally posted by Bruce70:

[QB] But for the turret example momentum will not give you a conclusive answer. However, in that case it makes more sense to consider the forces involved and hence the impulses as has been pointed out already.[QB]

In the turret case, we can approximate a perfectly inelastic collision if the shell penetrates and does not go out the back of the turret. If the shell breaks up or better yet, ricochets, then the momentum transfer can be greater (depending on the angle of ricochet), but if the shell doesn't penetrate, I don't think it would have much chance of blowing off the turret. Again, I am thinking that blowing the turret off requires something blowing up inside the tank, HE explosives or somesuch.

Originally posted by Bruce70:

[QB] And in fact it is probably easier to consider impulses in both cases. If the bullet accelerates over a distance of 1m and decelerates over a distance of 0.1m (lodging somewhere inside the body) then the firer will experience a force 10 times less than the target, but for 10 times longer... simple. I think that could easily be enough to make you lose balance if you weren't expecting it (in fact my grandfather was often knocked over by the recoil in his old age, let alone the impact), but not enough to send you flying through the air. No idea about a moose, but I would guess that it would barely notice it, if it wasn't for the damage caused. [QB]

I think you may be confusing distance with time in the example here. If the acceleration took place over 1sec and deceleration over .1sec, then the average force would be 10 times greater (though it would peak at a point higher then this since the force isn't distributed evenly in time). You example is roughly correct in it's conclusions, but is irrelevant. In each case, the time over which the force is transfered is practically instantaneous.

There is not enough momentum to send someone flying backward, but there is enough force to cause torque, tipping someone over, knocking them down. In the case of torque knockdown, how well the target is braced for the force is extremely important. This is why a prepared firer is far less prone to knockdown then a surprised, unprepared target unable to immediately try to recover due to the pain. Thus, the total angular impulse is the best factor when determining knockdown. This means that you have a much greater chance of knocking someone over is you hit in the head then in the torso.

[Bruce70]

"I think you may be confusing distance with time in the example here."

I was a little surprised to find out that it didn't matter whether you consider time or distance, the ratios come out the same. I thought that the t-squared in there would throw it off, but actually it doesn't.

[Neutrino 123]

I'm not exactly sure what you mean by the ratios, but there is no reason to assume that acceleration is constant in this scenario. In fact, quite to the contrary, one would expect the acceleration at the beginning of a gun tube to be signifigantly greater then at the end (the compressed gas has expanded by the time it has pushed the bullet to the end, thus exerting less pressure at that point). Meanwhile, the acceleration a bullet experiances when entering a person would vary wildly depending on what it happens to be going through in a given interval...