Space Lobsters and Gravity....

[Peter Cairns]

Rother,

Try this page, it's wikipedia's page on Momentum,

Oddly it's P=MV, Momentum = Mass x velocity, mirrors

my F=MA almost exactly.

Keep working peopel, together we'll get to the bottom of this.

http://en.wikipedia.org/wiki/Momentum

Peter.

[Rother]

Thanks Peter, but I've already been there and now I have a headache. :mad: It all makes sense but it doesn't make any sense.

[Cpl Steiner]

Peter,

That Wikipedia has entries for the strangest things, as can be seen in this link.

http://en.wikipedia.org/wiki/Combat_Mission

[c3k]

Rother,

Nice link. Now, having a more than passing knowledge of Newtonian physics equations, why did "g", the acceleration of gravity, come into ANY of those equations?

Thanks,

Ken

[flamingknives]

Originally posted by Rother:

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

Momentum is conserved in collisions. Any cartridge that would blow a person "clean off their feet" would do the same to the person firing. Movies are tripe in this regard - the impact of the bullet is similar to the impact of the gun on one's shoulder or hand.

Really? I'll use my rifle as an example just because I know the numbers. Bullet energy=300gr./2,700fps/4,850ft.lbs. of energy. Rifle recoil energy=63,000gr./16fps/36ft.lbs of energy. Those are pretty rough numbers but close enough for this. You're failing to take into consideration is the rifle weighing more than 200 times what the bullet weighs. Believe me when I say I can knock a 1200lb moose literally right off it's hooves and yet remain standing. I will agree that the movies way over do it though. </font>

[flamingknives]

Originally posted by c3k:

Rother,

Nice link. Now, having a more than passing knowledge of Newtonian physics equations, why did "g", the acceleration of gravity, come into ANY of those equations?

Thanks,

Ken

Possibly for converting kg into N?

Although in this case it's supposed to be a subscript.

[acrashb]

Originally posted by Rother:

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

Momentum is conserved in collisions.

Really?

[...]You're failing to take into consideration is the rifle weighing more than 200 times what the bullet weighs.

[...]Believe me when I say I can knock a 1200lb moose literally right off it's hooves and yet remain standing. </font>

[Rother]

I really hate it when my lack of education and ignorance is so obvious to everyone but myself. Thanks for learning me something new for the day guys. Now I have to call an engineer friend and have him explain it so I can understand it. My headache is only getting worse here.

[Peter Cairns]

I am for metric every time, all that pounds to stone to ounces and feet, inches and yards is such a bloody pain in the arse...

Peter.

[Miska]

This is what i found from internet

The Attractive Force of Glass

Our hero stands innocently on the sidewalk as a sinister car approaches with a shotgun protruding from the window. Suddenly he sees it, but—blam— it's too late. He's blown violently off his feet and flies several feet backward through the nearest display window. Fortunately, he's wearing his bulletproof vest and survives.

If he were not on the sidewalk by a display window, then invariably he'd be blown into a rack of whisky bottles, a giant mirror, or some other large glass object. This happens so often that if we didn't know better we'd think Hollywood had discovered a new principle of physics: the attractive force of glass for shooting victims.

Hollywood apologists would explain that the hero was blown backwards by the force of the shotgun blast, and glass objects are in the way 98% of the time due entirely to random chance. Unfortunately the current laws of physics don't agree.

A load of buckshot hitting a vest can be considered an inelastic collision. This means that the kinetic energy of the victim with the buckshot stuck on his vest is less than the original kinetic energy of the buckshot before the collision. The "lost" kinetic energy is not really lost, it has just changed forms. Some of it becomes a shock wave in the victim that creates bruises and possibly cracked ribs. Some is converted to heat.

Even though kinetic energy is "lost" during the collision, momentum is not. The momentum of the victim is the same as the original momentum of the buckshot. So, the collision can be analyzed using conservation of momentum. This will let us estimate the backwards velocity of the shooting victim and judge whether he would indeed be thrown violently backwards.

To make the analysis we have to decide on some simplifying assumptions. As a rule of thumb, physicists and engineers (who should be considered applied physicists) generally start with the simplest reasonable calculation or model when analyzing whether an event will occur. They will also attempt to make assumptions which favor the event's occurrence. The reasoning is that if a simple model with favorable assumptions shows there could be no effect then there's no point in making a more rigorous model.

We'll make a simplifying assumption that there is no friction to impede the backward motion of the victim. This would favor the event's occurrence.

To calculate the momentum of an object we use the following equation:

p = mv

Where p is momentum, m is mass, and v is velocity.

Before the buckshot collides with the victim, the victim's momentum is zero, since he is not moving. This means that we only have to consider the momentum of the buckshot. For simplicity we will treat the buckshot as though it is a single object rather than calculating individual momentums for each pellet and adding them together. Both of these methods give the same result.

After the collision the victim and buckshot stick together and so, again, we only have to calculate the momentum of their combined mass. We'll use a subscript of 1 to indicate conditions before the collision and a subscript of 2 to indicate conditions after the collision. Hence:

p2 = p1

It's Not Newton's 3rd Law

Contrary to the explanations given in some venues, the fact that shooting victims are not thrown violently backwards by bullet impact forces cannot be explained using Newton's 3rd law. These explanations usually claim that the recoil force on the shooter is an action/reaction pair with the bullet impact force on the victim—simply not true.

Action/reaction pairs of forces are equal in magnitude and opposite in direction. They occur simultaneously. While the recoil and bullet impact forces are opposite in direction they do not occur simultaneously. The recoil force begins before the bullet strikes the target. It is generally lower in magnitude than the bullet impact force but lasts for a longer time.

By substitution:

m2v2 = m1v1

Solving for the velocity of the victim after the collision gives:

v2 = (m1/m2)v1

Note that the velocity of the victim is proportional to the buckshot's mass to victim's mass ratio. This ratio is going to be tiny. Using the following values: the mass of the man with buckshot stuck to his vest equals 80 kg and the mass of the buckshot alone equals 0.0318 kg with a velocity of 486 m/s, we obtain:

v2 = (0.0318 kg)/(80 kg)(486 m/s)

= 0.193 m/s

This is about 0.4 miles per hour. Keep in mind that humans can walk about 4 miles per hour. Since our model was set up with favorable assumptions, we have to conclude that shooting victims aren't going to be blown backwards through display windows by the force of a shotgun blast.

There's yet another way to view the problem. Conservation of momentum works for shooters as well as victims. In other words, recoil from firing a weapon will give a shooter backward momentum equal to the forward momentum of the bullet and hot gasses from burning gun powder exiting the gun barrel. (Note: buckshot will also include a light weight, fibrous wad placed between the powder and buckshot.) When the bullet strikes the victim he'll end up with only the momentum the bullet had immediately before striking. The magnitude of the victim's backwards momentum will be less than the magnitude of the shooter's backward momentum because the victim will not be hit by the firearm's hot gasses. Also, thanks to air resistance, the bullet will be moving more slowly and have less momentum than when it first exited the gun barrel. If the recoil from discharging a firearm is insufficient to throw the shooter backwards through the nearest window then the bullet also will not throw the victim backwards through the nearest window.

There is one other possible mechanism for being blown through a window: involuntary muscle contraction. The victim could be so stunned by being shot that he involuntarily jumps backwards. Since we haven't run this experiment, and have no desire to do so, we can't totally rule it out, but it does seem unlikely.

-----------------------------------------------

And here is the website where i found it

http://www.intuitor.com/moviephysics/

[Peter Cairns]

So as I see it, if it won't knock the firer down, then it won't knock the target down, but does it mean that on the moon with one sixth gravity, the recoil and impact feel six times as strong.

Peter.

[dan/california]

There was a great Mythbusters on a lot of these bullet impact issues and how wrong Hollywood has them. They took an approximately man weight pig(dead), put a completely bullet proof somthing around it to ensure that the rounds didn't blow thru and was thus guaranteed to transfer all of it's momentum to the target, and hung it on a straight bar. They then proceeded to shoot it with every thing up to a Barret 50 cal and they may have used one of those too. End result, the pig barely twitches much less blows back. All caught on high speed camera ect., very good stuff.

[FAI]

Originally posted by dan/california:

There was a great Mythbusters on a lot of these bullet impact issues and how wrong Hollywood has them. They took an approximately man weight pig(dead), put a completely bullet proof somthing around it to ensure that the rounds didn't blow thru and was thus guaranteed to transfer all of it's momentum to the target, and hung it on a straight bar. They then proceeded to shoot it with every thing up to a Barret 50 cal and they may have used one of those too. End result, the pig barely twitches much less blows back. All caught on high speed camera ect., very good stuff.

There's something (wearable) that can stop a 50 cal Barret at point blank range?

:eek:

[Bruce70]

Actually momentum and energy are conserved. You have to solve both equations to find out what's happening. And, since the bullet spins and will probably tumble on impact, you have to consider angular momentum as well. And then there is deformation. BTW, F=ma is only useful if you know what 'a' is, ie how long does it take the bullet to stop, or over what distance.

Anyway as stated above you get the general idea by just considering the energies involved and the bottom line is that the bullet does not knock the moose over, the moose achieves that all by itself.

[flamingknives]

Originally posted by FAI:

</font><blockquote>quote:</font><hr />Originally posted by dan/california:

There was a great Mythbusters on a lot of these bullet impact issues and how wrong Hollywood has them. They took an approximately man weight pig(dead), put a completely bullet proof somthing around it to ensure that the rounds didn't blow thru and was thus guaranteed to transfer all of it's momentum to the target, and hung it on a straight bar. They then proceeded to shoot it with every thing up to a Barret 50 cal and they may have used one of those too. End result, the pig barely twitches much less blows back. All caught on high speed camera ect., very good stuff.

There's something (wearable) that can stop a 50 cal Barret at point blank range?

:eek: </font>

[Peter Cairns]

Bruce70,

I use 1,000m/s for a, as the muzzle velocity of a basic M-16 is 1090m/s and the barrel less than a meter, so although the energy level obviously drops with distance, 1000m/s is a good ball park figure.

You don't need precision to establish principle.

Peter.

[John D Salt]

Originally posted by dan/california:

There was a great Mythbusters on a lot of these bullet impact issues and how wrong Hollywood has them. They took an approximately man weight pig(dead), put a completely bullet proof somthing around it to ensure that the rounds didn't blow thru and was thus guaranteed to transfer all of it's momentum to the target, and hung it on a straight bar. They then proceeded to shoot it with every thing up to a Barret 50 cal and they may have used one of those too. End result, the pig barely twitches much less blows back. All caught on high speed camera ect., very good stuff.

And no doubt these folks can also explain why I have seen a single 7.62mm round hitting a water-filled 50-gallon oil drum knock it right over (it was explained to us before the demonstration that a 50-gal drum full of water was about the same mass and density as a human target).

Muscular spasm in the oil-drum, perhaps? Or was it surprised?

I look forward to the amateur physicists explaining why a 122mm round can knock the turret off a target tank without knocking the turret off the firing tank.

I think the word "impulse" might need to feature more in the physical explanations before they become entirely convincing.

All the best,

John.

[Bruce70]

Originally posted by Peter Cairns:

Bruce70,

I use 1,000m/s for a, as the muzzle velocity of a basic M-16 is 1090m/s and the barrel less than a meter, so although the energy level obviously drops with distance, 1000m/s is a good ball park figure.

You don't need precision to establish principle.

Peter.

Yes, that's fine. But I was talking about the deceleration at the other end. F=ma is useless for that, too many factors to consider, as per John's post about impulse above.

John: You are right about the impulse being important, of course. The projectile accelerates over the entire length of the barrel, but decelerates (mostly) over just the thickness of the armour, therefore larger forces are involved. But I have no idea how that applies to moose and it also doesn't take into account deformation of the projectile and heaps of other factors. In short, unless the bullet loses most of it's velocity penetrating only a few inches of the moose's body, I do not believe that it is the bullet that knocks over the moose. I refer you to the Mythbusters pig shooting episode

[ January 29, 2006, 06:40 PM: Message edited by: Bruce70 ]

[Neutrino 123]

Originally posted by John D Salt:

And no doubt these folks can also explain why I have seen a single 7.62mm round hitting a water-filled 50-gallon oil drum knock it right over (it was explained to us before the demonstration that a 50-gal drum full of water was about the same mass and density as a human target).

Muscular spasm in the oil-drum, perhaps? Or was it surprised?

I look forward to the amateur physicists explaining why a 122mm round can knock the turret off a target tank without knocking the turret off the firing tank.

I think the word "impulse" might need to feature more in the physical explanations before they become entirely convincing.

All the best,

John.

Impulse basically describes transfer of momentum, and is not needed in the simple models presented here.

To knock someone down, you need to consider torque and angular momentum. To knock someone backward parallel to the ground, one considers momentum, but that is different then knocking someone down. In a frictionless environment, even a tiny slow bullet is enough to tip someone over, and gravity does the rest. During the second after the bullet impact, the target is probably not going to be trying to stabilize itself.

I would guess that the reason good guns can blow the turret off of certain tanks would be detonation of explosives in the turret, giving the turret upward momentum.

[Peter Cairns]

John,

The simple reason for the tank round example, is that tank guns have a complex mechanism for absorbing shock through the recoil system and control of the gasses from firing.

Tank turrets on the other hand are solid objects designed to withstand impact and penetration.

Therefore when a gun is fired the forces are dispersed and controlled , while at the other end, when the turret armour doesn't give (i.e. penetration) or the shot deflect, then the next weakest point gives, the turret ring.

Fact is that if for some reason the recoil mechanism on an M1A2 gave on firing, there is a good chance that the gun would rip back through the turret, or just like it's target, the turret ring would give.

As to the barrel of water, when it's empty I bet the bullet goes straight through, so it's almost certainly down to the effect of the shockwave in a liquid medium.

As the bullet enters the force creates a pressure wave that redistributes the weight of the water rapidly to the back of the drum, this makes it unstable, and hey presto it tips.

I once saw a slow motion version of this with a water storage tank.

The team who built it didn't quite level the platform, so the water went to that side, which increased the weight, which pushed it down, which meant more water to that side, which increased the weight, etc etc .

It took it a couple of days for people to notice the slight shift, it was really noticable when they inspected it, and the whole thing collapsed before the repair team arrived.

As to whether the way in which the dispersal of force as a shockwave might cause a moose to fly, by altering it's center of gravity through the movement of internal fluids or soft tissue, I doubt it would be a factor, but it is why the oil drum tips.

Peter.

[Caesar]

Is that 122mm shell solid shot or does it have an explosive component. If it is a straight momentum calculation, for it to make sense it would need to be a solid projectile that hit straight on. I don't think in that case you can use anything so simple as a straight momentum calculation.

A 50 gallon barrel would weigh either 190 or 225 kg (depending on which gallon measure you use) which is a mighty large man. I assume the barrel was not full which might make the water surge idea make sense.

[Drusus]

Originally posted by Peter Cairns:

Bruce70,

I use 1,000m/s for a, as the muzzle velocity of a basic M-16 is 1090m/s and the barrel less than a meter, so although the energy level obviously drops with distance, 1000m/s is a good ball park figure.

You don't need precision to establish principle.

Either I didn't understand something, or you have your numbers mixed up. If the acceleration of the bullet is 1000 m*s^-2 (the unit is meters per seconds^2) then it would take a second for the bullet to reach 1000 m/s speed. I don't think the bullet stays in the barrel that long...

Some calculations to get the force, I suppose that the acceleration is a constant over the length of the barrel, which is 1m and the end velocity is 1000 m/s. And we have

a = v/t

x = 1/2 a t^2

f = m a.

f/m = v/t (using a=f/m).

1/t=f/(mv) and thus t = mv/f.

x = 1/2 * (f/m)*(mv/f)^2

= 1/2 * m v^2/f

f = 1/2 * m v^2/x

= 1/2 * 0.02 * 1000*1000 / 1 = 10000 N.

This is quite a large force, but one should remember that it is applied only a short time (0.002s) If we apply this force to the shooter, he will get an acceleration of 100 m/s (assuming 100kg shooter) but it will be applied only for 0.002s, which means total speed of 0.2 m/s. In reality some of the force is translated to the gun going upwards. And a 20g bullet is really quite heavy. The acceleration of the bullet is 500000 m*s^-2.

BTW I can only hope my calculations are correct, it has been some time I last did this kind of calculations...

Quick addition: You can have the same end result for the speed of the shooter by applying the conservation of momentum to the shooting.

0.02 kg * 1000 m/s = 100 * 0.2 m/s, thus I think my calculations are actually correct.

What happens to the energy? Most of it goes into deformation of the bullet and most of all in the deformation of the target. If the happening would be totally elastic, (a perfect bounce with no deformation whatsoever) then one should apply the conservation of energy together with conservation of momentum to get the resulting speed. In this situation the energy is also conserved (it always is) but it is transferred to non-kinetic energy.

[ January 30, 2006, 05:07 AM: Message edited by: Drusus ]

[Peter Cairns]

Drusus,

I probably put it badly so i'll try to make it clearer.

As acceleration in the equation F=MA, is measured in Msec, I just took the muzzle velocity of 1090msec, as a good figure for the acceleration on the assumption that it wouldn't have lost that much in the 1st metre.

1,000msec is a good ball park figure to use for the speed of the round. If as you say the force is compressed in to 0.02 of a second (1/50) then you could be right to use 50 times the force.

Which would make it, F= 0.004*50,000 = 200n.

It's odd that as the calculation is in neutons, metres per sec and kilogrammes I an using the weight of the projectile only which I get as 4 grams, some people are using 63grains ( which is the same thing) and I don't know where the figure of 250 comes from,

Again It's hard when people use different measures of imperial and metric.

I found this which might interest people. For 5.56x45mm Nato.

Velocity in feet-per-second (Muzzle, 100, 200, 300, 400, 500 yards)

5.56mm NATO (63 grain): 3200, 2862, 2521, 2198, 1900, 1612

7.62 NATO (168 grain): 2700, 2513, 2333, 2161, 1996, 1839

.30-06 (150 grain): 2700, 2473, 2257, 2052, 1859, 1664

.270 Win. (150 grain): 3000, 2804, 2613, 2429, 2253, 2084

Kinetic Energy in Ft-Pounds (Muzzle, 100, 200, 300, 400, 500 yards)

5.56mm NATO (63 grains @ 3200 FPS): 1432, 1146, 889, 676, 505, 364

7.62 NATO (168 grains @ 2700 FPS): 2719, 2355, 2030, 1742, 1486, 1261

.30-06 (150 grains @ 2700 FPS): 2428, 2037, 1697, 1403, 1150, 922

.270 Win. (150 grains @ 3000 FPS): 2997, 2618, 2273, 1965, 1690, 1446

Bullet Drop from line of bore (at 100, 200, 300, 400, 500 yards)

5.56mm NATO (63 grains @ 3200 FPS): -1.76, -7.79, -19.31, -38.06, -66.51

7.62 NATO (168 grains @ 2700 FPS): -2.37, -10.23, -24.47, -46.14, -76.53

.30-06 (150 grains @ 2700 FPS): -2.41, -10.51, -25.42, -48.60, -81.86

.270 Win. (150 grains @ 3000 FPS): -1.90, -8.22, -19.63, -36.92, -61.05

Bullet Path in Inches at Zero Range of 300 yards (100, 200, 300, 400, 500)

5.56mm NATO (63 grain @ 3200 FPS): +3.68, +4.58, 0.00, -11.81, -33.33

7.62 NATO (168 grain @ 2700 FPS): +4.79, +5.58, 0.00, -13.02, -34.75

.30-06 (150 grain @ 2700 FPS): +5.06, +5.94, 0.00, -14.20, -38.49

.270 Win. (150 grain @ 3000 FPS): +3.36, +4.36, 0.00, -10.25, -27.33

Like much of the net, it uses US ft/pounds, which makes it difficult to relate to metric.

Peter.

[acrashb]

Originally posted by Bruce70:

Actually momentum and energy are conserved. You have to solve both equations to find out what's happening.

No, kinetic energy is not conserved in collisions. Some kinetic energy will translate to the target, the rest will get transformed in many different ways: heat, new surfaces, deformation, etc.

Total energy is of course conserved, but we're interested here in kinetic only. Potential energy, heat energy, etc. are not in play.

Your later comment about angular momentum is true, but so small as to be irrelevant. At most, in some handguns with a large bore diameter, ones hand will twist a bit as the bullet spins up. At the target end, all the spinning does is to create weird bullet paths due to a gyroscopic effect called precession.

The only thing that matters in terms of "blowing someone off their feet" is momentum, and man-portable shoulder-fired guns can't do it, period, because momentum is conserved in collisions. I suppose you could build a gun that would knock both the target and the shooter off their feet, but I won't be the guy to test-fire it

Man-portable rocket launchers and other recoilless devices are a different matter, as anyone who caught a panzerschreck in the chest could attest to... in a seance.

[ January 30, 2006, 09:34 AM: Message edited by: acrashb ]

[Drusus]

Peter,

First, acceleration is measured in m/(s*s). And as velocity is acceleration*time, you can't just put velocity in the place of acceleration, unless the time in question is 1 second. I say the force is there for 0.002 seconds. And thus the acceleration is 500000 m/s^2. And the force is 10 kN with 0.02 kg bullet. With 0.004 kg bullet it is 2 kN.

While the energy doesn't matter in the knocking down effect, the energy of the bullet is the thing that makes the damage. Modern guns have high velocities with small bullets, and thus they get large amounts of energy delivered with small recoil. If you double the speed of the bullet but make it half the weight, the momentum of the bullet remains the same, but the energy is doubled. If you manage to get a lot of angular energy to the bullet, still better, as this only causes the gun to twist a bit.