Recoil energy vs. instantaneous recoil (felt recoil)?
350132
I'm not exactly a weapons expert, but I believe total recoil energy is the amount of recoil produced by the rearward push of a bullet of a certain weight propelled by a certain amount of gunpowder. Felt recoil is the amount of recoil that is transfered to the shooter after some of the total recoil has been dissipated by the weight of the rifle, the muzzle brake (if the rifle has one), and the rifle's gas operating system (again, if the rifle has one).
797135
Bound's hubby here:\n\nMany of the answers contained herein are good answers. If you dissect an automobile muffler, there is very little conceptual difference from a firearms' silencer (or more correctly, suppressor). \n\nThe firearm's silencer contains a perforated barrel extension, that allows the propellant gases to flow into the expansion chamber formed by the barrel extension and the outer shell. Between the barrel extension and the outer shell is a medium, typically, steel wool. As the gasses flow into the expansion chamber the gases flow around/through the steel wool, cooling the gasses, and muting the crack of the shot (to a degree). Silencers will work with both sonic and subsonic ammunition ... it is that the silencer is MORE EFFECTIVE with subsonic ammunition. \n\nWhat many people have neglected to state, is the silencer's effect on the external ballistics of the projectile. Typically, the bleed-off of the propellant gasses slows the velocity of the projectile, reducing both its muzzle velocity and its muzzle energy! A firearm shot with both a silencer on and off, will hit in different places, and different effects on the object being shot.\n\nGood luck!
802288
the impuls (m*v) is conserved, so the impulse in one direction;\n0.150*20*sin(30deg)\nEquals the impulse in the other direction\n50*v_recoil\nthe recoil velocity is thus 0.03 m/s
827118
We can't use conservation of energy because this is an inelastic collision. Conservation of momentum does hold, so\n\nm1*u1 + m2*u2 = (m1+m2)*v\n\nu2 is zero. So \n\nv = m1*u1/(m1+m2) = .24 kg*u1/2.85 kg = .084*u1\n\nThe projectile's incident kinetic energy \nKEp = (1/2)*m1*u1^2 = 0.12*u1^2 joules\nThe final kinetic energy \nKEf = (1/2)*(m1+m2)*v^2 = 1.42*(.084*u1)^2 = .0100*u1^2 joules.\n\nPercentage = 100%*KEf/KEp
361801
I use a .243 myself.  It's light and there's not much recoil, but travels well and will put one down.
834102
1)  I'm not entirely sure.  It may just be the preference of that publisher, or just that the quantities being compared were not measured more accurately than to two places.\n\n2)  The coeffecient of friction is basically a measure of how well two surfaces slide against one another.  Pressing down simply increases the force, it does not necessarily change the structure of the surfaces in contact.\n\n3)  The coefficient of friction has no units because it is a ratio.  If you were to write out the units being compared in a fraction, say 2 Newtons / 4 Newtons, the units would cancel out when divided.
1067939
Yes, they are essentially kinetic energy rounds. A quick glance at your periodic table of elements will show you that Uranium is quite heavy. DU rounds are far heavier than typical ordinance and have far greater penetrating power. Note that thier momentum and kinetic energy are the same as for a conventional round fired from the same gun (increased mass is made up for by thier decreased velocity), however, when you run the conservation of momentum equations for the bullet impacting a target, the DU round fares slightly better.\n\nAlso, DU is an environmental hazard, not becuse it may be radioactive (it isn't), but because it is naturally toxic and deletrious for the human body, and it readily forms uranium salts when it comes into contact with the air. Most DU materials are coated with a layer of steel to prevent this.
919525
Vibration:  ω = √(k/m) = √(100/4) = 5.0 rad/sec\n\nV at neutral point = Vo = Aω = 2*5.0 = 10 m/sec\n\nx-direction momentum: This momentum will be the same before and after dropping the 6 on the 4 →\n6*0 + 4*5 = (6+4)*v →  v = 20/10 = 2 m/s\n\nThe new KE = .5mv² = .5*10*2² = 20 Nm ; This will equal the new spring energy:\n\n20Nm = .5kA² → A = √(40/100) = .6324 m, so the change in amplitude is\na)  2-.6324 = 1.3676 m\n\nb) New period: T = 2π√(m/k) = 2π√(10/100) = 1.987 sec\n....Old period:  To = 2π√(4/100) = 1.2566 sec\n....Difference = .73 sec greater\n\nc)  KE1 = .5*4*10² = 200 Nm.  From (above) KE2 = 20 Nm\n....so, energy change is -180 Nm\n\nd) The energy loss is released as heat between the (momentarily) sliding surfaces of the 2 objects when they contact.
879121
.50AE has 60% more muzzle energy than a .44 magnum.\nBOOM!\n(.357 is just a pup compared to the .44)
142208
A viscous damper opposes motion with a force proportional to speed, thus dissipating mechanical energy. This contrasts with a spring, which opposes displacement with a force proportional to deflection and stores the energy required to deflect it, returning the energy to the vibrating system as it returns to the undeflected state. Viscous energy dissipation reduces the amplitude of a vibration.
255887
Over all in combat the AK, it can take a hell of a lot of abuse. Trust me here, I have used both.  The M-16 A2 does well but can not handle the same environment.
59800
There are two issues at hand.  One is energy.  The other is momentum transfer.  \n\nHere are some muzzle energy for factory loads of the two rounds.\n\n30-30 factory loads:\n150 grain round - 2,390 fps - 1,902 ft. lbs energy\n170 grain round - 2,200 fps - 1,827 ft. lbs energy\n\n7mm factory loads:\n140 grain round - 3,400 fps - 3,593 ft. lbs energy\n160 grain round - 3,200 fps - 3,637 ft. lbs energy\n\nThe 7mm clearly has the uupper hand here.  However, that energy needs to be delivered to the deer to be effective.  Or more correctly, the momentum of the round needs to be transfered to the deer. \n\nThis is where, for example, hollow points are preferable to full metal jackets.  The hollow point will cause the bullet to mushroom giving it better momentum transfer properties (and therefore knockdown power).  FMJ rounds tend to go straight through causing somewhat less damage because of less momentum transfer.\n\nYou want a careful balance here.  Too much energy in too small an animal will make things, let us say, much messier.  All you want is a clean take.
1074554
If you have a piece of wire and you increase its mechanical energy, for example by winding it into a spring and then compressing that spring, you can wonder what happens to that energy when the spring is dissolved in acid.\n\nWhen the energy is stored in the spring, it is because atoms have been pushed to a place which is away from their natural lowest-energy position.  You could say that by squeezing the spring, you are pushing the nuclei in the metal coil to be closer than they naturally want to be.  Because they are in a higher energy configuration than natural unstressed metal, when they are separated by the action of the acid, the atoms end up with more energy than they do in the unstressed case.  This additional energy manifests itself as heat.\n\nSo, in summary, if you took two identical springs, compressed one of them and left the other one relaxed, and dissolved them in identical acid baths, the acid bath which holds the stressed spring would end up hotter as the energy formerly stored in the spring is released as heat.
393875
The way I see it.... u calculate its impulse with the ground.... mass1*velocity1 (pre-shock)=m2*v2 (after shock) then caculate it's potential energy by newton's law of gravity using v2....(god I'm being a GEEK!)
881471
Bound's hubby here:\n\nA bull barrel is typically a barrel of larger diameter than a typical sporter barrel ... bull barrels are about 1" in diameter. Rifles that typically have "bull barrels" are varmint rifles and target rifles. These barrels serve two major fuctions ... \n\nFirst, bull barrels are stiffer and resist the harmonic vibrations generated by a bullet streaking down the bore ... thereby generating a higher degree of accuracy (both bullet on target and grouping). \n\nSecond, a bull barrel adds more mass to a rifle (generally making the rifle muzzle heavy) which helps minimize movement of the rifle and sighting errors. \n\nI know that the bull barrel on my Remington 700 Varmint generates a fair amount of heat in 1000 yard matches (generally 25 shots fired in a 30 minute period), and the bull barrel does not contribute to heat reduction! This is why most target shooters that shoot bull barrels have mirage bands running from muzzle to receiver to reduce the barrel heat generated mirage!\n\nActually, most "bull barrelled" rifles are not that much heavier (perhaps 2-3 pounds) and make accurated longer range hunting rifles. \n\nIt is not that barrels warp when hot, it is the vibration of the barrels. Thinner barrels vibrate more, causing more movement of the barrel which will displace shot placement!\n\nGood luck!
794091
I'd suggest you first try the site listed below where you can find a downloadable spreadsheet that predicts velocity. You can deduce the author's equations from it. It's actually a time-stepping simulation rather than analytical equations. However, a bit of theory wouldn't hurt either. First, to keep it simple, let's assume the pressure is constant for T, the lesser of the time the valve is open and the time the projectile remains in the barrel. This is actually not a good assumption since many spudguns have pressure vessels that are smaller than or not much bigger than the barrel volume. Anyway, during T the spud is subjected to a force F=P*barrel area = P*pi*d^2/4. The net force accelerating the spud is F - friction force, so its acceleration A=(F-Ffriction)/M and its velocity v=A*T. I don't know the practice on valving. It may be that open valve time is always less than barrel transit time, but if not, you can compute the distance S that the spud would travel during the open valve interval T; S=0.5*A*T^2. If S>l, then T is the time spent in the barrel, which is SQRT(2*l/A), and you should recompute v. \nAnother assumption would be that each shot entirely empties the pressure tank. In this case we simplify by disregarding pipe friction, thermodynamic losses (like failure to fully expand due to cooling) and any other losses you can think of, and assume the potential energy of the compressed air is converted to kinetic energy. The potential energy PE = 0.5*P*vol (where vol is tank volume = L*pi*D^2/4). The kinetic energy of the spud is 0.5*M*v^2. Thus v=SQRT(P*vol/M).\nAbout other questions, 1 psi = 6894.75729 Pascals, and speed of sound =340.29 m/sec.\nNote that with what you have given in your question, you have a lot of units conversion to do.\nEDIT: See friction stuff added above. The spreadsheet includes friction effects.
785358
It depends on what caliber you're using, and how far the target is.\nLike I told someone on another question. If you're going by what the manufacturer says, that's wrong. Manufactures ALWAYS \nEXAGGERATES what the gun actually shoots. It's called "marketing." They know if they say a gun will shoot 1000fps,  people with NO experience with airguns will buy that gun. They also use the lightest pellets to get those numbers, and when using light pellets you loose accuracy.\n\nDo you have a Chronograph that you have used to test the velocity of your gun? That is the only way to tell what your gun\nactually shoots. This is what you need if you want to test velocity:\nhttp://www.straightshooters.com/navagationpages/precisionequipmentselections.html\n\nAlso, what is the weight of the pellets you're going to use?\nThat also makes a diffrence.\nIf I were you I would use at least a .22 pellet, not a .177.\nYou're just going to piss it off.