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Recoil or no recoil? (Patreon)

Published:
2016-10-12 05:29:23
Imported:
2021-10

Content

This has been a really confusing experiment. I think I have it figured out, but that means that I am just now filming the explanation portion of the video. (for the 2nd time)

You guys want to take a stab at explaining why the gun shot the way it did? I'll post the video sometime tomorrow so you can check your answers. Hint: nether Nighthalkinlight or I where completely correct.

Files

gun with without

I shoot a gun once with a free recoil and once with the gun braced againsed the mountain. this is the video without the explanation, that part is taking me a looong time because this is actually a pretty good head scratcher.

Comments

x9x9x9x9x9

I am thinking the recoil doesn't matter because the explosion is basically happening in the center of 2 object and its going to exert the same amount of energy on the objects either way. the braced one the objects around it just absorbed the energy while in the other one the gun itself took it by flying backwards. but it does still seem counter intuitive that the bullets are the same speed. I have no idea if this is right this is just my simple guess at 1am.

Anonymous
>>2756158

This is also what I'm thinking, they are just pushing against each other. In both scenarios the explosion provides equal energy against all surfaces, the walls of the tube, the back of the tube and the back of the projectile. When the no recoil gun is fired, the energy released is absorbed by the wall, no further action occurs. However when the recoil is allowed the energy pushes the tube back at the same speed as the projectile was shot at (assuming equal mass) and the explosive energy is transferred into kinetic. In both cases the same amount of energy is applied to both masses therefore the projectile does not change due to recoil.

dancer42

BECAUSE THE BARREL IS RECOILING IT WILL ACT AS A SLIGHTLY SHORTER BARREL HAVING A VERY SMALL REDUCTION IN THE EXIT VELOCITY . THE CHAMBER PRESSURE WILL HAVE A BY A VERY SMALL AMOUNT SHORTER TIME TO ACCELERATE THE PROJECTILE ,THE DIFFERENCE WILL BE VERY SMALL.

dancer42

I really liked the demo hope to see it again with the chronograph.

Anonymous

I think there actually has to be a difference in the speed with this rather big projectile to gun ratio. That's because the energy released by the black powder is the same but the Kinetic energy you get is twice as big if you assume, that the projectile and the gun have the same mass and therefore move with the same speed which they have to do because of the conservation of momentum law. But you have to keep in mind, that forces do not imply energy. Energy is distance*Force so if the gun applies a force to the iron-thingy and the clay it doesn't transfer any energy as long as it doesn't move (although there is some deformation). And a problem with the speed measuring is that kinetic energy is equal to (mass*speed^2)/2 this means that if you could prevent any movement of the gun and direct all the energy in one direction (to the projectile) you would get two times the kinetic energy in the project which isn't the same as two times the speed. I don't know how fast the brass slug went but if it was around for example 20 m/s it would have had 400 J if you double that energy you get 800 J with translates to 28 m/s. So if you double the energy you get a square root of 2 times higher speed. So that makes it harder to measure. And with common projectile to gun ratios the difference it is surely negligible

Anonymous

My opinion on this : To me, pressure build-up is not the issue here. Since it's an explosion, we can approximate it as being instantaneous, and due to inertia, I think it happen before any part (bullet or gun) has had time to move. I think that, just after the explosion, the pressure is roughly the same in both cases. I believe the difference between the 2 experiments may come from the way the pressure decrease. One factor for pressure decrease is the expansion of the "explosion chamber" (between the log and the gun). In one case it expends slower (relative to the pressure, but at the beginning, we assume it's the same), because only the "log" side is moving : this may lead to a difference between experiments. But there are other factors to take into account. Heat dissipation from the gases will also decrease the pressure, and if it is the main factor in the pressure decrease, there will not be a major difference between the experiments. I think it would be very nice to design another experiment that allows us to plot the temperature of the gases before the bullet has completely exited the gun. Using that and the size of the cavity (that we can estimate from the position of the log and gun), we could deduce which it the most important factor in the pressure decrease, and the predict what to expect. (I hope you all can understand what I'm trying to say : I'm not a native English speaker)