June 22nd, 2012, 07:44 AM
A gun is fixed on the sidewall in a box. The muzzle points towards X direction.
There is a disk in the box and it can rotate freely around the axis Y .
The trajectory of the bullet is one of tangent line of the disk.
After shot, the bullet moves ahead and embeds into the disk.
Will the box move towards -X direction?
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It will NOT move because the momentum is conserved. All other complicated processes can be forgetted/ignored because nothing in the world *however complicated it seems* can violate the simple law of momentum conservation. -So the box will not move.
How about the disk?
Won't it rotate？why？
If it will rotate, do you think its angular momentum is lonely or not?
The disk will rotate, but so will the box in the opposite direction. Essentially what happens is this:Originally Posted by Emdrive
1. System starts with zero linear and angular momentum.
2. Bullet fires in x, direction. Box recoils in -x direction to maintain momentum.
3. Bullet hits disk, imparting some angular and linear momentum. The linear momentum transfer cancels out the recoil of the box. Since the axis of the disk is attached to a point of the box that is not inline with the bullet path, this also imparts a torque to the box, causing it to rotate in the opposite direction of the disk, which cancels out the angular momentum of the disk, leaving the system with zero angular momentum.
The main point is that the total momentum of the system both before and after will be zero.
I think that the box will move towards -X direction and the disk will rotate clockwise.
But it seems that the result will violate the Inertia law.
Can we explain it like this way: only particles, rigid and solid object apply to the Inertia law?
He is exactly right in what he has said here. There a few minor details which may occur like details on the box weight vrs the recoil and such but what he says is 100% true.I've used nail guns for 35+ years.The disk will rotate, but so will the box in the opposite direction. Essentially what happens is this:
1. System starts with zero linear and angular momentum.
2. Bullet fires in x, direction. Box recoils in -x direction to maintain momentum.
3. Bullet hits disk, imparting some angular and linear momentum. The linear momentum transfer cancels out the recoil of the box. Since the axis of the disk is attached to a point of the box that is not inline with the bullet path, this also imparts a torque to the box, causing it to rotate in the opposite direction of the disk, which cancels out the angular momentum of the disk, leaving the system with zero angular momentum.
The main point is that the total momentum of the system both before and after will be zero.
Nail guns have less power behind their driving force, than a gun.
But the motions are the same.
If what you think will happen violates the inertia law, then what you think will happen, is not what will happen.
The inertia law always applies. Objects which are not particles or are not solid and rigid objects, can be analyzed as assemblies of particles.
Even a bowl of soup follows the laws of physics. Momentum is conserved, see post #4.
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When a kid plays the water roller, there is no external force in a horizontal direction. The gravity will be mentioned .
Somebody even thanks the water for making his roller moving .
Last edited by Emdrive; June 23rd, 2012 at 07:26 PM.
Stand firmly at the head of a boat with a gun in your hand. Now fire towards right direction.
Obviously the boat will rotate anti-clockwise. NOthing violates any physics laws here.
But think about it carefully:
The part of angular momentum involve the bullet relating to the center of the earth? Will it remain unchanged?
Another item of angular momentum involve the bullet relating to the center of the sun? and around the center of galaxy? Will they all keep constant forever?
When accounting for the angular momentum of the boat, bullet, and water, for the boat-bullet-water system, you will find that the total angular momentum is zero. It actually doesn't matter with respect to what point you take the angular momentum around. The fact that the laws of physics remain unchanged regardless of what direction you are facing guarantees this.
The boat spins around , the angular momentum of the boat's centroid relating to the center of earth does not change.
But the bullet's angular momentum relating to the earth actually do change!——you choose to ignore it ?
In the same time the shot can't change the earth's angular momentum relating to the sun, but the bullet's angular momentum relating to the sun does change!
A local theory shouldn't be regarded as an absolute truth all over the world espacially in the infinite universe.
I'm sorry I prefer the truth to theories.
Last edited by Emdrive; June 23rd, 2012 at 10:02 PM.
The angular momentum of the system (boat and bullet) remains constant with respect to the Earth. Just like with the box and wheel, the firing of the gun will impart some movement to the boat with respect to the center of the Earth. If you take the angular momentum of this movement with respect to the center of the Earth, you find that it cancels out the angular momentum of the bullet with respect to the same point. (of course, friction between the boat and water will transfer this to the Earth itself, which then will itself become part of the system)Stand firmly at the head of a boat with a gun in your hand. Now fire towards right direction.
Obviously the boat will rotate anti-clockwise. NOthing violates any physics laws here.
But think about it carefully:
The part of angular momentum involve the bullet relating to the center of the earth? Will it remain unchanged?My last statement remains true for whatever point you may choose to measure the angular momentum from.
Another item of angular momentum involve the bullet relating to the center of the sun? and around the center of galaxy? Will they all keep constant forever?
Any time you come up with a scenario where it appears that you have lost or gained angular momentum, You have overlooked something.
When disk A revolves round axis Z and disk B revolves round axis Y , I have been wondering that wether I can add their angular momentums together or not?
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Last edited by Emdrive; June 24th, 2012 at 02:14 AM.
You can try build the system, and then you'll see that the box just jolted abit after the explosion but then it stopped/no-motion. What really happen is that our mind (including me) is not born to have an intuition for physic, so we can't use it to imagine something that we've never seen; eg: this box (so you must build it first). -The only way to be sure is for you to build it in real world or use computer-software to simulate it... else you'll just run into much more complicated reasoning and ridiculous long reduction-logics to prove something simple, (ie: "reduction logics" is a logics used to justify something using *only* mechanical movement, but it can extend until to the atomic level of individual atoms. No guarantee of success).
It won't move because that's what I learn in undergrad, from textbook. Because of momentum conservation.
You should test it first.
Last edited by msafwan; June 24th, 2012 at 06:05 AM.
About post #1, I can't ignore the rotation movement of the disk .
Once I put a toy electric ship in a sqaure basin with half water, then put the basin floating in the river.
I guess the basin will move .But it actually rotated slowly but uninterrupted.
In 2005,I built a swing with a shell and floor, also installed four wheels under the bottom--like a small van without engine.
Every body asserted that no one can make the van move straightly when he play swing in it.
The result was: when I sit on board of the swing and began to swing in rhythm , push-pulled the frame by hands.
The van moved ahead straightly with rhythm :fast,then slow;fast,slow……
Should I believe the theory or the experimental results?
Later when I discussed this in some forums , someone didn't trust its moving but refused to verify it by himself;someone stated that it must move otherwise it will violate the Inertia law . Those people included college students, doctors of physics, middle school teachers, even professors in university.
Which one I should believe ?
If your soup is driven by a pump and is forced to move in a spiral pipe with a plenty of small holes in its sidewall.
In the process of flowing more and more droplets of your soup will leaked out from those holes continuously.
At the exit end little soup left so did the momentum of the flowing beam.
The side pressure doesn't understand what is "ON STRIKE".
But the recoil force of pump will remain unchanged.
How about the closed container?Will it move or rotate?
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You should trust the experimental data, and you should study textbook because it is from the hardwork of countless hour of experiment. People 100 years ago are really good people, they performed alot of experiment to test alot of science: ranging from testing the theory of fluidic space (ether), radiation, motion, and chemistry... and even older generation tested with idea of turning stuff into gold, making potion and cures, and perform radical surgery like lobotomize living human brain, shock treatment, replacing brain and ect. -Reading textbook is really really worth it! save alot of time.
This prove that it doesn't move. It rotate.About post #1, I can't ignore the rotation movement of the disk .
Once I put a toy electric ship in a sqaure basin with half water, then put the basin floating in the river.
I guess the basin will move .But it actually rotated slowly but uninterrupted.
It prove that it doesn't move; it just rhythm like your body rhythm. 'Rhythm' is not 'moving'. -ie: Does a swing 'move'? no, a swing just 'swing!', swing doesn't move.In 2005,I built a swing with a shell and floor, also installed four wheels under the bottom--like a small van without engine.
Every body asserted that no one can make the van move straightly when he play swing in it.
The result was: when I sit on board of the swing and began to swing in rhythm , push-pulled the frame by hands.
The van moved ahead straightly with rhythm :fast,then slow;fast,slow……
*motion has different wording in physics because people always get confuse. Physic word is more accurate.
Last edited by msafwan; June 24th, 2012 at 11:21 AM.
I have no doubt that you could cause such a device to "creep" across a surface. I can do the same thing by standing on a four wheeled cart and just shifting my weight back and forth in the right way. However, neither of these violate conservation of momentum. They rely instead on the difference between static and non-static friction. Put simply, this means that the friction between surface that are not sliding against each other is greater than that for two surfaces that are. IOW, it takes more force to get something to start sliding than it does to keep it sliding. There is also some difference between sliding something slowly vs quickly.
In my example you take advantage of this by shifting your weight in one direction suddenly and sharply to generate enough force to overcome the static friction and shift the cart. you then shift your weight back more gently so as to create less than the static friction force so that the cart doesn't roll back. When you do this, you are transferring momentum to the Earth via the wheels.
With your device something similar occurs. As you swing you perform an action known as "pumping" were you lean back and extend your legs while swinging in one direction and lean forward and tuck your legs back when you go in the other. This is what allows you to start swinging and maintain your swing rate. This has the same effect as my shifting my weight back and forth. The force applied in one direction is greater than the other. There is also a difference in friction for the wheels when they turn slow vs. fast. When turning slow it is greater and thus more momentum is shifted to the Earth in one direction than in the other. Your forward creeping is a result of this uneven momentum transfer due to the varying friction, not to any violation of the laws of conservation. If you were to remove the friction from the set up entirely, you would not be able to cause your device to move forward.
Experimental results are only as good as the attention given to all the contributing factors when interpreting them.
Thank you.
But in which textbook can I find the answer of post #18？
I don't know the answer, sorry.Thank you.
But in which textbook can I find the answer of post #18？
Last edited by msafwan; June 25th, 2012 at 12:31 AM.
On post 17 : A swing installed at a small car is that ? To change position continuously all it takes is know how to swing properly. Each swing the car can jump a distance S to the floor then (and actually go through a stop between decellerating/accellerating and accellerating/decellerating to the floor. Accelleration decellerating is releative whatever direction you choose to accellerate to.
The thing is that a swing is not symmetric as the usual pendulum. It,s not very effective (walking goes faster) but still.
Unfortunately most universities where you study fysics have a billboard sign above the entrance : "leave you,re intuition outside".
[spelling corrected for clarity]
I have never seen such a sign. But it would be a good idea. I think a lot of the problems in this thread are due to Emdrive using intuition rather than maths.
Maybe you are right but I think that maths should operate after model .[spelling corrected for clarity]
I have never seen such a sign. But it would be a good idea. I think a lot of the problems in this thread are due to Emdrive using intuition rather than maths.
I'm trying another more intuitive experiment on electromagnetic force with only one direction. If obvious macro effect occurs there will be no ambiguity.
It all depends on how the holes are designed. You seem to believe that the water coming out of the holes will automatically be moving at a right angle to the pipe and towards the center of the coil. This is not the case. The water's natural tendency would be to hold on to the velocity it through the pipe at the time it left the hole. The water will come out at an angle. How this effects the entire set up depends on how the whole thing is balanced and how evenly the water flows from the holes.
If your soup is driven by a pump and is forced to move in a spiral pipe with a plenty of small holes in its sidewall.
In the process of flowing more and more droplets of your soup will leaked out from those holes continuously.
At the exit end little soup left so did the momentum of the flowing beam.
The side pressure doesn't understand what is "ON STRIKE".
But the recoil force of pump will remain unchanged.
How about the closed container?Will it move or rotate?
Attachment 809
You could overcome this by angling the holes in the opposite direction to force the water to come out at a right angle, but this would require the water to be deflected by the pipe which would put a force on the pipe that counters the recoil of the pump.
One thing that complicates this is that as the water flows through the pipe it loses pressure due to the loss through the holes. This means that water coming out of the first holes will have more effect than water out of later holes. You could compensate by decreasing the diameters of the holes as you go along.
If you arrange things such that the water has no net angular or linear momentum after it leaves the pipe, the pipe will not move or rotate, but in order to do this the water has to interact with the pipe in a way that would counteract the recoil of the pump.
This is somekind of homework I'm sure....
Where can you find such homework?
If both container A and B will not move or rotate the theory seems to be kid's Silly putty.
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Suppose I am on a boat. When I walk from the back to the front and then stop what has happened ?
The balance of the boat + me has changed. The massmiddle line for the system (technical sense all parts together) was more to the back at first and walking to the front I moved this balance line forward. If the boat (that,s not + me) would still be at the same spot again this would mean the technical system ( me + the boat) has moved forward. To keep this at same spot the boat has to move backward a little to compensate that I ran forward. Then I have not done work to the system of me and the boat.
Moderator note:
Since it appears that the OP's main purpose here is to claim that conservation momentum can be violated (as long as you are tricky enough), I believe that is time that this thread found a more fitting home. Moving to Pseudo.
Just asking for clarity :
With a box and a gun fixed at one end.... the gunpowder explodes. Does it shoot only the bullit or will it also shoot the box including the disk opposite direktion ?
Then - after the struck - can the bullit bring the box back ?
If not has box moved visually to the floor after the shot and struck ?
Last edited by Ghrasp; June 26th, 2012 at 08:15 AM.
Yes, firing the bullet will cause a recoil that will move the box. When the bullet strikes the far side it will move it back again.
In principle. In practice there may be complications because of the air in the box, the friction of the surface, etc. This may lead to some (apparent) departure from the ideal case. But if you are able to take all these factors into account, you will find that energy and momentum are fully conserved.
Slight correction. It will come to a stop, with the box having been displaced slightly, but the bullet on the other side of the box and the center of mass of the box+bullet system at its original location.
Ah yes. Good point. Thinking about it in a low friction environment makes that clear. Thank you.
Plus the mass of the gunpowder needs to be involved. The gunpowder has to be considered to shoot here not the gun because the gun is fixed as one with the box.
The smoke is not where the gunpowder was. This is not much less mass then a bullit.
But more important for considering is to realize that the explosion changes the momentum for the system.
A struck in the opposite wall can then be considered as a collision with zero elasticity.
But no matter how elastic this type of collision (no additional energy involved) it can,t change the momentum nor energy for this system (total angular plus linear ).
At least not according to this first law. Reason is both preservation of momentum and energy. (offcourse a temperature raise can be considered momentum also).
Any change of momentum would also imply a change of energy. Thus imply free energy.
Just a collision can,t do that as it doesn,t add energy by itself.
Where I have my doubts with collisions as with this disk is that the relation between angular and linear momentum/ energy is not necessarily preserved.
The total energy must be preserved (the bullit disk collision also doesn,t change the total energy) and the disk rotates / bullit is deflected (angled) without friction it keeps rotating.
When the bullit goes linear before and after the disk is angular momentum but also the reflection angle of the bullit shows angular momentum.
To me part of linear energy has been tranfered into angular/rotational energy then (but total energy preserved) Part of linear momentum has become angular momentum then.
Otherwise this system has gained rotational energy without a loss of linear energy. And when it lost linear energy it lost linear momentum. But the momentum is not lost ...the gain of angular momentum compensates that.
Preservation of momentum (I think) is just not saying preservation of linear and angular momentum to each other.
When the angular momentum is - artificially with math tricks - translated to linear it skips space to using a cirkle as reference to make angular linear to that cirkle. That,s not allowed I think.
No. If the fan is blowing air around a closed loop, it exerts no force on the outside world. The reaction force against the fan blades is exactly balanced by the force of the air blowing onto the interior of the tube. This is the beauty of conservation of momentum. We don't need any hydraulic or aerodynamic calculations to prove it.
That all depends on how much powder you have put into your cartridge. The mass of the powder counts more in the recoil calculation than the mass of the bullet, because the powder gases exit the muzzle faster than the bullet. I believe a weighting factor of 1.7 or thereabouts is used, although strictly speaking this only applies to cartridges in the 30-06 class. However, this is usually plugged into the calculation for all sorts of cartridges, because people are lazy.
No, it doesn't.But more important for considering is to realize that the explosion changes the momentum for the system.
No. Temperature does not affect momentum.A struck in the opposite wall can then be considered as a collision with zero elasticity.
But no matter how elastic this type of collision (no additional energy involved) it can,t change the momentum nor energy for this system (total angular plus linear ).
At least not according to this first law. Reason is both preservation of momentum and energy. (offcourse a temperature raise can be considered momentum also).
Wrong. Momentum can change with no change in energy. For example when something bounces off the wall and recoils at the same speed.Any change of momentum would also imply a change of energy. Thus imply free energy.
What are you trying to say? You are not making sense.Just a collision can,t do that as it doesn,t add energy by itself.
Where I have my doubts with collisions as with this disk is that the relation between angular and linear momentum/ energy is not necessarily preserved.
The total energy must be preserved (the bullit disk collision also doesn,t change the total energy) and the disk rotates / bullit is deflected (angled) without friction it keeps rotating.
When the bullit goes linear before and after the disk is angular momentum but also the reflection angle of the bullit shows angular momentum.
Linear momentum always has angular momentum with respect to some axis of rotation. It does not "become" angular momentum.To me part of linear energy has been tranfered into angular/rotational energy then (but total energy preserved) Part of linear momentum has become angular momentum then.
You should take some classes in physics before posting your foolish comments.Otherwise this system has gained rotational energy without a loss of linear energy. And when it lost linear energy it lost linear momentum. But the momentum is not lost ...the gain of angular momentum compensates that.
Preservation of momentum (I think) is just not saying preservation of linear and angular momentum to each other.
When the angular momentum is - artificially with math tricks - translated to linear it skips space to using a cirkle as reference to make angular linear to that cirkle. That,s not allowed I think.
When the fan blows air around the loop, the container will rotate in one direction as the air inside the container rotates the opposite direction. Angular momentum is preserved because the angular momentum of the fan/container is balanced by the equal and opposite angular momentum of the fluid inside. If two identical loops flowing the opposite direction are attached to one another, nothing moves because the torque applied by each loop at the point of attachment balances out the other loop.
Yes, but... will this work?
Emdrive, are you just going to keep posting pictures of systems you don't understand? Or do you have a serious point to make?
If cat,s would throw dice for this there would be no cats.
It seems that we lived in different logical world!
I'll take that as 'yes' and 'no'.
Trash?
That all depends on how much powder you have put into your cartridge. The mass of the powder counts more in the recoil calculation than the mass of the bullet, because the powder gases exit the muzzle faster than the bullet. I believe a weighting factor of 1.7 or thereabouts is used, although strictly speaking this only applies to cartridges in the 30-06 class. However, this is usually plugged into the calculation for all sorts of cartridges, because people are lazy.
The quantity makes no principal difference offcourse. The powder smoke just doesn,t recoil when the momentums of the gun and the bullit are in balance. To what could it recoil ? The bullit ór the gun (+ box aso).
Recoil would mean an imbalance of momentums between the bullit and gun (+..) and when there is balance what gives the additional momentum to the gunpowder (other then gunpowder plus further content of the expanding space between bullit and gun with/after ignition?.
Be aware that the massbalance for the gun on itself may be at one side of the gunpowder but the gun+box+ disk and all can have the massbalance in front of the barrel opening.
The gunpowder also has a massbalance somewhere plus the system as a whole. So it starts with four G,B,P plus total (G+B+P)
If these are considered massmiddleplanes these planes can be seen as continued through and ouside the barrel and box.
From P (powder) the momentums Pg and Pb must balance. But also Ppowder must be zero.
A recoil wouldn,t keep it to the reference P(powder) but choose the side of the gun .. Then the bullit and powder must have same momentum as the gun......Where would the energy come from then. That would make it a collision instead of a shot.
Last edited by Ghrasp; June 29th, 2012 at 06:21 AM.
Why would you write something like that? You said "The smoke is not where the gunpowder was. This is not much less mass then a bullit." If the amount of powder placed in the cartridge is much less mass than the bullet, then it does make a difference to what you said. How can it be otherwise?
I have no idea what you are trying to say here. Momentum of the powder gases is treated exactly the same as momentum of the bullet. However, it is a gas, and so it does not have a velocity that can be easily measured. A good approximation for the 30-06 cartridge is to take the mass of the powder, and multiply it by the 1.7 times the muzzle velocity of the bullet to find the momentum.
The powder smoke just doesn,t recoil when the momentums of the gun and the bullit are in balance. To what could it recoil ? The bullit ór the gun (+ box aso).
Recoil would mean an imbalance of momentums between the bullit and gun (+..) and when there is balance what gives the additional momentum to the gunpowder (other then gunpowder plus further content of the expanding space between bullit and gun with/after ignition?.
Be aware that the massbalance for the gun on itself may be at one side of the gunpowder but the gun+box+ disk and all can have the massbalance in front of the barrel opening.
The gunpowder also has a massbalance somewhere plus the system as a whole. So it starts with four G,B,P plus total (G+B+P)
If these are considered massmiddleplanes these planes can be seen as continued through and ouside the barrel and box.
From P (powder) the momentums Pg and Pb must balance. But also Ppowder must be zero.
A recoil wouldn,t keep it to the reference P(powder) but choose the side of the gun .. Then the bullit and powder must have same momentum as the gun......Where would the energy come from then. That would make it a collision instead of a shot.
First if it makes a difference then how can you assume it,s not much. Op's aim is not too shoot a bullit here but moving a box or system. He can use as much gunpowder as he wants.
He can also make the ignition come from the bullit,s side or start it in the center of the gunpowder or maybe from heating all at one.
If it,s a fuell oxidation process in a carengine it takes place in the whole cilinder chamber and burns from where the ignition takes place.
That,s what I was pointing at (in my poor english) with the smoke won,t be where the gunpowder was. it,s a process of expansion. Then it,s not allowed to give it a momentum also. At least not speaking of it as "the whole of it".
Pick any particle inside a fuell chamber all other particles move away from it. But then this particular particle would be reference.
If this is close to the gun it sees all other particles accellerate pushing the bullit. If it,s a particle chosen near the bullit it would see the content accellerate away from the bullit.
So none of them is a good choice ....at most a convenient choice (instinctive maybe choosing the gun). That,s not a fysics choice but a preference. I,m looking for preference choice for fysics.
And this is not a gun but an experiment that uses a (potential) gun for different purpose then guns are made for. Otherwise I wouldn,t even participate in this thread.
Then lets try to make a different choice and find a plane or brane that can funktion as zero reference for the momentums to the one dimensional direktion of such a shot.
It needs such a point zero for classic fysics to define an accelleration. Energy - accelleration - momentum. That,s the logic order.
Apart from the shot there are no further accellerations when all collisions would be 100% elastic.
A billiard that has 100 % elastic collisions ..would be a perpetuum mobile but has no accellerations. If it had it would be more then a perpetuum mobile and raise the question..From what energy ? Who interrupted in the shot from the player ?
So somehow somewhere you have to make a difference between just a collision and an initiation for the momentum.
In dutch we use (and I learned to use and think with ...) the word impulse when it,s an initiation to make things move (billiardplayer or ignigition+ gunpowder + oxygen). Once things move it,s about momentum. But this momentum stays related to the impulse and where it took place.
I assumed nothing. The first thing I wrote was "That all depends on how much powder you have put into your cartridge." Do you know what the word "depends" means?It doesn't matter where the ignition starts. The only thing that matters is where the powder gases go (hint: out the end of the barrel) and how fast.He can also make the ignition come from the bullit,s side or start it in the center of the gunpowder or maybe from heating all at one.
Clearly you do not understand conservation of momentum, and how it is used to solve physics problems. You should really learn these things before posting things that make you look foolish.
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