Reactionless drive critique.

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21 comments, last by Lucidium 18 years, 8 months ago
Ok, i've got an idea for a reactionless drive. This shouldn't work, but i think it does, therefore i need to learn something about physics badly. The idea You have a tube, with two magnetised weights, and two 'buffers' (to help stop the weights from rebounding when they hit the end of the tube). Ascii art

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Now, you accelerate each of the weights using the same force (and since there the same mass, they accelerate at the same rate). Because they are the same, there is no net change in velocity for the engine. Now the weight on the left just keeps going and hits the buffer at the end. The weight on the right, however, turns most of its velocity into electricity, before coming to a pretty gentle stop on its buffer. Now because the left weight is moving faster (much, much faster) then the one on the right, when they hit, he engine moves to the left. Now, the electromagnet in the middle pulls both of the weights back to the center. (same mass, same but oppisite distance away, so the forces cancels out). I've thought about this for a while, and there doesn't seem to be anything which would cause problems, and yet, this shouldn't work. Can you find the problem? From, Nice coder
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Yes. As the magnet on the right converts its energy to electricity, it'll encounter resistance in the form of the reaction force due to the induced field. That will balance the inertia.
Quote:Original post by Syphoon
Yes. As the magnet on the right converts its energy to electricity, it'll encounter resistance in the form of the reaction force due to the induced field. That will balance the inertia.


So, where does the energy thats in the electricity come from?

From,
Nice coder
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Hirr, I'm tired and got that wrong. No, you will get movement. Same way you can sit on a swivel chair and rotate yourself without having to touch anything. Of course the problem is when you try to draw the weights back. The speed doesn't matter, unlike with the swivel chair, because there's no external friction to leverage your drawback against.
Yes, but both weights are on oposite ends of two arms, they are both motionless, and they both have the same weight.

When you pull then both together, the forces cancel out, so wheres the problem?

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Nice coder
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Quote:Original post by Nice Coder
The weight on the right, however, turns most of its velocity into electricity, before coming to a pretty gentle stop on its buffer.


As you draw the weights back you will get the same effect - but in an opposite direction. So if the effect of the induced field has the effect that you say it has, then it'll have the same effect on the way back - so your weight's wont' meet in the centre - they'll be a distance to the right of centre (or they won't meet - the weight on the right won't return to the centre).

This also means that when returning to the centre the weight on the left will have a higher velocity than the weight on the right - which would mean that if they did meet the drive would probably move to the right a similar amount to which it had moved to the left in the initial motion.
Quote:Original post by Anonymous Poster
Quote:Original post by Nice Coder
The weight on the right, however, turns most of its velocity into electricity, before coming to a pretty gentle stop on its buffer.


As you draw the weights back you will get the same effect - but in an opposite direction. So if the effect of the induced field has the effect that you say it has, then it'll have the same effect on the way back - so your weight's wont' meet in the centre - they'll be a distance to the right of centre (or they won't meet - the weight on the right won't return to the centre).

This also means that when returning to the centre the weight on the left will have a higher velocity than the weight on the right - which would mean that if they did meet the drive would probably move to the right a similar amount to which it had moved to the left in the initial motion.


Hmmm.

I could imagin changing the coil (ie. charging it), so that it has no effect. (ie. energising the field lines just above/below the magnet. then energising the one in front, using the energy of the one behind. You would have to top it up, tho...)

Or else, you could move the coil away. (little moter?)

Or you could energise the coil with a small field.
When you do that, the magnet will spin as it moves through the field, but it will have no net effect. (or very little).

If this is a problem, then you could use a spherical magnet (with one pole in its core, the other on its surface)

From,
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The force exerted on the right hand weight by the electromagnet will be balanced by a force on the tube/engine. Since all the forces are internal to the system and balanced, there will be no net acceleration.
You are exerting internal forces on a closed system. The positions of the individual
pieces will change, but the center of mass will not.
Kul - we know that - even the OP acknowledges that he is missing something... we would just like to point out what...

Quote:The weight on the right, however, turns most of its velocity into electricity, before coming to a pretty gentle stop on its buffer


You can't turn velocity into electricity. Since we are doing physics here, let's be precise. If you wish, you may slow down the one on the right in such a way that it's kinetic energy is converted into electrical energy, but in doing so you will exert a force on the object (without a force, it won't slow down). When you exert a force on the object, it will exert one on you. This is not optional. The transfer of momentum on striking the end will differ from that transferred at the other end by exactly the same amount as the amount transfered via the electro magnetic field.

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