Does matter and/or energy

[jjjjj]

Hmm...I can't think of any example where a CYCLIC system doesn't lose ANY matter/energy. For example, a pendulum is cyclic, but at the cost of friction. As for the Earth going around the Sun, well like I said I'm 14 so I don't know much, but I doubt gravity is an axiom.

[swansont]

They all will, because of the second law of thermodynamics. If a system is left to itself, entropy will not decrease, and this manifests itself as a loss of energy from a mechanical system.

[ninus maximus]

The sphere is in contact with the ground. It translates because there is friction at the point(s) of contact. No friction, no translational motion.

 

I'm dissagreeing with you swansont so I suppose I will be banned or suspended again , but anyway for the sake of Physics here goes.

 

The Sphere does not have a external force acting on it.

 

the sphere is a closed system.

 

and the friction is not a external force it is the result of the internal force.

[Sisyphus]

I'm dissagreeing with you swansont so I suppose I will be banned or suspended again , but anyway for the sake of Physics here goes.

 

The Sphere does not have a external force acting on it.

 

the sphere is a closed system.

 

and the friction is not a external force it is the result of the internal force.

 

The sphere exerts a force on the ground, and the ground exerts an equal and opposite force on the sphere. It is not a closed system.

[ninus maximus]

The sphere exerts a force on the ground, and the ground exerts an equal and opposite force on the sphere. It is not a closed system.

 

I dont want to argue , only dissagree if that is allowed.

 

In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.

 

the sphere is sealed , matter cannot be exchanged.

 

closed system

 

nothing comes out of the sphere and reacts with the spheres surroundings

so the sphere is an example of reactionless propulsion.

Edited July 6, 2010 by ninus maximus

[swansont]

I'm dissagreeing with you swansont so I suppose I will be banned or suspended again , but anyway for the sake of Physics here goes.

 

To be very clear, you were not suspended for disagreeing. You were suspended for how you conducted yourself

 

 

The Sphere does not have a external force acting on it.

 

the sphere is a closed system.

 

and the friction is not a external force it is the result of the internal force.

 

Bollocks. The earth is external to the sphere, not a part of it. QED.

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I dont want to argue , only dissagree if that is allowed.

 

In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.

 

the sphere is sealed , matter cannot be exchanged.

 

closed system

 

nothing comes out of the sphere and reacts with the spheres surroundings

so the sphere is an example of reactionless propulsion.

 

But this is not a thermodynamics problem.

[ninus maximus]

To be very clear, you were not suspended for disagreeing. You were suspended for how you conducted yourself

 

 

 

 

Bollocks. The earth is external to the sphere, not a part of it. QED.

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But this is not a thermodynamics problem.

 

I suppose that the above shows that you are abiding by the rules by giving me an example of how I should conduct myself.

 

 

Bollocks is a word of Anglo-Saxon origin, meaning "testicles". The word is often used figuratively in British English, as a noun to mean "nonsense", an expletive following a minor accident or misfortune, or an adjective to mean "poor quality" or "useless". Similarly, the common phrases "Bollocks to this!" or "That's a load of old bollocks " generally indicate contempt for a certain task, subject or opinion. Conversely, the word also figures in idiomatic phrases such as "the dog's bollocks" and "top bollock(s)", which usually refer to something which is admired, approved of or well-respected.

 

Bollocks

 

I was suspended for a week for doing what you just did , now will a moderator please perform their duty and warn swantsont and begin to count his violations to the forum rules , that is if the rules also apply to the moderators.

 

I think that what is good for the goose is also good for the gander.

 

now to continue.

 

the earth is external to the sphere , but the earth does not apply a force

to the sphere if the truck inside the sphere is not turning its wheels.

 

the sphere will not simply move without the trucks wheels turning.

 

so the force is comming from within the sphere.

 

not from the earth.

 

the force aplied by the trucks wheels are the action and the inside of the sphere provides the reaction force , and this force is transfered to the sphere as a action force and the earth provides the reaction force.

 

it is a closed system and according to the deffinition of a reactionless propulsion drive , it is a reactionless propulsion drive.

 

reactionless drive

 

but it may be that you have a link to a better deffinition of a reactionless drive that

you could share.

Edited July 6, 2010 by ninus maximus

[Sisyphus]

The truck applies a force to the sphere and vice versa, correct. So far, you've got the sphere spinning in place.

 

Now, in order to get the sphere to go anywhere, it has to apply a force to the ground (and vice versa). It is able to do this because of friction, which is also how the truck and the sphere can interact.

 

There is nothing "reactionless" going on.

[ninus maximus]

The truck applies a force to the sphere and vice versa, correct. So far, you've got the sphere spinning in place.

 

Now, in order to get the sphere to go anywhere, it has to apply a force to the ground (and vice versa). It is able to do this because of friction, which is also how the truck and the sphere can interact.

 

There is nothing "reactionless" going on.

 

I believe that the deffinition on the internet does not stipulate friction.

 

it only stipulates that it is

"not based around expulsion of fuel or reaction mass."

 

nothing comes out of the sphere , so the sphere qualifies as a reactionless drive.

 

there was a time when the sphere being a closed system would violate newtons laws because the sphere could move without a external force being applied to it.

 

this time has passed , but I remember it well.

 

back then , nothing could cross the boundary , work , energy , were bound inside a closed system.

 

now work and energy can pass the boundary , only matter is denied.

 

this too will change.

Edited July 6, 2010 by ninus maximus

[Sisyphus]

The reaction mass is the Earth itself. For the sphere to move forwards, the rest of the Earth has to be given an equal and opposite momentum (which is obviously way too small to notice given the mass difference).

 

Though if the sphere were a "reactionless drive," then you wouldn't need the sphere. Just drive the truck.

[swansont]

You should actually read the links you provide.

 

A reactionless drive or inertial propulsion engine (also reactionless thruster, reactionless engine, bootstrap drive, and inertia drive) is any form of pseudoscientific propulsion not based around expulsion of fuel or reaction mass.

 

The name comes from Newton's Third Law of Motion, usually expressed as: "For every action, there is an equal and opposite reaction." Such a drive would use a hypothetical form of thrust that does not require any outside force or net momentum exchange to produce linear motion, and therefore necessarily violates the conservation of momentum, a fundamental principle of all current understandings of physics. In addition it can be shown that conservation of energy is violated.

 

In spite of their physical impossibility, such devices have been often proposed in recent history.

 

 

An action-reaction force pair is present here — the earth exerts a force on the sphere, and the sphere exerts a force on the earth. And since you provided the link, do you agree this is pseudoscience and violates conservation of momentum and/or energy?

 

 

I was suspended for a week for doing what you just did

 

You were suspended for a week for calling people stupid, and other related behavior. It would behoove you to learn the distinction between posts and the people who make them.

[J.C.MacSwell]

I dont want to argue , only dissagree if that is allowed.

In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings.

the sphere is sealed , matter cannot be exchanged.

 

closed system

 

nothing comes out of the sphere and reacts with the spheres surroundings

so the sphere is an example of reactionless propulsion.

 

 

 

In thermodynamics, a closed system cannot exchange heat and work with its surroundings.

 

Edit: I see some different use of the term "closed system".

Edited July 7, 2010 by J.C.MacSwell

[PaulS1950]

Ninus Maximus,

if the truck were sitting on the ground without the sphere would your opinion be different?

The sphere is a counter "wheel" of the truck - the tires of the truck drive the sphere in the same way that the motor drives the tires. All of the reactions come from the magnetic interactions within the electric motor that is supplied with power from a battery. It is all a number of action/reactions that end at the friction of the gravity working on the sphere allowing movement through the motor to the gears to the tires to the sphere. An inertial drive must be able to move even when it is in micro gravity and a vacuum. Your toy truck could be made to move its tires and they could be made to move the shere but it would not have linear travel without the weighted contact with a surface to supply friction.

I may not completely grasp what an inertial engine is but I know enough about mechanics to know that you are just using a final reduction of speed by running wheels inside a sphere. The tires are just friction gears to the final wheel which is the sphere.

[ninus maximus]

Ninus Maximus,

if the truck were sitting on the ground without the sphere would your opinion be different?

The sphere is a counter "wheel" of the truck - the tires of the truck drive the sphere in the same way that the motor drives the tires. All of the reactions come from the magnetic interactions within the electric motor that is supplied with power from a battery. It is all a number of action/reactions that end at the friction of the gravity working on the sphere allowing movement through the motor to the gears to the tires to the sphere. An inertial drive must be able to move even when it is in micro gravity and a vacuum. Your toy truck could be made to move its tires and they could be made to move the shere but it would not have linear travel without the weighted contact with a surface to supply friction.

I may not completely grasp what an inertial engine is but I know enough about mechanics to know that you are just using a final reduction of speed by running wheels inside a sphere. The tires are just friction gears to the final wheel which is the sphere.

 

I personaly dont think that the sphere is a reactionless drive , it is the deffinition that causes me to say that it is.

 

I had a perfect example of a inertial drive in another thread but it was closed because the concept was not understood by those in this group and name calling and insults resulted from the lack of understanding.

 

that missunderstanding resulted from swantsonts use of the below equation that requires the

weight of the mass changing , but the weight of the mass does not change.

 

 

notice the M1-M2 in the above , this calcuates a change in velocity due to a change in mass , not a change in the masses velocity becauses it encounters a turnaround.

this equation can only be used to describe the change in velocity a mass will undergo if its mass

changes while it is moving.

I would like to discuss that concept but it seems that theres no point in trying.

 

this forum seems to restrict thought if the thought doesnt fall within the moderators realm of reality

be it reality or not.

Edited July 7, 2010 by ninus maximus

[Sisyphus]

I personaly dont think that the sphere is a reactionless drive , it is the deffinition that causes me to say that it is.

 

The definition says something that accelerates without reaction mass, i.e. violates Newton's third law. A truck pushes off the ground to move forward, therefore it doesn't qualify.

 

Perhaps the definition as given on Wikipedia is not very clear, because it isn't really a science article. It's an article about a pseudoscientific concept that sometimes pops up in works of fiction.

 

I had a perfect example of a inertial drive in another thread but it was closed because the concept was not understood by those in this group

 

Have you considered the possibility that it was not everyone else, but you who misunderstood something?

 

this forum seems to restrict thought if the thought doesnt fall within the moderators realm of reality

be it reality or not.

 

No, what we do is close flame wars that aren't going anywhere.

 

If you say something that is technically, demonstrably inaccurate and someone corrects it, is that really "restricting thought?" Is it persecution to be contradicted in anything you say?

[ninus maximus]

The definition says something that accelerates without reaction mass, i.e. violates Newton's third law. A truck pushes off the ground to move forward, therefore it doesn't qualify.

 

Perhaps the definition as given on Wikipedia is not very clear, because it isn't really a science article. It's an article about a pseudoscientific concept that sometimes pops up in works of fiction.

 

 

 

Have you considered the possibility that it was not everyone else, but you who misunderstood something?

 

 

 

No, what we do is close flame wars that aren't going anywhere.

 

If you say something that is technically, demonstrably inaccurate and someone corrects it, is that really "restricting thought?" Is it persecution to be contradicted in anything you say?

 

yes , the wiki deffinition is vague but it is all that we have to work with.

perhaps they need to clarify the deffinition to include a stipulation that stipulates that

any reactionless drive cannot react against matter because there is matter everywhere even in deep space.

 

this way there would never be a possiblity of ever achieving a 100% reactionless drive.

 

 

there isnt anything to misunderstand , its just basic physics.

 

the example I used was a force of 500N applied to a 100kg mass

and a 500N force applied to a 1000kg mass

 

one force pushing a mass and one force pushing another mass.

 

theres nothing "technically, demonstrably inaccurate " about that.

 

then the 100kg mass passes through a turnaround

 

theres nothing "technically, demonstrably inaccurate " about that.

 

the 100kg mass goes through the turn and applies a force to the 1000kg mass

causing the 1000kg mass to stop.

 

theres nothing "technically, demonstrably inaccurate " about that.

 

assuming no friction , and friction can actually be negligible if engineered

correctly , the 100kg mass would exit the turn at the same velocity it entered the turn only in the opposite direction.

 

theres nothing "technically, demonstrably inaccurate " about that.

 

in fact all of the above are technically, demonstrably accurate.

Edited July 7, 2010 by ninus maximus

[swansont]

We've been through this. Conservation of momentum is trivial to apply to this system, at each step. How about doing it? Last time you declined to do so.

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that missunderstanding resulted from swantsonts use of the below equation that requires the

weight of the mass changing , but the weight of the mass does not change.

 

 

notice the M1-M2 in the above , this calcuates a change in velocity due to a change in mass , not a change in the masses velocity becauses it encounters a turnaround.

this equation can only be used to describe the change in velocity a mass will undergo if its mass

changes while it is moving.

 

Absolutely untrue. m1 and m2 are unchanging masses that comprise the whole system. m1 is the mass that reverses direction, and m2 is the mass of everything else in the system (100 kg and 1000 kg, using your numbers from above). This is from a physics 101 derivation, available in any introductory physics textbook, in case you can't derive it yourself.

Edited July 7, 2010 by swansont
Consecutive posts merged.

[ninus maximus]

We've been through this. Conservation of momentum is trivial to apply to this system, at each step. How about doing it? Last time you declined to do so.

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Absolutely untrue. m1 and m2 are unchanging masses that comprise the whole system. m1 is the mass that reverses direction, and m2 is the mass of everything else in the system (100 kg and 1000 kg, using your numbers from above). This is from a physics 101 derivation, available in any introductory physics textbook, in case you can't derive it yourself.

 

I used your description of the equation and gave it a try.

 

m1 = 100kg

m2 = 1000kg

 

m1-m2 / m1+m2 *v1i = v1f

 

100-1000/100+1000 * 40 m/s = v1f

 

-900/1100 * 40 m/s = v1f

 

= -32.727272727272727272727272727273 m/s

 

am I to believe that the mass accelerates from 40 m/s to -32.72 m/s?

 

because that is the results from your equation.

 

for that to happen a force would need to first stop the mass , then accelerate the mass

in the opposite direction.

 

how exactly do you use that formula if the above is not correct?

why dont you work the equation for me , because I cant see any possibility of

that answer being correct.

Edited July 7, 2010 by ninus maximus

[insane_alien]

why can you not see that answer being correct? it looks fine to me.

[ninus maximus]

why can you not see that answer being correct? it looks fine to me.

 

Because it shows that the 100kg mass stops and then backtracks , there is no obstruction causing the mass to stop , and there is no additional force causing the mass to accelerate in the opposite direction.

 

if it looks fine to you , then perhaps you can explain what the result tells you.

 

what does v1f describe.

 

m1 or m2 or both?

 

is it the final velocity of m1 or m2?

 

if it does describe a mases velocity , how did you determine which masses velocity it describes.

 

only the velocity of m1 is given in the equation , so just exactly how does the equation know

what the velocity of m2 is?

 

one other point I would like you to touch on , there is no resistance force of any kind mentioned in

the equation , so in that case the mass would not even slow down.

 

but the results I get show the mass stops then goes backwards it backtracks the way it went into the turn , Im not saying its momentum changed Im saying that it never completed the turn , it reversed its momentum and headed backwards.

 

thats not possible.

 

could you elaborate a little on how you believe the answers to be correct?

Edited July 7, 2010 by ninus maximus

[insane_alien]

if it looks fine to you , then perhaps you can explain what the result tells you.

 

what does v1f describe.

 

the velocity of 1 after the interaction. you seem to understand this later so why are you asking?

 

m1 or m2 or both?

 

is it the final velocity of m1 or m2?

m1 as shown by the one.

 

if it does describe a mases velocity , how did you determine which masses velocity it describes.

generally you use a subscript in the mathematics, like 1 and 2 for different objects.

 

only the velocity of m1 is given in the equation , so just exactly how does the equation know

what the velocity of m2 is?

 

it doesn't. nor does it need to, the final velocity of 2 doesn't need to be calculated for the final velocity of 2 to be found. you can work it out from the available info if you want though.

 

could you elaborate a little on how you believe the answers to be correct?

 

its mathematically correct, and theoretically correct. it doesn't use advanced physics notions at all. infact, this physics has been in use for centuries, we'd have figured out if it was wrong by now. especially when we do stuff like experiments.

[ninus maximus]

the velocity of 1 after the interaction. you seem to understand this later so why are you asking?

 

m1 or m2 or both?

 

m1 as shown by the one.

 

generally you use a subscript in the mathematics, like 1 and 2 for different objects.

 

 

 

it doesn't. nor does it need to, the final velocity of 2 doesn't need to be calculated for the final velocity of 2 to be found. you can work it out from the available info if you want though.

 

 

 

its mathematically correct, and theoretically correct. it doesn't use advanced physics notions at all. infact, this physics has been in use for centuries, we'd have figured out if it was wrong by now. especially when we do stuff like experiments.

 

so you are saying that the mass m1 stops?

and that the mass m1 then accelerates in the opposite direction?

 

100kg * 40 m/s = 4000N

it could be stopped but it would require a resistive force to do so , if there were

enought resistance , and it would require 4000N resistive force to stop it in 1 second , but

where does it get the needed force to reverse?

ie...100kg * 32 m/s = 3200N

plus there is the resistive force that the mass met entering the turn so we need

to include that also , there is a resistive force of 4000N that the mass meets as it accelerates

backwards , so now the mass needs a force of 7200N in order to achieve 32 m/s in the opposite

direction as it backtracks.

 

it only had 4000N when it entered the turn , how did it gain 7200N by

entering the turn?

Edited July 8, 2010 by ninus maximus

[insane_alien]

so you are saying that the mass m1 stops?

and that the mass m1 then accelerates in the opposite direction?

 

this is typically what happens when something changes direction, so yes.

 

100kg * 40 m/s = 4000N

 

this is not how you calculate force. what you calculated there was momentum, not force.

 

it could be stopped but it would require a resistive force to do so , if there were

enought resistance , and it would require 4000N resistive force to stop it in 1 second , but

where does it get the needed force to reverse?

ie...100kg * 32 m/s = 3200N

 

the force comes from whatever its colliding with. again you have the forces all wrong.

 

it only had 4000N when it entered the turn , how did it gain 3200N by

entering the turn?

 

it didn't.

[ninus maximus]

this is typically what happens when something changes direction, so yes.

 

100kg * 40 m/s = 4000N

 

this is not how you calculate force. what you calculated there was momentum, not force.

 

the force comes from whatever its colliding with. again you have the forces all wrong.

 

 

 

it didn't.

 

when something changes directon it does not stop.

 

it changes direction , and that is all.

 

the mass does not stop it turns.

 

4000N resistive force to stop it in 1 second = 4000N-s whats the difference?

 

are you saying that the mass does not go through the turn?

[swansont]

I used your description of the equation and gave it a try.

 

m1 = 100kg

m2 = 1000kg

 

m1-m2 / m1+m2 *v1i = v1f

 

100-1000/100+1000 * 40 m/s = v1f

 

-900/1100 * 40 m/s = v1f

 

= -32.727272727272727272727272727273 m/s

 

am I to believe that the mass accelerates from 40 m/s to -32.72 m/s?

 

because that is the results from your equation.

 

for that to happen a force would need to first stop the mass , then accelerate the mass

in the opposite direction.

 

how exactly do you use that formula if the above is not correct?

why dont you work the equation for me , because I cant see any possibility of

that answer being correct.

 

 

Then how is it possible that your scenario is correct, in which the mass goes from 40 m/s to -40 m/s?

 

the 100kg mass would exit the turn at the same velocity it entered the turn only in the opposite direction.

 

That the mass changes direction is not something that should be in question here, since you've already agreed that it happens.

 

The point is that the speed isn't 40 m/s after it turns around, since some momentum has to be transferred to the 1000 kg mass. And that's why your scenario is impossible.