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Question

Asked by: Blaze
Subject: One shot linear coasting motion using barycenters
Question: Please see the attached link for a diagram of the device. There are no spinning masses in this device (no gyroscopes or flywheels).

http://min.us/i/MQjSwbXP7hs4

I believe this device should produce linear motion from a one time centrifugal force "thrust pulse" and then, if in a non gravity environment, coast indefinitely. When the paired masses (A & B and A' & B') collide all the energy of the perfectly inelastic collision is turned into heat and is assumed to dissipate in all directions equally. Before collision, the barycenter movements of the paired masses perfectly counteract each other and what is left is centrifugal force from the angular momentum of the masses.

Any thoughts?

regards,
Blaze
Date: 19 July 2014
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Answers (Ordered by Date)


Answer: Glenn Hawkins - 20/07/2014 05:41:10
 Hello Blaze,
It is strange that this should come up now when earlier tonight I wrote, but did not post a very similar example and explanation. Still earlier I posted a kind of alert of it for MD.

On the thread: Subject: Equal but not opposite 2
Glenn Hawkins - 13/07/2014 09:40:10

“I did tests yesterday that validate my mechanical conclusions, which are that acceleration can be had. This is definitive. The tests were so simple and yet, I had searched for a means of proof for years.

You are so close to creating acceleration. You just don't know it.”
……………………………………………………………………………………
I think this schematic you offer would create a one shot acceleration and coasting, but it is a far cry from what similar ideas are capable of producing. There are more efficient and more powerful ways to produce continuous thrust. I have tested them; not merely reasoned them. The thrust is inarguable real in that you can watch it happening and it is powerful. People are finally closing in. It will be thrilling to watch. Gyroscopes are not necessary, but you can use them. I just never thought it was possible without them.

Regards,
Glenn
I am on vacation. If no one gets it, in the coming months I will post the information, if I don’t build it.

Blaze: “. . .what is left is centrifugal force from the angular momentum of the masses.”
As well as I can reason without testing to substantiate, that is precisely: the source of force to thrust.


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Answer: Harry K. - 20/07/2014 08:10:50
 Hi Blaze,

I don't believe that this will work. After the collision of all involved masses the centrifugal forces will decrease to zero in a certain (short) time span but the same happens with the counter directed centripetal forces and thus all acting forces will be canceled out.

Regards,
Harald

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Answer: Glenn Hawkins - 20/07/2014 12:15:52
 I haven’t tested it, but the idea is that during the time of motion, centrifuge pulls. Centripetal force on the other hand is merely a resistance, as I’ve said. It generates no power, and when the source of power i.e. angular momentum ceases the resistance i.e. centripetal ceases. Centripetal holds. It does not pull back and it can not resist coasting. And even if it could, which it can’t, it could not resist coasting that has already been put into play at an early time and is occurring.

Perhaps this schematically presented idea here won’t work; I cannot know for certain, but I have made something somewhat similar work perfectly. What more can I say? I watched it!
Glenn,


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Answer: Blaze - 20/07/2014 16:53:58
 Thank you for your comments gentlemen.

Centrifugal force is generated by the angular motion of the moving masses.
Centripetal force is equal to centrifugal force and is generated by the resistance to acceleration of the platform. If we know the centrifugal force we can calculate the rate of acceleration of the platform from F=M*A.
Fcpetal/Mass of platform = Acceleration of platform

The platform should accelerate during the angular movement of the masses due to centrifugal force. The velocity of the platform just before the masses collide is found from V=A*t (assuming acceleration is constant, which is probably isn't but even if it isn't it would just require an integral to calculate the velocity). When the masses collide, both centrifugal and centripetal forces are cancelled out at the same rate but that should not affect the existing velocity of the platform. To stop the existing velocity of the platform would require a counter force or acceleration to the direction of movement of the platform. I don't see where that happens.

cheers,
Blaze

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Answer: Harry K. - 21/07/2014 07:33:48
 Hello Blaze,

I do not agree. Assumed the complete system (platform and pivoted masses) would be placed in space, i.e. without influence of gravity and friction, the pivoted masses would move the platform in counter direction in a manner, that the center of gravity of the whole system would remain at its origin place.

Your consideration about what might happen is a bit illogical as well. If the platform would move in the direction of your assumed direction of centrifugal force, "real" centripetal forces could not exist to cause "fictious" centrifugal forces!

Regards,
Harald

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Answer: Nitro - 21/07/2014 12:29:44
 Dear Blaze

I am sorry to say that I cannot see anything about the proposed device that will “get round” the third. I understand that the angular deceleration of the two masses in each pair movement will produce a force acting in the direction of the resultants of their centrifugal acceleration and their impacts which would indeed act to move the base in the direction shown. However, (there is always a bloody “however!) I believe that you will find that there is an exactly opposite force that acted in the reverse direction upon the base as the masses were initially accelerated.

This seems to be a variant of the proverbial “pendulum on a skateboard” experiment. It will move in one direction more than any other only if you put a ratchet on it – unfortunately you cannot effectively use a ratchet in a space craft without transferring the reverse motion as well as the forward motion to the spacecraft. Good for shaking the teeth out of the spacemen but not much else, I’m afraid.

Kind regards
NM

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Answer: Blaze - 22/07/2014 03:12:41
 Thank you for your further thoughts gentlemen. You may be correct but before I describe why I see it differently here are a few more facts that you may have missed at first glance.

The final angle of collision is not evenly spaced between the two starting points for a mass pair. It is actually 60 degrees from A and A'. You can see that the collision point is not evenly spaced between the starting points in the diagram and it was drawn this way on purpose. This is one of the requirements to keep the pivot movements forward and backwards equal due to the barycenter effect. Because of the non centered collision angle, the individual masses of the mass pairs (A & B and A' & B') will have to accelerate at different rates. These different rates of acceleration vary throughout the whole rotation about the pivot until collision occurs. However the final speed the instant before collision is exactly the same for A, A', B, and B'.

Nitro, because the pivot movement forward and backwards due to the barycenter effect is exactly balanced during acceleration of the masses, the platform does not move forwards or backwards during acceleration of the masses due the the barycenter effect. Any platform movement is due to the centrifugal forces generated by the masses rotating about the pivot points.

Harry, you are concerned that there will be no centripetal force if the platform moves. As I explained in my previous posting, the centripetal force is generated because the platform moves or more correctly stated, because it accelerates. This is a key piece to understand about the device. It is the Acceleration of the Mass of the platform that causes the resisting Force to centrifugal acceleration which is the centripetal acceleration. If the platform were attached to an infinite mass then it would not have to move to generate the centripetal force. However, since it is free to move (but not rotate like a binary star system) it has to accelerate to generate any resisting (centripetal) force.

regards,
Blaze


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Answer: Glenn Hawkins - 22/07/2014 16:38:07
 Two mules are chained together and pulling against one another in a tug-of-war. They pull a length of chain in the center apart. The length of chain falls still on the ground while the mules walk away. The chain was only a molecular structure holding two balanced and outward forces that were pulling against one another. If you remove one mule, the other mule walks away dragging the dead chain. Remove both mules and you are left with nothing but a slack chain lying uselessly and motionlessly in the dirt. The mules exert the power, not the center length of a chain.

I will entertain any countering argument given with logic and reason, but nothing for blanket statements drilled into children's minds to remain forever unexplained.

Glenn,

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Answer: Glenn Hawkins - 23/07/2014 00:47:52
 
The action on this platform design as it is suggested to cause inertial propulsion, has been reasoned in this way.

1) The method to force B to swing forward must be anchored onto the platform. The platform then will react backwards in equal mass times distance to the forward swing of the weight. The impact of B’s collision with A drives the platform forward into a STOP of platform motion. Mass has been displace inside the confines of a circle so to speak. All the action follows standard physics and nothing is new here.

2) The twins A + A, however are driven by a shared force of one against the other. The reaction onto the anchored platform is then countered by reaction against reaction. The platform itself is not moved sideways. The resulting centrifuge force caused by the angular momentum of A+A is free to pull the platform at dual and equal angles forward into coasting.

3) However, we cannot really believe that so simple a solution to cause inertial propulsion could have escaped notice for two thousand years of investigation. Based on the obvious of past trial and error I can assure you this will not work, but I do not know why the logic of it fails.

Toast to all drinking men of good cheer. We know this is not you Nitro. We know you do not indulge. It is that other wild man who looks like you. ;-)

Glenn,


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Answer: Blaze - 24/07/2014 02:21:49
 Hi Glenn. You have most of the functioning correct but are still missing some of it.

Both the A and B pairs cause the platform to move, the A pair forward, the B pair backwards. The A pair generates very, very little barycenter movement when starting out and the B pair generates a lot of barycenter movement when starting out. To keep the barycenter forward and backward movement of the platform the same amount from both the A and B pairs it is necessary to move them different amounts. For a small movement of the A pair it is necessary to move the B pair a very, very small amount. If the A pair moves 1 degree of rotation the B pair only moves 0.00873 degrees of rotation. This sounds extremely small but that is because the amount of barycenter movement from the A pair in that first degree of rotation is also extremely small.

Looking at it the other way around, if the B pair is moved 1 degree of rotation, the A pair has to move 10.7 degrees of rotation. When the B pair has moved 2 degrees, the A pair has moved 15.2 degrees. When the A and B pairs are very near the collision point (like a degree or less away) they both move very nearly the same distance as each other for that last little bit of rotation.

The collision of the A and B pairs does not move the platform in either direction or stop it from moving in either direction. This is because the speeds and “weights” of the masses are all exactly the same at the time of collision. The kinetic energy is all turned to heat, assuming a perfectly non elastic collision. If one of the masses was moving faster or was a bit more massive (not the case here) then the collision would move the platform one way or another.

Because the barycenter forward and backward motions are perfectly countering each other during the entire rotation of the masses, there is only centrifugal force pulling the platform forward.

Finally, if the concept in your point 3 regarding a simple solution to a problem going unnoticed for a long time were true, then wheel, the lever, the compound pulley, gears, timing pulleys or any other number of simple inventions for that matter, would have been invented much earlier than they were.

Cheers,
Blaze


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Answer: Glenn Hawkins - 24/07/2014 11:12:36
 Hi Blaze’
Your last paragraph got me interested in the subject of time and invention and so I looked up the following and I will shear it with you. These inventions came at the very dawning of of the study physics and mathematics and depicted information, but surely much earlier.

1) Wikipedia
The earliest well-dated depiction of a wheeled vehicle (here a wagon—four wheels, two axles) is on the Bronocice pot, a c. 3500 – 3350 BC clay pot excavated in a Funnelbeaker culture settlement in southern Poland

2) The Greek mathematician Archimedes was the first to describe levers in approximately 260 B.C

3) Archimedes screw, the compound pulley and the ... Archimedes of Syracuse invented the first compound pulleys 287 BC - 212 BC

4) Aristotle mentions gears around 330 BC, (wheel drives in windlasses). He said that the direction of rotation is reversed when one gear wheel drives another gear wheel.

5) Hero of Alexandria 10 – c. 70 AD) identified the pulley as one of six simple machines used to lift weights.[2] Pulleys are assembled to form a block and tackle in order to provide mechanical advantage to apply large forces.

Cheers,
Glenn


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Answer: Glenn Hawkins - 24/07/2014 11:42:31
 Hi Again Blaze,
Let us say you are correct in your assessments and explanations. Very good, but it doesn’t matter to me as much as it normally would, because last week and again last night I have been physically testing similar principles, not the same structure. The reason I am having trouble excepting the positive and observed test results is that the conditions are so simple, as are the theoretical schematic you offered, that I think my findings would have been discovered long before now. If that were not true, then we both are on to the great solution. It would explain why Nitro and MD have had a measure of success. Incidentally there is a much better way, (I can’t believe my eyes of what I have witnessed). The better way is somewhat more complicated, but the primary principle is similar, but not exact at all. In observation and theory it can produce contagious acceleration quite easily and substantially. I don’t trust what I see, what I measure, what I find.
Cheers,
Glenn


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Answer: Harry K. - 24/07/2014 12:43:44
 Hello Blaze,

Now I believe to know what you mean. The idea behind it is pretty good!
If you consider 1 pair of masses, e.g. A&B, the common center of mass is identical in the platform moving direction at the origin starting position and the final colliding position, assumed the angle in final position is 30° above horizontal axis. To collide at 30° mass A or A' must move twice the time as mass B or B'.

That means, if you could achieve to keep the center of mass constant during the movement of each mass pairs, your assertion would be correct because the center of mass of the complete assembly would not be changed.

Unfortunately the center of mass of the mass pairs A/B or A'/B' moves from the starting position upwards and backwards again to the colliding position in a sinusoidal like manner. This causes in return a downward movement of the platform.
But again, if the center of mass would be constant for each both mass pairs during their movement, your findings would be correct.

Regards,
Harald


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Answer: Glenn Hawkins - 24/07/2014 20:42:17
 You are on to something. Stick with it, Blaze.

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Answer: Glenn Hawkins - 29/07/2014 17:04:30
 Hello All,

MD, I was wrong and there is no reason to see my false discovery.

I've done more theory and tested them. My 'great discovery' that I called it, and stated did not trust it, is a fluke. Friction can act in the most baffling ways-- if you are innovating enough to design so much functioning testing. I found that it was rotational friction that was responsible for the thrust. Equal and opposite was in full play. The thing just can't work.

Therefore Blaze, your schematic will not work for the same reason.

Centrifuge and momentum on another note. This is for those capable of imagination and following described mental pictures.

I cut a section of hollow graphite fishing rod. I dissembled a toy gyroscope; removing the bell housing, leaving only the wheel and shaft. I inserted one side of the shaft into an opening of the hollow graphite piece and tilted the piece upward so that the gyro would not slide out. I managed to spin the gyro up then holding the piece, I begin rotating my body. I tested several different things similar, actually. Each time centrifuge pulled the gyroscope out of the opening in the hollow graphite piece and flung it several feet.

TO SATISFY YOURSELF YOU CAN TRY THIS experiment this way or your own such way The conclusion is that in forced precession centrifuge is present and therefore momentum to cause centrifuge is present. There seems to be a full measure of centrifuge whether the gyroscope is spinning or not.

The cause of all the confusion that in forced precession causes the gyro to fold over, twisting the same as in gimbals rings. The torque from this presses oppositely down on the hub area and the gyro lifts itself as a lever curving upwards. As the wheel must move inward in order to move upward, it was assumed the centrifuge was absent. It is not at all, but the folding torque can be strong enough to overpower centrifuge.

Another fallacy. As the he gyroscope rises, the diameter of its rotation decreases, therefore you see the same effect as the ice-skater bring her arms in to rotate faster. Faster at a smaller diameter of rotation does not mean less resistance is occurring at faster speeds.

I have one more designed to check out and if it doesn't work I will have gone through everything I dreamed up over the years. I will have tried everything.

Cheers, Glenn


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Answer: Blaze - 29/07/2014 19:36:24
 Hi Glenn. "There seems to be a full measure of centrifuge whether the gyroscope is spinning or not."

When it comes to me, you are preaching to the choir. Here are a few of my thoughts from a previous postings on other threads:

"The incorrect assumption that nearly everybody on this forum seems to make is that a gyro has no centrifugal force when precessing and therefore there will be no barycenter for the gyro and the tower."

"All centrifugal force, centripetal force, reactions at the pivot when starting the gyro, etc. etc. etc. apply to gyros whether they are overhanging or not, whether they are precessing at full speed, over full speed, or under full speed, and all of these forces act fully and to the same degree on spinning or non spinning masses."

Regarding my schematic in this thread, it REQUIRES that "equal and opposite" are in full play and that all centrifugal and centripetal forces are present.

One thing I like to say regarding gyros is:
"Thinking out of the box will only get you to the point that propulsion using gyros may be possible. To actually come up with a solution you have to think out of the tesseract."

cheers,
Blaze






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Answer: Glenn Hawkins - 30/07/2014 03:21:33
 Hi Blaze,
Well, you can't get more adamant than that. Yes it is all true,
Cheers,
Glenn

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Answer: Glenn Hawkins - 30/07/2014 15:13:55
 Above I wrote: ". . .faster at a smaller diameter of rotation does not mean less resistance is occurring at faster speeds."

I think this is right only with not spinning mater, but I think is is wrong in forced rotation of spinning flywheels. In fact I think the faster the rotation of spinning flywheels the more resistance they would produce.

I will shut up for a while now. Good luck everybody.

Glenn,

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Answer: Blaze - 06/09/2014 20:37:51
 Hi Harry.
You said "Unfortunately the center of mass of the mass pairs A/B or A'/B' moves from the starting position upwards and backwards again to the colliding position in a sinusoidal like manner."

Could you explain how you see this happening? I have been puzzling over this for a while and I don't see that happening but it doesn't mean that I am correct. Please explain further.

Thanks,
Blaze

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Answer: Harry K. - 07/09/2014 17:48:32
 Hello Blaze,

I had to think again about my statement you quoted and I cannot remember the reason why the platform should move in a sinusoidal manner in vertical direction. I'm getting old!

Assumed that both mass pairs collide at an angle of 30 degrees above horizontal plane and assumed that masses A/A' rotate downwards at double speed than masses B/B* rotate upwards, the barycenter of massrs A/B respectively A'/B' should remain at constant vertical position. Only the horizontal component of the barycenter will move from the orign position in outside direction to the colliding point.

Based on my stated assumptions the complete platform with the 4 weights should move upwards (if you look on your sketch) based on the acting centrifugal forces during rotation of the 4 masses.

That means that I believe it could work if I made no mistakes in thinking. However, I have doubts because this mechanism would be to easy. But good solutions are always easy... ;-)

Did you made some experiments in this matter? I would be curious about the results.

Regards,
Harald

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Answer: Harry K. - 07/09/2014 17:48:32
 Hello Blaze,

I had to think again about my statement you quoted and I cannot remember the reason why the platform should move in a sinusoidal manner in vertical direction. I'm getting old!

Assumed that both mass pairs collide at an angle of 30 degrees above horizontal plane and assumed that masses A/A' rotate downwards at double speed than masses B/B* rotate upwards, the barycenter of massrs A/B respectively A'/B' should remain at constant vertical position. Only the horizontal component of the barycenter will move from the orign position in outside direction to the colliding point.

Based on my stated assumptions the complete platform with the 4 weights should move upwards (if you look on your sketch) based on the acting centrifugal forces during rotation of the 4 masses.

That means that I believe it could work if I made no mistakes in thinking. However, I have doubts because this mechanism would be to easy. But good solutions are always easy... ;-)

Did you made some experiments in this matter? I would be curious about the results.

Regards,
Harald

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Answer: Harry K. - 07/09/2014 17:49:03
 Hello Blaze,

I had to think again about my statement you quoted and I cannot remember the reason why the platform should move in a sinusoidal manner in vertical direction. I'm getting old!

Assumed that both mass pairs collide at an angle of 30 degrees above horizontal plane and assumed that masses A/A' rotate downwards at double speed than masses B/B* rotate upwards, the barycenter of massrs A/B respectively A'/B' should remain at constant vertical position. Only the horizontal component of the barycenter will move from the orign position in outside direction to the colliding point.

Based on my stated assumptions the complete platform with the 4 weights should move upwards (if you look on your sketch) based on the acting centrifugal forces during rotation of the 4 masses.

That means that I believe it could work if I made no mistakes in thinking. However, I have doubts because this mechanism would be to easy. But good solutions are always easy... ;-)

Did you made some experiments in this matter? I would be curious about the results.

Regards,
Harald

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Answer: Harry K. - 07/09/2014 17:51:47
 Sorry for the multiple postings. This was not untended.

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Answer: Blaze - 07/09/2014 20:13:14
 Hi Harry.
This mechanism is not as easy as it first appears. It requires a constantly changing speed for both the A and B arms (it is not as simple as the A arms moving twice the speed of the B arms), a proper understanding of how the barycenter actions can be counteracted, the understanding of the proper angle of collision and its importance, and of course, some how getting a non elastic collision (I am thinking that lead weights might work fairly well).

The easiest way to keep the A and B arms in sync would be to use gears but because of the constantly changing rotational speeds of the arms in relation to each other, the gears would have to be a complicated spiral gear set. I have been investigating how to make these at a reasonable cost and may have come up with a way of doing it by cutting the shapes out of plate aluminum and attaching a flexible gear rack the the machined aluminum plates. To provide the rotational motive force for the A/B arms I am considering using a spring. The spring would have to be quite strong to generate enough rotational movement because the device would only move due to centrifugal acceleration which is a squared function of rotational speed. If the rotational speed is too low the thrust could only be a few ounces or maybe even less.

Those are my thoughts so far.

cheers,
Blaze

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Answer: Harry K. - 07/09/2014 21:27:05
 Hi Blaze,

It's not clear to me why the A/B arms should change their rotational speed during the cycle of their collision? To my understanding, the upper weight A has to run with twice the speed (better: with twice of angular velocity) as weight B to ensure that the each barycenter of the 2 weights A/B and A'/B' remains at its vertical position. What is the reason to vary the speeds?

I also think that the collision is not necessarily important because the movement of the whole platform is the cause of the acting centrifugal forces of the weights during their rotation but not the cause of the collision itself? The collision will stop only the cycle which has to be reseted in any useful form to start a new cycle.

Or do I miss something important?
Just my 2 cents.

Regards,
Harald

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Answer: Blaze - 07/09/2014 22:45:34
 Hi Harry. The average speed of the A pair is twice the speed of the B pair. However, to keep the barycenter movement the same for both the A pair and the B pair it is necessary to vary at least one of the A or B pairs, the other can be a constant speed.

The A pair generates very, very little barycenter movement when starting out and the B pair generates a lot of barycenter movement when starting out. To keep the barycenter forward and backward movement of the platform the same amount from both the A and B pairs it is necessary to move them different amounts. For a small movement of the A pair it is necessary to move the B pair a very, very small amount. If the A pair moves 1 degree of rotation the B pair only moves 0.00873 degrees of rotation. This sounds extremely small but that is because the amount of barycenter movement from the A pair in that first degree of rotation is also extremely small.

Looking at it the other way around, if the B pair is moved 1 degree of rotation, the A pair has to move 10.7 degrees of rotation. When the B pair has moved 2 degrees, the A pair has moved 15.2 degrees. When the A and B pairs are very near the collision point (like a degree or less away) they both move very nearly the same distance as each other for that last little bit of rotation.

So from the above descriptions you can see that at least one of the pairs has to have a varying rotational speed to keep the barycenter movements equal.

I am not sure what you mean by the first sentence in the second paragraph of your response but you are correct that "The collision will stop only the cycle which has to be reseted in any useful form to start a new cycle".

This device is not intended to produce continuous propulsion. If the cycle were to continue past the collision point the arms would eventually have to be stopped when the B arms got to the starting position of the A arms (pointing straight up in the diagram). This means that the A arms would be pointing sideways on the diagram and stopping the arms then would likely also stop all the forward motion that the device just gained.

regards,
Blaze



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Answer: Harry K. - 08/09/2014 11:04:20
 Hello Blaze,

I did remember about the reason why each barycenter is moving in a sinusoidal manner. I introduced this in a geometrical way by drawing the path of the masses in Autocad. I considered only each common barycenter of the 2 masses A/B in steps of 5 respectively 10 degrees. That means that I didn't considered the acting centrifugal forces, because before the centrifugal forces can be released, a stable barycenter must be present to deliver the necessary centripetal forces.

Now I understand why you intend to vary one speed namely to achieve a stable barycenter.

I made a PDF of this drawing and if you are interested in I will send it to you. Unfortunately I do not find your email address because I currently do not have access to my computer.

Regards,
Harald

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Answer: Blaze - 09/09/2014 03:01:49
 Hi Harry. If the pdf doesn't take into account the varying speeds and therefore distances of the arm pairs then it won't be much good and I have already done the math on a spreadsheet for one degree increments.

Now, the really interesting thing about this device is that it doesn't break any of Newton's laws to produce movement beyond it own dimensions but it does break the conventional thinking of a closed system, that is, if you perceive this device to actually be a closed system. One could easily argue that it is shedding energy from the collision in the form of heat and therefore it is not a closed system. Even if you consider the fact that emitting energy in one direction to generate propulsion is in fact a viable concept, there is still the issue that the energy (heat) emitted from this device moves off in all directions equally and therefore can produce no thrust. All the thrust from this device (even though it is only a one shot device) is due to centrifugal acceleration and it ejects no mass. This device does not have to shed mass to get movement.

cheers,
Blaze

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Answer: Harry K. - 09/09/2014 10:19:12
 Dear Blaze,

You wrote:

"Hi Harry.
You said "Unfortunately the center of mass of the mass pairs A/B or A'/B' moves from the starting position upwards and backwards again to the colliding position in a sinusoidal like manner."

Could you explain how you see this happening?..."

My drawing which I have prepared again, does explain how I see this happening. Thus I'm wondering why you are not interested to look at this drawing. Anyway, it doesn't matter.

My drawing shows that the common barycenter of each 2 masses move from its starting position in Y-direction to a maximum value, when the masses moved half of their angle, i.e. 30 deg. for mass A and 15 deg. for mass B and afterwards backwards to its orign position. That means the complete platform would move in 1 cycle forward and backward in Y-direction. The horzontal components are canceled out by both mass systems, thus there is only an oscillating movement in Y-direction.

At this stage nothing has been won because the platform only moves forward and backward. Centrifugal forces cannot act because centripetal forces cannot be achieved because of the moving, "unstable" barycenter.

It is not clear to me how you want to achieve a "stable" (relativ) non moving barycenter by changing one or more rotational speeds of the masses?
If you change one speed the masses will not collide at 30 deg which is mandatory.

By the way the heat energy caused by the collision of the masses is caused by the input energy to accelerate the masses.

Unfortunately I do not yet believe that a mass displacement could be achieved.

Regards,
Harald

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Answer: Blaze - 09/09/2014 19:24:20
 Hi Harry. Well I can see that you have completely missed the whole point.

You are absolutely correct that if one of the masses moves at twice the speed of the other then it cannot work. I have never claimed that one mass moves at twice the speed of the other.

What I have said is that if one mass moves at a constant speed then the other mass has to move at a varying speed but they are both moving the same speed when the collision occurs. One could vary the speeds of both masses but that would be unnecessarily complicated.

You are also absolutely correct that if one of the masses moves 15 degrees and the other moves 30 degrees then it cannot work. I have never claimed that either. When one mass has moved 15 degrees the other is far from 30 degrees.

regards,
Blaze

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Answer: Harry K. - 09/09/2014 21:18:19
 Hi Blaze,

I wish you much success with your ideas.

In the moment I'm very busy with my own ideas, also not based on gyro dynamics but based on centrifugal forces as well. However not related in any form to your one shot experiment.

I only came in here because you asked me a question. I gave my answer and thus t did my job. I'm not too much interested in one shot devices because they are senseless for me. But this is only my own humble opinion.
Anyway, good luck!

Regards,
Harald

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Answer: Blaze - 10/09/2014 19:31:09
 Thanks for your input Harry. It is unfortunate that I couldn't explain my concept well enough to get the message across.

Good luck in your projects.

Regards,
Blaze

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Answer: Blaze - 14/09/2014 21:02:05
 Ok, it seems that no one gets the picture about how this is supposed to work so I have made another picture that is a graph of movement of Arm B vs Arm A. Unfortunately the picture came out a little grainy from the conversion to pdf from Excel but it is still legible.

You will notice that the graph displays a non linear curve of the arm rotations in relation to each other. The rotation (and therefore the speed) of one arm is NOT a simple multiple of the other which is the point that seems to have gotten lost somewhere in my previous explanations.

To further exemplify this point, you will also see that for the first 5 degrees of arm A rotation, Arm B rotates an extremely small amount but for the last 5 degrees of rotation the two arms move almost the same amount.

These non linear arm rotations are required to keep the center of mass of the platform stationary (see the the original link http://min.us/i/MQjSwbXP7hs4). The center of mass remains stationary because for every incremental LINEAR distance that the mass of arm A backwards, the mass of arm B moves forwards the same incremental LINEAR distance simultaneously. Since the center of mass remains stationary, the platform will be pulled forward by centrifugal force and after the perfectly non elastic collision it would continue to coast indefinitely in a non gravity environment in space.

Again, as explained in previous postings in this thread, this motion is generated without breaking any of Newton's Laws even though there is no mass ejected from the platform.

The graph is in the link below.
http://min.us/i/u5pEKnA8PCQs

Hopefully that will clear things up a bit.

cheers,
Blaze

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