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Author Topic: Artificial Gravity Power  (Read 15178 times)

nybtorque

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Artificial Gravity Power
« on: October 30, 2013, 12:44:34 PM »
Hi all!


I posted this in the mechanics forum, but maybe it's better suited here?


http://www.overunity.com/13557/double-pendulum-power/60/#.UnDwyCiizxg


Regards NT


-----------------------------


One way to make my concept (http://www.scribd.com/doc/146232946/Double-Pendulum-Power-AC-Power-from-a-Mechanical-Oscillator)  easier to grasp, is to think of the inner pendulum mass as in constant free fall.


It's not in free fall with the acceleration G in one direction (down…) but it's in free fall using an oscillating artificial gravity (varying in strength and direction with the frequency of oscillation) provided by the centrifugal force of the outer pendulum. And, the nice part is that this artificial gravity is free, as the system is set in rotation.


It's analogous to free fall, and as I pointed out; power has no direction.


The general critique is that if energy is taken out of the system it has to be replaced. And this certainly is valid, and applies to the kinetic energy in the system, as in friction. But kinetic energy is a function of velocity, not acceleration.


So, all we have to do, is learn how to utilize this power.


And, what do we do when we want to utilize gravity for power?


Of course we use a generator to capture it… Like in a hydropower plant. If we let the water fall with no resistance - no power, and if we build a dam - no power (but lots of static forces…) BUT, for all resistances in between we capture the potential energy as E=mgh.


The same applies in my concept because we're dealing with potential energy from artificial gravity - If the mass is fixed; only static forces, and if it is free to move (as in the double pendulum) no power… But, for all resistances in between… 


Here we even have the mass oscillating so we do not need to worry about the water getting up the hill again. Instead we get AC-power from the oscillations.


As Nicola Tesla said:
Quote
"If you want to find the secrets of the universe, think in terms of energy, frequency and vibration."

telecom

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Re: Artificial Gravity Power
« Reply #1 on: October 30, 2013, 06:26:04 PM »
Hi nybtorque,
I would like to ask you if my reasoning based on the power calculation makes any sense.

The inner pendulum have to overcome friction and other losses while activating the outer pendulum.
Lets say its a motor which consumes 100 w to oscillate an outer pendulum with an amplitude of 1 cm.
1000 w will approximately mean of lifting 100 kg by 1 meter in 1 second, in this case 100w will be equal 10 kg x 100 cm in 1 second, or 1000 kg x 1 cm /1 sec.

Lets say an outer pendulum makes 10 oscillation per second, which will be equal to 600 rpm for the inner pendulum.
Since an outer pendulum makes  2 maximal movement per the oscillation, to generate 100 w it will need to generate  the centrifugal force of 1000/20 = 50 kG ~500 N.
Then we can find the radius/ mass ratio for the outer pendulum by the formula for the centrifugal force, F = mV^2/r.
Does it make sense or I missed something?

nybtorque

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Re: Artificial Gravity Power
« Reply #2 on: October 31, 2013, 08:12:34 AM »
Hi nybtorque,
I would like to ask you if my reasoning based on the power calculation makes any sense.

The inner pendulum have to overcome friction and other losses while activating the outer pendulum.
Lets say its a motor which consumes 100 w to oscillate an outer pendulum with an amplitude of 1 cm.
1000 w will approximately mean of lifting 100 kg by 1 meter in 1 second, in this case 100w will be equal 10 kg x 100 cm in 1 second, or 1000 kg x 1 cm /1 sec.

Lets say an outer pendulum makes 10 oscillation per second, which will be equal to 600 rpm for the inner pendulum.
Since an outer pendulum makes  2 maximal movement per the oscillation, to generate 100 w it will need to generate  the centrifugal force of 1000/20 = 50 kG ~500 N.
Then we can find the radius/ mass ratio for the outer pendulum by the formula for the centrifugal force, F = mV^2/r.
Does it make sense or I missed something?


@Telecom


I'm not sure if you got the concept right. I'm using the outer pendulum to activate the inner pendulum. In my report I set the outer pendulum in motion at a certain rotational speed and/or angle depending on what I want to analyze. The accelerations and velocities of the two masses then interact in a complicated manner which I solve numerically (using Runge Kutta method, ie. some kind of FEM method).


Actually the dimensions of the pendulums are not that important, bigger, heavier and faster pendulums give more power. Design and construction issues are more relevant i believe since canceling vibrations and harvesting the power are the hardest problem to solve. There seem to be optimal load though, since no load gives no power and a big load makes the mass more or less stop, ie. low velocity = low power. I found that a load that is more or less equal to the pendulum mass is optimal, but might not be the most practical.


Regards NT

telecom

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Re: Artificial Gravity Power
« Reply #3 on: October 31, 2013, 10:44:22 AM »
Hi NT,
it should be red in reverse -  inner for the outer and vice versa.
What I wanted to say is that the outer pendulum is not really a pendulum,
but the body which can be described by the first law of Newton, and yet it produces a force acting on the inner pendulum. This force can be calculated as a centrifugal.
This force is due to a missing reaction which is caused by the floating axis of the
outer pendulum. Since the outer pendulum acts according to the first law, the turning
torque only should be sufficient to overcome the friction, no matter how big the speed and the mass are.
Regards

nybtorque

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Re: Artificial Gravity Power
« Reply #4 on: October 31, 2013, 05:31:58 PM »
Hi NT,
it should be red in reverse -  inner for the outer and vice versa.
What I wanted to say is that the outer pendulum is not really a pendulum,
but the body which can be described by the first law of Newton, and yet it produces a force acting on the inner pendulum. This force can be calculated as a centrifugal.
This force is due to a missing reaction which is caused by the floating axis of the
outer pendulum. Since the outer pendulum acts according to the first law, the turning
torque only should be sufficient to overcome the friction, no matter how big the speed and the mass are.
Regards


Ok. I get it. You're correct in that the torque needed is to overcome friction, but you also need torque to replace the kinetic energy taken out by the generator (reduction of pendulum velocity). This however is not the same that you have to replace all power taken out by the generator because a reduction of velocity actually increases acceleration more or less half of the time. This is the nature of oscillations and AC-power.

telecom

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Re: Artificial Gravity Power
« Reply #5 on: October 31, 2013, 06:01:15 PM »

Ok. I get it. You're correct in that the torque needed is to overcome friction, but you also need torque to replace the kinetic energy taken out by the generator (reduction of pendulum velocity). This however is not the same that you have to replace all power taken out by the generator because a reduction of velocity actually increases acceleration more or less half of the time. This is the nature of oscillations and AC-power.
Hi NT,
we need to look very carefully at the actual mechanics of the transfer, since an outer pendulum may not "see" the inner pendulum,
since we are taking the reaction of the axis which otherwise would be spent on the stressing of the support itself (applying
the tension).
regards!

nybtorque

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Re: Artificial Gravity Power
« Reply #6 on: November 01, 2013, 06:28:55 AM »
Hi NT,
we need to look very carefully at the actual mechanics of the transfer, since an outer pendulum may not "see" the inner pendulum,
since we are taking the reaction of the axis which otherwise would be spent on the stressing of the support itself (applying
the tension).
regards!


I believe this work is elegantly done already by Euler Lagrange and the the forces acting on the pendulums are described in equations (1) and (2) in my report. They cover the Newtonian laws.

telecom

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Re: Artificial Gravity Power
« Reply #7 on: November 01, 2013, 10:48:01 AM »

I believe this work is elegantly done already by Euler Lagrange and the the forces acting on the pendulums are described in equations (1) and (2) in my report. They cover the Newtonian laws.
Hi NT,

Their work was done for a different physical model where two pendulum are interconnected to each other.
In our case there is no physical , or mechanical interconnection between two pendulums.In our case we use a part of the reaction of the axis of the outer
pendulum to pendulum to oscillate the inner pendulum.
As I pointed out earlier, this is very important because outer pendulum in this case
works according to the 1st law of Newton in equilibrium.
There is no external force which would cause change in acceleration of the outer pendulum, according to the second law of Newton!
Another importance is the fact that it allows to transfer reaction into motion,
possible opening ways to create a reactionless motion.
It is very easy to apply a different set of formulas in our case to calculate the power produced and the period of the oscillations based upon the formula for a centrifugal force.
So, the power will be proportional to the rpm2,m,r of the outer pendulum.
Hope this helps.

nybtorque

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Re: Artificial Gravity Power
« Reply #8 on: November 04, 2013, 07:15:01 AM »
Hi NT,

Their work was done for a different physical model where two pendulum are interconnected to each other.
In our case there is no physical , or mechanical interconnection between two pendulums.In our case we use a part of the reaction of the axis of the outer
pendulum to pendulum to oscillate the inner pendulum.
As I pointed out earlier, this is very important because outer pendulum in this case
works according to the 1st law of Newton in equilibrium.
There is no external force which would cause change in acceleration of the outer pendulum, according to the second law of Newton!
Another importance is the fact that it allows to transfer reaction into motion,
possible opening ways to create a reactionless motion.
It is very easy to apply a different set of formulas in our case to calculate the power produced and the period of the oscillations based upon the formula for a centrifugal force.
So, the power will be proportional to the rpm2,m,r of the outer pendulum.
Hope this helps.


I'm unfortunately not following your reasoning. The way I see it, there is a mechanical connection and both the centrifugal force of the outer pendulum and the angular acceleration of both pendulum masses (second law of Newton) are part of the equation and force equilibrium at all points. That is the Euler Lagrange equations. As you say, the centrifugal force of the outer pendulum acts on the inner pendulum mass, which then accelerates. However this acceleration acts back on the outer pendulum according to Newtons second law, and that mass in turn accelerates. This is the way they interact.

telecom

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Re: Artificial Gravity Power
« Reply #9 on: November 04, 2013, 01:33:06 PM »

I'm unfortunately not following your reasoning. The way I see it, there is a mechanical connection and both the centrifugal force of the outer pendulum and the angular acceleration of both pendulum masses (second law of Newton) are part of the equation and force equilibrium at all points. That is the Euler Lagrange equations. As you say, the centrifugal force of the outer pendulum acts on the inner pendulum mass, which then accelerates. However this acceleration acts back on the outer pendulum according to Newtons second law, and that mass in turn accelerates. This is the way they interact.
First of all it acts back according to the third law.
Secondly, it acts back on the housing of the bearings. If the bearings are
of a good quality, than it absolutely is not affecting the rpm of the unbalanced disk,
the outer pendulum itself. Look at the Milkovich for example. The outer pendulum
keeps going as a normal pendulum with only friction losses, unaffected by what
the inner pendulum does (hitting an anvil as a hummer).
In the Lagrange case, the outer and the inner pendulum are interconnected by their
ends, which is a completely different case.
Regards

nybtorque

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Re: Artificial Gravity Power
« Reply #10 on: November 04, 2013, 02:34:44 PM »
First of all it acts back according to the third law.
Secondly, it acts back on the housing of the bearings. If the bearings are
of a good quality, than it absolutely is not affecting the rpm of the unbalanced disk,
the outer pendulum itself. Look at the Milkovich for example. The outer pendulum
keeps going as a normal pendulum with only friction losses, unaffected by what
the inner pendulum does (hitting an anvil as a hummer).
In the Lagrange case, the outer and the inner pendulum are interconnected by their
ends, which is a completely different case.
Regards


But when the inner pendulum acceleration acts back (Newton third law, as you say...) on the housing it will either pull or push the rotating pendulum mass so that it either accelerates or decelerates depending on the angle.


This is the case with Milkovic as well. I've done a simulation with a pendulum where each pendulum mass is 10kg, the inner pendulum lever is 1m and the outer lever 0,5m. In reality Milkovic uses a much heavier inner lever which will make the amplitude and variations smaller, but nevertheless existent.




telecom

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Re: Artificial Gravity Power
« Reply #11 on: November 04, 2013, 07:37:01 PM »

But when the inner pendulum acceleration acts back (Newton third law, as you say...) on the housing it will either pull or push the rotating pendulum mass so that it either accelerates or decelerates depending on the angle.


This is the case with Milkovic as well. I've done a simulation with a pendulum where each pendulum mass is 10kg, the inner pendulum lever is 1m and the outer lever 0,5m. In reality Milkovic uses a much heavier inner lever which will make the amplitude and variations smaller, but nevertheless existent.

It will not push the mass, it will push the bearing of the axis.
The problem is that your simulation doesn't reflect the reality because you are using the wrong physical model.
you started with the oranges, but applying them to the apples.
Take a look at the video video of the Milcovich pendulum - the period of oscillations is the same, not as in your simulation.
I learned in school that someone said the reality is the criteria for the truth.

nybtorque

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Re: Artificial Gravity Power
« Reply #12 on: November 05, 2013, 07:06:29 AM »
It will not push the mass, it will push the bearing of the axis.
The problem is that your simulation doesn't reflect the reality because you are using the wrong physical model.
you started with the oranges, but applying them to the apples.
Take a look at the video video of the Milcovich pendulum - the period of oscillations is the same, not as in your simulation.
I learned in school that someone said the reality is the criteria for the truth.


But, the period IS the same in my simulations... It's the amplitude that varies. This is kind of hard to verify in the Milkovic videos since he is pushing the pendulum each oscillation to overcome friction.  However it is easy to verify the uneven motions of the inner lever which is a result of the momentum transfer.

telecom

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Re: Artificial Gravity Power
« Reply #13 on: November 05, 2013, 05:45:51 PM »
Your point was that the inner pendulum affects the outer pendulum as per lagrange model. However,
in the current model, which is different from lagrange, an outer pendulum is not being affected by the inner one.
Take a look at Milcovic videos -  what happens when he stops pushing?
The pendulum oscillates unaffected.
http://www.youtube.com/watch?v=dvst47E5CvM
No matter how elegant are the equations of lagrange, they are not applicable.
You need to use different set of formulas, which is not very hard to derive from the equations for the centrifugal force.
This video confirms this:
http://www.veljkomilkovic.com/Video/Veljko_Milkovic_(video-3)_Fast_model.wmv


nybtorque

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Re: Artificial Gravity Power
« Reply #14 on: November 06, 2013, 11:02:45 AM »
Your point was that the inner pendulum affects the outer pendulum as per lagrange model. However,
in the current model, which is different from lagrange, an outer pendulum is not being affected by the inner one.
Take a look at Milcovic videos -  what happens when he stops pushing?
The pendulum oscillates unaffected.
http://www.youtube.com/watch?v=dvst47E5CvM
No matter how elegant are the equations of lagrange, they are not applicable.
You need to use different set of formulas, which is not very hard to derive from the equations for the centrifugal force.
This video confirms this:
http://www.veljkomilkovic.com/Video/Veljko_Milkovic_(video-3)_Fast_model.wmv





Unfortunately I don't understand your reason for the conclusion that Euler Lagrange is not applicable in the Milkovic case. Centrifugal force is a very important part of the Euler Lagrange model. Basically it's only a description of the Newtonian force equilibrium for each of the pendulum masses... As I see it there is no other way to do it. 


In my opinion the videos verify the model. The pendulum amplitude looks unaffected, but that is only because the amplitude of the inner lever (pump) is so much smaller and because of friction slowing it down pretty fast. However it is easy to se the uneven movements of the inner pendulum lever; i.e. one bigger move followed by a somewhat smaller, and so on... This is a result of the momentum transfer. Look at the graph below which verify this variation.


I would say that Euler Lagrange this is the best way to show overunity from the double pendulum on a theoretical level. The centrifugal force is essential as it creates the "artificial gravity" which makes it possible to to extract work, not from the kinetic energy input, but from the oscillating potential energy of the system. Milkovic and Feltenberger pendulum pumps are good examples of this concept.