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Author Topic: Double Pendulum Power  (Read 58605 times)

Offline vince

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Re: Double Pendulum Power
« Reply #60 on: October 26, 2013, 12:57:22 AM »
Hi Telecom


you are right , the numbers are high but it is because it is an old motor with bad bearings and all the bearings are recycled old bearings. The motor alone draws just over 3 amps with nothing on it at all. I was just trying to show relative numbers to different situations.  The  main inference here is that the draw does not go up appreciably when loading the pendulum.

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Re: Double Pendulum Power
« Reply #60 on: October 26, 2013, 12:57:22 AM »

Offline nybtorque

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Re: Double Pendulum Power
« Reply #61 on: October 27, 2013, 10:46:28 AM »
@NT

As this is your design, I do have some questions.   To design and build a proper machine that can be harnessed for testing I would appreciate your comments on the following questions.

Do the offset weights have to turn in opposite directions ( CW and CCW) I realize that they must me positioned correctly to each other to balance the centrifugal forces but can the same thing be done with them spinning in the same direction.  This would open the design up to sprockets and chains and be less dependent on gear size and weight.

Is there an optimum radius for the offset weight?

Is there an optimum radius for the inner pendulum?

Does the weight of the pendulum matter as much in a balanced system where one side offsets the other?

Do you have any thoughts on a possible generator for the system? You mention a dc motor but if it is a brushed motor it would only utilize a few bars of the commutator and their respective coils. A PM motor would be better but can it generate any real power with such little movement even though the frequency would be high. Im not sure how a conventional ac generator would react?

Thanks
Vince


@Vince


1. CCW or CW rotation doesn't matter as long as they are synchronized the right way.


2. There are no optimus radius for either of the pendulums. They do have impact on the power and amplitude on different aspects, but there is no optimization to done for other than for design reasons.


3. Larger weight gives larger output power since the centrifugal force is larger. The reason for two pendulums is to cancel out the forces that are not utilized for power and makes the machine vibrate. Basically, two pendulums gives twice the power.


4. Unfortunately I don't know enough about generators/motors to know which would be the best to utilize small angular amplitudes with hight frequence. The way I calculate it should be abie to output high voltage low current power, since voltage is correlated with acceleration and current with velocity.


Regards NT

Offline nybtorque

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Re: Double Pendulum Power
« Reply #62 on: October 30, 2013, 12:40:20 PM »
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.

Free Energy | searching for free energy and discussing free energy

Re: Double Pendulum Power
« Reply #62 on: October 30, 2013, 12:40:20 PM »
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Offline telecom

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Re: Double Pendulum Power
« Reply #63 on: October 30, 2013, 02:21:53 PM »
Hi nybtorque,
can you please post your paper as a pdf file. I have difficulty reading it on where it is now, as well as downloading.
But otherwise, very fascinated with your concept!
May be it will be beneficial to connect your oscillating ring to a pivoting arm, to increase the leverage?
Regards

Offline LibreEnergia

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Re: Double Pendulum Power
« Reply #64 on: October 31, 2013, 12:01:49 AM »
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:

You are going to find out shortly that this 'power' is simply an illusion of your flawed analysis.

Any attempt to extract it WILL slow the pendulum down. Once that happens the pendulum has less kinetic energy to exchange for potential energy and the system ends up in a lower potential state. That lower potential is subsequently exchanged for a lower amount of kinetic energy and so on oscillating back and forward until it stops. The total amount of energy that you can extract is equal to the amount of energy it was given to start.

It is ludicrous to analyse in terms of  power (the time derivative of energy) and expect that it can be magically extracted without lowering the total energy state of the system.

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Re: Double Pendulum Power
« Reply #64 on: October 31, 2013, 12:01:49 AM »
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Offline nybtorque

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Re: Double Pendulum Power
« Reply #65 on: October 31, 2013, 07:14:12 AM »
You are going to find out shortly that this 'power' is simply an illusion of your flawed analysis.

Any attempt to extract it WILL slow the pendulum down. Once that happens the pendulum has less kinetic energy to exchange for potential energy and the system ends up in a lower potential state. That lower potential is subsequently exchanged for a lower amount of kinetic energy and so on oscillating back and forward until it stops. The total amount of energy that you can extract is equal to the amount of energy it was given to start.

It is ludicrous to analyse in terms of  power (the time derivative of energy) and expect that it can be magically extracted without lowering the total energy state of the system.


@LibreEnergia


I value your feedback since it gives me direction on where to put my effort in explaining the line of thought. I would appreciate though if you could give me a less opinion based explanation to why it's "ludicrous" to integrate power over time and calculate work performed.


I do agree that if you take energy out the pendulum WILL SLOW DOWN and that kinetic energy WILL have to be replaced.


If we leave the energy extracting aspect, do you agree with the free fall/artificial gravity analogy? I.e the inner pendulum mass being in constant acceleration (varying in direction and size)?


Then, think about it this way. How much energy does it take to stop a 1 kg mass in free fall? Well, anything from 0 J to whatever depending on the speed it have reached due to gravity (E=mgh=mv^2/2). Then again, think about it as if you always reduce the acceleration 50% by a generator and book the energy difference. How much energy will you book? Well, E=mgh/2 of course....


As you can see, this example is only limited by "h". The mass will continue to accelerate, although at a slower rate, gain velocity, etc. All because of gravity. So, what if "h" is infinite and "g" can be decided by speed of rotation. How much energy can you extract?


I'm only proposing a non-linear model of this example. The concept is used for high-G training already, although at lower rates...


Regards NT

Offline nybtorque

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Re: Double Pendulum Power
« Reply #66 on: October 31, 2013, 07:27:05 AM »
Hi nybtorque,
can you please post your paper as a pdf file. I have difficulty reading it on where it is now, as well as downloading.
But otherwise, very fascinated with your concept!
May be it will be beneficial to connect your oscillating ring to a pivoting arm, to increase the leverage?
Regards


I'll try to post it below.


Pivoting arms of different length is part of the concept. It could be beneficial depending on what you want to achieve. If you want angular amplitude short arm is best, if you want linear amplitude a long arm is better. The power will be the same though.




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Re: Double Pendulum Power
« Reply #66 on: October 31, 2013, 07:27:05 AM »
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Offline LibreEnergia

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Re: Double Pendulum Power
« Reply #67 on: October 31, 2013, 09:34:50 AM »

I do agree that if you take energy out the pendulum WILL SLOW DOWN and that kinetic energy WILL have to be replaced.


That is all you need to realise.

You cannot use free fall of gravity to replace the energy lost. For that to occur the system would need to have gravitational potential to allow it to fall, but the maximum of potential reduces at exactly the same rate as the energy is dissipated either by a generator or by friction.

Build a simulation that has a damping force proportional to the angular velocity at the pivots and you'll see that.


Offline nybtorque

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Re: Double Pendulum Power
« Reply #68 on: October 31, 2013, 03:34:20 PM »
That is all you need to realise.

You cannot use free fall of gravity to replace the energy lost. For that to occur the system would need to have gravitational potential to allow it to fall, but the maximum of potential reduces at exactly the same rate as the energy is dissipated either by a generator or by friction.

Build a simulation that has a damping force proportional to the angular velocity at the pivots and you'll see that.


Well, that is not really ALL I have to realize!! Of course friction in the pivots will make the pendulum stop eventually. Friction in the outer pendulum needs so be replaced continually. With the inner pendulum it's a bit more complicated. You need to consider that friction is in the opposite direction of velocity, but that acceleration is more or less in the opposite direction of velocity half of the time as well, so friction is actually more or less neutral to power. This is the reason why the potential does not reduce in the same rate as the kinetic energy. Friction actually has direction (DC), but power has not and an if you replace friction with a generator you will get AC-power.






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Re: Double Pendulum Power
« Reply #68 on: October 31, 2013, 03:34:20 PM »
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Offline telecom

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Re: Double Pendulum Power
« Reply #69 on: October 31, 2013, 05:51:46 PM »
Hi NT,
I'm not sure this statement is correct:
The interdependencies between the two pendulums continuously exchanging
kinetic energy as they oscillate is complex. The two pendulums transfer
momentum between themselves in both directions as described by the Euler
Lagrange equations earlier. If we try to force the outer pendulum into a
certain speed or movement, we will no doubt disturb the inner pendulum as
it transfer momentum back. Constant rotation of the outer pendulum will
simply not do.

The way I see it, the outer pendulum simply doesn't "see" the inner pendulum.
It keeps rotating in equilibrium (less friction) according to the first law of motion:
2. Rotational equilibrium:When body is not rotating at all or its rotating at constant rate it is said to be in rotational equilibrium. This is Newton's first law of motion,equilibrium.

To activate the inner pendulum with use the reaction of the axis of the outer pendulum by allowing the axis to move freely in one direction, being attached to the pivot of the inner pendulum.
What do u think?
Regards!

Offline nybtorque

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Re: Double Pendulum Power
« Reply #70 on: October 31, 2013, 06:01:27 PM »
Hi NT,
I'm not sure this statement is correct:
The interdependencies between the two pendulums continuously exchanging
kinetic energy as they oscillate is complex. The two pendulums transfer
momentum between themselves in both directions as described by the Euler
Lagrange equations earlier. If we try to force the outer pendulum into a
certain speed or movement, we will no doubt disturb the inner pendulum as
it transfer momentum back. Constant rotation of the outer pendulum will
simply not do.

The way I see it, the outer pendulum simply doesn't "see" the inner pendulum.
It keeps rotating in equilibrium (less friction) according to the first law of motion:
2. Rotational equilibrium:When body is not rotating at all or its rotating at constant rate it is said to be in rotational equilibrium. This is Newton's first law of motion,equilibrium.

To activate the inner pendulum with use the reaction of the axis of the outer pendulum by allowing the axis to move freely in one direction, being attached to the pivot of the inner pendulum.
What do u think?
Regards!


I do think the statement is correct. Because the centrifugal force of the outer pendulum acts on the inner pendulum so that i in turn accelerates, and this creates a force acting back on the outer pendulum mass which either accelerates decelerates it.... and on... This is the reason that the solution of Euler Lagrange equation is quite complex.




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Re: Double Pendulum Power
« Reply #70 on: October 31, 2013, 06:01:27 PM »
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Offline telecom

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Re: Double Pendulum Power
« Reply #71 on: October 31, 2013, 06:23:35 PM »

I do think the statement is correct. Because the centrifugal force of the outer pendulum acts on the inner pendulum so that i in turn accelerates, and this creates a force acting back on the outer pendulum mass which either accelerates decelerates it.... and on... This is the reason that the solution of Euler Lagrange equation is quite complex.
Lets remove the motor 1 on fig. 5 and place two motors coaxial to the weights 2 and 3.
What is the mechanism of slowing down the above motors with weights attached?
The reaction of the centrifugal forces is only able to push back on the bearings holding the shafts of the motors, IMHO.

Offline nybtorque

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Re: Double Pendulum Power
« Reply #72 on: October 31, 2013, 07:12:18 PM »
Lets remove the motor 1 on fig. 5 and place two motors coaxial to the weights 2 and 3.
What is the mechanism of slowing down the above motors with weights attached?
The reaction of the centrifugal forces is only able to push back on the bearings holding the shafts of the motors, IMHO.


Yes, and by pushing back on the bearings it either accelerates or decelerate the pendulum mass depending on the angle. This is actually kind of interesting as a way to extract power because it would generate power spikes in the driving motor. I don't know how to make it useful but it would be interesting to analyze.

Offline telecom

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Re: Double Pendulum Power
« Reply #73 on: October 31, 2013, 07:26:23 PM »

Yes, and by pushing back on the bearings it either accelerates or decelerate the pendulum mass depending on the angle. This is actually kind of interesting as a way to extract power because it would generate power spikes in the driving motor. I don't know how to make it useful but it would be interesting to analyze.
What is the mechanism of action?

Offline nybtorque

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Re: Double Pendulum Power
« Reply #74 on: November 01, 2013, 09:55:45 AM »
That is all you need to realise.

You cannot use free fall of gravity to replace the energy lost. For that to occur the system would need to have gravitational potential to allow it to fall, but the maximum of potential reduces at exactly the same rate as the energy is dissipated either by a generator or by friction.

Build a simulation that has a damping force proportional to the angular velocity at the pivots and you'll see that.


I have another analogy for you. Think of it as an high-frequency, high voltage(acceleration), low current(velocity) circuit 90 degrees out of phase. If you put a resistive load on it you will not get much heat because P=i^2 * R. It's the same with kinetic energy turning into friction heat.


However if you put an inductive load it should be different because P=L * i * i', where i' = di/dt, is and analogy to acceleration. This is why I expect high-voltage AC-output from a generator with an oscillating input torque.

 

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