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Author Topic: Double Pendulum Power  (Read 57175 times)

Offline nybtorque

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Re: Double Pendulum Power
« Reply #15 on: June 20, 2013, 05:59:38 PM »
Vidar,
Yes, that is the double pendulum I'm refering to.
 
 
Quote
Maybe you could build a small prototype and take a video of it while playing with it. Because I'm not sure if I follow you on this

 Well, I would like to, but I'm not really in a position to do it right now. The easiest example is the bicycle test, which is covered before in this thread:
 http://www.overunity.com/12119/centripetal-force-yealds-over-unity/15/#.UcMIHhpvmUk
 
If we play with my equation [13] : P(work)=(2*f^3*m(load)*m(pend)^2*r^2*pi^4*sin(pi/4)^2)/(m(pend)+m(load)+m(machine))^2
and som data that resembles the bicycle. For example:
 
f=10Hz, m(load)=10kg, m(machine)=10kg, m(pend)=1kg, r=0,3m
 
We get P(work)= 195W
 
Actually this kind of make sense since that is the about the amount muscle power a man kan sustain for some period of time. (Lift 20 kg, 1m, once a second...) And this is exactly what it feels like trying to keep the bicycle fixed to the ground while pedaling. It's hard work...

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Re: Double Pendulum Power
« Reply #15 on: June 20, 2013, 05:59:38 PM »

Offline Low-Q

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Re: Double Pendulum Power
« Reply #16 on: June 21, 2013, 10:17:35 AM »
Vidar,
Yes, that is the double pendulum I'm refering to.
 
 
 Well, I would like to, but I'm not really in a position to do it right now. The easiest example is the bicycle test, which is covered before in this thread:
 http://www.overunity.com/12119/centripetal-force-yealds-over-unity/15/#.UcMIHhpvmUk
 
If we play with my equation [13] : P(work)=(2*f^3*m(load)*m(pend)^2*r^2*pi^4*sin(pi/4)^2)/(m(pend)+m(load)+m(machine))^2
and som data that resembles the bicycle. For example:
 
f=10Hz, m(load)=10kg, m(machine)=10kg, m(pend)=1kg, r=0,3m
 
We get P(work)= 195W
 
Actually this kind of make sense since that is the about the amount muscle power a man kan sustain for some period of time. (Lift 20 kg, 1m, once a second...) And this is exactly what it feels like trying to keep the bicycle fixed to the ground while pedaling. It's hard work...
The bicycle experiment can be compared with riding a bicycle on a very bumpy road. If there is no dampers and no suspension on the bicycle, it require less energy to ride, but if you put on dampers it requires more energy. Some of the energy you put into the ride will convert into heat in the dampers. It's like riding on a hard surface vs muddy surface. Cars with very stiff or no suspension is more fuel efficient than cars with soft dampers. The fuel efficiency is not depending on the suspension which is a steel spring that has very little loss.

If you take out energy from the bicycle in your experiment, the mass that is put on one side of the wheel will cause retardation of the wheel so you must put in extra energy to sustain the RPM of the wheel.
You can see this more clear if you have an electric motor with an imbalanced flywheel on it. The motor will start to viberate if you hold it in your hand, and the RPM drops. But if you fix the motor to the table or the ground, the RPM rises. Your hand isn't a perfect suspension, but also an effective damper. Tha damping cause reduction of RPM, and the motor must be fed with more energy to sustain RPM.

Further, you can translate this experiment into the double pendulum. So I guess there is a missing part in your equation. The missing part is most probably a variable that depends on the load/friction you put into the system.

I have done an experiment with a pendulum that partially consist of a soft steel spring with very little loss. Mass is added on the bottom of the spring. If the spring isn't damped the pendulum sustain its swing for quite some time. If I put whool inside the spring, the pendulum stops relatively quick. That is because some of the kinetic energy that i represented by the stretch/bounce in the spring is converted to heat in the whool because it is no longer sync between the bounce and the period of the pendulum, but some delay.

Vidar

Offline nybtorque

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Re: Double Pendulum Power
« Reply #17 on: June 23, 2013, 06:48:17 PM »
Vidar,
 
Quote
The bicycle experiment can be compared with riding a bicycle on a very bumpy road. If there is no dampers and no suspension on the bicycle, it require less energy to ride, but if you put on dampers it requires more energy. Some of the energy you put into the ride will convert into heat in the dampers. It's like riding on a hard surface vs muddy surface. Cars with very stiff or no suspension is more fuel efficient than cars with soft dampers. The fuel efficiency is not depending on the suspension which is a steel spring that has very little loss.

I guess you could compare the feeling, and I agree that friction in any system generates heat. However, I do not see the analogy with a double pendulum.

 
Quote
If you take out energy from the bicycle in your experiment, the mass that is put on one side of the wheel will cause retardation of the wheel so you must put in extra energy to sustain the RPM of the wheel.
 
Correct. But I do not propose to take any energy out of the rotating wheel, only from the fixture point (center of the inner pendulum). And this is important. Because, for the system to work it has to be fixed to the ground in one direction (through the inner pendulum arm), so that it can use the ground as a "counterweight".

 
Quote
You can see this more clear if you have an electric motor with an imbalanced flywheel on it. The motor will start to viberate if you hold it in your hand, and the RPM drops. But if you fix the motor to the table or the ground, the RPM rises. Your hand isn't a perfect suspension, but also an effective damper. Tha damping cause reduction of RPM, and the motor must be fed with more energy to sustain RPM.

Exactly. What you do here is that you release the motor from the ground completely.Then the motor with the unbalanced flywheel will start to behave "complicated". Basically it would like to rotate freely around its combined center of mass. RPM will of course be reduced. But since the ground as a counterweight is an essential part of the theory, it is not valid as an example of a double pendulum.

 
Quote
Further, you can translate this experiment into the double pendulum. So I guess there is a missing part in your equation. The missing part is most probably a variable that depends on the load/friction you put into the system.

No, as I explained, you can not translate it to a double pendulum when you dissconnect the system from the ground. As I said, if you do that, you are not analyzing the problem correctly. It is not difficult to put friction in the equation as a counter momentum in the both centers of rotation. Load is already taken into account by the masses of the pendulums.

 
Quote
I have done an experiment with a pendulum that partially consist of a soft steel spring with very little loss. Mass is added on the bottom of the spring. If the spring isn't damped the pendulum sustain its swing for quite some time. If I put whool inside the spring, the pendulum stops relatively quick. That is because some of the kinetic energy that i represented by the stretch/bounce in the spring is converted to heat in the whool because it is no longer sync between the bounce and the period of the pendulum, but some delay
 
I'm not sure I understand completely, but it seams that it would behave like friction where the motor (outer pendulum center) is. This would of course have a direct effect on the RPM and input power needed. I do not see how it is related to the double pendulum though.
 
/NT

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Re: Double Pendulum Power
« Reply #17 on: June 23, 2013, 06:48:17 PM »
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Offline nybtorque

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Re: Double Pendulum Power
« Reply #18 on: June 26, 2013, 09:53:41 AM »
Now I am confused.  If the axis of the motor and driving cogwheel as the axis of arm D and the generator is the same axis, then wheels B and C cannot rotate.
In your drawing you used a dotted line, probably for indicating the center line (this is what I meant in my question on the fixed axle above) and you used cylindrically shaped shafts for both the input motor and the output generator axle, there is no any continuos and cylindrical axle lines in your drawing to connect the motor and the generator axles. 
The best would be to clear this setup if you make another drawing which leaves no questions.
You used a circular arrow for the input motor to indicate its rotational direction and you drew a two-way arrow for both arm D and the generator shaft: how can it be if you now has said you use the same axis?

Gyula

Gyula,
 
Here is a somewhat more detailed drawing of the proposed machinery. Efficiency depends on maximizing pendulum mass and minimizing the mass of the oscillating machine parts (cogwheels, bearings and levers).
 
 

Offline Low-Q

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Re: Double Pendulum Power
« Reply #19 on: June 26, 2013, 02:17:37 PM »
I understand the design now. This might be very funny as a seesaw in the kindergarden :-))
Well, back to the serious part:
Have you analyzed the inner workings if you load the seesaw with the generator?


At first glance I cannot point out anything that cannot work, but I have a feeling that if the seesaw is loaded there will occour a delay between its motion and the position of the weights, OR that the motor will work hardest when the generator isn't loaded at all, and not work at all if the generator is short circuit (The seesaw stops).


I can make a simple demonstration with your initial drawing, but replace the lower pendulum with an inbalanced flywheel. What I imagine is that if the actual pendulum is massless, the weight in the flywheel will not longer gain momentum as it spins up, since the flywheel spins the weight it self will not longer follow a circular path, but more or less going stright up and down. This will effectively stop the rotation (Due to lack of momentum)


Vidar

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Re: Double Pendulum Power
« Reply #19 on: June 26, 2013, 02:17:37 PM »
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Offline nybtorque

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Re: Double Pendulum Power
« Reply #20 on: June 26, 2013, 03:48:49 PM »
Vidar,
Thankyou for showing interest.  Well, at 3000 rpm it'll be a hell of a tickle for the kids... :D
 
I believe you are correct in some respects. At low RPMs the motor will rotate the "whole thing" since the centrifugal forces will not be big enough to overcome friction and bring the levers to oscillate. The same thing will happen if the pendulum mass are to low compared the the mass of the cogwheels, levers, etc.
 
To be clear; the output levers should not rotate but oscillate. According to my calculation an oscillating amplitude of about 40-50% of the pendulum radius is ideal (more or less as I've drawn the arrows in the pic).
 
If you short the generator the amplitude of oscillation will be really small generating little power. If there is no load, amplitude will be large, but of course; no power. So there is a maximum power somewhere in between depending on the design (masses, radius and frequency)
 
Physically there should be no delay (not exactly sure what you mean here); the forces are always what the are sort of... You say that the pendulum masses will not follow a circular path. This is correct. When the machine is oscillation I think they will move more or less horisontal depending on the amplitude, since upward acceleration is maximized when the position is at it's lowest point and vv.
 
So, there are some optimizing to be done with the different masses. Ideally the any mass that is not unbalanced  should be minimized and the pendulum mass should be maximized. That is what I mean by using lightweight cogwheels and levers. This enables the system to take more relative load from the generator. The double pendulum equations states that ideally m(load)+m(machine)=m(pendulum) is optimal.
 
But,that said. As soon as you get the system to oscillate from the centrifugal forces, you should be able to get "overunity". Allthough it is not overunity per se, since we are utilizing the "ground" as counterweight in the oscillations.
 
It will be really interesting to see a demo!
Regards NT

Offline nybtorque

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Re: Double Pendulum Power - AC power from Mech. Oscillator
« Reply #21 on: October 22, 2013, 12:00:49 PM »
Hi again!


I've updated my report on a mechanical oscillator (double pendulum). In the previous version I made too many simplification and ignored the feedback momentum to the driving pendulum. I'm now using the Runge Kutta method to solve the equations numerically instead. By doing this it is possible to simulate all kind of setups. As an example I use the Milkovic Pendulum in the report. Please have a look. All possible feedback is appreciated!


http://www.scribd.com/doc/146232946/Double-Pendulum-Power-AC-Power-from-a-Mechanical-Oscillator


Regards NT

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Re: Double Pendulum Power - AC power from Mech. Oscillator
« Reply #21 on: October 22, 2013, 12:00:49 PM »
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Offline TechStuf

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Re: Double Pendulum Power
« Reply #22 on: October 22, 2013, 05:37:29 PM »
Keep going....it is quite clear that more energy is produced in these systems than is necessary to keep them going.  An example:

http://www.youtube.com/watch?v=vjVQWG7xUQk

Even the layman can see that, at least in the preceding example, much more energy is produced than is required to put the device back at the starting point/energy.
 

TS

Offline LibreEnergia

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Re: Double Pendulum Power - AC power from Mech. Oscillator
« Reply #23 on: October 23, 2013, 01:46:50 AM »
Hi again!


I've updated my report on a mechanical oscillator (double pendulum). In the previous version I made too many simplification and ignored the feedback momentum to the driving pendulum. I'm now using the Runge Kutta method to solve the equations numerically instead. By doing this it is possible to simulate all kind of setups. As an example I use the Milkovic Pendulum in the report. Please have a look. All possible feedback is appreciated!


http://www.scribd.com/doc/146232946/Double-Pendulum-Power-AC-Power-from-a-Mechanical-Oscillator


Regards NT


I think you are missing something. 

Sure energy is being transferred between the two pendulums and alternately causes one or the other to accelerate.  However I guarantee you the sum of all the potential and kinetic energies in the system remains constant. (or decreasing slowly due to friction).

As soon as you try to extract energy from one location the system will very quickly come to a stop as there is no excess being generated.




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Re: Double Pendulum Power - AC power from Mech. Oscillator
« Reply #23 on: October 23, 2013, 01:46:50 AM »
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Offline TechStuf

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Re: Double Pendulum Power
« Reply #24 on: October 23, 2013, 03:22:33 AM »
Nybtorque, they "guaranteed" the wright brothers never flew.  They "guaranteed" the atom would never be split....many of the "guarantees" of the high minded, are about as useful as a wet paper sack.

PHDs.....Piled Higher and Deeper.

And where is the pile sitting?

Squarely on top of the heads of those who blindly follow them.

That, I "guarantee".
 

 
TS
 

Offline LibreEnergia

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Re: Double Pendulum Power
« Reply #25 on: October 23, 2013, 03:46:03 AM »
Nybtorque, they "guaranteed" the wright brothers never flew.  They "guaranteed" the atom would never be split....many of the "guarantees" of the high minded, are about as useful as a wet paper sack.

PHDs.....Piled Higher and Deeper.

And where is the pile sitting?

Squarely on top of the heads of those who blindly follow them.

That, I "guarantee".
 

 
TS


Ok then, I'll do the math that shows this to be so, and post it here.

In the meantime it shouldn't be too hard to connect a small generator to this and prove me wrong, should it :)
Your time starts now.

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Re: Double Pendulum Power
« Reply #25 on: October 23, 2013, 03:46:03 AM »
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Offline TechStuf

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Re: Double Pendulum Power
« Reply #26 on: October 23, 2013, 07:23:49 AM »
LOL.  Do the math on a double pendulum.....Do ALL that math.....if you can.  Math is a tool.  And man too often employs it like a blunt instrument.   In so doing, which then, is the bigger tool?  Look at what he's done with the wheel, the wing, and his other weapons.  It is certainly no coincidence that he assumes too much and therefore misses much more.

 
TS
 

Offline nybtorque

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Re: Double Pendulum Power - AC power from Mech. Oscillator
« Reply #27 on: October 23, 2013, 04:47:09 PM »

I think you are missing something. 

Sure energy is being transferred between the two pendulums and alternately causes one or the other to accelerate.  However I guarantee you the sum of all the potential and kinetic energies in the system remains constant. (or decreasing slowly due to friction).

As soon as you try to extract energy from one location the system will very quickly come to a stop as there is no excess being generated.


You're correct. Kinetic and potential energy remains constant at all times. This is accounted for by using Euler Lagrange. What you and most other seems to miss is that the pendulums actually performs real work constantly; accelerating/decelerating mass; i.e. E(work)= m * a  * s. This is the work/power I calculate in the report, and propose might be partially productive (by replacing the mass by a resistive load or pumping water). And this does not violate the energy equilibrium.


The easiest way to comprehend the function is to think of the inner pendulum mass as water which is moved back and forth and replaced by more water as we go; i.e. a pump… (as in the case of Milkovic, Feltenberger and others)


So, how is this possible? Well, I suppose it has to do with the centrifugal force acting on the fixture resulting from swinging/rotating a pendulum. This force is there at all times and is a function of speed of rotation squared. So even though the kinetic energy is constant (constant speed) of pendulum, we get a force acting on a fixture or a mass as in the case of the double pendulum. And if that mass is free to move the force will accelerate/decelerate the mass, i.e work is performed. It's the power of vibrations. (it's a little bit more complicated of course, and thats the reason for solving the Eular Lagrange equations numerically)


No doubt my model can be validated/falsified by putting a double pendulum on a generator and start swinging. I would suggest replacing as much as possible of the inner pendulum mass with a resistive load on the generator (which can be analyzed) and have a heavy enough outer pendulum to get good results. Even better if the outer pendulum can be set in rotation before the inner pendulum is released to act on the generator. 


I've used my model on the High Perfomance Double Pendulum in the youtube clip above and if we assume the pendulums mass is 1kg each and the levers are 0.4m each and that the operator sets them i motion at a 2 Hz speed from a straight up position (86 J is needed to do this). Then, if we replace the inner pendulum mass with a resistive load and assume no losses, the output would be 21 W . So within 5 seconds we have a COP>1.


Regards NT

Offline TechStuf

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Re: Double Pendulum Power
« Reply #28 on: October 23, 2013, 07:27:09 PM »
Quote

 

 
No doubt my model can be validated/falsified by putting a double pendulum on
a generator and start swinging. I would suggest replacing as much as possible of
the inner pendulum mass with a resistive load on the generator (which can be
analyzed) and have a heavy enough outer pendulum to get good results. Even
better if the outer pendulum can be set in rotation before the inner pendulum is
released to act on the generator. 


I've used my model on the High Perfomance Double Pendulum in the youtube clip above and if we assume the pendulums mass is 1kg each and the levers are 0.4m each and that the operator sets them i motion at a 2 Hz speed from a straight up position (86 J is needed to do this). Then, if we replace the inner pendulum mass with a resistive load and assume no losses, the output would be 21 W . So within 5 seconds we have a COP>1.

In just five seconds?  That would be impressive!  Generators are now common, both linear and rotary, that are in the high nineties for efficiency.  So even at 10 seconds, or heck, say the entirety of it's cycle until dead stop.....

A "smidge" more than necessary to start all over again.

Bessler's way is cool, no doubt, but with the tech of today, following his entire path is unnecessary.
 
 
TS

Offline TinselKoala

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Re: Double Pendulum Power
« Reply #29 on: October 23, 2013, 09:07:48 PM »
Your "model" can be falsified simply by observing that the "double pendulum", more commonly known as a Chaotic Pendulum, always comes to a stop.

Consider this: If you had really sticky bearings the thing would swing a few times and slow down and stop. Right? SO decrease the friction a little bit. Now it takes a bit longer to run down and stop. Reduce friction even more, even put it in a vacuum. It takes a bit longer... but it still stops.

Why? Because the friction eventually dissipates _all_ the energy you put in with your initial starting impulse... and nothing comes in from anywhere to replace it.

If there were _any_ excess energy in the system, by reducing friction to some arbitrarily small value... a value that is less than the magical "incoming" or created power... it would not stop swinging.  But it always does. Therefore... there is no extra energy, no excess power coming in.

ANY load you put on the system will make it come to a stop faster.


The physics simulation "Phun" or "Algodoo" even comes with a couple of Chaotic Pendulums as example scenes. If you think that the chaotic pendulum cannot be modeled mathematically... how does Phun do it, by smoke and mirrors?

 

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