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Overunity Machines Forum



Double Pendulum Power

Started by nybtorque, June 10, 2013, 01:03:21 PM

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nybtorque

Vidar,
Yes, that is the double pendulum I'm refering to.
 
QuoteMaybe 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...

Low-Q

Quote from: nybtorque on June 20, 2013, 11:59:38 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

nybtorque

Vidar,
 
QuoteThe 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.

QuoteIf 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".

QuoteYou 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.

QuoteFurther, 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.

QuoteI 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

nybtorque

Quote from: gyulasun on June 14, 2013, 06:13:18 PM
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).


Low-Q

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