Lets consider the experiment below. A bottle opener and a garlic press, connected with a thin thread through a ring. Hold the ring between your fingers and spin the bottle opener around to reach an equilibrium where the the centrifugal force of the rotating bottle opener keeps the garlic press in a constant position.
If we assume that the mass of the garlic press is 200g and the bottle opener mass is 50g and the radius of rotation is 20cm we need a rotational speed of 14 rad/s, i.e. 2.25 Hz.
To keep this rotational speed and equilibrium we need 0.2J of kinetic energy invested in the rotation.
This is all fine. But lets say we increase the rotational speed a little. What will happen?
Of course the garlic press will begin to move upwards since the centrifugal force will become larger than the gravitational force pulling the garlic press down... Actually, in theory an infinitesimal increase of kinetic energy (a one time energy input) in rotation of the bottle opener will be enough for a continuos movement (acceleration) against gravity for the garlic press and thereby an continuos increase in potential energy, i.e. infinite overunity...
Enjoy!
NT
Quote from: nybtorque on November 07, 2013, 09:21:56 AM
Lets consider the experiment below. A bottle opener and a garlic press, connected with a thin thread through a ring. Hold the ring between your fingers and spin the bottle opener around to reach an equilibrium where the the centrifugal force of the rotating bottle opener keeps the garlic press in a constant position.
If we assume that the mass of the garlic press is 200g and the bottle opener mass is 50g and the radius of rotation is 20cm we need a rotational speed of 14 rad/s, i.e. 2.25 Hz.
To keep this rotational speed and equilibrium we need 0.2J of kinetic energy invested in the rotation.
This is all fine. But lets say we increase the rotational speed a little. What will happen?
Of course the garlic press will begin to move upwards since the centrifugal force will become larger than the gravitational force pulling the garlic press down... Actually, in theory an infinitesimal increase of kinetic energy (a one time energy input) in rotation of the bottle opener will be enough for a continuos movement (acceleration) against gravity for the garlic press and thereby an continuos increase in potential energy, i.e. infinite overunity...
Enjoy!
NT
Hi,
can you please draw a diagram of your experiment since I have a hard following of what should be rotated and what should rise
against the gravity . Also, how you have calculated the forces, using a centrifugal force equation or Lagrange?
Regards.
Quote from: nybtorque on November 07, 2013, 09:21:56 AM
...snip... To keep this rotational speed and equilibrium we need 0.2J of kinetic energy invested in the rotation.
IINM .. the 0.2 joules of energy invested to maintain equilibrium of forces is the amount of energy inputted to overcome windage [drag forces] of the bottle opener & rotating wire ?!
N.B. for a constant velocity [rpm] if we know the drag force [proportional to the square of the speed] & the circumference of the circle at radius 20cm then f x d = joules of energy expended - since the drag force is a function of air mass [e.g. Force in Newtons = Cd.1/2(air density).velocity^2.frontal area] then according to Newton's Laws we must input muscle energy equivalent to the energy lost to air drag.
This is all fine. But lets say we increase the rotational speed a little. What will happen?
Of course the garlic press will begin to move upwards since the centrifugal force will become larger than the gravitational force pulling the garlic press down... Actually, in theory an infinitesimal increase of kinetic energy (a one time energy input) in rotation of the bottle opener will be enough for a continuous movement (acceleration) against gravity for the garlic press and thereby an continuous increase in potential energy, i.e. infinite overunity...
Force [N's] = mass x acceleration
Centrifugal force = Centripetal force = m.v^2/r
Gravity force = Weight force = m.g where 'g' = 9.81 m/s^2
1. A small increase in muscle input energy will break the equilibrium of forces & increase rpm & velocity of the bottle opener [bo] - this will also increase air drag proportionately since it is a factor of air velocity squared.
2. the Cf's acting on the bo will increase by the square of the velocity so that the Cf is greater than the weight force of the suspended mass.
3. since Cf's [mv^2/r] > mg the suspended mass will rise upwards gaining Pe [mgh in joules].
4. the radius of rotation will increase to greater than 20 cm.
5. the Cf will reduce because of a greater radius.
Questions:
a. does the Potential Energy [Pe] gained by the suspended mass ever exceed the excess muscle energy input required to maintain the new & higher rpm ?
b. is there ever a situation using a mechanical assembly of any sort where Pe gained is greater than the input energy ?
Take a look at this torque increasing drive wheel, as the blue disk rotates with all the attached gears and weights, the weights will want to move outward.
The outward movement rotates the center gear with increased torque dependent on the RPM of the disk and the value of the weights.
The system cannot produce back torque or it would be able to do work from gravity and this is impossible, but the increase in torque could be valuable if used correctly.
If an electric motor was used to turn the disk then the weights would simply fly outward and the center gear would rotate the same speed as the disk, but as a load on the center gear increases, like from a generator under load, the weights would want to rotate back inward to the 90 degree position as shown. This applies no additional load on the driving motor.
Now, if the generator was connected to a load like the grid, where the load could sink very large currents, a circuit could be built to keep the weights at the 90 degree point by varying the load on the generator. The motor is now free to spin the disk faster and faster, further increasing the output without any additional load but with huge additional output.
Centrifugal force put to work.
Hi Lumen,
I presume the blue disk is rotating ACW?
Quote from: telecom on November 07, 2013, 08:30:51 PM
Hi Lumen,
I presume the blue disk is rotating ACW?
Sorry,
I ment CCW.
Quote from: telecom on November 07, 2013, 08:30:51 PM
Hi Lumen,
I presume the blue disk is rotating ACW?
It could rotate either direction but it makes sense to rotate this setup ACW because of where the weights are.
I have done some simulation testing and this is at least a very interesting configuration in that the torque on the output is based solely on the weights, without them there is no output torque.
I plan to build a small test setup with some small DC motors to test the idea.
This concept of yours looks very similar to UE.
http://www.universalengines.com.au/how-ue-works
Please keep us posted on the progress.
I agree it is almost the same, but his is overly complicated in that he is trying to time out the imbalance.
(actually after reading some of his work, I believe it's the same principal)
The same thing can be done even easier than I have shown using just four sprockets and some chain on the wheel. (+ the weights)
I have been an engineer for over 30 years and though I am sure this can't work, I don't see why!
The torque on the output is totally decoupled from the drive. If this was not so than a similar setup could be use to build a gravity wheel.
It's like the weights can be continuously falling as the wheel turns, driving the output, and the faster it turns is like stronger gravity.
If anyone knows why this does not work, then speak up so I don't need to build it!
Quote from: fletcher on November 07, 2013, 04:04:12 PM
IINM .. the 0.2 joules of energy invested to maintain equilibrium of forces is the amount of energy inputted to overcome windage [drag forces] of the bottle opener & rotating wire ?!
N.B. for a constant velocity [rpm] if we know the drag force [proportional to the square of the speed] & the circumference of the circle at radius 20cm then f x d = joules of energy expended - since the drag force is a function of air mass [e.g. Force in Newtons = Cd.1/2(air density).velocity^2.frontal area] then according to Newton's Laws we must input muscle energy equivalent to the energy lost to air drag.
The 0.2 J of kinetic energy E=( m * w^2 * r^2 ) / 2, where w is angular velocity (rad/s), i.e. (0.05 * 0.2^2 * 14.1^2 ) / 2 =0.2 J. I assume no friction from windage.
Quote from: fletcher on November 07, 2013, 04:04:12 PM
Force [N's] = mass x acceleration
Centrifugal force = Centripetal force = m.v^2/r
Gravity force = Weight force = m.g where 'g' = 9.81 m/s^2
1. A small increase in muscle input energy will break the equilibrium of forces & increase rpm & velocity of the bottle opener [bo] - this will also increase air drag proportionately since it is a factor of air velocity squared.
2. the Cf's acting on the bo will increase by the square of the velocity so that the Cf is greater than the weight force of the suspended mass.
3. since Cf's [mv^2/r] > mg the suspended mass will rise upwards gaining Pe [mgh in joules].
4. the radius of rotation will increase to greater than 20 cm.
5. the Cf will reduce because of a greater radius.
Questions:
a. does the Potential Energy [Pe] gained by the suspended mass ever exceed the excess muscle energy input required to maintain the new & higher rpm ?
b. is there ever a situation using a mechanical assembly of any sort where Pe gained is greater than the input energy ?
It is easy to calculate an example to answer the questions. If we start with my example of equilibrium, where 0.2 J of kinetic energy is invested for force equilibrium. Then, lets say we invest another 0.2 J in the rotation using finger and arm muscles, total kinetic energy is now 0.4 J. What will happen?
As Fletcher stated RPM will increase, the radius of the rotating BO will increase and the GP will move upwards in search for a new force equilibrium. This will be found sooner or later because energy is proportional to radius squared and force to radius (non squared).
Now we have two unkowns, w (new rpm) and r (new radius), but we have two equations as well:
(1)The new kinetic energy of our system: E(new) = w(new)^2 * r(new)^2 * m(bo) / 2 = 0.4 J and
(2)The new force equilibrium: w(new)^2 * r(new) * m(bo) = m(gp) * g
=> r(new) = 2*E(new) / ( m(gp) * g ) = 0.4 m
The adding of 0.2 J in kinetic energy actually lifted the m(gp) 0.2 m (since the radius increased from 0.2m->0.4m). How much potential energy is that?
Well, by lifting the garlic press 0.2 m we get 0.2kg*g*0.2m = 0.4 J...
The result is by investing another 0.2 J of kinetic energy in the rotation we get TWICE back in gained potential energy. This is done by using the "artificial" gravity caused by rotation (centrifugal/centripetal force). I believe pumping water the same way is even more interesting since we do not have to deal with increased radius and rpm...
I'm afraid Viktor Schauberger beat us all to the punch bowl.
http://www.youtube.com/watch?v=GctFKIp8uXI
And look what he got to show for it, even way back then. Before the hydra dared dream of commandeering local space.
Your computations mean little on the page. Especially in this day and age. Should one manage a real go of things....
He should be aware that the hydra doesn't like competition. And in a world where the truth, what remains, is often but a lie undiscovered, if such were allowed to proliferate, it could only be with the beast's blessing.
http://www.rense.com/general54/babalc.htm
Beware of geeks bearing gifts.
TS
Quote from: lumen on November 07, 2013, 11:05:55 PM
I agree it is almost the same, but his is overly complicated in that he is trying to time out the imbalance.
(actually after reading some of his work, I believe it's the same principal)
The same thing can be done even easier than I have shown using just four sprockets and some chain on the wheel. (+ the weights)
I have been an engineer for over 30 years and though I am sure this can't work, I don't see why!
The torque on the output is totally decoupled from the drive. If this was not so than a similar setup could be use to build a gravity wheel.
It's like the weights can be continuously falling as the wheel turns, driving the output, and the faster it turns is like stronger gravity.
If anyone knows why this does not work, then speak up so I don't need to build it!
Hi Lumen,
how do you transmit the original torque from the blue wheel to the central gear
in your machine? This is not very clear to me..
Also, can you elaborate on a simpler design with four sprockets + weights you mentioned?
Thank you on advance!
Quote from: telecom on November 08, 2013, 06:05:37 AM
Hi Lumen,
how do you transmit the original torque from the blue wheel to the central gear
in your machine? This is not very clear to me..
Also, can you elaborate on a simpler design with four sprockets + weights you mentioned?
Thank you on advance!
The original torque is never transferred to the output gear so at very low RPM or no weights, there is no output torque.
To make an easier version you would make the center and two outer gears as sprockets with the center sprocket a double sprocket, then run a chain from one center sprocket to one outer sprocket and another chain from the center to the other outer sprocket.
It's also easier to understand if you think of it running horizontally to avoid the effects of gravity.
@nybtorque: I believe pumping water is not the same because you must constantly accelerate new mass as new water enters, and that will require additional work.
Quote from: lumen on November 08, 2013, 11:05:56 AM
The original torque is never transferred to the output gear so at very low RPM or no weights, there is no output torque.
To make an easier version you would make the center and two outer gears as sprockets with the center sprocket a double sprocket, then run a chain from one center sprocket to one outer sprocket and another chain from the center to the other outer sprocket.
It's also easier to understand if you think of it running horizontally to avoid the effects of gravity.
@nybtorque: I believe pumping water is not the same because you must constantly accelerate new mass as new water enters, and that will require additional work.
I think I understand - an input only accelerate the weights towards the required speed.
This model with the sprockets is quite easy to make and try.
Regards.
I want to say that even though the chain configuration appears to achieve the same goal, there may be important differences between the chain and gear configurations in how the centrifugal forces are utilized.
I think I know what you mean - the chain is not rigid enough to transfer the torque positively.
Quote from: lumen on November 08, 2013, 11:05:56 AM
@nybtorque: I believe pumping water is not the same because you must constantly accelerate new mass as new water enters, and that will require additional work.
Yes, but that is exactly what the centrifugal force of the rotating "garden hose" is doing!
Consider replacing the the thread and bottle opener with a hose filled with water and rotate it... Lets say its an 3/4" hose and its 2m long and you swing it above your head at 96 rpm (10 rad/s). That equals 0.57 kg of water, and with the center of the water mass rotating with a 1m radius, we get the force Fc=10^2*0.57*1=57N and kinetic energy Ek=10^2*1^2*0.57/2=28.5J.
So what mass of water do you need to reach equilibrium with the rest of the hose hanging down at the center of the rotation? Well, to get m*g=57N you need 5.7kg of water to keep the rotating water from escaping through the open end of the hose... Thats a 20m long hose hanging straight down...
But, what if the it is only 2m (suction head) from the swinging hose above your head to the water bucket you're trying to pump? What will happen?
The way I see it, there will be a net force of 51.3 N acting on the rotating water wanting it to escape. That would translate to an average pressure of 180 kPa in the 3/4" hose.
Of course new water will need to be accelerated at all times. With a mass of 0.57 kg rotating, it will accelerate by the centrifugal force at 90 m/s2 (F=ma) from 0 velocity at the center of equilibrium to an escape velocity of 9.5 m/s at the open end of the hose. That equals a flow of about 2.7 l/s and a pumping power of 487W... And this is just by keep rotation at the same rpm, i.e. no extra kinetic energy have to be added since the mass, radius and angular velocity is unchanged.
Quote from: nybtorque on November 07, 2013, 09:21:56 AM
Lets consider the experiment below. A bottle opener and a garlic press, connected with a thin thread through a ring. Hold the ring between your fingers and spin the bottle opener around to reach an equilibrium where the the centrifugal force of the rotating bottle opener keeps the garlic press in a constant position.
If we assume that the mass of the garlic press is 200g and the bottle opener mass is 50g and the radius of rotation is 20cm we need a rotational speed of 14 rad/s, i.e. 2.25 Hz.
To keep this rotational speed and equilibrium we need 0.2J of kinetic energy invested in the rotation.
This is all fine. But lets say we increase the rotational speed a little. What will happen?
Of course the garlic press will begin to move upwards since the centrifugal force will become larger than the gravitational force pulling the garlic press down... Actually, in theory an infinitesimal increase of kinetic energy (a one time energy input) in rotation of the bottle opener will be enough for a continuos movement (acceleration) against gravity for the garlic press and thereby an continuos increase in potential energy, i.e. infinite overunity...
Enjoy!
Your analysis is incorrect.
In the equilibrium position the gravitational potential energy of the garlic press would equal the energy of the rotating bottle opener. If you increase the rotation speed by applying work the garlic press moves up increasing its gravitation potential energy.
AND the radius of the rotation of the bottle opener increases.
If you do the sums you will find that the increase in in GPE = the increase in rotational kinetic energy = work supplied to the system and the system finds a new equilibrium point. No excess of energy is left over and that is what would be required if it were 'OU'.
Quote from: LibreEnergia on November 10, 2013, 05:15:02 PM
Your analysis is incorrect.
In the equilibrium position the gravitational potential energy of the garlic press would equal the energy of the rotating bottle opener. If you increase the rotation speed by applying work the garlic press moves up increasing its gravitation potential energy.
AND the radius of the rotation of the bottle opener increases.
If you do the sums you will find that the increase in in GPE = the increase in rotational kinetic energy = work supplied to the system and the system finds a new equilibrium point. No excess of energy is left over and that is what would be required if it were 'OU'.
Well, that is exactly what I described in a later post. However, I came to a different conclusion. So please show me your numbers for exactly what happens with the potential energy of the garlic press if you increase the kinetic energy of the bottle opener with a specific amount... Only stating that my analysis is incorrect doesn't do it for me unless you show me where! It should be easy enough.
Clue: you need to explain it using the following equations;
w2 * mbo * r = mgp * g (the force equilibrium regardless of
w and
r) and
Ekin = 1/2 * w2 * r2 * mbo (kinetic energy, regardless of
w and
r) and
Epot = mgp * g * (rafter-rbefore) (potential energy after the increase in
Ek which will result in a new equilibrium with a new
w and
r)
Quote from: lumen on November 07, 2013, 07:35:18 PM
Take a look at this torque increasing drive wheel, as the blue disk rotates with all the attached gears and weights, the weights will want to move outward.
The outward movement rotates the center gear with increased torque dependent on the RPM of the disk and the value of the weights.
The system cannot produce back torque or it would be able to do work from gravity and this is impossible, but the increase in torque could be valuable if used correctly.
If an electric motor was used to turn the disk then the weights would simply fly outward and the center gear would rotate the same speed as the disk, but as a load on the center gear increases, like from a generator under load, the weights would want to rotate back inward to the 90 degree position as shown. This applies no additional load on the driving motor.
Now, if the generator was connected to a load like the grid, where the load could sink very large currents, a circuit could be built to keep the weights at the 90 degree point by varying the load on the generator. The motor is now free to spin the disk faster and faster, further increasing the output without any additional load but with huge additional output.
Centrifugal force put to work.
It's good to be optimistic, but the only way to know what you said is true or not is to build and experiment. If you want I can help.
Quote from: broli on November 11, 2013, 04:21:51 PM
It's good to be optimistic, but the only way to know what you said is true or not is to build and experiment. If you want I can help.
I seen some problems in the previous design so I made some changes and now......OU?
It's very simple and is OU!
Well I suppose I might be a bit optimistic but now none of the back torque on the generator will be felt in the main drive wheel.
If torque had any effect on the wheel, then one could simply put a spring to apply torque on the outer sprockets and the wheel would turn forever which is impossible.
So even though the generators will exert back torque into the outer sprocket, that torque is simply applied to the center stationary sprocket.
In effect, the wheel could be turned by hand and generate huge amounts of power, or a tiny motor could generate thousands of watts of output.
I believe the problem in the previous design is in the fact that the rotation of the large disk is also the device that rotates the smaller working components. This means that resistance to rotation in the working components also results in resistance to rotation of the large disk.
In the new design, the working outer sprocket never rotates and is essentially a stationary platform moving in a circle. The artificial gravity generated then is moving in a circle around this stationary platform. If the direction of gravity on earth was rotating, one could simply build an offset wheel and generate power. Luckily, Earths gravity does not rotate because other things would suck, but now you see the point.
Because of the one to one sprocket gearing, applying any amount of torque on the outer sprocket, no torque is applied to the large disk. So in effect, no back torque can be see in this generating system.
Does anyone see anything different?
Or is that it, Game Over!
The sprockets cannot be completely independent of the big wheel can they, unless I'm misunderstanding the concept. To me it looks like the countertorque of the generator will be projected on the big wheel. It's just like someone grabbing the chain with his hands, the whole thing will grind to a halt.
Quote from: broli on November 12, 2013, 02:59:20 PM
The sprockets cannot be completely independent of the big wheel can they, unless I'm misunderstanding the concept. To me it looks like the countertorque of the generator will be projected on the big wheel. It's just like someone grabbing the chain with his hands, the whole thing will grind to a halt.
That's what makes this so interesting!
You need to ask yourself if there is any way the large wheel would turn by applying a torque to the outer sprocket when the inner sprocket is mounted stationary (not on the big wheel).
The answer is yes, only if the two sprockets are of different sizes. When the inner and outer sprockets are the same size, then it is impossible to make the big wheel rotate by applying any torque to the outer sprocket when the center sprocket is mounted stationary.
If the big wheel can see no torque from the outer sprocket, then what is there that could apply any back torque on the big wheel when generator loading occurs?
I don't know!...... That's why I'm asking.
The only thing the big wheel would see is the mass of the generators and weights.
By torqueing the outer sprocket the only thing the occurs is the compression of the solid mass through the centerlines of the sprockets and the stretching of the chain at the outer edge of the sprockets.
Do you think that there might be a force vector that does not pass directly through the center of the big wheel when torque is applied to the outer sprocket?
If the big wheel could rotate from a torque applied on the outer sprocket then we could build a sensitive wheel and place magnets on the outer sprocket, then because the outer sprockets do not rotate, the earths magnetic field could apply a constant torque and the wheel would run for nothing!
I believe the problem in making any of these devices work as imagined, depends on which components the torque is applied or more so, the reference from which the torque is applied.
Torque applied to the outer sprocket from the external or stationary frame cannot make the large wheel rotate, but torque applied from the frame of the large wheel can rotate the wheel.
It seems the answer is somewhere in this type of wheel, locked between gravity and centrifugal force.
lumen
I like the version on you post on Reply #12 best. But it won't work as I have played with this before. But there are other roads you should look at for secondary uses. I have seen people use this approach for mechanical anti gravity as well.
It is not always our journeys end but what we learn on the way.
Alan
AB
Yes I like that one too but I believe it's no different than directly driving the center sprocket with the big wheel.
It would be the same if the outer sprockets were bolted directly to the big wheel.
So now we can use two Gyro wheels to maintain position for half the generator and drive it with centrifugal force, both independent of the rotating wheel.
I added a short video of operation.
Quote from: lumen on November 13, 2013, 06:41:19 PM
AB
Yes I like that one too but I believe it's no different than directly driving the center sprocket with the big wheel.
It would be the same if the outer sprockets were bolted directly to the big wheel.
So now we can use two Gyro wheels to maintain position for half the generator and drive it with centrifugal force, both independent of the rotating wheel.
I added a short video of operation.
lumen
But what is going to power the Gyros? Many things can work as a Gyro, but something has to power them. From your video you can use more of a pendulum instead for it will not require power.
I have an even better way!
2 wheels connected like a steam engine for the stationary drive.
Nope! The stationary frame of reference must be from a point on the main wheel just like the other source, the centrifugal weights.
Any external source will apply work against the wheel.
The Gyro's are the way...... unless the torque against the gyro's will cause them to slow faster and require more power.
I need to look into this but it's centrifugal force operating the gyro's also, so it could be that the faster they spin the less they will slow down at a given torque.
It's possible that the steam engine parallel connecting rod idea could work if the wheels rotated in opposite directions.
This would require some type of sliding or extending rod, but it would apply the resistance to rotation of one wheel to assist the other and this might cancel the counter torque against the wheel.
Though the torque transferred into the connecting arm works against the wheels rotation for half the rotation, it works with the wheels rotation for the second half. This should cancel the reverse torque problem with a net resistance of zero.
One could also add several more generators to each wheel and provide huge gains in output for nearly zero input.
You can also watch the movie if you have a need to feel sick!
Hi Lumen,
well done !
In the Generator 3 test simulation are these just weights there on the discs or do you still use Gyros ?
Does it selfaccelerate as in the video ?
What simulation software did you use and could you post the simulation file for others to play with it ?
What were your friction settings in the simulation software ?
Many thanks.
Regards, Stefan.
Stefan
The idea here is to find a method to turn a generator by using external forces so the drag of the generator from loading is never returned to the driving motor.
The gyros in the generator 3 drawing have been removed as the stationary generator support force, and replaced with the connecting rod to achieve the same goal. I am not fully convinced of the independence of the connecting rod to prevent torque from the generators from moving into the driving wheels and increasing the load on the driving motor. The use of gyros totally prevents this but may have some other problems.
The simulation is only motion simulation and does not measure any values from the generators or torque loading on the wheels so a prototype will need to be built to do further testing.
The video is run with the gravity direction towards the face of the wheels and with some friction on the weights to simulate the generator loading.