Harvesting energy from the earth rotation

[Simon Bridge]

Those aren't references supporting what you say - they are just references to you saying it...

The PE calc looks like the difference between gravitational PE and Centrifugal effect.

If you just fired something straight up, in a rotating reference frame, it does not go straight. Put it on a radial track, though, and push it straight up, then the rotation gives you some help ... though you lose energy from the thing rotating. Now your rotating reference frame no longer has a constant angular speed wrt an inertial frame - did you take this into account?

The further you run the mass up the track, the more energy you draw off from the rotation. This is the same effect as an ice-skater pushing their arms out to slow their spin. At some stage, you'll slow to a stop.

At the end of the track, above geostationary, the mass seems to accelerate madly away for a bit - but isn't that just a classical relativity effect? The mass just heads off at a constant velocity in the inertial frame.

There is a common mistake people make when they consider rotational gravity - they may have someone start out in the "low g" section of a cylindrical habitat moving slowly radially and then figure that since gravity depends on the distance from the center, that the person experiences an increasing gravitational pull as they move outwards.

 

[mfb]

Those aren't references supporting what you say - they are just references to you saying it...

Those are calculations supporting what I say.

Now your rotating reference frame no longer has a constant angular speed wrt an inertial frame - did you take this into account?

The slowdown of Earth is negligible for any reasonable mass lifted with a space elevator. I did not take it into account, for anything below ~10^18 kg the effect is smaller than 0.1% (the rotational energy of Earth is the same as the kinetic energy of 7*10^21kg, moving around Earth with 100.000km radius once a day like a giant slingshot)

The further you run the mass up the track, the more energy you draw off from the rotation. This is the same effect as an ice-skater pushing their arms out to slow their spin. At some stage, you'll slow to a stop.

Right. You certainly do not want that to happen, and stop the device if the days become too long. The geostationary orbit will move upwards a bit.

At the end of the track, above geostationary, the mass seems to accelerate madly away for a bit - but isn't that just a classical relativity effect? The mass just heads off at a constant velocity in the inertial frame.

This is just Newtonian mechanics in rotating systems.

There is a common mistake people make when they consider rotational gravity - they may have someone start out in the "low g" section of a cylindrical habitat moving slowly radially and then figure that since gravity depends on the distance from the center, that the person experiences an increasing gravitational pull as they move outwards.

If they keep stationary relative to the rotating cylinder (and their mass is small compared to the cylinder), this is true. If they are in free fall, it is wrong. Conclusion: You need a space elevator as rotating "floor", otherwise it will not work.

 

[Simon Bridge]

The slowdown of Earth is negligible for any reasonable mass lifted with a space elevator.

Not if you want to to go to arbitrarily high kinetic energy as per your previous posts.

Right. You certainly do not want that to happen, and stop the device if the days become too long. The geostationary orbit will move upwards a bit.

If, however, you return mass by the same route, you can recover the lost energy.

If they keep stationary relative to the rotating cylinder (and their mass is small compared to the cylinder), this is true. If they are in free fall, it is wrong. Conclusion: You need a space elevator as rotating "floor", otherwise it will not work.

You have to modify the equation for an elevator frame that is not classically rigid. It will flex as the mass travels up.

In general you don't expect the beanstalk-type elevator to be rigid ... or even to go up strictly radially. From Earth, carbon-nanofibers are not strong enough ... though I have seen calculations that make such a thing more feasible for, say, Mars.

We often neglect small reaction in our calculations - eg. when we jump, we neglect the recoil of the earth. The trouble comes when doing the same for large distances and energies. I'm not so much concerned about your ideas but that someone less versed in physics will read what you have written and think "Wow: free energy!" I seem to attract people so this is a kind of self defense ;)

The last set of calculations I saw had a space elevator go twice the distance to geostationary ... 71,400km. I couldn't help thinking of tidal drag... and what would happen if it broke! I have Endless Frontier VolII, Jerry Pournelle (ed). as a lay-accessible optimistic take on the possibilities.

 

[mfb]

Not if you want to to go to arbitrarily high kinetic energy as per your previous posts.

Well, a real space elevator would have a finite length anyway - even if you ignore all engineering issues, at 350,000km it would crash into moon. You can gain energy if the cable extends to >150,000km if my calculation is right.
My old posts have an error - the energy should be 1/2 ω^2 r^2 instead of 1/3 ω r^3

If, however, you return mass by the same route, you can recover the lost energy.

Mass at geostationary orbit is perfect - you can gain energy by lowering or raising it.

You have to modify the equation for an elevator frame that is not classically rigid. It will flex as the mass travels up.

I neglected this, too, with the assumption that most mass of the elevator is required to support the elevator, and only a small fraction of its mass can be transported on it.

We often neglect small reaction in our calculations - eg. when we jump, we neglect the recoil of the earth. The trouble comes when doing the same for large distances and energies. I'm not so much concerned about your ideas but that someone less versed in physics will read what you have written and think "Wow: free energy!" I seem to attract people so this is a kind of self defense ;)

It is not free energy, but energy which is available in large amounts. This is similar to fusion power, for example - I think that fusion is way better than slowing earth, but that is not the topic here.

 

[K^2]

The fact that your equation gives arbitrary amount of energy should immediately make you suspicious. Earth has finite amount of energy available. Seems like a thing to take into account. Let's throw angular momentum conservation in.

\left(I+mR^2\right)\omega = \left(I + mr^2\right)\omega '

Using ω' in equation for effective potential.

U = -\frac{MG}{r} -\frac{\omega '^2 r^2}{2} = -\frac{MG}{r} - \left(\frac{I+mR^2}{I+mr^2}\right)^2 \frac{\omega^2 r^2}{2}

Note that this potential goes to 0 at r->inf, which is the sort of thing you'd expect. Rotation just doesn't matter if you take it to infinity.

For m<<M, this potential has 6 extreme points, 3 of which are on positive real axis, tow of which correspond to equilibria. First is unstable equilibrium at ~42,000 km, corresponding to geostat. Assuming Earth can be modeled as homogeneous sphere for purposes of moment of inertia estimate, the second equilibrium point is stable and is located about 10^16 km above Earth. At that point, the potential is at about -6x10^28/m J, which means, you can extract up to about 1/4 of Earth's total rotational energy.

 

[Simon Bridge]

Mass at geostationary orbit is perfect - you can gain energy by lowering or raising it.

Nooooo ... by that argument you can set a mass going up and down in a cycle and eventually stop the Earth rotating. Remember where the energy comes from, and small numbers add up to big ones.

I neglected this, too, with the assumption that most mass of the elevator is required to support the elevator, and only a small fraction of its mass can be transported on it.

You are hoping that the energy transferred to this very small mass, from the Earth, via the elevator is also small compared with the transverse rigidity (on this scale the tower is more like a rope - stretchy and flexy) ... but then, right now, such a thing would need magic materials just to exist.

 

[mfb]

The fact that your equation gives arbitrary amount of energy should immediately make you suspicious.

It should simply show you the limits of the equation.

the second equilibrium point is stable and is located about 10^16 km above Earth.

=1057ly. Sounds a bit... impractical :D.
If you plan to lift significant fractions of Earth's mass, the gravitational energy needs a modification, too.

Mass at geostationary orbit is perfect - you can gain energy by lowering or raising it.

Nooooo ... by that argument you can set a mass going up and down in a cycle and eventually stop the Earth rotating.

No. Geostationary orbit is just a local maximum of the potential (for m<<m_earth). If you balance a pen on its tip you can gain energy by letting it fall in every direction. Does that imply any possible cycle?

 

[K^2]

The entire proposal isn't practical, so that's a moot point. The question was, can energy be gained using this method in principle. If you don't factor in conservation of angular momentum, you end up with a result that diverges to infinity, which is wrong and obviously doesn't tell you about whether the actual result is a net gain or net loss. For a sufficiently massive object with sufficiently low angular velocity, the second equilibrium will actually be at higher potential than surface potential, despite your formula still saying you can gather energy from it.

As far as accounting for Earth's loss of mass, my formula does account for it. Starting mass of Earth is M+m.

And if we are talking about practical limitations, there is no material out of which you can build the elevator like this. Nothing will provide sufficient resistance to bending over such distance. The way a space elevator works around that is by having an anchor point slightly above geostat to provide tension. Placing an anchor for your method would eat up more energy than you can harness.

 

[Simon Bridge]

No. Geostationary orbit is just a local maximum of the potential (for m<<m_earth). If you balance a pen on its tip you can gain energy by letting it fall in every direction. Does that imply any possible cycle?

Ah now I get you ... you mean: starting from geostationary you gain kinetic energy in either direction. Excuse me.

 

[Simon Bridge]

Placing an anchor for your method would eat up more energy than you can harness.

Isn't that, if you got the anchor mass from the Earth?... though I understood that the elevator idea is not really for generating energy but to get easier subsequent access to space.

Also - wouldn't such a cable try to lie along an equipotential - so it would have to be tethered at both ends? (Otherwise we are describing a long thin satellite which is not only in geostationary orbit but must rotate to always present one end to the Earth.)

I'm reading mfb's proposal as "having paid the enormous cost of getting one in the first place - we can use the space elevator to draw off energy from the Earth rotation ..." which would satisfy OPs question - sort of.

 

[mfb]

I'm reading mfb's proposal as "having paid the enormous cost of getting one in the first place - we can use the space elevator to draw off energy from the Earth rotation ..." which would satisfy OPs question - sort of.

This.

Building a space elevator purely as power plant is highly impractical. But if that structure is ever built for cheap space launches, it would be possible to use it as power plant, too (engineering issues are a different thing). The first cable would extend to some point a bit above GEO to keep the whole thing in place, afterwards you can extend it outwards with material lifted on the space elevator itself.

For a sufficiently massive object with sufficiently low angular velocity, the second equilibrium will actually be at higher potential than surface potential, despite your formula still saying you can gather energy from it.

I think you have to extend the equation a bit in the general case, as the center of mass and rotation shifts, too. Assuming I of Earth already contains the space elevator (or neglecting its mass):
The initial center of mass relative to the geometric center of Earth is Rm/(M+m), assuming that the mass to lift just lies around on the surface. It becomes important once you lift your mass over large distances as done in your calculation:

The center of rotation will be at x=r m/(M+m), this gives a total moment of inertia of I + Mx^2 + m(r-x)^2 (neglecting momentum of inertia of m).

Using the reduced mass \mu=\frac{Mm}{M+m}, the r-dependent angular momentum becomes
\left(I+M\frac{r^2m^2}{(M+m)^2}+m\frac{r^2M^2}{(M+m)^2}\right)\omega(r)<br /> = (I+\mu r^2)\omega(r)Short:
(I+\mu R^2)\omega_0 = (I+\mu r^2)\omega(r)
Solving this for ω(r) gives
\omega(r)=\omega_0\frac{I+\mu R^2}{I+\mu r^2}

The total potential+kinetic energy at distance r now is given by
U<br /> =\frac{-GMm}{r} + \frac{1}{2}(I+\mu r^2)\omega_0^2 \left(\frac{I+\mu R^2}{I+\mu r^2}\right)^2<br /> =\frac{-GMm}{r} + \frac{\omega_0^2(I+\mu R^2)^2}{2}\frac{1}{I+\mu r^2}<br />

Fixing m to some value and introducing positive constants a,b and c, the general shape is
U=\frac{-a}{r} + \frac{b}{c+r^2}
Derivative:
U&#039;=\frac{a}{r^2} + \frac{-2br}{(c+r^2)^2}
Solving for U'=0:
a(c+r^2)^2 = 2br^3
Small angular momentum or high mass of Earth gives small b or large a, while c is constant (given by the geometry of earth), therefore the equation will not have any solution.
Ok, I can confirm your result.

And if we are talking about practical limitations, there is no material out of which you can build the elevator like this. Nothing will provide sufficient resistance to bending over such distance.

Bending is significant only if you want to lift a lot of mass at the same time.

 

[Simon Bridge]

Bending is significant only if you want to lift a lot of mass at the same time.

That is just the definition of "a lot of mass" in this context.
(Of course, still need to define "significant".)

How much mass were you thinking of lifting in one go?
Apollo CSM was 30Tonnes or so. How much lateral rigidity would be needed? Ballpark figure?

Of course you need not go that way - you could imagine a structure that flexes like a kite-string with an anchor "out there" maintaining tension. You can still run mass up and down that sort of thing. But it changes the equations.

 

[mfb]

Ballpark figure: Lifting 100 tons with 100m/s requires an initial power of 10MW and gives a Coriolis force of F_c = 2m \vec{v}\times\vec{\omega} = 1300N. This would probably bend the space elevator a little bit within some meters of its length.
I found this image for illustration - a 1°-bend in the cable would require an internal tension of ~60*1300N = 80kN, or the gravitational force of 8 tons. I would expect that a space elevator maintains more minimal tension as safety margin (especially when weights of 100 tons have to be lifted).

 

[Simon Bridge]

Pournell's beanstalk was 2x geostationary long ... 1deg is a deflection of over 625km at 1x geo. (picturing the deflection as a kink in the line). Unless the cable stretched, this would pull the midpoint in by 5.4km.

Pournell's (and others) solution was to transport much smaller masses ... 100s of kg to 1T at a time ... in a chain of buckets. Ship components and do your assembly in orbit.

However - I think this topic is just about done to death.

 

[johnmn3]

Supposing we had a gyroscope with 100% mechanical efficiency -- say it is in a vacuum chamber, suspended in place by some frame of superconductors.

Also suppose this gyroscope has magnets on it, facing outward from the walls of the spinning surface (where the white paint on a car's white-wall tires would be), such that we can externally apply force against the axis of the gyroscope's spin, without causing friction.

Now, it is not a stretch to imagine how, as the Earth rotates, the gyroscope's magnets will apply force against a set of magnets connected to a motor that is planted on the earth.

I guess the question is, does applying force against the gyroscope's axis of rotation cause the rotation to slow down? Even if the spin is 100% mechanically efficient?

 

[Simon Bridge]

Welcome to PF;
What is keeping the frame from moving?

 

[johnmn3]

Thanks. Pleasure to be here. The frame would be secured to an object on the earth. It would have a gear system that puts pressure on the gyro as it tries to turn. Would the rotation slow down if the gyro were pushed against its spinning axis?

 

[mfb]

I don't understand the details of your setup (sketch?), but you cannot gain energy with it:
Reducing the angular velocity of Earth requires an increased angular velocity of the gyro (angular momentum conservation). But that violates energy conservation unless you use some external energy source to do so.

 

[johnmn3]

Sure, let me pull out my handy dandy Galaxy Note.

Attached is a PNG of the sketch. A shows the gyro on the Earth's equator. In B you see the gyro inside its frame, attached to the gears (I'm no artist). The frame is a vacuum sealed enclosure, using magnets (and/or superconductors) to 1) let this internal disc spin freely, while 2) allow the application of frictionless force against the walls of the spinning disc, in order to turn the gears.

C shows how the gyro attempts to maintain a constant orientation as the Earth spins.

D shows how the gears apply resistance to the relative spin of the gyro.

The question is, does applying force against spinning axis of the gyro add to or take away from the energy in the spinning axis.

I know there's an input axis and an output axis that I don't fully understand yet. But if the spinning axis can gain and lose energy via the input and output axis, I would argue one could position the generator such that the Earth's turn pushes into the input axis. Then the generator could simply bleed energy from the spin, using coils on the frame and the magnets on the disc.

 

[johnmn3]

Looks like these guys have already put some thought into it: http://www.google.com/patents/US5313850

 

[johnmn3]

Here's a businessweek article on these guys in 1995: http://www.businessweek.com/stories/1995-08-27/as-the-world-turns-free-energy

 

[johnmn3]

According to this research, graphene flakes can spin up to 60,000,000 rpm. http://www.newscientist.com/article/dn19514-levitating-graphene-is-fastestspinning-object-ever.html

Couple that with research showing that strained graphene can generate pseudo-magnetic fields in excess of 300 teslas: http://newscenter.lbl.gov/news-releases/2010/07/29/graphene-under-strain/

Graphene sounds like a good material for this.

A nanoscale gyro might have more optimal friction characteristics as well: http://www.nepachemistry.com/2011/01/macroscopic-and-microscopic-molecular.html

 

[LURCH]

Supposing we had a gyroscope with 100% mechanical efficiency -- say it is in a vacuum chamber, suspended in place by some frame of superconductors.

Also suppose this gyroscope has magnets on it, facing outward from the walls of the spinning surface (where the white paint on a car's white-wall tires would be), such that we can externally apply force against the axis of the gyroscope's spin, without causing friction.

Now, it is not a stretch to imagine how, as the Earth rotates, the gyroscope's magnets will apply force against a set of magnets connected to a motor that is planted on the earth.

I guess the question is, does applying force against the gyroscope's axis of rotation cause the rotation to slow down? Even if the spin is 100% mechanically efficient?

Yes, the magnetic field acting against the gyro will impart some drag. Eventually, this drag will slow the gyro's rotation to a stop (relative to the magnets). The amount of energy that can be extracted during this process will be slightly less than the amount it took to get the gyro spinning in the first place.

 

[johnmn3]

People keep thinking this a free energy scheme. It is not.

Just imagine a bike wheel spinning on a string. Now add another small wheel to the rim (perpendicular to the big wheel). Spin the big wheel really fast. Now spin the small wheel really fast. The small wheel will start to slow down the spin of the big wheel -- their angular momentums are fighting each other. The small wheel is taking energy away from the big wheel.

Now put the small wheel on a gimbal and now it can maintain the small wheels orientation freely, letting the big wheel spin on its own accord. Now the small wheel is not taking energy away from the big wheel.

Now put a gear box on the small wheel's gimbal and let the large wheel compete with the small wheel with half the force it normally would exhibit without the gimbal. Now the small wheel is being forced to spin at half the speed of the big wheel. This takes energy away from the big wheel but feeds that energy into the gear box.

Does that make sense?

 

[Simon Bridge]

Thanks. Pleasure to be here. The frame would be secured to an object on the earth. It would have a gear system that puts pressure on the gyro as it tries to turn. Would the rotation slow down if the gyro were pushed against its spinning axis?

Well ... yes.
The problem with these sorts of projects is getting the thing harvesting the energy to take it from the Earth and not the Gyro. In practice you end up only getting the energy out that you put in when you set the gyro spinning.

Here it is the frame turning wrt the gyro axis that is the motion you want to get energy from.
You had to put energy in by spinning the gyro in the first place - in practice you'll have to keep boosting this but work out the plausibility by assuming it does not run down by itself.

The next bit will be where you put the magnet and where the generator coils - don't get bogged down in the details of what kind of magnets and conductors you will use: that's for later. You need them placed to harvest the Earth's rotational energy ... eg. if you tried it with a mechanical gear-chain driving an electric generator, you'd have trouble placing the generator so it is not also rotating with the earth.

People keep thinking this a free energy scheme. It is not.

No - it is just massively impractical though well tried.
Usually people forget about where the generator has to go, or they fail to account for the energy used spinning up the gyro in the first place.

Also have a look at this demo:
http://www.lhup.edu/~dsimanek/museum/advanced.htm
... you seem to have that setup ... generator attached to support-frame, magnet on the gyro-frame.
The PHYS-L discussion in question is here, and still available.
Of particular note is this observation: a steel disk 1m thick and 3m in radius spinning at 10 cycles/s (!?). Would have a max (ideal) power output of ~100mW and cost you some 2GJ to spin it up.

You can have a go tweaking the figures:

$$P_{max} = \tau_g \omega_s = \frac{\omega_s}{2}\frac{dL_g}{dt} = \frac{1}{2}\omega_e L_g = \frac{1}{4}I_g\omega_g\omega_e^2$$ ... where:
$\tau_g$ is the torque on the gyro (under load),
$\omega_s$ is the angular velocity of the shaft wrt Earth,
$L_g$ is the angular momentum of the gyro,
$\omega_e$ is the angular velocity of the Earth, and
$\omega_g$ is the angular velocity of the gyro about it's spin axis.

 

[mfb]

Just imagine a bike wheel spinning on a string. Now add another small wheel to the rim (perpendicular to the big wheel). Spin the big wheel really fast. Now spin the small wheel really fast. The small wheel will start to slow down the spin of the big wheel -- their angular momentums are fighting each other. The small wheel is taking energy away from the big wheel.

To get any increase of the speed of the small wheel, you need external interaction (friction or whatever) - something you do not have in the earth+gyro system. You can spin up the big wheel and slow the smaller one, but that is not useful.


You can design anything you want, it does not matter. Just with energy and momentum conservation, it is possible to show that your setup does not work in an isolated system. To make it work, you need at least one of those 4:
- an external interaction: This can be the moon, or launching things to space, or whatever
- a violation of energy conservation
- a violation of linear momentum conservation
- a violation of angular momentum conservation

It can be shown (this is a mathematical proof) that energy conservation follows from time-independent physical laws, and momentum conservation from space- and orientation independence of the physical laws. Those can be studied on a very fundamental level. If you do not find any violation there (certainly worth a Nobel prize), your concept cannot work unless you go to space.

 

[Simon Bridge]

To get any increase of the speed of the small wheel, you need external interaction (friction or whatever) - something you do not have in the earth+gyro system. You can spin up the big wheel and slow the smaller one, but that is not useful.

The big-wheel+small-wheel thing puzzled me since it has little to do with the setup being considered with the variations on the gyrogenerator.

The equivalent with the big wheel being the Earth would be just to run the small wheel along the ground... if you could get into a position where the Earth is turning under you then you could just hold a small wheel to the Earth to generate power: the trick is getting in that position to start with.

However - I thought that the generator being considered was like the one in the link I supplied (the apparent 1/day precession of a gyroscope being used to generate power). This one does not violate any conservation laws that I know of - it is just very very impractical.

People get excited over it for much the same reasons they go starry eyed of perpetual motion or over-unity machines: important stuff got left out. Lots of attention gets lavished on supermagnets and conductors and high-tech materials, just like with pmm, then the proponent points out that this is not a "free energy" machine... yet the basics of how much energy is available vs invested gets overlooked.

 

[mfb]

if you could get into a position where the Earth is turning under you then you could just hold a small wheel to the Earth to generate power: the trick is getting in that position to start with.

To get in such a position and keep it (even with the force required to accelerate the small wheel), you need some external anchor.

it is just very very impractical.

It is not just impractical, it is a pure waste of useful power. You can simply use the flywheel in the conventional way as energy storage and get the same result, but with a better efficiency.

 

[Buckleymanor]

To get in such a position and keep it (even with the force required to accelerate the small wheel), you need some external anchor.

Why do you need an external anchor.Is it not the proposal of using a gyro to be able to use it as an anchor.The giro is set spinning say on it's side at the north pole, if it's left spinning for a day it will appear to make one complete rotation.It is not rotateing it's just obeying Newton's laws in as much that it wan't to move in one direction once the giro is spinning.
I'ts the Earth that is rotateing beneath it.
The giro stays put like an anchor.

 

[willem2]

Why do you need an external anchor.Is it not the proposal of using a gyro to be able to use it as an anchor.The giro is set spinning say on it's side at the north pole, if it's left spinning for a day it will appear to make one complete rotation.It is not rotateing it's just obeying Newton's laws in as much that it wan't to move in one direction once the giro is spinning.
I'ts the Earth that is rotateing beneath it.
The giro stays put like an anchor.

This thread has really been dead since post #7.
If you wan't to get energy out of the Earth's rotation you have to slow it down.
If you want to slow down the earth, you have to give some of its angular momentum to something else, which in this case can only be the anchor. For this reason the anchor will have to start spinnng with the earth.

 

[mfb]

Why do you need an external anchor.

Please read the quoted text in my post, my comment was related to a different setup.

This thread has really been dead since post #7.

So true...

 

[D H]

This thread has really been dead since post #7.

Pretty close to that. (But you did miss the option of transferring angular momentum on objects sent from the Earth.)

This thread is closed.

johnmn3, please read our rules.