Why is no one trying to create a UFO-like engine out of superconductors that can levitate in Earth's magnetic field?

[TheVat]

1 hour ago, exchemist said:

 It is said that eddy currents will be triggered in the superconductor which will form a perfect mirror image of the magnet, with like poles adjacent, so a repulsive force is generated. 

 

How does something a few meters across form a mirror image of Earth's magnetic field?  

[Genady]

10 minutes ago, exchemist said:

Can’t be that or the  magnetic pressure would be a function of how the field was applied.

If the currents are such that the force is not sufficient to hold the superconductor in place (levitating), the superconductor will start falling into the field. This will cause the currents to strengthen ... etc. until the forces balance.  -- ?

[studiot]

1 hour ago, TheVat said:

"We Also Walk Dogs" 

- by RAH, yes.  

Thanks. +1

I remember

"We'll spank the baby and feed the cat"

and now you mention it, 'we also walk dogs'

[swansont]

2 hours ago, exchemist said:

Is a superconductor different, and if so why would that be?

It’s a quantum mechanical phenomenon.

[Kassander]

On 4/5/2023 at 7:37 PM, exchemist said:

To be honest I think to make progress in understanding this topic we should forget ChatGPT and have one of our physicists talk us through the Meissner-Ochsenfeld effect a bit. ChatGPT is basically as thick as mince and just plagiarises stuff it looks up on the web that it hopes is relevant, based on some algorithm. There's no reason to expect it to  be able to do this stuff properly. But does it give you references for where it gets its formulae from? If we can read those sources we might get somewhere.

Meanwhile, I've had another look at the Wiki article, which gives a remarkably simple formula for something called the "magnetic pressure" that a magnetic field exerts on a superconductor. This is Pmag  = B²/μ₀, where P is force per unit area at the superconductor/field interface, in Pascals, B in Tesla. Here's the link: https://en.wikipedia.org/wiki/Magnetic_levitation

What happens if you plug in the numbers for the Earth's field? 

 

 

  

I also think it would be better if we try to do it ourselves. I have asked ChatGPT multiple times, and this is the only useful answer I have received:

Here is the link to the Wikipedia article on the London Equations:

https://en.wikipedia.org/wiki/London_equations

When I input the numbers for Earth's magnetic field into the formula from your link, this is what I get:

(0.00005²T)/(2(4π*(10^-7))) = 0.0009947Pa

Now, when I use the formula for buoyancy in liquid water, I get:

B = pVg | 0.0009947Pa * 523.599m³ * 9.81 = 5.11N

5.11N is a very low number - you could lift just about 0.5 kg, and the weight of the construction has not yet been subtracted. So this may be the reason why no one uses this concept, if the calculation is correct. However, I am not sure if it is appropriate to use the formula for buoyancy in liquid water in this particular case. ChatGPT mentioned the London Equations, which are also related to the Meissner–Ochsenfeld Effect:

https://en.wikipedia.org/wiki/Meissner_effect

I also think that ChatGPT chose a Niobium-Titanium (Nb-Ti) superconductor because it is a Type-II superconductor, which can be used for "flux pinning", this would generate a much greater force than just the magnetic levitation, but no formulas are mentioned:

https://en.wikipedia.org/wiki/Flux_pinning

I believe that with this Wikipedia article, I have found the key to what I was previously looking for. It mentions that “the ability to fix the superconductor in space can be used as a damping device, like a spring. This idea has been proposed for isolating vibrations for parts in satellites.” This is a pretty similar idea to my suggestion of using it in a spacecraft for deceleration.

On 4/5/2023 at 10:14 PM, Phi for All said:

!

Moderator Note

Very good. We've established that ChatGPT is NOT a good tool for science discussion. Can we continue to discuss the concept you mention in your OP, or is it dependent on the linguistic AI?

 

No, an AI is not necessarily necessary. I think we are already close to a solution at this point.

Edited April 6, 2023 by Kassander

[exchemist]

9 hours ago, Kassander said:

I also think it would be better if we try to do it ourselves. I have asked ChatGPT multiple times, and this is the only useful answer I have received:

Here is the link to the Wikipedia article on the London Equations:

https://en.wikipedia.org/wiki/London_equations

When I input the numbers for Earth's magnetic field into the formula from your link, this is what I get:

(0.00005²T)/(2(4π*(10^-7))) = 0.0009947Pa

Now, when I use the formula for buoyancy in liquid water, I get:

B = pVg | 0.0009947Pa * 523.599m³ * 9.81 = 5.11N

5.11N is a very low number - you could lift just about 0.5 kg, and the weight of the construction has not yet been subtracted. So this may be the reason why no one uses this concept, if the calculation is correct. However, I am not sure if it is appropriate to use the formula for buoyancy in liquid water in this particular case. ChatGPT mentioned the London Equations, which are also related to the Meissner–Ochsenfeld Effect:

https://en.wikipedia.org/wiki/Meissner_effect

I also think that ChatGPT chose a Niobium-Titanium (Nb-Ti) superconductor because it is a Type-II superconductor, which can be used for "flux pinning", this would generate a much greater force than just the magnetic levitation, but no formulas are mentioned:

https://en.wikipedia.org/wiki/Flux_pinning

I believe that with this Wikipedia article, I have found the key to what I was previously looking for. It mentions that “the ability to fix the superconductor in space can be used as a damping device, like a spring. This idea has been proposed for isolating vibrations for parts in satellites.” This is a pretty similar idea to my suggestion of using it in a spacecraft for deceleration.

No, an AI is not necessarily necessary. I think we are already close to a solution at this point.

What has buoyancy in water got to do with this problem? 

[Kassander]

4 hours ago, exchemist said:

What has buoyancy in water got to do with this problem? 

The formula for the magnetic pressure you asked me to calculate gave out 0.0009947Pa. Pa or Pascal is the unit for pressure. If you multiply it with the volume of any object and g you should get the buoyancy/upthrust, not just in liquid water but in general I assume. Was not that what you wanted to know?

[exchemist]

4 minutes ago, Kassander said:

The formula for the magnetic pressure you asked me to calculate gave out 0.0009947Pa. Pa or Pascal is the unit for pressure. If you multiply it with the volume of any object and g you should get the buoyancy/upthrust, not just in liquid water but in general I assume. Was not that what you wanted to know?

I don't think this scenario is like buoyancy in a fluid. I would think the relevant dimension of the superconductor is the area it presents to the magnetic field. But I may be wrong - this is outside my area of knowledge.   

[Genady]

deleted, wrong quotation

Edited April 7, 2023 by Genady

[exchemist]

25 minutes ago, Genady said:

The dimensions don't match. The unit of pressure is [force]/[length²], volume is [length³], g is [length]/[time²]. Multiply:

[force]/[length²]x[length³]x[length]/[time²] = [force]x[length²]/[time²]

which is not a force/buoyancy/upthrust.

Eh? I think you must be quoting @Kassander. 

[Genady]

1 hour ago, Kassander said:

pressure. If you multiply it with the volume of any object and g you should get the buoyancy/upthrust, not just in liquid water but in general I assume

The dimensions don't match. The unit of pressure is [force]/[length²], volume is [length³], g is [length]/[time²]. Multiply:

[force]/[length²]x[length³]x[length]/[time²] = [force]x[length²]/[time²]

which is not a force/buoyancy/upthrust.

 

16 minutes ago, exchemist said:

Eh? I think you must be quoting @Kassander. 

Sorry for that. Fixed.

[Kassander]

1 hour ago, Genady said:

The dimensions don't match. The unit of pressure is [force]/[length²], volume is [length³], g is [length]/[time²]. Multiply:

[force]/[length²]x[length³]x[length]/[time²] = [force]x[length²]/[time²]

which is not a force/buoyancy/upthrust.

 

Sorry for that. Fixed.

Thank you for your answer. I´m not good at the usage of formulas. The same goes for the conversion of units. But according to the following Wiki-Article, which was previously linked by exchemist and was the base of my calculation, the result should be in pascals:

https://en.wikipedia.org/wiki/Magnetic_levitation

To avoid any further mistakes, I now used this online calculator:

https://www.omnicalculator.com/physics/buoyancy

When I now fill in p=0.0009947Pa and V=523.599m³ I get B=5.108N and a Mass of displaced fluid of 0.5208kg. Which is what I calculated bevor. Or got I something wrong?

[Genady]

35 minutes ago, Kassander said:

When I now fill in p=0.0009947Pa and V=523.599m³ I get B=5.108N

How do you get this? 

How come B is in N?

36 minutes ago, Kassander said:

Mass of displaced fluid of 0.5208kg

How do you get this?

What fluid?

[TheVat]

I found this helpful, though the equations didn't paste too well. (from StackExchange)

The lift generated by magnetic field B on a superconductor of area S is:

 

F=B2S/2μ0

disregarding lateral forces and assuming superconducting cylinder (or similar shape) with area S at the top and bottom and height h, we need three forces to remain in the equilibrium: magnetic pressure on top, bottom and gravity force:

Fb−Ft=Fg

denoting density of the superconductor as ρ, Earth' gravity as g and magnetic field at the top and bottom of the object as B_t and B_b, we have

 

1/2μ0(B2b−B2t)=ρgh

assuming the vertical rate of change of magnetic field is nearly constant and denoting the average magnetic field as B, we have

 

−B* dB/dz=μ0ρg

Compare with diamagnetic levitation (superconductor's magnetic susceptibility is -1).

Now, Earth magnetic field is between 25 to 65 μT. For the derivative I have found this survey from British Columbia with upper point on the scale being 2.161 nT/m. Assuming this to be the maximum for vertical derivative we get the required density of 1.1394e-08 kg/m³. For comparison air density at the sea level at 15C is around 1.275 kg/m³, so required density is 8 orders of magnitude smaller.

Even assuming a very high vertical derivative where B goes from its maximum 65 μT to 0 on 1 m of height results in density required of 0.00034272 kg/m³.

[Kassander]

54 minutes ago, Genady said:

How do you get this? 

How come B is in N?

How do you get this?

What fluid?

“How do you get this?”

Here again the calculation of the magnetic pressure in pascal according to the formula:

(0.00005²T)/(2(4π*(10^-7))) = 0.0009947Pa

 

“How come B is in N?”

There are two “B”. The first one seen in the formula for magnetic pressure is not in n Newton but in Tesla. So I filled in 0.00005T for earth’s magnetic field. The second “B” in the formula to calculate the buoyancy stands for buoyancy. Buoyancy is of course a force and that’s why it´s in “N” for Newton.

“How do you get this?”

Exactly as I said, I filled in p=0.0009947Pa and V=523.599m³ in this online calculator:

https://www.omnicalculator.com/physics/buoyancy

This calculator give than the mentioned result´s out B=5.108N and a Mass of displaced fluid of 0.5208kg.

“What fluid?”

The mentioned online calculator is made for buoyancy calculations in water. That´s why the programmer called it “Mass of displaced fluid”. However I think it would be proper to use this for our calculation.

[Kassander]

1 hour ago, TheVat said:

I found this helpful, though the equations didn't paste too well. (from StackExchange)

The lift generated by magnetic field B on a superconductor of area S is:

 

F=B2S/2μ0

disregarding lateral forces and assuming superconducting cylinder (or similar shape) with area S at the top and bottom and height h, we need three forces to remain in the equilibrium: magnetic pressure on top, bottom and gravity force:

Fb−Ft=Fg

denoting density of the superconductor as ρ, Earth' gravity as g and magnetic field at the top and bottom of the object as B_t and B_b, we have

 

1/2μ0(B2b−B2t)=ρgh

assuming the vertical rate of change of magnetic field is nearly constant and denoting the average magnetic field as B, we have

 

−B* dB/dz=μ0ρg

Compare with diamagnetic levitation (superconductor's magnetic susceptibility is -1).

Now, Earth magnetic field is between 25 to 65 μT. For the derivative I have found this survey from British Columbia with upper point on the scale being 2.161 nT/m. Assuming this to be the maximum for vertical derivative we get the required density of 1.1394e-08 kg/m³. For comparison air density at the sea level at 15C is around 1.275 kg/m³, so required density is 8 orders of magnitude smaller.

Even assuming a very high vertical derivative where B goes from its maximum 65 μT to 0 on 1 m of height results in density required of 0.00034272 kg/m³.

Wow, thanks’ a lot! This seems to be the answer why a type-I superconductor would never get enough uplift for any of my suggested usages. Exchemist gave me another formula yesterday and we also calculated a very small uplift of 0,5kg. But we were not sure if the formulas we used where appropriate in this case. Anyway, it seems to stabilize that an extension of the volume is not enough to increase the uplift capacity to a useful amount.

So the only remaining question is if a flux pinned type-II superconductor would also not provide enough uplift in earth’s magnetic field for a practical usage. Or is the flux pinning-effect already part of your calculations? To be honest, I don´t understand every part of what you have calculated 😉. But I think ChatGPT had used the flux pinning-effect in his calculation. It choose the Niobium–titanium (Nb-Ti) superconductor probably because this is a type-II superconductor. According to all online sources I can find the flux pinning get stronger when the superconducting layer gets thinner. But I can´t find any formula for that.

5 hours ago, exchemist said:

I don't think this scenario is like buoyancy in a fluid. I would think the relevant dimension of the superconductor is the area it presents to the magnetic field. But I may be wrong - this is outside my area of knowledge.   

“I would think the relevant dimension of the superconductor is the area it presents to the magnetic field.”

Since the volume is part of the equation the area should be considered. Anyway, TheVat calculated the lift capacity in detail. The result is pretty close to the 0,5kg. So increasing the volume and using a hollow superconductor is not the solution to get enough lift capacity in earth’s magnetic field for practical usage. The only remaining question is if the flux pinning-effect also not provides a significant increase in lift capacity. But I think we can assume that this is the case because we would see for sure many technical uses if this would be possible.

[MigL]

Been busy the last couple of days with work.
This is the resource I used for the Meissner effect

Meissner effect for superconductors (gsu.edu)

and the embedded link to the London equation

Characteristic Lengths in Superconductors (gsu.edu)

Haven't searched through any of the embedded links for generated forces, but you guys seem to have found solutions to the problem already.
One solution is not to use ChatGPT.

[Kassander]

2 hours ago, MigL said:

Been busy the last couple of days with work.
This is the resource I used for the Meissner effect

Meissner effect for superconductors (gsu.edu)

and the embedded link to the London equation

Characteristic Lengths in Superconductors (gsu.edu)

Haven't searched through any of the embedded links for generated forces, but you guys seem to have found solutions to the problem already.
One solution is not to use ChatGPT.

Thanks, yes ChatGPT seems to cause more trouble than it helps. I think we can conclude that my basic suggestion to increase the lift capacity of a superconductor by increasing it´s volume make not enough difference that any practical usage in earth’s magnetic field is possible. But I am not sure if ChatGPT probably tried to use a Niobium–titanium (Nb-Ti) superconductor because it´s a type-II superconductor. This leads to a so called flux pinning-effect witch seems to be much stronger than ordinary magnetic levitation. Wikipedia says “On a simple 76 millimeter diameter, 1-micrometer thick disk, next to a magnetic field of 28 kA/m, there are approximately 100 billion flux tubes that hold 70,000 times the superconductor's weight.”

https://en.wikipedia.org/wiki/Flux_pinning

I´ve converted the 28kA/m with the converter in the following link to Tesla and it gave out 0.0351858T.

https://maurermagnetic.com/en/demagnetizing/technology/convert-magnetic-units/

These 0.0351858T are about 704 times bigger than the 0.00005T of earth’s magnetic field. But because the type-II superconductor could carry 70,000 times the superconductor´s weight it should still be possible to use this in the magnetic field of the earth. However, sins there are no technical usages known, I would assume that there are other problems that prevent such usages.