[championoftruth (OP)]

Space Elevators at the current level of materiel science are not possible due insufficient strength of the tether.
Also very expensive.

Why not use super powerful Magnetic Fields?

Helmholtz Coils when a current is passed thru them show a uniform field between the two.

2 coils a meter in diameter each will have a roughly uniform field between them if the coils are separated by a meter.

Now if the one of the coils is 500 kilometers wide and lying flat and the other coil is say a 1/2 a kilometre wide
either with a current passing thru it or permanent magnets it can be repelled into space up to a distance of about 500 km into space.


One coil flat on the ground and the other coil being the ship/payload.

 F=B I L.

What you would do is have is a huge square or round grid (500 miles by 500 miles)  in the Sahara or a desert in Utah with thousands of amps of amps flowing through it while the craft is going up.

The force would be F= B  I  L . 

The craft which will be a smaller version of the grid on the Earth (plus payload) will have the opposite magnetic field and repel itself away. (Or use permanent opposite polarity magnets).

The large size of the grid in the desert means the field will be nearly uniform 500 km into space.

Also no moving parts. Will cost only 0.01% of the space elevator or less.

E.g. The electric or magnetic field between 2 large parallel sheets of metal is roughly uniform hence similar for electric or magnetic fields. (Helmholtz coils).

The reason for the 500 km size of the grid is that the inverse square law will not operate and the field will be roughly constant/uniform between the ground coil and spacecraft coil.

Using giant Helmholtz coils for the magnetic fields. One 500 km helmholtz coil lying flat in the desert.The other smaller one about a 1/2 km wide with payload will be repelled into orbit at least 400 km to 500 km into space.

( Or use electric fields which will not require huge currents but huge voltages).

[alancalverd]

thousands of amps of amps

Good physics. Now let's do the engineering.

Please calculate how many thousands of amps for a payload of, say, 1 kilogram.  Then assuming a 100 deg C temperature rise in a copper coil, the voltage required to achieve this in your coil of how many turns of what thickness of material.  The numbers you need are all in your textbooks.

[syhprum]

The coils would of course be room temperature superconductors and the power generated by cheap fusion reactors both of which will be available in the very near future

[championoftruth (OP)]

thousands of amps of amps

Good physics. Now let's do the engineering.

Please calculate how many thousands of amps for a payload of, say, 1 kilogram.  Then assuming a 100 deg C temperature rise in a copper coil, the voltage required to achieve this in your coil of how many turns of what thickness of material.  The numbers you need are all in your textbooks.

Can't post links for some reason. See Wiki.

Description

A Helmholtz pair consists of two identical circular magnetic coils that are placed symmetrically along a common axis, one on each side of the experimental area, and separated by a distance h {\displaystyle h} h equal to the radius R {\displaystyle R} R of the coil. Each coil carries an equal electric current in the same direction.[1]

Magnetic field of two Helmholtz coils:
Configuration of two Helmholtz coils to produce a uniform magnetic field
To setup a Helmholtz coil two similar coils with radius R are placed in the same distance R. When the coils are so connected that the current through the coils flows in the same direction, the Helmholtz coils produce a region with a nearly uniform magnetic field.

The magnetic field at center of the coils with N wire windings is proportional to current through coils:

B=μ0⋅8⋅I⋅N√125⋅R I = coil current, μ0 = vacuum permeability, N = windings, R = radius and distance of coils

The magnetic field of the Helmholtz coil used on the next pages, depending on the coil current I, is:B≈7,48⋅10−4TA⋅II = coil current, μ0=4π⋅10−7NA2, N = 124 windings, R = 14,9 cm

[alancalverd]

I know the physics. I am in the business of capital project engineering. Please give me the numbers for your proposal, not for a 15 cm diameter coil.

[evan_au]

the inverse square law will not operate and the field will be roughly constant/uniform

The inverse square law only applies to a monopole, like an electric charge.

So far as we know, there is no such thing as a magnetic monopole (although physicists have worked on theories that allow for them).

The problem with a magnetic dipole is that there is a North end and a South end, generally fairly close together.
- Once you get away from the magnet, the strength of the magnetic field tends to drop as an inverse cube
- The maximum useful distance is around the diameter of the loop, so I agree that a 500km loop would be a good size to get you above the atmosphere, if you could power it (and not black out entire states due to the power drain or electromagnetic interference...)

PS: How would you fix the orientation your 1kg payload so that it didn't just flip over to align with the external magnetic field from the loop?

See: https://en.wikipedia.org/wiki/Magnetic_monopole

[alancalverd]

But a small superconductor would levitate quite nicely in a symmetric nonuniform field. I have shareholders and bankers waiting. We just need the numbers.

[chiralSPO]

Why go through all the trouble of using magnets when pure electrostatics can do it? And there is no problem making an electrostatic monopole!

You wanna get 1 kg into space? Just put 1 Coulomb of charge on it, and 10 Coulombs of charge on your launch pad (both positive or both negative, so they repel). It will still be able to overcome gravity at an altitude (separation) of 100 km.

What could go wrong?

Lemme just crunch some numbers...

[championoftruth (OP)]

Why go through all the trouble of using magnets when pure electrostatics can do it? And there is no problem making an electrostatic monopole!

You wanna get 1 kg into space? Just put 1 Coulomb of charge on it, and 10 Coulombs of charge on your launch pad (both positive or both negative, so they repel). It will still be able to overcome gravity at an altitude (separation) of 100 km.

What could go wrong?

Lemme just crunch some numbers...

You are correct .i did number crunching and the answers were awful . requiring thousands of amps to flow thousands of coils for not much space.


My next solution was actually using electrostatics and i had it written but you preempted me...

The field between two parallel plates is uniform if the distance between the two is similar to the size of the plates...

I rechecked where outer space is and :-

A common definition of space is known as the Kármán Line, an imaginary boundary 100 kilometers (62 miles) above mean sea level.

So this means we need a flat conductor on the ground of about 100 km square..

[chiralSPO]

There are still some significant impracticalities with the electrostatic approach, unfortunately....

If you calculate the electric field around the "flat conductor on the ground" you will find that it likely exceeds the breakdown voltages of air (best case scenario is that you accidentally build a lightning machine). https://hypertextbook.com/facts/2000/AliceHong.shtml

[evan_au]

So this means we need a flat conductor on the ground of about 100 km square..

That's not the biggest problem....

The field between two parallel plates

You need to find some way of building and suspending another flat plate of 100 km square directly above the one on the ground.
- This implies an amazing amount of space capability to launch the upper plate
- And it is not in a stable orbit, so it will crash and burn
- and anti-gravity too, to stop the upper plate crashing down on the lower plate, as they will attract each other strongly).

And then you need some way of generating and maintaining an enormous voltage between said plates.

A space elevator is looking more and more attractive....

[championoftruth (OP)]

So this means we need a flat conductor on the ground of about 100 km square..

That's not the biggest problem....

The field between two parallel plates

You need to find some way of building and suspending another flat plate of 100 km square directly above the one on the ground.
- This implies an amazing amount of space capability to launch the upper plate
- And it is not in a stable orbit, so it will crash and burn
- and anti-gravity too, to stop the upper plate crashing down on the lower plate, as they will attract each other strongly).

And then you need some way of generating and maintaining an enormous voltage between said plates.

A space elevator is looking more and more attractive....

No second upper plate. The payload vehicle will have a pusher plate of say 50 meters square which will be also negatively charged or same polarity as ground plate.

Have you heard of electret microphones? The material has a permanent charge.

[alancalverd]

No problem!

The lower plate can be at any potential relative to the ground since all that matters is that the charges on both plates have the same sign.

So we use the surface of the earth as a reference and just bolt the "launch" plate to it. That will obviate any lightning around the edges.

Now to get a repulsive force we need to induce the same sign charge on the "lift" plate, so we just place it in contact with the launch plate, and away it goes!

The principle has been applied for thousands of years in magic carpets, UFOs and metal airplanes, which obviously can't fly by any other means (Archimedes)

Have a great Christmas.

[championoftruth (OP)]

So this means we need a flat conductor on the ground of about 100 km square..

That's not the biggest problem....

The field between two parallel plates

You need to find some way of building and suspending another flat plate of 100 km square directly above the one on the ground.
- This implies an amazing amount of space capability to launch the upper plate
- And it is not in a stable orbit, so it will crash and burn
- and anti-gravity too, to stop the upper plate crashing down on the lower plate, as they will attract each other strongly).

And then you need some way of generating and maintaining an enormous voltage between said plates.

A space elevator is looking more and more attractive....

No second plate is required as the payload will be negatively charged.

The idea is that once it about 100 miles above it will fire onboard motors to obtain orbital velocity friction free..

no upper plate.

high voltages will only be maintained on the ground plate and the payload...

[championoftruth (OP)]

No problem!

The lower plate can be at any potential relative to the ground since all that matters is that the charges on both plates have the same sign.

So we use the surface of the earth as a reference and just bolt the "launch" plate to it. That will obviate any lightning around the edges.

Now to get a repulsive force we need to induce the same sign charge on the "lift" plate, so we just place it in contact with the launch plate, and away it goes!

The principle has been applied for thousands of years in magic carpets, UFOs and metal airplanes, which obviously can't fly by any other means (Archimedes)

Have a great Christmas.

i dont see why mention magic carpets as this based on science.  Felect = k • Q1 • Q2 / d2

We know the field at location qqqq due to dQ\text dQdQstart text, d, end text, Q; it's the definition of the field created by a point charge,
dE=14πϵ0dQℓ 2\text dE = \dfrac{1}{4\pi\epsilon_0} \dfrac{\text dQ}{\ell\,^2}dE=4πϵ0​1​ℓ2dQ​

The electric field near an infinite plane is,
E=σ2ϵ0\large E = \dfrac{\sigma}{2\epsilon_0} E=2ϵ0​σ​E, equals, start fraction, sigma, divided by, 2, \epsilon, start subscript, 0, end subscript, end fraction   newtons/coulomb\;\text{newtons/coulomb}newtons/coulombstart text, n, e, w, t, o, n, s, slash, c, o, u, l, o, m, b, end text

E=sigma /2Eo

Lower case "sigma" represents the amount of charge per area of the plane, in units of coulombs/meter^2. This parameter is called the "charge density".

Conclusion
This the electric field (the force on a unit positive charge) near a plane. Amazingly, the field expression contains no distance term, so the field from a plane does not fall off with distance! For this imagined infinite plane of charge, it doesn't matter if you are one millimeter or one kilometer away from the plane, the electric field is the same.
This example was for an infinite plane of charge. In the physical world there is no such thing, but the result applies remarkably well to real planes, as long as the plane is large compared to aaaa and the location is not too close to the edge of the plane.
Review
Using the notion of an electric field, the analysis technique is,

    Charge gives rise to an electric field.
    The electric field acts locally on a test charge.

Summarizing the three electric field examples worked out so far,
The field due to a   falls off at
point charge   1/r21/r^21/r21, slash, r, squared
line of charge   1/r11/r^11/r11, slash, r, start superscript, 1, end superscript
plane of charge   1/r01/r^01/r01, slash, r, start superscript, 0, end superscript
These three charge configurations are a useful toolkit for predicting electric field in lots of practical situations.

[Bored chemist]

Please calculate the force on a spherical craft 10 metres in diameter when it is two hundred  metres above a large flat plane  and when the potential difference is a million volts.
For extra credit, please calculate the potential to which this  10 metre diameter craft can be charged before the air round it is ionised by the field gradient.

Also calculate the up-thrust due to the density of air (per Archimedes principle)

[alancalverd]

This example was for an infinite plane of charge. In the physical world there is no such thing, but the result applies remarkably well to real planes, as long as the plane is large compared to aaaa and the location is not too close to the edge of the plane.

Generally, "large" means that the radius of the plane is at least 10 times the separation from the test charge. So if we want to launch a very small craft to 100 km, we need a driving plane 2000 km wide. That's one heck of a construction project.

[championoftruth (OP)]

This example was for an infinite plane of charge. In the physical world there is no such thing, but the result applies remarkably well to real planes, as long as the plane is large compared to aaaa and the location is not too close to the edge of the plane.

Generally, "large" means that the radius of the plane is at least 10 times the separation from the test charge. So if we want to launch a very small craft to 100 km, we need a driving plane 2000 km wide. That's one heck of a construction project.

incorrect. for a 100 km plane you can go up a 100km.
https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge

also you dont need construct much.

you can use a crater shaped water filled lake as one of the planes and charge it with electrons. making it repulsive. same with the payload.

[championoftruth (OP)]

Please calculate the force on a spherical craft 10 metres in diameter when it is two hundred  metres above a large flat plane  and when the potential difference is a million volts.
For extra credit, please calculate the potential to which this  10 metre diameter craft can be charged before the air round it is ionised by the field gradient.

Also calculate the up-thrust due to the density of air (per Archimedes principle)

i will soon as i can't find my scientific calculator at the moment.

Maybe someone else does.

[alancalverd]

incorrect. for a 100 km plane you can go up a 100km.
https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge [nofollow]

There is quite a difference between "infinite"  and 100 km, especially if you are 100 km away from the supposed "infinte" surface. Try reading before quoting.


Anyway, suppose we can indeed spray the Sahara with aluminum. Not a big deal since it won't be carrying much current, so can be a few microns thick. What equipment are you going to use to charge the foil and the magic carpet?