Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: championoftruth on 22/12/2020 13:44:38
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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).
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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.
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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
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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
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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.
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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
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But a small superconductor would levitate quite nicely in a symmetric nonuniform field. I have shareholders and bankers waiting. We just need the numbers.
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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...
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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..
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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
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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....
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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.
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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.
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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...
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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=4πϵ01ℓ2dQ
The electric field near an infinite plane is,
E=σ2ϵ0\large E = \dfrac{\sigma}{2\epsilon_0} E=2ϵ0σE, 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.
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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)
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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.
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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.
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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.
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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?
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a crater shaped water filled lake
That's about as good a ground connection as you could wish for. How do you plan to insulate it?
i will soon as i can't find my scientific calculator at the moment.
Even a rough calculation would do- just keep track of the zeros.
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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?
Equipment is not an issue. All we are doing is giving it an excess of electron. you are correct a thin layer of aluminum and charged to a high negative voltage. Electret materials can have CHARGES frozen in place.
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Equipment is not an issue
You can not be sure of that until you do the maths.
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Equipment is not an issue.
Good to hear it. So what equipment will you be using?
Please don't waste time with handwaving. I have engineers and investors standing by. We just need your specification or preferably a lead to the relevant manufacturer's catalog, and you will be credited and rewarded as the true and first inventor of the system that launches 10 kg to 100 km.
We usually start such projects with a £10,000 limited company, in which you as inventor hold 51% of the 100,000 enterprise shares. You choose the name of the company and I will do the rest when I receive your cheque for £5100.
Per elektron ad astra.
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Equipment is not an issue.
Good to hear it. So what equipment will you be using?
Please don't waste time with handwaving. I have engineers and investors standing by. We just need your specification or preferably a lead to the relevant manufacturer's catalog, and you will be credited and rewarded as the true and first inventor of the system that launches 10 kg to 100 km.
We usually start such projects with a £10,000 limited company, in which you as inventor hold 51% of the 100,000 enterprise shares. You choose the name of the company and I will do the rest when I receive your cheque for £5100.
Per elektron ad astra.
Surely it would be best for you to get your engineers to do the calculations instead of relying on ME?
This way you can be confident that your MANY trusted engineers verify and confirm it independently of each other.
Look what happened at NASA to the Hubble when they outsourced it to a company which miscalculated the curvature of the mirror.
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If you don't have the courage of your own calculations, you shouldn't make suggestions you can't support. But you said "Equipment is not an issue" so you must know where to buy what we need.
It's a well-known fact that rocket science is easy (Newtonian mechanics and the spontaneous combustion of the simplest chemicals - hydrogen and oxygen) but rocket engineering is expensive and difficult. Now I'm offering to raise the money and find the personnel to do the difficult bit, using the equipment you have stated as "not an issue". Commercial confidentiality, professional project management.....it's all there, and I've built much more complicated things in the past.
Happy to build a 1 km prototype in a field, which will increase your share value a hundredfold even before we go for the big one.
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Look what happened at NASA to the Hubble when they outsourced it to a company which miscalculated the curvature of the mirror.
They calculated just fine. They screwed up the engineering.
But calculation is cheap- so get on with it.
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Look what happened at NASA to the Hubble when they outsourced it to a company which miscalculated the curvature of the mirror.
They calculated just fine. They screwed up the engineering.
But calculation is cheap- so get on with it.
The equation is
E= Sigma/2e Newtons/coulomb..Note that distance is NOT a factor
Sigma is the charge density
The SI unit is C/m–1. (ii) Surface charge density; σ=qA, where, q is the charge and A is the area of the surface.
e = 8.854 187 8128 x 10-12 F m-1 Vacuum permittivity
Note FORCE =ELECTRIC FIELD/Q
q = the charge
You can put in the numbers while i try to find my scientific calculator
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You can put in the numbers while i try to find my scientific calculator
you can just enter your expressions as search terms into the google search bar and it will act as a scientific calculator.
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The equation is
E= Sigma/2e Newtons/coulomb..Note that distance is NOT a factor
distance is not a factor when an infinite plane is used, and this equation is still a reasonable approximation when the distance from the plate is much smaller than the size of the plate. If you want 100 km to be much smaller than the plate, be prepared for an expensive project!
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Also, quick question:
Why don't charged clouds in thunderstorms ever launch anything into space?
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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)
If you want something done properly, do it yourself...
OK
The spherical craft acts as a capacitor.
We can calculate the capacitance as shown here
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html
C= 4 pi e0 R
R= 5 metres
Pi =3.141
e0 = 8.8541878128×10^−12
That's 5.5 *10^-10 farads.
At a million volts it will have a charge Q= CV so that's 5.5*10^-4 Coulombs.
And we need to see how strongly that's repelled.
The force between a charged body and a large plane is the same as the force between two charged bodies with the same charge, and twice as far apart as the distance from the charge to the plane.
This guy explains it about 5 minutes in.
It's as if the charge sees its "mirror image" reflected in the plane.
The charge is 200 metres above the plane.
So the force is the same as having two charges 400 metres apart.
OK that's just Coulombs law.
F= K Q1 Q2 /R^2
OK, I'm lazy, so I'm going to find an on-line calculator for that.
https://www.omnicalculator.com/physics/coulombs-law
and stuff the numbers in.
It gives me a force of 0.01404305 Newtons.
About one gram's worth of force.
I admit I have cheated slightly, I used the formula for a charge and an uncharged (earthed) plane.
But you are going to charge the plane electrode. I think that increases the force by a factor of 4.
So that's 4gf
Now, for the second part of the exercise- the up thrust due to density.
I said I didn't need precision; I'm going to pretend that the sphere is a cube.
Well, the up thrust is bigger than that on a cube half as big as the sphere.
So it's bigger then the buoyancy of a 5 metre cube.
And that's the density of air times the volume of the cube
about 1.2 Kg/m3 times 5*5*5. That's about 150 Kgf (or 150000gf)
So the craft would work better if it was a balloon by a factor of about 150000/4
(Nobody tell Alan; you might still get £10K of investment out of him.)
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Don't forget, the force reduces as you get higher (as the inverse square of the distance).
In space- say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of micrograms by the time you were reaching space.
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Thank you for your concern, BC, but whilst waiting for the lad to respond I put the money on a horse instead. The great thing about horses is that they have a very good sense of things like energy, speed and distance, and make fairly reliable business partners compared with some humans. 3:1 ain't a bad return for a few minutes' work. Might do some physics tomorrow.
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Don't forget, the force reduces as you get higher (as the inverse square of the distance).
In space- say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of micrograms by the time you were reaching space.
inverse square law does not apply. The calculation is wrong.
see the diagram and derivation below:-
https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge
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We have been through this before. Literate correspondents will have read the introduction in that reference, my reply #19, and chiral's response # 29. I won't bother to say it again 'cos there's none so blind as them as will not look.
For numerate readers and anyone aspiring to electrostatic engineering, the near-field approximation is fairly good if the separation d is less than the radius r of the plane. From d = r to d = 10r, the force falls off roughly as 1/r, tending to 1/d^2 beyond that. We use similar approximations for radioactive contamination.
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Don't forget, the force reduces as you get higher (as the inverse square of the distance).
In space- say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of micrograms by the time you were reaching space.
inverse square law does not apply. The calculation is wrong.
see the diagram and derivation below:-
https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge
Did you see the bit where it said "This example was for an infinite plane of charge. In the physical world there is no such thing," ?
That page is great- if you know what sigma is.
It's the charge density on the plane, but how do you figure that?
Anyway, let's have a look at a slightly different way of considering it.
Imagine that we have two plane electrodes- one on the ground and one "a long way up".
If we put a voltage across the two plates the a charged object- like our ship will be repelled by one plate and attracted ot the other.
In that case the force is independent of the height.
It's awkward to build the top electrode, but there's an easy solution; we can use the ionosphere.
It's (very roughly) 100 KM up so that's close enough to space.
Again let's imagine that we can put a million volts across the two plates.
And let's assume that our craft starts near the bottom plate and is attracted to the top one.
How much force is on the craft?
Well, if we move the ship from the bottom to the top then we alter the electrical potential by a million volts.
And the ship has (as before) a charge of about 5.5*10^-4 Coulombs.
So the energy transferred to the ship is the product of those
That's 5.5 * 10^2 Joules. (About the energy you would get from a ounce of low Calorie cola).
OK, we need to convert that to a force.
Well that force acts over a distance of 100 Km and in doing so it transfers 550J of energy
Energy is force times distance so we can divide the energy (550J) By the distance (100,000 M) to get the force.
That's a force of about 0.0055 Newtons
Which is close enough to the answer I gave before (0.014N).
I could get the "right" answer by fiddling with the distance if I wanted.
It's an interesting thought- this pair of plates looks a lot like a capacitor.
Let's consider a single square metre of it and work out the capacitance
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html
That will have a capacitance of E0 times the area divided by the distance.
So that's 1/100000 times 8.8541878128×10^−12
And that's not very big.
8.8541878128×10^−17 Farads
OK that's the capacitance of each square metre of the pair of plates.
And, if you charge that to a million volts each square metre will acquire a charge (Q=CV) of
8.8541878128×10^−11 Coulombs.
Some of you might be wondering why I did that.
Well, as far as I can tell, that's the mystical "sigma" in the equation in the Kahn academy page.
"The total charge on the plane is of course infinity, but the useful parameter is the amount of charge per area, the charge density: sigma "
And what I just calculated was the charge density.
So let's bung the numbers into the equation.
E=sigma/2 e0
Well sigma is 8.8541878128×10^−11 Coulombs per square metre.
Divide that by 2 times e0
8.8541878128×10^−11 / 8.8541878128×10^−12 gives us 10. It's interesting to consider the units here. 10 volts per metre because we have a million volts and 100,000 metres of spacing- I could have taken the short cut, but the OP wouldn't have believed me.
And the formula says to halve that.
So the electric field near this plane is 5 volts per metre (Which is obviously different from 10; the difference is that using a second electrode doubles the field strength).
OK, so what can we do with that?
Well, it's a field gradient.
It has units of Newtons per Coulomb.
So, we should be able to multiply it by the charge on our ship, and get the force on the ship.
We know the charge; it's 5.5*10^-4 Coulombs
And so we can multiply that by 5 and get 0.00275 Newtons. (as I said, adding a second plate doubles the force)
About a quarter of a gram.
My apologies; when I said
I think that increases the force by a factor of 4.
it looks like I should have divided, rather than multiplying.
Alan's investment in a horse is 16 times more sound than we thought it was.
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So, we should be able to multiply it by the charge on our ship, and get the force on the ship.
We know the charge; it's 5.5*10^-4 Coulombs
And so we can multiply that by 5 and get 0.00275 Newtons. (as I said, adding a second plate doubles the force)
About a quarter of a gram.
My apologies; when I said
I think that increases the force by a factor of 4.
it looks like I should have divided, rather than multiplying.
Alan's investment in a horse is 16 times more sound than we thought it was.
I think your calculations need to be reviewed as your figures are far too low.
I have an ionizor in the house and yesterday i connected a 3 gram metal plate to one of the 4 outputs and when i put hand near it it moved towards my hand quite strongly. The force was more than a a gram.
This ionizer only uses 6 kilovolts!!!
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I think your calculations need to be reviewed as your figures are far too low.
Feel free to try.
But I did three different calculations and got practically the same response.
The force was more than a a gram.
Did you measure this or are you saying I'm wrong, based on your ignorant guess?
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It all depends on "near". From 100 km distance you'd have difficulty even seeing a 5 kV ioniser, never mind detecting its (far) field.
The quantitative study of electrostatics began in 1770 and was pretty much completed and verified by experiment within 20 years.
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It all depends on "near". From 100 km distance you'd have difficulty even seeing a 5 kV ioniser, never mind detecting its (far) field.
The quantitative study of electrostatics began in 1770 and was pretty much completed and verified by experiment within 20 years.
You could say the same for any field and implies a cessation of progress...
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I think your calculations need to be reviewed as your figures are far too low.
Feel free to try.
But I did three different calculations and got practically the same response.
The force was more than a a gram.
Did you measure this or are you saying I'm wrong, based on your ignorant guess?
Well this guy is using 200kv and that forces looks more than 0.0026 grams.
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Don't forget, the force reduces as you get higher (as the inverse square of the distance).
In space- say150 Km up- the force will be (150000/200)^2 times weaker
So what was about 4 gram's worth of force (at 200 M) would be reduced to about a couple of micrograms by the time you were reaching space.
inverse square law does not apply. The calculation is wrong.
see the diagram and derivation below:-
https://www.khanacademy.org/science/electrical-engineering/ee-electrostatics/ee-fields-potential-voltage/a/ee-plane-of-charge
Did you see the bit where it said "This example was for an infinite plane of charge. In the physical world there is no such thing," ?
That page is great- if you know what sigma is.
It's the charge density on the plane, but how do you figure that?
Anyway, let's have a look at a slightly different way of considering it.
Imagine that we have two plane electrodes- one on the ground and one "a long way up".
If we put a voltage across the two plates the a charged object- like our ship will be repelled by one plate and attracted ot the other.
In that case the force is independent of the height.
It's awkward to build the top electrode, but there's an easy solution; we can use the ionosphere.
It's (very roughly) 100 KM up so that's close enough to space.
Again let's imagine that we can put a million volts across the two plates.
And let's assume that our craft starts near the bottom plate and is attracted to the top one.
How much force is on the craft?
Well, if we move the ship from the bottom to the top then we alter the electrical potential by a million volts.
And the ship has (as before) a charge of about 5.5*10^-4 Coulombs.
So the energy transferred to the ship is the product of those
That's 5.5 * 10^2 Joules. (About the energy you would get from a ounce of low Calorie cola).
OK, we need to convert that to a force.
Well that force acts over a distance of 100 Km and in doing so it transfers 550J of energy
Energy is force times distance so we can divide the energy (550J) By the distance (100,000 M) to get the force.
That's a force of about 0.0055 Newtons
Which is close enough to the answer I gave before (0.014N).
I could get the "right" answer by fiddling with the distance if I wanted.
It's an interesting thought- this pair of plates looks a lot like a capacitor.
Let's consider a single square metre of it and work out the capacitance
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html
That will have a capacitance of E0 times the area divided by the distance.
So that's 1/100000 times 8.8541878128×10^−12
And that's not very big.
8.8541878128×10^−17 Farads
OK that's the capacitance of each square metre of the pair of plates.
And, if you charge that to a million volts each square metre will acquire a charge (Q=CV) of
8.8541878128×10^−11 Coulombs.
Some of you might be wondering why I did that.
Well, as far as I can tell, that's the mystical "sigma" in the equation in the Kahn academy page.
"The total charge on the plane is of course infinity, but the useful parameter is the amount of charge per area, the charge density: sigma "
And what I just calculated was the charge density.
So let's bung the numbers into the equation.
E=sigma/2 e0
Well sigma is 8.8541878128×10^−11 Coulombs per square metre.
Divide that by 2 times e0
8.8541878128×10^−11 / 8.8541878128×10^−12 gives us 10. It's interesting to consider the units here. 10 volts per metre because we have a million volts and 100,000 metres of spacing- I could have taken the short cut, but the OP wouldn't have believed me.
And the formula says to halve that.
So the electric field near this plane is 5 volts per metre (Which is obviously different from 10; the difference is that using a second electrode doubles the field strength).
OK, so what can we do with that?
Well, it's a field gradient.
It has units of Newtons per Coulomb.
So, we should be able to multiply it by the charge on our ship, and get the force on the ship.
We know the charge; it's 5.5*10^-4 Coulombs
And so we can multiply that by 5 and get 0.00275 Newtons. (as I said, adding a second plate doubles the force)
About a quarter of a gram.
My apologies; when I said
I think that increases the force by a factor of 4.
it looks like I should have divided, rather than multiplying.
Alan's investment in a horse is 16 times more sound than we thought it was.
force more than 0.25 grams and he only using 50000 volts!
This proves your calculations are in error or your assumptions in error or a misplaced decimal point.
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https://en.wikipedia.org/wiki/Inverse-square_law#Electrostatics
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Quote from: championoftruth on 24/01/2021 12:31:25: "The force was more than a a gram."
Did you measure this or are you saying I'm wrong, based on your ignorant guess?
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It all depends on "near". From 100 km distance you'd have difficulty even seeing a 5 kV ioniser, never mind detecting its (far) field.
The quantitative study of electrostatics began in 1770 and was pretty much completed and verified by experiment within 20 years.
You could say the same for any field and implies a cessation of progress...
There has been plenty of progress in electrostatic technology since 1770, but no change in the laws of physics for the last 13.8 billion years..
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When you increase the voltage... it moves down
With 50,000 Volts in (say) 0.5 meters = 100,000 V/m, you can levitate a small aluminium foil shape.
- When you increase the voltage, it doesn't go as high.
- Extrapolating the two data points provided, you should be able to launch aluminium foil into orbit with zero volts...
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Exactly so. If you put two flat plates in contact and share a negative charge between them, they are both at the same potential but repel one another. Remember the gold leaf electroscope? So you don't need a high voltage source to propel yourself into space!
Where's the tongue-in-cheek emoticon?
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Exactly so. If you put two flat plates in contact and share a negative charge between them, they are both at the same potential but repel one another. Remember the gold leaf electroscope? So you don't need a high voltage source to propel yourself into space!
Where's the tongue-in-cheek emoticon?
Doesn't the gold-leaf electroscope only work because it uses very thin metal foil.
If you tried to make an electroscope with thick metal sheets, say 6-inches thick, the sheets wouldn't move at all.
They would be kept still by their inertial mass.
But - could a spaceship be built of thin metal foil? The foil would be pressurised internally against the vacuum of outer-space. That would keep the ship's foil-hull, rigid and taut. It would be as light, and efficient, as a rubber balloon.
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Exactly so. If you put two flat plates in contact and share a negative charge between them, they are both at the same potential but repel one another. Remember the gold leaf electroscope? So you don't need a high voltage source to propel yourself into space!
Where's the tongue-in-cheek emoticon?
Doesn't the gold-leaf electroscope only work because it uses very thin metal foil.
If you tried to make an electroscope with thick metal sheets, say 6-inches thick, the sheets wouldn't move at all.
They would be kept still by their inertial mass.
But - could a spaceship be built of thin metal foil? The foil would be pressurised internally against the vacuum of outer-space. That would keep the ship's foil-hull, rigid and taut. It would be as light, and efficient, as a rubber balloon.
Depends on how much charge you give it the higher the charge the greater the force. f=q1q2/r2 x Eo
in a flat plane the charge density is the key.
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Doesn't the gold-leaf electroscope only work because it uses very thin metal foil.
It works because it is light.
You could just about use aluminised Mylar or something. Though the "usual" grade is heavier (per unit area) than gold leaf.
Depends on how much charge you give it
Yes, that's why I asked this
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.
You don't seem to have answered it.
Did you get distracted by nonsense where you forgot about the inverse square law?
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from: charles1948 on 29/01/2021 00:24:24
Doesn't the gold-leaf electroscope only work because it uses very thin metal foil.
It works because it is light.
But the fun bit is that, whatever material you use, both leaves are at the same potential. Therefore you do not need a high voltage source to drive your space elevator - the elements repel each other at zero potential difference!
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from: charles1948 on 29/01/2021 00:24:24
Doesn't the gold-leaf electroscope only work because it uses very thin metal foil.
It works because it is light.
But the fun bit is that, whatever material you use, both leaves are at the same potential. Therefore you do not need a high voltage source to drive your space elevator - the elements repel each other at zero potential difference!
You can use an electroscope as a crude way to measure voltage. The higher the potential, the more the leaves diverge.
With a potential of zero, you get a divergence of zero.
Sadly, it's only in championoftruth's fantasy world where we could launch a ship for free (or, indeed, where the idea could be any use at all).
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The higher the potential,
with respect to what? the more the leaves diverge.
Two electrons in deep space, with no third body reference potential, will repel each other.
F = keq1q2/r2 - no mention of any potential, just charge
Indeed if there is no third body within a distance D >> r, they will fly apart indefinitely.
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with respect to what?
With respect to the potential it would have without a charge.
You can measure potential by looking at the work done bringing a small test charge from infinity to the object.
You can calculate the potential of, for example, an isolated sphere by calculating the capacitance and then combining that information with the charge (which you can usually find by counting electrons and protons).
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So I have two metal plates in contact. Being conductors, they share charge so they both have the same surface electron density. So they fly apart. Which is why Champion asserted that the power source isn't a problem.
Or more to the point, they don't.
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I think I saw something involving a canon once?
How to get your payload to the moon I think?
=
Eh, sorry, remembering some more. It involved a cannon, not a canon.