Can we fly in air

[Boerseun]

I suppose if you create a platform out of some ridiculously light material like aerogel, and find some really light-weight magnets (I don't know what you'll make them out of) and park one of them at each corner of the platform with the proper pole pointing downwards and you can somehow prevent the whole thing from flipping over so that the attracting poles are pointing at the Earth, then, presumably, you could have a magnetically levitating flying machine. It won't take any cargo or passengers, though, but it'll definitely be kinda neat.

[Janus]

 

So, a total east-west current of 64 million amps is needed. It can be carried in many or a few conductors – for example, a 1 m * 0.2 m cross-section of 200000 0.001 m insulated wires each carrying 320 amps. Assuming an adequate cooling system, this could be achieved using an ordinary conductor, such as high-quality aluminum wire.

Whoa! 0.001m wire is 18 AWG which is generally rated at a max current of 24 amps. Getting it to carry 13 times as much will take some cooling system. And considering that your bundle of wires already are taking up 424Kg of the thousand kg of your ship, A cooling system would take up quite a bit more, and we haven't even considered the power plant yet.

 

The energy requirements of this aircraft would be modest – to move upward at 1 m/s would require just the mechanical power (10,000 W) plus the power due to resistance of the electrical circuit (assuming aluminum wire, a surprisingly low 6 W).

Surprisingly low, yes, and also completely wrong.

Take one of your wires. One meter of 18 AWG wire has a resistance of .021 Ohms. Using P = I^2R. we get a wattage of 320^2*.021 = 2150 watts per wire. You have 200000 individual wires, which gives a total of 430,080,000 watts just to push the current through the wires. This would mean that your cooling system would have to be able to disipate this much heat and your power plant, this much power. Hard to squeeze into the 576 kg you have left for your craft.

 

I think I know where you went wrong with your wattage calculation. You figured that since wires in parallel reduce the total resistance of the circuit all you had to do was multiply this reduced resistance by the 320 amps squared to get the wattage. But this isn't how it works. If this were true, then having two light bulbs wired in parallel would use fewer kilowatts than one Light bulb. And since all the lights in your house are wired in parallel, turning on all the lights in the house would save more on your electric bill than just burning one. But this doesn't happen. Each light bulb is on a different branch of a parallel circuit, and each carries its own current. To get the total current supplied by the power plant, you have to add the currents of each branch together. You can then get the power supplied by multiplying this total current by the voltage of the source.

 

Another way of calculating the power usage would be like this:

 

Each wire has a resistance of .021 ohms and a current of 320 amps. Using Ohm's Law, this comes out to a voltage drop of 6.72 volts across each wire and this is the voltage the power plant supplies. To figure out how much power the power plant must provide, we use P=E^2/R. Since the power plant sees the resistance of all the wires in parallel, R is going to be very, very small, which drives the wattage up.

[CraigD]

Whoa! 0.001m wire is 18 AWG which is generally rated at a max current of 24 amps. Getting it to carry 13 times as much will take some cooling system. And considering that your bundle of wires already are taking up 424Kg of the thousand kg of your ship, A cooling system would take up quite a bit more, and we haven't even considered the power plant yet.

This matches what my intuition told me, and what I set out to show, when I began my first post in this tread. Based on my experience with the ampacity guidelines that have come with the various kinds of wire I’ve bought, I was expecting to support the conclusion that, yes, a vehicle propelled the interaction of a current with Earth’s magnetic field is in principle possible, but would require much lower resistance than any material other than a superconductor, and that all known superconductors would cease superconducting in the presence of the magnetic field such a current would produce.

Surprisingly low, yes, and also completely wrong.

With Janus’s ever-reliable direction, I’ve found my error in calculating the resistance of the aluminum wire coil in my tentative design.

 

Recall that the 1000 kg vehicle requires a force of about 1000 kg * Acceleration of gravity = 10000 N (kg*m/s/s) to hover.

 

F = B * I * L

, where F is force, B is magnetic flux density, I is current, and L is length.

Solving for F = 10000 kg*m/s/s, B= 0.00003125 T (kg*A*s*s), gives

I * L = 320000000 A*m

, and for a 200000 wires each 5 m long,

I = 320 A

 

R = p * L / A

, where p is resistivity R is resistance, and A is cross sectional area of the conductor.

Solving for p = 2.8*10^-8 ohm * m (the resistivity of aluminum), L = 10^6 m, and A = Pi * r^2 = Pi * .0005^2 = ~ 7.85 *10^-7 m^2,

R = 2.8*10^-8 * 10^6 / 7.85 *10^-7 = ~ 36000 ohm

 

(here lay the error in my previous calculation – I omitted the division by 7.85 *10^-7 m^2, understating resistance by a factor of roughly a million!)

 

P = I^2*R

, where P is power.

Solving with the above,

P = 320^2 *36000 = 3,686,400,000 W

 

This is even worse that the 430,080,000 W Janus calculated, and unfeasible using any present day engine of which I’m aware. :(

 

Working from the above equations, and introducing an additional one,

L = V / A

, where V is the volume of the unshielded part of the conductor, I get:

P = F^2 * p / B^2 * V

. Notice that A disappears from the equation. It doesn’t matter if the conductor consists of many wires, or a single big one (though it would matter considerable when it comes to cooling it!).

 

F = M * g

, where M is the mass of the vehicle, g the acceleration of gravity, so

P = M^2 * g^2 * p / B^2 * V

 

Ignoring parts of the vehicle other than the conductor,

M = d * V

, where d is density of the conductor, so

P = V * d^2 * g^2 * p / B^2

 

So power can be reduced by decreasing the volume of the conductor, it’s density, or its resistivity, or increasing the magnetic flux of the Earth. The only one of these factors that isn’t a non-manipulable property of Earth or nature is volume, but as volume is reduced, the vehicle becomes smaller, making it less able to contain a power source. :shrug:

 

I think my flying mag-lev car is doomed, until a material capable of superconducting in an intense magnetic flux is available. I suspect increased understanding of the physics of superconductivity will show this to be prohibited. So it’s back to helicopter, airplanes, and rockets. :)

 

It’s amazing what false excitement one little failure to divide by a million can generate. :D

[Jay-qu]

This matches what my intuition told me, and what I set out to show, when I began my first post in this tread. Based on my experience with the ampacity guidelines ...

 

maybe you need to read my guide on basic electronics :)