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 Adventures in EngineeringThe wanderings of a modern ronin.

 Date: 2006-12-23 22:42 Subject: In which Ben does the math on a hoverboard. Public Mood: innumeracy ensues Music: Chemlab (The Magnetic Field Remixes) - 21st Century

The equation for the magnetic force on a wire is:

F = BLI

B is the magnetic field, in Teslas
L is the length of the wire, in meters
I is the current, in amps

The earth's magnetic field is about .3-.6 Gauss. There are 10,000 Gauss in a Tesla. Let's be optimistic and call it .5 Gauss. Converting to Teslas, B = .00005

Let's say the wire is one meter long, because it makes the math easier. L = 1.

How much force do we need? Let's say a 200 lb person, call it 90 kg. Gravity pulls them down with an acceleration of 9.8 m/s^2. F = ma, so F = 90 * 9.8, or 882 Newtons.

882 = I * 1 * .00005

I = 882 / .00005

I = 17,640,000 amps

A little more than seventeen and a half million amps.

To get some sense of scale, most circuit breakers for houses are about 200 amps. Your whole house, everything in it, all the lights going full blast, stereo, fridge, washer/dryer, everything, takes less than 200 amps. (And requires a cable about 3/4 of an inch across.) Doing the math, we see that this is the electrical equivalent of 88,200 houses going full bore.

In terms of energy, P = I^2V, so this comes to 3.7 x 10^16 Joules per second. In more familiar terms, this about the same energy as you would produce if you exploded 40 million tons of TNT. Per second.

As your lawyer, I advise you to stick with Inductrack. ;]

 User: Date: 2006-12-24 12:06 (UTC) Subject: (no subject)
Why just 1 meter? If you up it to 1000 m (spooled, somehow) it's only 17,640A! Easy!

Or else wait to see if someone solves the final unified theory.

 User: Date: 2006-12-24 13:33 (UTC) Subject: (no subject)
P=IV=I^2R , not I^2V. Assuming that you use a piece of copper 1m long with 1cm x 1cm cross section, its resistance will be .00017 Ohm, so the dissipated power will be 5.3 x 10^10W, or 53GW. The voltage will be "only" 3000V. Still quite unrealistic (it's impossible to remove this amount of heat from such a small piece of copper), but about a million times less than your estimate.

If you use a coil that Nick had described (it will be 1m high cylinder with 30cm diameter and walls 10cm thick, not counting the insulation), you will have R=Rorig*1000, I=Iorig/1000, thus reducing the dissipated power 1000 times to 53MW or about 2200 "houses".

Of course, there will be another problem that Earth's magnetic field's direction will only make such a hoverboard possible in the regions close to Earth's magnetic poles. On most of the Earth surface you will accelerate horizontally or at some small angle to horizon instead of hovering. I think, ionizing the air and accelerating it in some electric field, or with a combination of electric and magnetic field will be a much more realistic design. And of course, you can always use that energy to compress the air, or heat it in a jet engine instead of burning fuel in it.

 User: Date: 2006-12-24 19:59 (UTC) Subject: Whoops!
Thanks for the correction on power!

As far as using a coil goes, the actual formula for force on a charge in a magnetic field is F = qv x B. In other words, it's a cross product. So any segment of wire parallel to the field lines contributes no force. Using a coil is not going to help. Only the tiny segments of wire in the coil that are nearly perpendicular to the earth's magnetic field will contribute significant lift. Also, the current flowing in the other side of the coil is in the opposite direction. So it would tend to either cancel out the total force, or else try and make the coil flip itself upwards. (Maybe some triangular kind of coil?)

It's true that there will be a relatively strong magnetic field inside the coil, but I couldn't find any equations that imply having a strong magnetic field will help you. I'd like to be wrong here, but I'm not sure I am. Solenoids and relays don't levitate the circuit boards they're attached to, and electromagnets don't float...

 User: Date: 2006-12-24 23:16 (UTC) Subject: Re: Whoops!

Right -- if you have a small enough coil in a field that can be assumed to be constant and parallel, all you can do is to make it align with the field, so if you place one side of the coil on the ground, the other one side will rise.

This also applies to the original "single-wire" version because you will have to somehow complete the circuit. You will have some current going in the opposite direction through the field, thus creating the force in the opposite direction, so if you won't place your power source on the ground, this force will be applied to your device as well. If your power source is a generator or battery, current has to go through the whole loop, thus making it impossible to raise the power source if it is a part of your device. If it's a charged capacitor, you have a small distance where the charges don't travel, however magnetic field produced by the change in electric field will compensate for that.

So the whole thing can only work if either:

1. Part of the circuit is fixed on the ground.
2. The circuit is large enough to cover the area where Earth's magnetic field is not uniform.
3. You are in the region where Earth's magnetic field is non-uniform enough for the device to work.

First case is probably not what it is supposed to be for -- if it's possible to place a part of the device on the ground, why not also place a coil, magnet or passive loop there, and make a much more efficient device without crazy power requirements?

Second thing for most of the Earth's surface would require a truly giant structure. A loop suspended on a space elevator column may work, but then you would already have a space elevator. A loop around the Earth will also work, though unless the loop is made of a superconductor, the amount of energy necessary to power it would be unsafe to produce or release.

The third case (what is actually part of the second one) is what may make it work around the Earth's magnetic poles, or magnetic anomalies, though it would still require a very large device, with no hope for any practical use.

 User: Date: 2006-12-26 20:05 (UTC) Subject: (no subject)
The only things I can see working with current physics is a ducted fan or small monopropellant sources... Both have similar power issues. Though high speed fans have the advantage of gyroscopic stabilization to help you balance.
Would have to be one heck of a power source, maybe an in-fan gas turbine to spin it up.

 User: Date: 2006-12-26 21:15 (UTC) Subject: (no subject)
Sounds like a helicopter...

 User: Date: 2006-12-26 21:29 (UTC) Subject: (no subject)
Not quite, since it could operate on ground effect it needs somewhat less power than a helicopter.

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