Talk:Inductrack

Latest comment: 6 years ago by InternetArchiveBot in topic External links modified (January 2018)

Stationary cars

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I would remove this section. It's almost certainly someone's "wishful thinking" fantasy rather than anything the researchers announced. "No moving parts friction"? Doesn't that section start out by saying it would use a moving track? There's no benefit to keeping it levitated and there's even a major disadvantage to getting rid of the wheels. The wheels are part of the fail-safe design of Inductrack. Sometime, somewhere, if it were to be built, there would be some kind of system failure and probably not at this fantastical "moving track" area, at which point it would need the wheels to settle down on. If there were no wheels, it would grind to a stop on the track like an airliner without landing gear. 24.58.246.19 08:09, 30 December 2006 (UTC)Reply

energy consumption

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How much of the kinetic energy in my train is converted to heat by the electric resistance of the loops? How fast does it slow down ? And what kind of perpetuum mobile do i get when i use superconducting loops? ;-) --Maxus96 (talk) 21:33, 16 March 2010 (UTC) A superconductor uses principles similar to permanent magnet. Tell me, based on *knowledge*, how you expect a permanent magnet to run down. In fact quantum mechanics has been invented to explain why atoms don't run down! But superconductors approach perpetual motion more than helium atoms in that they maintain electric current on a macroscopic scale. Yet this is not a perpetual motion machine of the second kind because one needs energy input in order to shift the magnetic force so the train does not just hover a fixed distance from the electromagnet. The article on perpetual motion has described attempts to convert magnetic energy into motion without dissipation. Tell me, is it more realistic to make a device completely 'fail-safe' than use permanent magnetism? And moreover, how can 'conventional roads' have magnets or rails built in?24.184.234.24 (talk) 02:03, 26 September 2010 (UTC)LeucineZipperReply

How long will the induction persist?

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OK, so permanent magnets induce a current that levitates a train in a superconducting coil. But won't they lose some of their magnetism in the process? Otherwise it seems like so many perpetual motion machines that turn magnetic field into a boundless source of motion. 24.184.234.24 (talk) 21:41, 2 October 2010 (UTC)LeucineZipperReply

No, not at all, the maglev takes significant power, which is taken from the forward motion of the train. The generated currents create resistive losses in the coils, and an inevitable effect of the resistance creates a slight phase shift on the magnetic field that puts the magnetic field slightly ahead of the train, and this forces the train backwards (although mostly upwards). So it costs energy to give the levitation effect; if there was no resistance in the coils (they were superconducting) then the magnetic field would be (in principle) right under the train and the lift would cost nothing. The magnets themselves are neodymium magnets and are magnetically very robust and wouldn't degauss in normal use. If you were using ferrites though, all bets would be off.Rememberway (talk) 19:38, 30 November 2010 (UTC)Reply

Description

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I believe the primary inventor, Richard Post, has died.

Here is a preliminary answer to Citation Needed:

v*x = w*t B.x=B.0*sin(w*t) B.y=B.0*cos(w*t) L=0.0001 henry C=1 farad R.L = 1 ohm R.C = 100 ohm V.L=L*(dB/dt) sqrt(1/LC) = 10 Hz NeFeB 1 in cubes lamba = 4 in v = 40 in/sec ~B = B.0 * 10 Hz V = data collected I = induced current

Questions:

What is the induced current I? What is the induced average lift? What is the induced average drag?

Answers:

RL track: "The Inductrack Approach..."

RLC track: "Spring 2013" (working title) NVCC-Annadale SPS Team 4



Doug Goncz (talk) 12:47, 3 November 2012 (UTC)Reply

Patent - Inductrack III

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I've added the Inductrack III, but I've munged the patent link. It doesn't generate the right query to retrieve the patent application on espacenet so I've commented it out. This is a direct link to the patent, 2010064929 A1. If anyone can see what I've stuffed up, feel free to fix and uncomment. Thanks -- PaulxSA (talk) 23:49, 21 May 2013 (UTC)Reply

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