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Hyperloop, Waves, and Unexciting Reality

From: andrew cooke <andrew@...>

Date: Sun, 28 Jul 2013 20:18:55 -0400

There's been some discussion on HN and blogs about Hyperloop:

http://charlesalexander2013.wordpress.com/2013/07/18/hyperloop-riding-sounds-density-peak-to-exploit-the-drag-equation/
http://conscienceofanentrepreneur.blogspot.com/2013/07/hyperloop-lets-you-travel-on-resonant.html
https://news.ycombinator.com/item?id=6117114

However, people don't seem to be thinking the physics through.  It's not as
simple, obvious, or right, as it seems (as far as I can tell).  I may be
wrong, of course, but here's my take:


First, the general idea is that there is a closed loop, filled with air, and
that some kind of cabin "rides" a "wave" around the loop.  That's the vision,
which is certainly intuitively attractive.

However, it seems to be ignoring some practical details that typically come
out of the maths in problems like this (here's a good place for a disclaimer:
what follows is just "general knowledge" from studying simple physical systems
as a student - I have a degree in physics (more or less) - and it may be that
I am using too simple an approximation, or have forgotten something...)


First, lets consider a single, traveling wavefront.  Imagine standing in the
loop and shouting in one direction.  The sound wave travels round the loop and
eventually you hear it coming from behind you (assuming you shouted hard
enough).

That's cool, and it's what seems to be being discussed in the first link
above.  But it's not a steady state solution.  If you used that in a real
system then you'd need something making a huge shout for each journey.  And
the shout would die out as it traveled round the loop.  That doesn't sound
very efficient.


There's an obvious fix to this: generate a series of shouts that are timed so
that each shout coincides with a previous one.  In other words, there's an
exact number of "shout wavelengths" round the loop.  This gives a steady state
solution that's likely much more efficient: you've got a resonant system and
you need less energy to keep it going.

But it won't work.

This solution is called a "standing wave".  And it's called "standing" for a
reason.  The waves don't travel.  Or, if you like, there are two equal but
opposite sets of waves, traveling in opposite directions.  The two views are
equivalent.  And that means that there's no moving wavefront to "ride" (or,
equivalently, that there's an equal and opposite moving wavefront that screws
you up).


The second link recognises this and hopes that there is some other solution.
And it's likely that there is.  It's likely that you can simply add a constant
velocity to all the air particles.  So instead of them being, on average, in
one place (but vibrating backwards and forwards), they are also moving round
the loop.

If do that, then you rotate the high pressure nodes in the standing wave
pattern.  So there's a pressure wave to ride.  But you're not really gaining
much, because the "win" in the first link no longer exists.

The first link hoped to avoid the costs of moving at high speed relative to
the surrounding air.  But in our current solution the air is moving on average
too.  So you might as well not bother with the waves at all.  Just move the
air round the loop in a steady stream, at the same speed as the cabin.

And that works fine but (1) isn't as cool and (2) implies lots of losses from
the wind against the walls of the tunnel.


So I don't see a consistent, useful, efficient physically reasonable argument
for wave riding.

Andrew

Wrong!

From: andrew cooke <andrew@...>

Date: Mon, 29 Jul 2013 21:35:42 -0400

A comment on HN (link above) pretty much convinced me I'm wrong on this.  I
think I am biased by solving problems with fixed nodes.  There should be a
solution that has moving waves and stationary (on average) air.  Andrew

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