Re: Stability of an O'Neill Cylinder


James A. Donald,

We're pretty clearly in strong agreement here that single-cylinder
colonies are death traps, and the only question is how quickly everyone
dies following the first computer crash. Our disagreement is whether
the destruction lasts for the whole movie, or just a special-effects
sequence.

My posts give an upper bound based on energy dissipation from fluid
sloshing. Your posts give a lower bound based on rapid dissipation of
energy due to structural flexing. Somewhere in between is the answer.


Narrowing this range is a valid intellectual exercise that can be
pursued without the vitriol thrown by people who can't believe that a
design that O'Neill never proposed might have problems.

In article <jqd973hnvghbpivmdqt86gohu4j1gfbgea@xxxxxxx>, James A.
Donald <jamesd@xxxxxxxxxxx> wrote:

James A. Donald:
It would tumbling be end for end in about ten
minutes or half an hour.

"David M. Palmer"

The yield strain of steel is about 0.1%, and the
designers would have probably used a safety factor of
about 2. This probably implies that the energy
transfered into and out of strain with each wobble,
shortly before the station breaks up, is comparable to
that fraction of the wobble kinetic energy. Thus even
if a large fraction of that energy is lost each cycle,
its contribution to 1/Q is going to be small compared
to that of the fluids.

So to summarize, the Q of the wobble is about 100
dominated by the sloshing of the open water.

The Q of the aluminum and the dirt being infinite?

A Q of 100 is more like that of piano wire.

The dirt does not have to break loose and fly around to
cause dissipation. It will dissipate energy merely by
being shaken. We could neglect the rather low Q of
aluminum if the natural vibrational frequencies of the
aluminum are much higher than the wobble period, because
we could treat the aluminum as entirely rigid

Metals don't have Q, structures do. Tap an aluminum rod on its end and
it will sing for minutes at kHz frequencies. Tap an empty coke can and
you get a dull thwock. (Agreed, different metals do have different
characteristics that give structures made from them different Q.)

The Q of the colony to vibration would definitely be much less than
100, and would probably be small enough that you could treat it as
zero-elasticity. This would be true for any competently designed
structure of this sort, so we can both ignore the red herring of
whether it is made of steel from a NiFe asteroid, or Al tossed up by
mass drivers on the Moon. It would not be true if the colony were a
solid NiFe asteroid (or an Al asteroid if such existed) into which
humans had dug a few tunnels, but that's not what we are talking about.

So, we are in agreement that the Q for structural flexing is low, and
we may as well set it to Q_flex=0 for the sake of argument--all energy
that goes into flexing the structure is immediately dissipated. The
structure is completely inelastic.

The cylinder's flexing in response to wobble will be significant.

However, elasticity and rigidity are two different things. The flexing
will be significant, but it will not be the primary channel of wobble
energy. Instead, I believe that the kinetic energy of the wobble will
be much more than the flexural energy.

Consider the analogous case of a pendulum bob on a free-swinging
damped-elastic string. If you start it oscillating up and down like a
mass on a spring, it will bob up and down with a decay rate given by
the Q_extension of the string.

However, if you set it to swinging, the swing will not damp out with
that Q. If the spring-constant of the string is such that suspending
the weight of the mass on the string extends it by 1%, then the energy
that transfers into and out of the string extension is only 1% of that
which converts between the kinetic and potential energy of the pendulum
motion with each swing. (I haven't worked it out in detail, so there
are probably near-unity factors I am ignoring.)

In that case the effective Q_swing_extension is 100x Q_extension [*],
since the amount of energy available for dissipation is only 1% of the
total energy.
[*] approximately for Q_extension > 1. If Q_extension is 0, then
Q_swing_extension ~ 100. These details are not important
to this argument.

The O'Neill colony will be strong enough to support itself under 1 g
with an extension comparable to the yield strength of the engineering
materials, or about 0.1% . This will also be about the magnitude of
the fraction of the wobble energy that will go into flexure with each
wobble. For Q_flexure = 0, all of it is dissipated; for Q_flexure = 1,
only (1-1/e) is. Either way, it gives a Q_wobble_flexure of ~1000.

So with
Q_wobble_water_slosh ~100,
Q_wobble_air_slosh ~ 1000,
Q_wobble_flexure ~ 1000
so you are first killed by liquids, and if somebody prevented that by
confining the water unsloshably, you get killed by gases and solids
contributing about equally.

And just to round out the elements with a representative of fire, the
mirror panels are large, non-rigid objects that are going to contribute
much more dissipation than the flexure of the rigid can is. It might
even exceed the water's dissipation.

In any case, everyone dies, near enough.

Let us suppose, however, the Q really was 100. Well
yippee. For Island three, we may then conclude that the
instability doubles every three hours, so instead of
half an hour to live, you have a couple of days to live.
Ooh, how wrong I was, A couple days before everyone
gets smacked apart, instead of half an hour. Well that
puts all my worries to rest.

Totally agree.

--
David M. Palmer dmpalmer@xxxxxxxxx (formerly @clark.net, @ematic.com)
.

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