Larry Niven Fan Forum???
Neil Williams
in science fiction. It would appear from an initial survey of the
postings that it is little more than a Larry Niven fan club. Now I like
Niven's work, I enjoyed it when I first read it, but come on folks...he
isn't the only "hard science" sf writer. Not to mention the time warp
feel of this newsgroup...there has been some "hard science" sf written
since the seventies (at least half the posts are not bitching and whining
about the "new wave" which is all that's needed to make the time warp
complete).
Neil
<well now you've stirred the hornet's nest>
--
Stewart Robert Hinsley
ac...@freenet.hamilton.on.ca "Neil Williams" writes:
It's partly sampling error; Niven science discussions come in cycles, and
there a cycle in progress at the moment.
I have a hypothesis as to why Niven gets discussed more than say Clement
or Anderson, or any other 'hard SF' writer; it's that he's hard enough
that you don't just discount his science, but soft enough that it's easy
to poke holes in it. (The way I'd put it is that Larry Niven's writing
is focused on science and technology, but that he doesn't let reality
spoil a neat story idea.)
Anyway, better a Larry Niven fan club than Relativity flame wars :-)
--
Stewart Robert Hinsley The adequate is the enemy of the good
Erik Max Francis
> One would think that this newsgroup was on the net to discuss the science
> in science fiction. It would appear from an initial survey of the
> postings that it is little more than a Larry Niven fan club. Now I like
> Niven's work, I enjoyed it when I first read it, but come on folks...he
> isn't the only "hard science" sf writer.
Fine. Then post about someone else and start a new thread. What are you
complaining about?
--
Erik Max Francis, &tSftDotIotE. && m...@alcyone.darkside.com || m...@alcyone.com
San Jose, California, U.S.A. && 37 20 07 N 121 53 38 W && the 4th R is respect
H.3`S,3,P,3$S,#$Q,C`Q,3,P,3$S,#$Q,3`Q,3,P,C$Q,#(Q.#`-"C`- && 1love && folasade
Omnia quia sunt, lumina sunt. && GIGO Omega Psi && http://www.alcyone.com/max/
"Out from his breast/his soul went to seek/the doom of the just." -- _Beowulf_
steve hix
:One would think that this newsgroup was on the net to discuss the science
:postings that it is little more than a Larry Niven fan club. Now I like
:Niven's work, I enjoyed it when I first read it, but come on folks...he
:since the seventies (at least half the posts are not bitching and whining
:about the "new wave" which is all that's needed to make the time warp
:complete).
New to the neighborhood, Neil?
The Niven threads are, at most, a couple of weeks old, with probably only
a few more, at most, to run.
If you want to talk about something else, start up a new thread.
It's easy.
Tony Buckland
>Neil Williams wrote:
|> One would think that this newsgroup was on the net to discuss the science
|> in science fiction. It would appear from an initial survey of the
|> postings that it is little more than a Larry Niven fan club. Now I like
|> Niven's work, I enjoyed it when I first read it, but come on folks...he
|> isn't the only "hard science" sf writer.
>complaining about?
Just as America's moral universe is partitioned into those who
think the main point is liberty, to be cautiously diminished
only in cases of necessity, and those who think the main point
is to obey a moral code, with liberty to be permitted
cautiously when they are sure there is no danger; so the net
universe is partitioned into those who skip what doesn't
interest them, and those who think nothing that doesn't
interest them should be there to offend them and waste their
time.
the ROYster-Meister
On Thu, 22 Feb 1996, Stewart Robert Hinsley wrote:
> In article <4gget1$l...@main.freenet.hamilton.on.ca>
> ac...@freenet.hamilton.on.ca "Neil Williams" writes:
>
> > One would think that this newsgroup was on the net to discuss the science
> > in science fiction. It would appear from an initial survey of the
> > postings that it is little more than a Larry Niven fan club. Now I like
> > Niven's work, I enjoyed it when I first read it, but come on folks...he
> > isn't the only "hard science" sf writer. Not to mention the time warp
> > feel of this newsgroup...there has been some "hard science" sf written
> > since the seventies (at least half the posts are not bitching and whining
> > about the "new wave" which is all that's needed to make the time warp
> > complete).
> >
> [snip]
>
> I have a hypothesis as to why Niven gets discussed more than say Clement
> or Anderson, or any other 'hard SF' writer; it's that he's hard enough
> that you don't just discount his science, but soft enough that it's easy
> to poke holes in it. (The way I'd put it is that Larry Niven's writing
> is focused on science and technology, but that he doesn't let reality
> spoil a neat story idea.)
>
>
Hmmmm.... Niven himself said something to that effect once. Keep
the science good, but the story is the thing -- is why there are
inconsistencies in some of the Known Space stuff.
Speaking of science one can poke holes in, anybody have comments
about the "Smoke Ring" concept? It's kool, but I haven't got the shmarts
to judge it on, and sometimes the whole thing doesn't seem believeable.
------------------------------------------------------------------------------
-- the ROYster-meister + wil...@Peak.org --
one of God's >peculiar< people
"But you are a chosen race, a royal priesthood, God's peculiar people."
-- the Apostle Peter (KJV)
------------------------------------------------------------------------------
Erik Max Francis
> Speaking of science one can poke holes in, anybody have comments
> about the "Smoke Ring" concept? It's kool, but I haven't got the shmarts
> to judge it on, and sometimes the whole thing doesn't seem believeable.
A torus of gas would have troubles maintaining its consistency; after all,
it's just a cloud of gas, and it would easily dissipate (certainly over
astronomical timescales). The idea is that Goldblatt's World replenishes the
Smoke Ring, but this cannot last forever.
Furthermore, the composition of the gases in the Smoke Ring makes one wonder.
The gases are breathable to humans, which means that they're roughly 75% N2
and 25% O2. So what are the gases on Goldblatt's World? If it's a gas giant,
which is I believe what it was supposed to be, then where does the N and O
come from? Gas giants are composed primarily of H and He for a reason:
Massive worlds are composed of low-molecular-weight gases because their
massive gravitational field and generally their cold exosphere temperatures
allow this to happen. So why isn't the Smoke Ring composed mostly of H and
He? What preferential process allows the N and O to be retained while the
lighter elements bleed off, and quite rapidly? Gravity prevents the leaking
of atmospheres off worlds; what's to prevent them from leaking out of a gas
torus in orbit around a neutron star?
It's interesting that you brought this up, since I was interested about this
myself.
Alice Holt
Erik Max Francis says :
>A torus of gas would have troubles maintaining its consistency; after all,
>it's just a cloud of gas, and it would easily dissipate (certainly over
>astronomical timescales). The idea is that Goldblatt's World replenishes the
>Smoke Ring, but this cannot last forever.
What lasts for ever?
[Explanation that gas giants consist mostly of H and He deleted]
>What preferential process allows the N and O to be retained while the
>lighter elements bleed off, and quite rapidly? Gravity prevents the leaking
>of atmospheres off worlds; what's to prevent them from leaking out of a gas
>torus in orbit around a neutron star?
A system like this - a large mass orbiting close to a much bigger mass,
produces an orbital path between the two that is very stable - where
the gravitational fields of the gas giant and the neutron star ballance
each other out.
The smoke ring is going to continuously lose a large portion of it's gas
to space, but as you say it is also being continualy replenished. Such a
system will inevitably reach an equilibrium state. The point of equilibrium
will depend on the masses of the neutron star, Goldblatt's World and their
relative possitions.
Very light gasses, such as H and He, will bleed off first just as they do on
earth. In fact, most of the ring will still be composed of these gasses, but
they will disipate out rapidly into the vast halo around the core of the
ring. The high pressure core will consist of the heaviest gasses, such as N,
O2 and CO2, which will disperse far more slowly.
All this is explained in the book.
Simon
Erik Max Francis
> What lasts for ever?
Well, of course. But I meant a rather short forever.
> A system like this - a large mass orbiting close to a much bigger mass,
> produces an orbital path between the two that is very stable - where
> the gravitational fields of the gas giant and the neutron star ballance
> each other out.
I wasn't questing the orbital dynamics of the Voy-Goldblatt pair; that's
obvious. The question is the stability of the Smoke Ring itself, which is
_not_ stable. It's a cloud of warm gas in orbit around a neutron star dense
enough to create 100 kPa of atmospheric pressure. The gases must be diffusing
into the vacuum at a ferocious rate.
> The smoke ring is going to continuously lose a large portion of it's gas
> to space, but as you say it is also being continualy replenished.
By Goldblatt's World, which cannot have an infinite supply, particularly
considering that if it's a typical gas giant, it has far more H and He than N
and O; so where are the H and He going? They aren't going through the Smoke
Ring.
> Such a
> system will inevitably reach an equilibrium state. The point of equilibrium
> will depend on the masses of the neutron star, Goldblatt's World and their
> relative possitions.
An equilibrium state? The equilibrium state is that Goldblatt's World
atmospheric pressure drops to the atmospheric pressure of a very wispy
interstellar gas cloud, and so does the Smoke Ring. Gravity has little to
participate in, here. The question is one of gas diffusion. Nothing is
keeping gases from diffusing away from the Smoke Ring; the only thing
counterbalancing that effect is the leeching of gases from Goldblatt's World.
> Very light gasses, such as H and He, will bleed off first just as they do on
> earth.
But in the Smoke Ring, they should be being injected at enormous rates, since
gas giants have far more lighter elements than they do heavier ones (that's why
they're classified as gas giants; they can retain the lighter elements). So H
and He should be flowing through the Smoke Ring in enormous amounts. Where are
they?
Jeff Suzuki
: allow this to happen. So why isn't the Smoke Ring composed mostly of H and
: He? What preferential process allows the N and O to be retained while the
: lighter elements bleed off, and quite rapidly?
Surely you could ask the same question of the Earth: when it started,
its atmosphere had to be mostly H and He (since, after all, most of
the primordial nebula was H and He). The H and He bled off
preferrentially, leaving behind the N and O.
What I'd wonder about, though, is how you got so _much_ N and O to
support a breathable atmosphere over such an enormous volume.
Jeffs
the ROYster-Meister
On Sat, 24 Feb 1996, Erik Max Francis wrote:
> the ROYster-Meister wrote:
>
> > Speaking of science one can poke holes in, anybody have comments
> > about the "Smoke Ring" concept? It's kool, but I haven't got the shmarts
> > to judge it on, and sometimes the whole thing doesn't seem believeable.
>
> A torus of gas would have troubles maintaining its consistency; after all,
> it's just a cloud of gas, and it would easily dissipate (certainly over
> astronomical timescales). The idea is that Goldblatt's World replenishes the
> Smoke Ring, but this cannot last forever.
>
Yah, but I'm not entirely clear on how that would work....
> Furthermore, the composition of the gases in the Smoke Ring makes one wonder.
> The gases are breathable to humans, which means that they're roughly 75% N2
> and 25% O2. So what are the gases on Goldblatt's World? If it's a gas giant,
> which is I believe what it was supposed to be, then where does the N and O
> come from? Gas giants are composed primarily of H and He for a reason:
> Massive worlds are composed of low-molecular-weight gases because their
> massive gravitational field and generally their cold exosphere temperatures
> allow this to happen. So why isn't the Smoke Ring composed mostly of H and
> He? What preferential process allows the N and O to be retained while the
> lighter elements bleed off, and quite rapidly? Gravity prevents the leaking
> of atmospheres off worlds; what's to prevent them from leaking out of a gas
> torus in orbit around a neutron star?
>
I got the impression that the lighter gases got blown away by the
equivalent of a solar wind, but I couldn't cite anything from the books
to support that. I did always wonder if maybe you wouldn't have to
construct such a set-up, 'cause it didn't seem like such a thing would
occur naturally.
Timothy J. Miller
N> One would think that this newsgroup was on the net to discuss the
N> science in science fiction. It would appear from an initial survey of
N> the postings that it is little more than a Larry Niven fan club. Now
N> I like Niven's work, I enjoyed it when I first read it, but come on
N> folks...he isn't the only "hard science" sf writer. Not to mention
N> the time warp feel of this newsgroup...there has been some "hard
N> science" sf written since the seventies (at least half the posts are
N> not bitching and whining about the "new wave" which is all that's
N> needed to make the time warp complete).
Would you like some cheese with your whine?
-- Cerebus <tmi...@ims.advantis.com>
"I wonder if I can restart the 'Deckard was the 6th replicant' thread. 8)"
Curtis Bartley
On the subject of the Smoke Ring: The diagram of the system in one (or
both books) describes a radius of about 25,000 km to the Smoke Ring
median. I can't remember if this figure also pops up in the text. The
neutron star has about 1/2 the mass of the Sun, and the orbital period of
the Smoke Ring ranges from about 2 to 6 hours (this is in the text).
I did some BOTE calculations (well, actually I used my trusty HP
calculator) and determined that these figures had to be way off. (For a
point of comparison, a geosynchronous satellite orbits the Earth at about
(I think) 22,300 km radius in 24 hours, which is way too close to the
above figures for them to be right.
For a radius of 250,000 km the orbital periods seemed to work out, and
furthermore the gravity gradient at the ends of the integral trees worked
out about right (about 1/10 g at the ends of a 40 km tree as I recall).
I could chalk this up to just an error in the diagram (but where would the
fun be in that?) but in several different places Niven asserts a habitable
volume of 30 times the volume of Earth, which strikes me as being far too
small for a 250,000 km radius.
Anybody care to check these figures?
-- Curtis Bartley (cbar...@macromedia.com)
Erik Max Francis
> I got the impression that the lighter gases got blown away by the
> equivalent of a solar wind, but I couldn't cite anything from the books
> to support that. I did always wonder if maybe you wouldn't have to
> construct such a set-up, 'cause it didn't seem like such a thing would
> occur naturally.
But the problem is, the H and He would still have to get through the Smoke
Ring, so you'd end up having ridiculously high levels of H and He in the Smoke
Ring. Remember that a gas giant is _defined_ as being able to retain lighter
elements, and due to element ratios, a gas giant would contain far, far more H
and He than N and O (for example).
So where is all the H and He? Furthermore, what levels would be noticeable?
(Inability to breathe, or sparks tending to cause big problems?) I imagine
that due to the sheer volume of H and He pumping through the Smoke Ring, we
would have noticed something as astute readers (i.e., they're breathing; they
shouldn't be able to be).
Erik Max Francis
> Surely you could ask the same question of the Earth: when it started,
> its atmosphere had to be mostly H and He (since, after all, most of
> the primordial nebula was H and He). The H and He bled off
> preferrentially, leaving behind the N and O.
It's a completely different situation with the Smoke Ring. At first, the Earth
was mostly H and He. Earth's meek gravitational field (compared to a gas
giant's, anyway) allowed the lighter elements to escape from the exosphere, and
thus it retained only the heavier gases in its atmosphere. After the vast
majority of the H and He were gone, the Earth's atmospheric composition settled
down to an equilibrium (at least when it comes to these considerations!).
The Smoke Ring would have atmospheric leaking with a vengeance (nothing's going
to prevent gas from escaping into vacuum from the edge of the Smoke Ring, and
_boy does it want to_), but we have a gas giant which is feeding the whole
system: Goldblatt's World. But since it _is_ a gas giant, that means it is
able to retain light elements such as H and He, and gas giants are singularly
good at retaining them very well. So even in the ideal case, the gas feeding
the Smoke Ring from Goldblatt's is _nearly all_ H and He. Now the H and He can
still preferentially escape due to their low molecular mass, but that means
that we have ridiculous amounts of H and He pouring through the Smoke Ring.
> What I'd wonder about, though, is how you got so _much_ N and O to
> support a breathable atmosphere over such an enormous volume.
Well, that's another question: How the Smoke Ring got started in the first
place, and what makes it even marginally stable, at least in the short run.
Aaron P Teske
Excerpts from netnews.rec.arts.sf.science: 26-Feb-96 Re: Smoke Ring (was
Larry N.. by the ROYster-Meister@PEAK
> equivalent of a solar wind, but I couldn't cite anything from the books
> to support that.
I think you're right, though I don't have either of the books with me to
examine.
>I did always wonder if maybe you wouldn't have to
> construct such a set-up, 'cause it didn't seem like such a thing would
> occur naturally.
On the other hand, I *do* have _N-Space_ here with me, and in his notes
on "The Kiteman" he says that what got him started on the idea was
Titan: Titan isn't massive enough to hold onto its (comparatively)
massive atmosphere, so that atmosphere has been slipping away... into
orbit around Saturn. Some of the gas, of course, escapes anyway, but
most of it remains in orbit, eventaully to wind up back in Titan's
atmosphere, only to leak away again... lather, rinse, repeat, and you
have a smaller-scale version of the Smoke Ring.
Whether the actual Smoke Ring could exist or not, I dunno. Niven admits
that the neutron star has to be given time to spin down before life
could start to develop, and in that time all (or, at least, too much) of
the gas could have escaped into space. But, IMHO, it is still a great
story, and a fantastic, er, world....
Aaron Teske
Mithr...@cmu.edu
Dave Greenbaum
closest orbit that two neighboring bodies can share without one body being
torn apart by tidal effects of the second.
In _The Smoke Ring_ and the _Integral Trees_(SR & IT hereafter), Niven
describes Goldblatt's World as having fallen within the Roche Limit of
the neutron star - the tides were tearing apart the world, sending atmosphere
into different orbits from the planet - that's how gas torii form.
Remember the terrific tidal density of the neutron star, and that the gas
from the atmosphere of the gas giant remains in orbit around the neutron star.
So, theoretically the SR could exist. My problem is that the incredible
magnetic field densities in close (<250,000 kilometers) would disassociate
most molecules, and the SR should exist as an ionized gas cloud.
But, perhaps there is some kind of ozone layer or magnetic effect caused by
the cloud or the planet that shields the ecology of the SR.
Dave Greenbaum
--
Marriage has many pains, but chastity hath no pleasures.
Samuel Johnson
Erik Max Francis
> So, theoretically the SR could exist. My problem is that the incredible
> magnetic field densities in close (<250,000 kilometers) would disassociate
> most molecules, and the SR should exist as an ionized gas cloud.
I don't think anyone's disputing that you can have toroids of gas surrounding
objects, distended from nearby worlds by tidal influences; after all, Jupiter
and Io are a good example. The questions are 1. how long would this
configuration last, 2. how could you get anywhere near an atmospheric pressure
of 100 kPa, 3. how in the world would you get a human-breathable composition,
and 4. how in the world can you have life with a neutron star just a few
hundred thousand kilometers away?
Jeff Suzuki
: to prevent gas from escaping into vacuum from the edge of the Smoke Ring, and
: _boy does it want to_), but we have a gas giant which is feeding the whole
: system: Goldblatt's World. But since it _is_ a gas giant, that means it is
: able to retain light elements such as H and He, and gas giants are singularly
: good at retaining them very well. So even in the ideal case, the gas feeding
: the Smoke Ring from Goldblatt's is _nearly all_ H and He.
Hmmm...I think Niven does indicate that the region near Goldblatt's is
uninhabitable, though the reason is never stated; one might assume
this has to do with a higher concentration of H and He near
Godlblatt's.
Maybe Goldblatt's is unusual. If it was a survivor of the supernova
that produced the neutron star, it might be that it lost most of its
H/He in the process. (After all, the same process that deprived the
Earth of most of its H and He would still be at work)
Jeffs
DAVID WAYNE MC KEE
: > Speaking of science one can poke holes in, anybody have comments
: > about the "Smoke Ring" concept? It's kool, but I haven't got the shmarts
: > to judge it on, and sometimes the whole thing doesn't seem believeable.
: A torus of gas would have troubles maintaining its consistency; after all,
: it's just a cloud of gas, and it would easily dissipate (certainly over
: astronomical timescales). The idea is that Goldblatt's World replenishes
: the Smoke Ring, but this cannot last forever.
I can address this question.
The short answer is that each, individual molecule of the gas torus is in
orbit around the neutron star, and it's position is constrained by the
usual dynamics (i.e. energy and angular momentum).
The slightly longer answer (first order approximation) takes into account
the fact that the molecules interact. Consider a small volume, V, of gas.
Assume for the moment that it is nearly in equalibrium with it's
surroundings, and that it is moving along the orbit of Goldblatt's world
with the same speed as the planet. The volume is at the minimum of an
effective potential due to the gravitational attraction of the neutron
star, and its own angular momentum (the same dynamics that hold any body
in orbit), and is acted on by some small pressure differential.
The point is that the effective orbital potential acts as a restoring
force against the pressure differential, so our unit of gas isn't going
anywhere.
This analysis applies to any unit of gas in the cloud. (Recall that is
conservation of angular momentum will not allow the whole cloud to just
stop and fall in). So we have a suprisingly stable system.
(I would guess that the dissapation mechanism is gas interactions which
effectively give one molecule lots of energy, taking it out, and another
very little energy, taking it in. Analysis of the pressure differentials
in terms of the Boltzman distribution to find the rate of dissapation is
left as an exercise for the student.)
: Furthermore, the composition of the gases in the Smoke Ring makes one
: wonder.
Err... Niven's desire to have a breathable atmosphere? I can't answer this one.
: Gravity prevents the leaking
: of atmospheres off worlds; what's to prevent them from leaking out of a gas
: torus in orbit around a neutron star?
Conservation of angular momentum. See the above.
--
David McKee | Graduate students are not "immediate" type people. We
dmc...@nmsu.edu | are experts at delayed gratification. We have to be.
| That, or nuts. --"Gary T. Ward" <gw...@U.Arizona.EDU>
| in alt.grad-student.tenured
the ROYster-Meister
On Mon, 26 Feb 1996, Curtis Bartley wrote:
>
> On the subject of the Smoke Ring: The diagram of the system in one (or
> both books) describes a radius of about 25,000 km to the Smoke Ring
> median. I can't remember if this figure also pops up in the text. The
> neutron star has about 1/2 the mass of the Sun, and the orbital period of
> the Smoke Ring ranges from about 2 to 6 hours (this is in the text).
>
1/2 a solar mass strikes me as a bit too small for a neutron
star... anybody have some figures on that? My handy little
desk-reference doesn't get that detailed (it never does, with the fun stuff).
Erik Max Francis
> Erik Max Francis made the same comment. However, that looks like one of
> the least problems with Niven's numbers for the Smoke Ring.
True. My comment was that all neutron stars whose masses have been directly
measured are consistent with 1.2-1.6 masses solar; but all equations of state
for neutron star matter (at least ones that I've seen) certainly _allow_ 0.5
mass solar neutron stars. The question is merely one of how they'd form.
As you say, though, this seems to be way down on the list of complaints.
Eric E Tolle
>On 26 Feb 1996, Jeff Suzuki wrote:
>>
>> What I'd wonder about, though, is how you got so _much_ N and O to
>> support a breathable atmosphere over such an enormous volume.
> (that'll answer everything....)
Oh, of course....you mean....
IT WAS ALL DONE BY THE OUTSIDERS!!!!
Y'Know, your right- it _does_ make sense. The question is- why? Is
it a Mana-collection field?
:')
Eric Tolle unde...@mcl.ucsb.edu
"An' then Chi...@little.com, he come scramblin outta the terminal room
screaming "The system's crashing! The system's crashing!"
-Uncle RAMus, 'Tales for Cyberpsychotic Children'
Erik Max Francis
> Beowulf Shaeffer's ship should have left with an end-over-end spin.
> (Of course, since he was in the middle of the ship, he'd be
> fine...unable to reach any controls, but fine...)
The periastron was _one mile_ above the surface of the neutron star. The tides
alone would have killed him; and do you know what kind of spin the ship would
have left with?
> At the distance of the smoke ring from the neutron star, you shouldn't
> have any (gravitational-type) problems, though.
> Goldblatt's and the gas molecules in the smoke ring are both in orbit
> around the neutron star, so there wouldn't be as much wind as you
> might suppose.
But the inner portions are orbiting faster than the outer ones, so you'll get a
differential wind. That's what he was talking about.
Erik Max Francis
> Yah, but I'm not entirely clear on how that would work....
Goldblatt's replenishment is not so much the problem; the real problem is
simply that so much of the Smoke Ring would be escaping away to space so
rapidly that it would really take a considerably amount of Goldblatt's
atmosphere to keep the process going. I suspect the "slow dissipation" would
be more like a steady, strong wind.
> I got the impression that the lighter gases got blown away by the
> equivalent of a solar wind, but I couldn't cite anything from the books
> to support that. I did always wonder if maybe you wouldn't have to
> construct such a set-up, 'cause it didn't seem like such a thing would
> occur naturally.
Well, the idea for _how_ the gases dissipate is clear: The gases on the outer
edge of the gas torus are adjacent to vacuum, and in such low pressure, they
naturally just float away (or get whisked away is more like it). Lighter gases
are more susceptible to this process, since at the same temperature, the
lighter gases mass less and will have a greater speed. (This is why H and He
are only largely prominent on gas giants, whereas the terrestrial planets are
comparatively H- and He-poor.)
But here's the real problem. Okay, so let's say the lighter elements are being
selectively dissipated at the edge of the gas torus; that's fine. But the
lighter elements are being injected into the Smoke Ring by Goldblatt's world.
Goldblatt's world is a gas giant, and gas giants have the characteristic of
being very massive, and therefore retaining heavy atmospheres of chiefly light
elements. Now the Smoke Ring's atmosphere breathable to humans, which means
that it must be mostly N and O. But much more H and He than N and O are being
injected into the Smoke Ring from Goldblatt's world, so where is the H and He?
It's still got to make its way to the outer edge of the gas torus where it can
be preferentially dissipated over the N and O (but probably not by all that
much, when it comes down to it).
Even in equilibrium, the Smoke Ring should have very high levels of H and He.
Likely it wouldn't be breathable; certainly you wouldn't want to strike a match
in it.
Erik Max Francis
> But the mass for voy these numbers imply is about 1e-5 solar masses.
> About 5 earth masses; way, way WAY too small to be a neutron star.
> My conclusion: the distance numbers are all too small by a factor
> of about 46.4 (call it a bit less than 50). (The periods given could
> also be about 300 times too large, but then the tree would orbit in
> about 1 minute, and the tides on a 100km tree would be about 10g...)
Yes. I made a quick calculation that indicated that either 1. the mass of Voy
was completely wrong (someone quoted it as 0.5 masses solar) or 2. the Smoke
Ring is _way_ too close. There's no way around it; the numbers Niven gives are
just plain wrong.
He might have had in mind a much less massive neutron star, but neutron stars
aren't stable below ~0.1 masses solar. This has been known since the sixties,
so Niven would have no excuse if he got this worng. (It's worth nothing, once
again, that all neutron star masses are consistent with 1.2-1.6 masses solar --
just what you'd expect from white dwarfs being pushed over the Chandrasekhar
limit.)
> Note that tides (to a close approximation) are entirely determined by the
> period, so exploding the orbits to more reasonable size doesn't affect
> the tidal calculation. It *does* seem a bit low, though.
Eh? Tides are proportional to the gravitational gradient, which in Newtonian
gravitation is the inverse cube. If you keep the mass the same and increase
your distance by a factor of 50, then the tidal effects are going to be 50^3 =
125 000 times less.
> Parameters:
>
> earth r=1.5e11,t=3.155693e7 -> k(sun)=1.337962745e+20
> (book value) -> k(sun)=1.32613297e20
> neutron star= 1.4 solar masses =1.856586158e+20
Early in this thread someone indicated that Voy had a mass of about 0.5 solar.
I didn't check this, but even then the mass is still way too high to explain
the long orbital period.
Erik Max Francis
> In astronomy there is a concept known as the Roche limit, which defines the
> closest orbit that two neighboring bodies can share without one body being
> torn apart by tidal effects of the second.
>
> In _The Smoke Ring_ and the _Integral Trees_(SR & IT hereafter), Niven
> describes Goldblatt's World as having fallen within the Roche Limit of
> the neutron star - the tides were tearing apart the world, sending atmosphere
> into different orbits from the planet - that's how gas torii form.
> Remember the terrific tidal density of the neutron star, and that the gas
> from the atmosphere of the gas giant remains in orbit around the neutron star.
>
> So, theoretically the SR could exist. My problem is that the incredible
> magnetic field densities in close (<250,000 kilometers) would disassociate
> most molecules, and the SR should exist as an ionized gas cloud.
>
> But, perhaps there is some kind of ozone layer or magnetic effect caused by
> the cloud or the planet that shields the ecology of the SR.
The problem I'm having beliving the Smoke Ring is not the formation of the gas
torus; this makes perfect sense and is actually a well-known phenomenon (take Io
around Jupiter). My question is twofold: 1. How in the world could you get
this sucker to maintain ~100 kPa of pressure in its densest portion stably for
any length of time? The higher the internal pressure the more likely the gas
torus will bleed away to space. 2. How in the world do you get all these heavy
elements (C, N, O) from a gas giant, which is basically defined upon its
exception propensity to keep light elements (H, He)?
Erik Max Francis
> Well, taking those numbers as gospel, we can derive the mass
> of the primary, and then figure the radii at 2 and 10 hours.
> (Messy formulae and such appended.) I get this table:
>
> radius(meters) period(sec) tides((m/s^2)/m)
> tree 2.60e7 19440 2.09e-7
> inner edge 1.34e7 7200 1.52e-6
> outer edge 3.92e7 36000 6.09e-8
I used the same figures, made a mistake and think I've figured out where Niven
went wrong.
I'll demonstrate this quickly by deriving and calculating the orbital radius,
given the orbital period. Early in the thread, someone cited Voy's mass as 0.5
masses solar, although they didn't indicate where that appeared in the book (I
can't find it in any obvious places). Let's assume that this is correct.
First, we have an orbit (we're presumably circular here) when centripetal
acceleration a equals gravitational acceleration g:
a = g.
Knowing a = omega^2 r = 4 pi^2 r/T^2 (omega is angular speed, r is radius, T is
orbital period) and g = F/m = G M/r^2 (G is universal constant of gravitation,
M is Voy mass), and setting these equal, we have:
4 pi^2 r/T^2 = G M/r^2.
We can now quickly solve for T and find:
r = [G M T^2/(4 pi^2)]^(1/3).
(For those that are paying attention, this is merely Kepler's third law.)
Calculate this for T = 5.4 h and M = 1.0 x 10^30 kg and you find that r = 8.6 x
10^8 m, or 33 times too big. At the radius quoted and a mass of Voy of 0.5
masses solar, we have an orbital period of 1.7 min -- way too fast.
I think I might have found out where Niven made his mistake (because the first
time through I made it myself). Let's say you dropped the square term off the
T value in the calculation would be
"r" = [G M T/(4 pi^2)]^(1/3).
(I write "r" in quotes since it's no longer a radius, of course.) If you do
the numbers using SI (or m.k.s.) you find that the "radius" comes to about 3.2
x 10^7 "m" for T = 5.4 h and M = 1.0 x 10^30 kg. This is surprisingly close to
2.6 x 10^7 m, which is the correct value. The other values for the inner and
outer limits come pretty close, too.
My suspicion is that someone (Niven himself, probably) just fudged the
calculation when they were doing it. But how come no one noticed it?
> But the mass for voy these numbers imply is about 1e-5 solar masses.
I find the same thing.
> Note that tides (to a close approximation) are entirely determined by the
> period, so exploding the orbits to more reasonable size doesn't affect
> the tidal calculation.
Er, watch yourself, there. Tides are proportional to the gradient of the
gravitational field; in an inverse square field they vary as the inverse cube.
So increase the distances by a factor of 50 and you decrease the tides by a
factor of 125 000.
> Sigh. Can somebody check my figures? Now that I'm about to post,
> I have a horrible, sinking feeling that I've made some error.
> Better run through it again; "Ho", "Ha", "Guard", "Turn", "Parry" ...
> --
> Wayne Throop throopw%sheol...@dg-rtp.dg.com
> thr...@aur.alcatel.com
>
> --
> Formulae:
>
> v=(2*p*r)/t velocity, given period t, radius r
> a=(v^2/r) acceleration given v & r
> k=(a*r^2) Gm of primary, given a & r
>
> k/r^2=v^2/r equate accelerations
> k/r=v^2 then
> v=(k/r)^(1/2) express velocity as radius
>
> t=2*p*r/v period, given r & v
> t=2*p*r/((k/r)^(1/2)) substitute v=(k/r)^(1/2)
>
> With these, we have related orbital period, radius, and mass of primary,
> and can solve for any one of them, given the others. Express that to
> my simple expression solver:
>
> e t-2*p*r/((k/r)^(1/2)) t r k k/r^2-k/(r+1)^2
> v p=2*asin1
> p k 1 1 1e30
> p r 1 1 1e20
> p t 1 1 1e20
> i 500
>
> Note that the k/r^2-k/(r+1)^2 calculates the tides for the given scenario.
Uh, what? Where does the 1 come from? (That's not dimensionally consistent,
you know.)
The proper formulation for a tidal strength is either the differential force or
acceleration (we'll use acceleration) over the two furthest points on the
object (in the radial direction), which we will call deltar:
tau o= |g(r) - g(r + deltar)|
o= M (1/r^2 - 1/(r + deltar)^2).
deltar must be a physical distance; it can't be a dimensionless number.
If you take the limit as deltar -> 0, you find that tau o= dg/dr, or M/r^3.
(Here the symbol `o=' means "is proportional to.")
Jeff Suzuki
: 2. How in the world do you get all these heavy elements (C, N, O)
: from a gas giant, which is basically defined upon its exception
: propensity to keep light elements (H, He)?
I think I have a solution: if Goldblatt's world started out very close
to its star, it would be at a much higher mean temperature, which
would give the H/He atoms more of a likelihood of escaping, enriching
its atmosphere in the heavier elements.
Jeffs
Leonard Erickson
> The problem I'm having beliving the Smoke Ring is not the formation of the ga
> torus; this makes perfect sense and is actually a well-known phenomenon (take
> around Jupiter). My question is twofold: 1. How in the world could you get
> this sucker to maintain ~100 kPa of pressure in its densest portion stably fo
> any length of time? The higher the internal pressure the more likely the gas
> torus will bleed away to space. 2. How in the world do you get all these he
> elements (C, N, O) from a gas giant, which is basically defined upon its
> exception propensity to keep light elements (H, He)?
The two questions actually answer each other.
Most of the H and He *have* escaped. Leaving behind the heavier
elements. Remember, just because the overwhelming majority of the atoms
composing a gas giant are H and He doesn't mean that it doesn't have
*huge* amounts of heavier elements.
So the question is if a maximally sized gas giant (brown dwarf?) would
survive the formation of the neutron star, and still have enough
heavier gases to sustain the torus for long enough for life to adapt.
Leonard Erickson (aka Shadow)
sha...@krypton.rain.com <--preferred
leo...@qiclab.scn.rain.com <--last resort
Erik Max Francis
> 1/2 a solar mass strikes me as a bit too small for a neutron
> star... anybody have some figures on that? My handy little
> desk-reference doesn't get that detailed (it never does, with the fun stuff).
All neutron stars whose masses have been measured are consistent with a mass
between 1.2 and 1.6 solar. (The Chandrasekhar limit is about 1.44 masses solar,
so this makes sense.)
The minimum and maximum stable masses depend upon the equation of state used to
describe the neutron star matter, and there is no "right" answer. For the
minimum mass, values of about 0.1 masses solar are common; for the maximum, 3 is
about right.
Neutron stars with a mass of 0.5 solar are allowed by theory, but they would be
pretty exceptional. (I posted a more detailed account of this a week or so
ago.)
--
Erik Max Francis &tSftDotIotE && http://www.alcyone.com/max && m...@alcyone.com
San Jose, California, U.S.A. && 37 20 07 N 121 53 38 W && the 4th R is respect
H.3`S,3,P,3$S,#$Q,C`Q,3,P,3$S,#$Q,3`Q,3,P,C$Q,#(Q.#`-"C`- && 1love && folasade
Omnia quia sunt, lumina sunt. && Dominion, GIGO, GOOGOL, Omega, Psi, Strategem
Leonard Erickson
> In article <P35ekD...@krypton.rain.com>, Leonard Erickson
> <sha...@krypton.rain.com> writes
> >
> >As a first approximation, the spin would be such that the forces due to
> >tides would be replaced by forces due to spin.
> The ship is long and relatively thin and has uneven distribution of mass
> inside (a GP#2 hull is long, with a narrow waist 2/3rds of the way
> along). It was hanging nose down to the neutron star (BVS#1 - a non
> rotating neutron star [IIRC]), because the tides were too strong for the
> gyros to overcome them. Why would the ship pick up a spin? Wouldn't those
> same tidal forces keep the nose pointing at the centre of mass of the
> star until the ship was far enough away for the gyros to become usefull
> again?
>
> >As a second approximation, it'd spin slower than that as the same tides
> >that sped it up would try to slow it down. But it'd still be high.
> The ship would whip around quite quick at closest approach, but those
> same tides would then slow its rotation again afterwards. Whatever
> angular momentum it gained during its fall, it would lose again during
> its climb (wouldn't it?). It would always be nose down to the star. It
> wouldn't complete a full 360 (unless its orbit did the same).
I'm not up on the details. But apparently due to the fact that the
orbit is hyperbolic, and due to the way the tidal forces interact with
the moment of inertia of the ship, it *will* pick up a spin.
I don't know the nitty-gritty details, and the calculations required
are out of my league. But folks who know what they are doing say that
it happens.
My "best guess" picture of it is like this:
As the ship approaches, the tidal forces try to make it line up
pointing at the star. But as the ship is in a hyperbolic orbit, that
direction *changes*, thus the ship *is* rotating, if only to keep
pointing at the star. So as it approaches periastron, the angular
momentum is increased. But once it is past, the tidal forces trying to
*brake* the rotation are *decreasing* faster than they can brake it.
Remember, tidal forces vary as the *cube* of the distance. And the
rotation induced therefore varies as well.
> Bey held himself in the inspection tube at the ships' centre of mass and
> used his arms and legs to press against the sides of the tube to keep
> himself in position and facing the front of the ship and therefore the
> star (would you face the other way and miss the view?) with the intent of
> reducing the tidal forces as much as possible. The tides at about 10cm
> would be the important ones.
Nope. The tides *increase* as you move away from the center of mass.
The tides at 10 cm are *nothing*. It's the tides at the points of his
body *farthest* from the center that count.
Also, beside the pull towards his head and feet, there's also a *push*
trying to compress him radially.
front head
| ^
v |
left-->body<--right body
^ |
| v
back feet
This is in addition to the pull towards "head" and "feet". And it may
kill him because even though it is half as strong as the pull (at the
same distance from the center) it's probably a lot stronger than Bey
can *push* at the walls.
So it would appear that rather than having his head-foot axis aligned
with the access tube, to survive he may have had to have his front-back
axis aligned that way. Then he could but more of his strength into
trying to push againt the walls.
Again, the tide is a pull of X away from the center of mass along axis
that points towards the star. It's also a *push* of X/2 at right angles
to this axis.
Erik Max Francis
> So the question is if a maximally sized gas giant (brown dwarf?) would
> survive the formation of the neutron star, and still have enough
> heavier gases to sustain the torus for long enough for life to adapt.
The question is not how come there's so much heavier elements. The question is
how come there's so of these elements _in comparison_ with H and He. You can
argue that H and He preferentially escape, that's fine; and it's true.
But the ligher elements don't teleport from Goldblatt to the edge of the gas
torus; they have to travel through the Smoke Ring to get there. In
equilibrium, there's going to be a flow of lighter elements out toward the gas
torus edge and off into space. Since there's so much more H and He than
heavier elements, _where are those lighter elements_? They're still migrating
through the Smoke Ring to preferentially escape. I'd wager that the atmosphere
would be unbreathable to humans; at the very least, you'd better not light any
matches.
Wayne Throop
:: Bey held himself in the inspection tube at the ships' centre of mass
:: and used his arms and legs to press against the sides of the tube to
:: keep himself in position and facing the front of the ship and
:: therefore the star (would you face the other way and miss the view?)
:: with the intent of reducing the tidal forces as much as possible.
:: The tides at about 10cm would be the important ones.
: sha...@krypton.rain.com (Leonard Erickson)
: Nope. The tides *increase* as you move away from the center of mass.
: The tides at 10 cm are *nothing*. It's the tides at the points of his
: body *farthest* from the center that count.
: Also, beside the pull towards his head and feet, there's also a *push*
: trying to compress him radially. [...]
: This is in addition to the pull towards "head" and "feet". And it may
: kill him because even though it is half as strong as the pull (at the
: same distance from the center) it's probably a lot stronger than Bey
: can *push* at the walls.
Bey was facing the star. Thus, the compression forces would have been
pushing on his head/feet, and the stretching forces pulling on his
chest/spine. Bey is skinny, and if his chest is 20cm (about 8in)
away from his spine (with center of mass halfway between), that would
make Steve's estimates for the primary tides about right. That is,
if Bey could keep himself centered with millimeter precision. But
let's go with the extreme-skinny notion, and see where it leads.
OK. The Gm of a 1.4 solar masses star is about 1.8e20 m^3/s^2.
So the tides at 14 km would be 1.3e8 m/s^2 per meter, and at 15 km
they would be 1.1e8 m/s^2 per meter. The 1-km-altitude flyby, then,
should have subjected Bey to better than 10 million gravities across
the 10 cm of his chest cavity. The compression forces on his head
and feet should (IIRC) be about ((1/2)*h)/r of the front/back tides,
or in this case about 60g.
But note, the escape velocity from there is about half-lightspeed.
So he's only experiencing those forces for a very small fraction of
a second. If we take his distance such that he spends a thousandth
of a second inside the given radius, we still have something like
10 thousand g per meter tides (less than a g compression tides),
meaning a thousand g tides for a thousandth of a second, with a
shockwave in the middle rising to a million g.
Basically, what we've got here is a classic, textbook case
of "smucker's syndrome".
And while I'm doodling around with this, an interesting point occurs to
me. Consider: when will Bey start to notice the tides? Just picking a
millionth of a g per meter, I get a third of a billion meters out, or
just about two minutes before peri-neutron-asteron. If we say he can
detect trillionth of a g per meter (the tides a spacecraft in LEO gets
from earth), that's still only 20 minutes or so warning. Going from the
portrayal, he didn't start to seriously pay attention until the
flywheels wouldn't let him raise his orbit, so let's just say something
like a tenth of a g across the whole spacecraft. That's more like a
thousadth of a g per meter, and at that point, he had only 12 seconds
warning, more or less. During those 12 seconds, he gets flung
from about 0.02 lightspeed to more than 0.5 lightspeed.
"MAN what a rush! Augh! Augh! Augh!"
--
Wayne Throop throopw%sheol...@dg-rtp.dg.com
thr...@aur.alcatel.com
--
Formulae (no relativistic effects considered):
escape velocity v=sqrt(2GM/r)
tide(neglect spin) a=(GM/r^2)-(GM/(r+1)^2)
Parameters
GM earth ~= 1e10 m^3/s^2
GM neutron star ~= 1.8e20
r earth ~= 6500km
r periNeutronAsteron ~= 15km
John Schilling
>Leonard Erickson wrote:
>> So the question is if a maximally sized gas giant (brown dwarf?) would
>> survive the formation of the neutron star, and still have enough
>> heavier gases to sustain the torus for long enough for life to adapt.
>The question is not how come there's so much heavier elements. The question
>is how come there's so of these elements _in comparison_ with H and He. You
>can argue that H and He preferentially escape, that's fine; and it's true.
>But the ligher elements don't teleport from Goldblatt to the edge of the gas
>torus; they have to travel through the Smoke Ring to get there. In
>equilibrium, there's going to be a flow of lighter elements out toward the gas
>torus edge and off into space. Since there's so much more H and He than
>heavier elements, _where are those lighter elements_? They're still migrating
>through the Smoke Ring to preferentially escape. I'd wager that the
>atmosphere would be unbreathable to humans; at the very least, you'd better
>not light any matches.
Except that airflow within the torus is almost guaranteed to generate some
sort of lighting equivilant, and so provide plenty of sparks. And for that
matter, the H2 + 0.5O2 -> H2O reaction will proceed just fine at room temp,
in the absence of sparks. The rate may be glacially slow by chemical
kinetics standards, but nonetheless fast by atmospheric-evolution standards.
So a flammable or explosive hydrogen/oxygen atmosphere is right out; the
minor component will be consumed as fast as it is introduced. At worst,
you get a region with periodic lighting-induced explosions near Goldblatt,
with a stable equilibrium elsewhere. The resulting water vapor migrates
to the edge of the torus, photodissociates, and the hydrogen leaves.
And in fact, there seems to be plenty of water and water vapor around.
Helium, we don't really have to worry about. Until someone makes it
into a movie, we can freely assume everyone speaks in a squeaky voice.
The real problem is that, for anything resembling a Jupiter-style (or
even Neptune-style) atmosphere as a source, the equilibrium composition
should be water vapor plus hydrogen, not water vapor plus oxygen.
I see two possible answers. First, preferential escaped of hydrogen
from Goldblatt could have stripped the bulk of that planet's hydrogen
atmosphere early in the system's evolution, leaving a gas giant remnant
with most of its original oxygen and water but little hydrogen. As
the torus is subsequently depleted, it is replenished by the now oxy-rich
atmosphere of what is left of Goldblatt. The torus shifts from hydrogen
with a touch of water, through nearly pure water, to oxygen with a touch
of water.
Or, perhaps the relative speed of hydrogen escape from the torus, as
compared to oxygen escape, is so great that even with a 90+% hydrogen
feed the equilibrium composition is mostly oxygen. This also requires
that the rate of escape from Goldblatt be extremely small compared to
the equilibration timescale of the torus itself, such that the vast
majority of the torus mass is due to the long-time equilibrium population,
with hydrogen from Goldblatt diffusing (as water vapor) towards the edge
being a small fraction of the total.
--
*John Schilling * "You can have Peace, *
*Member:AIAA,NRA,ACLU,SAS,LP * or you can have Freedom. *
*University of Southern California * Don't ever count on having both *
*Aerospace Engineering Department * at the same time." *
*schi...@spock.usc.edu * - Robert A. Heinlein *
*(213)-740-5311 or 747-2527 * Finger for PGP public key *
Erik Max Francis
> Not necessarily (though I don't have a calculator handy to check this
> out). Human bodies can survive pretty impressive forces over a
> sufficiently short period of time; if the time he spent in the
> "killing zone" was short enough (a hundredth of a second? a
> thousandth?) then he'd be able to survive (though probably with every
> bone in his body broken...)
The question is what we consider fatal. A recent discussion concluded that 200
gees will basically kill you (this is enough to dislodge your aorta).
Certainly a few thousand gee, no matter how short, is going to kill you.
According to "Neutron star," BVS-1 is 1.3 masses solar, and at closest approach
the _Skydiver_'s "nose was just seven miles from the center of BVS-1. The tail
was three hundred feet further out." This corresponds to a differential
acceleration between two freely floating objects at these points of about 5
_billion_ gee.
At this time, Beowulf is at the midpoint of the ship, stuck in a service tube;
presumably this means his body is perpendicular to the direction toward BVS-1.
Presuming that Beowulf's thickness is about a foot (he's a crashlander, after
all), this corresponds to a differential acceleration across his torso of a
whopping 7 _million_ gee. You can't tell me that magically he would have
survived this.
I mean, he came within one _mile_ of the surface of a neutron star. There is
just _no_ way you can tell me he survived.
Leonard Erickson
> And in fact, there seems to be plenty of water and water vapor around.
> Helium, we don't really have to worry about. Until someone makes it
> into a movie, we can freely assume everyone speaks in a squeaky voice.
Nope. We can't assume that. Besides squeaky voices, heli-ox atmospheres
are *notoriuos* for one other thing. They conduct heat almost as well
as water. That's why folks in undersea habitats need to dress so
warmly, and keep the temp so high. You can die of exposure in a heliox
atmosphere at "room temp" or even above.
This doesn't fit the Smoke Ring. So there isn't a significant helium
fraction, or staying warm would be a *major* consideration. Fires
wouldn't help much, because the atmosphere would make it hard to
concentrate the heat (rubbing sticks would be a *real* chore)
> I see two possible answers. First, preferential escaped of hydrogen
> from Goldblatt could have stripped the bulk of that planet's hydrogen
> atmosphere early in the system's evolution, leaving a gas giant remnant
> with most of its original oxygen and water but little hydrogen.
Don't forget that Goldblatt's has to have survived whatever created the
neutron star. And have been pretty close before that. So it would have
had a *much* elevated temperature for an extended period before the
formation, and if the formation involved any sort of "energetic" event
like a nova or supernova, it'd lose more from the shockwave.
Start with a "brown dwarf", and then heat it for a few billion years to
drive of H and He. Then hit it with a good shock. I wonder if the
remains would resemble Goldblatt's?
Drive off the free H, and you wind up with water, ammonia, and methane
(plus various other things in *much* smaller amounts). Allow a torus of
those to form and photo-dissociate at the edges, and you get nitrogen
and CO2. Add photosynthesizing life, and you get oxygen, water, CO2,
nitrogen, and the barest traces of ammonia (ammonia and methane both
disappear *rapidly* in an oxygen atmosphere).
So figure that what is being added from Goldblatt's is methane, ammonia
and water, and it all works out nicely.
Wayne Throop
: throopw%sheol...@dg-rtp.dg.com (Wayne Throop)
: And while I'm doodling around with this, an interesting point occurs
: to me. Consider: when will Bey start to notice the tides? Just
: picking a millionth of a g per meter, I get a third of a billion
: meters out, or just about two minutes before peri-neutron-asteron. If
: we say he can detect trillionth of a g per meter (the tides a
: spacecraft in LEO gets from earth), that's still only 20 minutes or so
: warning.
Unfortunately, that was based on a transcription error; I underestimated
earth's mass by several orders of magnitude. Plus, I may have made
some errors comparing distances-fallen. Let me try to do better.
Take earth's GM value to be about 3.98e14 m^3/s^2 (which I think it better
than the bogus 1e10 value I used before). This yields tides of 2.9e-6
m/s^2 per meter. That seems to correspond to about 5 minutes warning.
The 1e-4 m/s^2 per meter tides that the book portrays him noticing
would give him about 60 seconds warning.
Just barely plausible, I think.
In the last 10 seconds, he is slung from 0.01 lightspeed
to more than 0.50 lightspeed. Stunningly, the first 9 seconds
of the last ten only get him up to 0.02 lightspeed; in the last
SECOND, he's flung from 2 percent lightspeed to 50 percent lightspeed.
Whoooosh! (If I haven't blown my calculations again.)
--
Wayne Throop throopw%sheol...@dg-rtp.dg.com
thr...@aur.alcatel.com
--
Parameters used this time
k(sun) = 1.8e20 m^3/s^2
k(earth) = 3.98e14 m^3/s^2
r(earth) = 6.5e6 m
e f-(k/r^2-k/(r+1)^2) f k r
p k 1 1 1e30
p r 1 1 1e20
p f 0 0 1e6
Table of time/distance/tide/percent-lightspeed
0 1.5e+04 1.07e+08 0.516
1 9.34e+06 0.442 0.0208
2 1.48e+07 0.11 0.0165
3 1.95e+07 0.0489 0.0145
4 2.36e+07 0.0274 0.0132
5 2.74e+07 0.0175 0.0122
6 3.09e+07 0.0121 0.0115
7 3.43e+07 0.00891 0.011
8 3.75e+07 0.00681 0.0105
9 4.06e+07 0.00537 0.0101
10 4.36e+07 0.00434 0.00976
20 6.95e+07 0.00107 0.00782
30 9.14e+07 0.000471 0.00688
40 1.11e+08 0.000262 0.00629
50 1.29e+08 0.000166 0.00587
60 1.46e+08 0.000115 0.00556
70 1.63e+08 8.35e-05 0.0053
80 1.78e+08 6.35e-05 0.0051
90 1.93e+08 4.98e-05 0.00492
100 2.08e+08 4.01e-05 0.00477
200 3.36e+08 9.45e-06 0.00393
300 4.48e+08 4.01e-06 0.00353
400 5.5e+08 2.16e-06 0.00329
500 6.46e+08 1.34e-06 0.00312
600 7.38e+08 8.97e-07 0.003
700 8.26e+08 6.39e-07 0.0029
800 9.12e+08 4.75e-07 0.00282
900 9.95e+08 3.65e-07 0.00275
1000 1.08e+09 2.88e-07 0.0027
Leonard Erickson
The way I recall the story, he had his long axis aligned with the tubes
axis. But I suspect it doesn't make much difference.
> OK. The Gm of a 1.4 solar masses star is about 1.8e20 m^3/s^2.
> So the tides at 14 km would be 1.3e8 m/s^2 per meter, and at 15 km
> they would be 1.1e8 m/s^2 per meter. The 1-km-altitude flyby, then,
> should have subjected Bey to better than 10 million gravities across
> the 10 cm of his chest cavity. The compression forces on his head
> and feet should (IIRC) be about ((1/2)*h)/r of the front/back tides,
> or in this case about 60g.
Well, for one thing you forget that the "compression" force is along
*all* directions that are at right angles to the radial axis that the
"stretching" forces act on. So it is not only trying to bring his
breastbone and spine together, but it's also trying to compress the
*sides* of his chest together. And the "compression" forces are *one
half* of the "stretching" forces.
Also, you are (incorrectly) assuming that the tides are *just* the
difference in g-force between the two points. They aren't. Those are
the forces that would act on a body *hovering* at those distances. The
tidal forces are caused by the fact that point at anyplace other than
the center of mass would "like" to be travelling in a different
*orbit*.
The formula for tidal force is:
a=(GM/R^3)*l
a = compression acceleration (ie at right angles to the radius vector)
1/2 stretching accel (along the radius vector)
G = gravitational constant
M = mass of body causing the tides
R = distance between center of mass of two bodies
l = distance between center of mass of body experiencing tide, and point
at which tide is measured.
So. You give GM=1.8e20 m^3/s^2. And R as 15 km. So R^3 is 3.375e12 m^3.
GM/R^3 is 5.33e7 s^-2. l varies...
l a
------ ------------
10 cm 5.33e6 m/s^2 (544 thousand g)
1 m 5.33e7 m/s^2 (5.4 million g)
So, not only can't Bey survive a 1 km approach, but nothing *but* the
hull can! The entire contents of the hull should be stretched and
compressed until it is a long, thin rod running the length of the ship...
> But note, the escape velocity from there is about half-lightspeed.
> So he's only experiencing those forces for a very small fraction of
> a second. If we take his distance such that he spends a thousandth
> of a second inside the given radius, we still have something like
> 10 thousand g per meter tides (less than a g compression tides),
> meaning a thousand g tides for a thousandth of a second, with a
> shockwave in the middle rising to a million g.
Aside from the errors in computing the tidal forces, you are also
making several assumptions about the orbit.
> Basically, what we've got here is a classic, textbook case
> of "smucker's syndrome".
*This*, I agree with!
> And while I'm doodling around with this, an interesting point occurs to
> me. Consider: when will Bey start to notice the tides?
Assuming he can detect an acceleration of 1 meter/sec^2 at 1/2 the length
of the ship. Hmmm.
a= .05 (remember, it's the *stretching* force we are interested in),
l= 10? (I can't recal the size of the ship, let's call it 20 metters long)
GM we have, we're solving for R.
.05=(1.8e20/R^3)*10
.5=1.8e20/R^3
1=3.6e20/R^3
R^3=3.6e20
R=(3.6e20)^1/3
R=7.1e6
So he can detect the pull at 7 thousand km.
> Formulae (no relativistic effects considered):
>
> escape velocity v=sqrt(2GM/r)
Correct.
> tide(neglect spin) a=(GM/r^2)-(GM/(r+1)^2)
Nope! See above.
Wayne Throop
error Niven (or whomever) made in doing the Smoke Ring calculations.
(See <314365ED...@alcyone.com> for details.)
Some further miscelania.
:: throopw%sheol...@dg-rtp.dg.com (Wayne Throop)
:: Note that tides (to a close approximation) are entirely determined by
:: the period, so exploding the orbits to more reasonable size doesn't
:: affect the tidal calculation.
: Erik Max Francis <m...@alcyone.com>
: Er, watch yourself, there. Tides are proportional to the gradient of
: the gravitational field; in an inverse square field they vary as the
: inverse cube. So increase the distances by a factor of 50 and you
: decrease the tides by a factor of 125 000.
Right, but if you increase the *distances* by a factor of 50, you must
also increase the *velocity* by a factor of 50 to keep the same period.
Since a=v^2/r (or w^2*r with constant w), this implies that the
centripetal acceleration needed for a circular orbit just went up by the
same factor. Since a=GmM/r^2, we have three factors of r to make up,
which means the mass of the primary must go up by a cubed factor.
Thus, while Eric is exactly right that the gradient for a fixed mass
drops off as inverse cube, the gradient for a fixed *orbital* *period*
is constant, because the required mass to keep that orbit exactly
compensates (to a close approximation... if the object is large compared
to the radius to the primary, this breaks down).
:: Note that the k/r^2-k/(r+1)^2 calculates the tides for the given scenario.
: Uh, what? Where does the 1 come from? (That's not dimensionally
: consistent, you know.)
: The proper formulation for a tidal strength is either the differential
: force or acceleration (we'll use acceleration) over the two furthest
: points on the object (in the radial direction), which we will call
: deltar:
: tau o= |g(r) - g(r + deltar)|
: o= M (1/r^2 - 1/(r + deltar)^2).
: deltar must be a physical distance; it can't be a dimensionless
: number. If you take the limit as deltar -> 0, you find that tau o=
: dg/dr, or M/r^3. (Here the symbol `o=' means "is proportional to.")
Exactly right. But I just did a quick and dirty approximation by taking
a delta(r) of 1 meter, instead of doing it symbolically and taking the
limit. That's what makes it dimentionally consistent: that (r+1) IS
(r+delta(r)).... I arbitrarily chose delta(r) to be 1 meter, so that
the answer pops out in terms that can be interpreted as
acceleration-per-meter. I can then approximate the tides on the ends of
trees of various lengths quickly by simply multiplying by this factor.
As I say, a quick and dirty approximation, but dimentionally legitimate.
John Schilling
>schi...@spock.usc.edu (John Schilling) writes:
>> And in fact, there seems to be plenty of water and water vapor around.
>> Helium, we don't really have to worry about. Until someone makes it
>> into a movie, we can freely assume everyone speaks in a squeaky voice.
>Nope. We can't assume that. Besides squeaky voices, heli-ox atmospheres
>are *notoriuos* for one other thing. They conduct heat almost as well
>as water. That's why folks in undersea habitats need to dress so
>warmly, and keep the temp so high. You can die of exposure in a heliox
>atmosphere at "room temp" or even above.
Depends on how far above. I doubt that exposure/hypothermia would be a
problem at 310K regardless of how efficient the atmosphere was at conducting
heat. In fact, a high-conductivity atmosphere would *improve* things at
temperatures greater than that of the human body - less risk of heat
exhaustion and heat stroke.
>This doesn't fit the Smoke Ring. So there isn't a significant helium
>fraction, or staying warm would be a *major* consideration. Fires
>wouldn't help much, because the atmosphere would make it hard to
>concentrate the heat (rubbing sticks would be a *real* chore)
As I don't recall any exact temperatures given in the "Smoke Ring" novels,
we can always assume that the core of the torus is warm enough to allow
humans to live in comfort even in a high-conductivity atmosphere. That
this may be ten degrees or so warmer than the corresponding condition on
Earth, is not a qualitative change and so doesn't contradict the described
environment.
>> I see two possible answers. First, preferential escaped of hydrogen
>> from Goldblatt could have stripped the bulk of that planet's hydrogen
>> atmosphere early in the system's evolution, leaving a gas giant remnant
>> with most of its original oxygen and water but little hydrogen.
>Don't forget that Goldblatt's has to have survived whatever created the
>neutron star. And have been pretty close before that. So it would have
>had a *much* elevated temperature for an extended period before the
>formation, and if the formation involved any sort of "energetic" event
>like a nova or supernova, it'd lose more from the shockwave.
>Start with a "brown dwarf", and then heat it for a few billion years to
>drive of H and He. Then hit it with a good shock. I wonder if the
>remains would resemble Goldblatt's?
>Drive off the free H, and you wind up with water, ammonia, and methane
>(plus various other things in *much* smaller amounts). Allow a torus of
>those to form and photo-dissociate at the edges, and you get nitrogen
>and CO2. Add photosynthesizing life, and you get oxygen, water, CO2,
>nitrogen, and the barest traces of ammonia (ammonia and methane both
>disappear *rapidly* in an oxygen atmosphere).
>So figure that what is being added from Goldblatt's is methane, ammonia
>and water, and it all works out nicely.
I hadn't considered this, actually. But yes, the neutron-star formation
process could well have driven the evolution of Goldblatt in the right
direction for a subsequent oxygen-atmosphere torus.
And since we now have pretty good evidence that, yes, planets *can* survive
to orbit a neutron star, the prospects for the Smoke Ring look somewhat
better. You still need an unlikely set of coincidences for a human-friendly
environment in any particular pulsar/gas giant system, but now at least it
looks like we may have a substantial number of such systems to work with.
Wayne Throop
:: The 1-km-altitude flyby, then, should have subjected Bey to better
:: than 10 million gravities across the 10 cm of his chest cavity. The
:: compression forces on his head and feet should (IIRC) be about
:: ((1/2)*h)/r of the front/back tides, or in this case about 60g.
: sha...@krypton.rain.com (Leonard Erickson)
: Well, for one thing you forget that the "compression" force is along
: *all* directions that are at right angles to the radial axis that the
: "stretching" forces act on.
I didn't forget it. True, I didn't mention it. But I didn't forget it.
The forces side-to-side would be different than those on the head/feet,
simply because the distance is different. I needed to specify that the
numbers I gave were for the head/feet. I didn't give numbers for
side-to-side, since he was already jam.
: Also, you are (incorrectly) assuming that the tides are *just* the
: difference in g-force between the two points. They aren't. Those are
: the forces that would act on a body *hovering* at those distances.
: The tidal forces are caused by the fact that point at anyplace other
: than the center of mass would "like" to be travelling in a different
: *orbit*.
Nope. The (gravitational) tides *are* simply the difference in g-force.
The effects of "wanting to be in a differen orbit" are moot.
Now, tidally locked objects in near-circular orbits also experience
a spin component. But that's another issue, one I was explicitly
ignoring (not "incorrectly assuming") in this case.
: The formula for tidal force is: a=(GM/R^3)*l
Nope. That's 2GMl/r^3, I'm pretty sure. It's just a calculation of
differential gravitational acceleration, and it gives nearly the same
answer I got. Which isn't surprising, since it's derivable from the
formula I used, under the assumption that l<<r.
But note I said "10 million gravities across the 10 cm of his chest cavity".
I got confused between m/s^2 and gravities; that should be 1 million, not 10.
Still quite fatal.
: So, not only can't Bey survive a 1 km approach, but nothing *but* the
: hull can! The entire contents of the hull should be stretched and
: compressed until it is a long, thin rod running the length of the
: ship...
Why a rod? Sure, there are compression forces (of 60g/m or so)
but they won't form a rod against a million gravities/meter
for-and-aft-ward. We end up with a puddle of metal and gore
in the bow and stern.
Further, note, that million gravities/meter only persists for
a microsecond or so. It'd come on like an incredibly strong shockwave.
So perhaps neither rod nor puddle: think "shrapnel". You end up with
floating gore-coated shards.
: Assuming he can detect an acceleration of 1 meter/sec^2 at 1/2 the
: length of the ship. Hmmm. a= .05 (remember, it's the *stretching*
: force we are interested in), l= 10? (I can't recal the size of the
: ship, let's call it 20 metters long) GM we have, we're solving for R.
Um... we're talking about Bey, strapped in at the nose, dropping
something, and having it go 1m/s^2 (a tenth-g). Whence this .05, then?
The "a" we're looking for is 1.0 per half-ship-length, not 0.05.
: .05=(1.8e20/R^3)*10
: .5=1.8e20/R^3
: 1=3.6e20/R^3
: R^3=3.6e20
: R=(3.6e20)^1/3
: R=7.1e6
: So he can detect the pull at 7 thousand km.
Also note, the second line should be dividing both sides by 10. That
should yield .005 on the lhs, not .5; carrying on from there (and
forgetting about disputed factors of 2 (ie, GM/r^3 vs 2GM/r^3)), we get
r=((1.8e20/1.0)*10)^(1/3)=1.2e7 meters, or a bit more than 12 thousand
kilometers. If we use 7.1e6, we get a=(1.8e20/7.1e6^3)*10=5.03 m/s,
or about a half-g, not a tenth-g.
ANYhow, 12,000 kilometers is 2 seconds warning.
7,000 kilometers is a tad less than 1 second warning.
I was presuming the ship was perhaps ten times longer, and Bey could
detect 1/100g at the nose. Thats more like
r=((1.8e20/0.1)*100)^(1/3)=5.6e7 or 56 thousand kilometers.
This would give something like 15 seconds warning.
If he can detect .001g (.01m/s) tide over a 100 meter distance,
r=((1.8e20/0.01)*100)^(1/3)=1.2e8m, or close to 60 seconds warning.
Leonard Erickson
> : Also, you are (incorrectly) assuming that the tides are *just* the
> : difference in g-force between the two points. They aren't. Those are
> : the forces that would act on a body *hovering* at those distances.
> : The tidal forces are caused by the fact that point at anyplace other
> : than the center of mass would "like" to be travelling in a different
> : *orbit*.
>
> Nope. The (gravitational) tides *are* simply the difference in g-force.
> The effects of "wanting to be in a differen orbit" are moot.
> Now, tidally locked objects in near-circular orbits also experience
> a spin component. But that's another issue, one I was explicitly
> ignoring (not "incorrectly assuming") in this case.
Wrong. Read Dr. Forward's article on neutralizing tidal forces (for the
purpose of getting "real" zero g while in orbit). The tidal forces
*are* due to orbital considerations as well.
> : The formula for tidal force is: a=(GM/R^3)*l
>
> Nope. That's 2GMl/r^3, I'm pretty sure. It's just a calculation of
> differential gravitational acceleration, and it gives nearly the same
> answer I got. Which isn't surprising, since it's derivable from the
> formula I used, under the assumption that l<<r.
You didn't read the rest of it. The "a" is the *compression*
acceleration. The "stretching" aceleration is 2a.
See the aforementioned article (page 162 in the paperback edition of
"Indistinguishable From Magic).
> But note I said "10 million gravities across the 10 cm of his chest cavity".
> I got confused between m/s^2 and gravities; that should be 1 million, not 10.
> Still quite fatal.
>
> : So, not only can't Bey survive a 1 km approach, but nothing *but* the
> : hull can! The entire contents of the hull should be stretched and
> : compressed until it is a long, thin rod running the length of the
> : ship...
>
> Why a rod? Sure, there are compression forces (of 60g/m or so)
> but they won't form a rod against a million gravities/meter
> for-and-aft-ward. We end up with a puddle of metal and gore
> in the bow and stern.
Sorry, but the compression forces are *not* any measly 60g/m. At any
given distance they are *one half* the stretching forces. So youyou've
got o million gees trying to pull twoards the nose and tail, and a half
million trying to *push* towards the nose-tail line.
> : Assuming he can detect an acceleration of 1 meter/sec^2 at 1/2 the
> : length of the ship. Hmmm. a= .05 (remember, it's the *stretching*
> : force we are interested in), l= 10? (I can't recal the size of the
> : ship, let's call it 20 metters long) GM we have, we're solving for R.
>
> Um... we're talking about Bey, strapped in at the nose, dropping
> something, and having it go 1m/s^2 (a tenth-g). Whence this .05, then?
> The "a" we're looking for is 1.0 per half-ship-length, not 0.05.
Sorry. It should be 0.5. Remember, the strteching force is 2a, not a!.
Here's a re-calc, given the 300 foot (call it 100 meters) length of the
ship someone else quoted:
.5=(1.8e20/R^3)*50
.01=1.8e20/R^3
1=1.8e22/R^3
R^3=1.8e22
R=(1.8e22)^1/3
R=26.2e6
So he can detect the pull at 26 thousand km.
Erik Max Francis
> Facinating: I'd say Eric has indeed stumbled on the
> error Niven (or whomever) made in doing the Smoke Ring calculations.
> (See <314365ED...@alcyone.com> for details.)
This makes sense. After all, it's clear that his figures are _quite_ wrong, so
it was probably some systematic (and probably trivial; Niven knows his stuff,
he just made a mistake somewhere and never noticed it) goof with the equations.
> Thus, while Eric is exactly right that the gradient for a fixed mass
> drops off as inverse cube, the gradient for a fixed *orbital* *period*
> is constant, because the required mass to keep that orbit exactly
> compensates (to a close approximation... if the object is large compared
> to the radius to the primary, this breaks down).
Eh? How do arrive at this conclusion? Take it from the beginning:
For a (circular) orbit, centripetal acceleration a equals gravitational
acceleration, and we use this to derive Kepler's second law (even though here
we only derive it for circular orbits):
a = g.
a = omega^2 r [omega, angular speed; r, orbital radius] and g = G M/r^2 (G,
gravitational constant; M, mass of primary (Voy)], so
omega^2 = G M/r^2.
We can write omega = 2 pi/T [T, orbital period], and get
4 pi^2 r/T^2 = G M/r^2,
and associate terms to find
4 pi^2/T^2 = G M/r^3.
People should notice that this is Kepler's second law "upside-down." Now we
have a nice equation relating period T, mass of primary M, and radius r.
Now we find the expression for the gravitational grradient, grad g. In
one-dimension (which is what we're dealing with), this is simply dg/dr, and we
find that
grad g = -2 G M/r^3.
We take our modified form of Kepler's law above, noting that G M/r^3 appears in
both the equations here, and substitute:
grad g = -2 (4 pi^2/T^2)
= -8 pi^2/T^2.
That is, the gravitational gradient is proportional to the inverse square of
the period of the orbit. That's a far cry from "approximately constant."
(Unless, of course, you were considering the period only over a very short
interval, such that grad g[T_1} ~= grad g[T_2]. But then that works for any
function, so it doesn't really tell us much; and for the variances in orbital
periods that we've been encountering, this _definitely_ doesn't hold.)
> Exactly right. But I just did a quick and dirty approximation by taking
> a delta(r) of 1 meter, instead of doing it symbolically and taking the
> limit. That's what makes it dimentionally consistent: that (r+1) IS
> (r+delta(r))....
Well, that's fine. But 1 is not the same as 1 m. (r + 1 m) is dimensionally
consistent; (r + 1) is not.
--
Erik Max Francis &tSftDotIotE && http://www.alcyone.com/max && m...@alcyone.com
San Jose, California, U.S.A. && 37 20 07 N 121 53 38 W && the 4th R is respect
H.3`S,3,P,3$S,#$Q,C`Q,3,P,3$S,#$Q,3`Q,3,P,C$Q,#(Q.#`-"C`- && 1love && folasade
Omnia quia sunt, lumina sunt. && Dominion, GIGO, GOOGOL, Omega, Psi, Strategem
Triple Quadrophenic
Treitel) says...
>
>To my surprise and delight, Erik Max Francis <m...@alcyone.com> wrote:
>
>>According to "Neutron star," BVS-1 is 1.3 masses solar, and at closest
approach
>>the _Skydiver_'s "nose was just seven miles from the center of BVS-1.
>>
>
>uh, ketchup.
>
Center != surface on planet Earth where the rest of us come from.
--
-- BEGIN NVGP SIGNATURE Version 0.000001
Frank J Hollis, Mass Spectroscopy, SmithKline Beecham, Welwyn, UK
Frank_H...@sbphrd.com or fj...@tutor.open.ac.uk
All Opinions My Own (So My Employer Tells Me)
Wayne Throop
:: while Erik is exactly right that the gradient for a fixed mass drops
:: off as inverse cube, the gradient for a fixed *orbital* *period* is
: Erik Max Francis <m...@alcyone.com>
: Eh? How do arrive at this conclusion? Take it from the beginning:
: grad g = -2 G M/r^3.
: We take our modified form of Kepler's law above, noting that G M/r^3
: appears in both the equations here, and substitute:
: grad g = -2 (4 pi^2/T^2)
: = -8 pi^2/T^2.
: That is, the gravitational gradient is proportional to the inverse square of
: the period of the orbit. That's a far cry from "approximately constant."
Yes. Tides are proportional to inverse square of the orbital period.
Therefore, Erik has just proven that, given a constant orbital period,
the tides are likewise constant. You will note that any dependence
on mass or distance has fallen out: once you know the orbital period,
you know the tide, EVEN IF you don't know the radius. Just as I said.
Note well: I said FOR A FIXED ORBITAL PERIOD, tides are (also) fixed.
You can vary r and M around all you want, but since orbital period
T = sqrt( 4pi^2r^3/GM )
you must balance the effects of r^3 and M on tides, to keep constant
orbital period T. Also just as I said.
Thanks to Erik for confirming what I said more rigorously.
Wayne Throop
::: you are (incorrectly) assuming that the tides are *just* the
::: difference in g-force between the two points. They aren't. Those
::: are the forces that would act on a body *hovering* at those
::: distances. The tidal forces are caused by the fact that point at
::: anyplace other than the center of mass would "like" to be travelling
::: in a different *orbit*.
:: throopw%sheol...@dg-rtp.dg.com (Wayne Throop)
:: The (gravitational) tides *are* simply the difference in g-force.
: sha...@krypton.rain.com (Leonard Erickson)
: Wrong. Read Dr. Forward's article on neutralizing tidal forces (for
: the purpose of getting "real" zero g while in orbit). The tidal
: forces *are* due to orbital considerations as well.
OK. I've read it. Forward explains it in terms of orbits, then says
There is an alternate way of looking at the same effect that uses
the concept of gravity gradients, or the change of the gravity field
of the [primary] with distance. --- Forward, IfM, p 159
Since it is an ALTERNATE view of the SAME EFFECT, then using a formula
that captures the gravity gradient is exactly sufficient to account
for the entire effect. And indeed, the formula Forward gives is
exactly the gradient of g, derivable from the formula I used
to start with. [1]
In fact, he then says (on p 160) "This gravity gradient or differential
acceleration effect is better known to you as the tidal force."
I think it's clear that Forward agrees with me;
tides are differential gravity.
:: Why a rod? Sure, there are compression forces (of 60g/m or so) but
:: they won't form a rod against a million gravities/meter
:: for-and-aft-ward. We end up with a puddle of metal and gore in the
:: bow and stern.
: Sorry, but the compression forces are *not* any measly 60g/m. At any
: given distance they are *one half* the stretching forces. So
: you've got a million gees trying to pull twoards the nose and tail,
: and a half million trying to *push* towards the nose-tail line.
Ah, I see; I was indeed miscomputing the compression; as Forward
points out (and is clear from doing differential gravity via vector addition)
it's half the radial tide. [2]
So, at 15km distance, stretched by 2GM/r^3 = 10 million g/m,
squished by GM/r^3 = 5 million g/m.
Nevertheless, the question remains... "why a rod?".
Sure, there are huge compression forces, but the ends of the rod are still
being pulled at 10 million g/m; it will still break, and form puddles in the
fore and aft of the ship. The puddles are heaped up a bit more than I
expected, but puddles fore and aft nevertheless. Not a rod, no way, no how.
Especially consider: as the compression forces squish centerward, they
lose the factor of "l" in their formula (it gets smaller), but as the
streching forces pull, they *increase* their force (l gets larger).
Again, this implies puddles, not a rod.
:: Whence this [0.5], then? The "a" we're looking for is [1.0 m/s^2] per
:: half-ship-length, not [0.5].
: It should be 0.5. Remember, the strteching force is 2a, not a!.
Ah, I see. Instead of doing acceleration-bey-can-detect = formula,
you do half-acceleration-bey-can-detect = formula-for-half-tides.
An odd, but correct, approach, methinks.
So, yes, if Bey's ship is 100 meters long, that's 1.0=50*2GM/r^3=1.8e22/r^3,
or r=(1.8e22)^(1/3)=2.62e7. That gives something under 5 seconds warning.
--
Wayne Throop throopw%sheol...@dg-rtp.dg.com
thr...@aur.alcatel.com
--
[1] Sloppy derivation, under presumption that l<<r
If we start with GM/r^2 - GM/(r+l)^2 for radial tides, one can derive
(GM(r^2+2rl+l^2) - GMr^2) / r^2(r+l)^2
GM(2rl+l^2) / r^2(r+l)^2
(l<<r) -> 2GMrl/r^4
(2GM/r^3)*l
which is exactly Forward's formula for radial tides.
[2] Basically, we're off-center by distance l, thus
the centering force is going to be the sin(a) times the gravity
(where a is the angle subtended by l from the center of mass of
the primary, which for small angles we can do via l/r, so that's
GM/r^2 * l/r = (GM/r^3)*l, which is exactly Forward's formula
for transverse tide.
Erik Max Francis
> Thanks to Erik for confirming what I said more rigorously.
Ah. So you were saying that tides are a unique function of the orbital
period; the other orbital factors do not enter into it. Well, I'd certainly
agree with that.
(Actually, that statement is not even true in general, but we're only
concerning ourselves with elliptical orbits.)
--
Erik Max Francis &tSftDotIotE && http://www.alcyone.com/max && m...@alcyone.com
San Jose, California, U.S.A. && 37 20 07 N 121 53 38 W && the 4th R is respect
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Leonard Erickson
> :: Whence this [0.5], then? The "a" we're looking for is [1.0 m/s^2] per
> :: half-ship-length, not [0.5].
> : It should be 0.5. Remember, the strteching force is 2a, not a!.
>
> Ah, I see. Instead of doing acceleration-bey-can-detect = formula,
> you do half-acceleration-bey-can-detect = formula-for-half-tides.
> An odd, but correct, approach, methinks.
Remember, Bey can detect the "compression" forces too. In fact, I rather
suspect that they are part of what *starts* the ship turning. Figure
that it is oriented not *quite* tangent to the orbit. So the
"compression" forces start the turn! Once it gets past 45 degrees, then
the "stretch" forces will speed things up. It might be interesting
doing a plot of the forces with the ship in various orientations.
> So, yes, if Bey's ship is 100 meters long, that's 1.0=50*2GM/r^3=1.8e22/r^3,
> or r=(1.8e22)^(1/3)=2.62e7. That gives something under 5 seconds warning.
Or in other words, we get "Bey puree". :-)
Pity.