Hi wulf
The question, as posed, may be interpreted in several ways.
I am posting this answer based on simplest interpretation in the hope
that this answer will help clarify the terms and physics involved.
I will also provide reference to websites which deal with these issues.
This may lead to more specific, more narow question.
Numerous discussions of these issues and data are to be found under this:
SEARCH TERM : space elevator
Space Elevator (popular overview, feasibility, cost)
http://www.space.com/businesstechnology/technology/space_elevator_020327-1.html
Space Elevator is a cable (whose center of gravity is ) in the
geosynchronous orbit. It extends up from that elevation to a
counterweight, and down, usually all the way the surface (where it can
be tethered). It can be made from a single wall nanotube, multiwall
tube or from from many nanotubes glued or sintered together.A single
nanotube can be made with any diameter. It is just more dificult to
make them lerger.
The thickness of the elevator cable will not be uniform (some
elevations expose cable to damage from micrometeorites and need to be
thicker)
That (the meterorite damage) is more serious factor determining
thickness of the cable just the ability to withstand the tension (for
practical case).
Here are two sites which describe the cable design considerations:
Cable design and production - micrometeorite damage - counterweight
http://www.isr.us/Downloads/niac_pdf/chapter2.html
Self-Assembling Nanotubes Make Space Elevators Possible
... the path of meteoroids and space debris, a ... manufacturing
techniques for single-walled
carbon nanotubes. ... technologies might render the elevator obsolete before ...
www.popularmechanics.com/science/ space/2002/7/going_up/print.phtml
And few more references are here:
http://en.wikipedia.org/wiki/Space_elevator
So, the conclusions are: The weight of such a cable is close to zero.
Depending on details of construction, (making the lower end thicker,
counterweight larger) we can make the weight a bit negative (so that
it will pull up at the thether). That will help stability.
The other design, (positive weight) would not be a good design, since
nanotubes are good in tension, not in buckling strength.
Here we use term 'weight' in technical sense = a total force of
gravity, integrated over the whole body.
Top part (above GEO) pulls up, the bottom part pulls down. Result is
zero, in the same sense in which a satelite in free orbit has zero
weight and objects in orbiting shuttle are weightless.
Perhaps you mean mass of the cable: that would be length * crosssection * D ,
where D, density, depending on the type of nanotube, is comparable
to that of diamond - about 2267 kg/m3,
http://en.wikipedia.org/wiki/Carbon_nanotube
http://en.wikipedia.org/wiki/Carbon
The GEO is at r=42 Mm (Mega-meters) above the center of the Earth as
calculated here:
http://scienceworld.wolfram.com/physics/GeostationaryOrbit.html
The velocity v there is (omega * r) = 40e6/86e3 = about .5 km/s
You specify velocity of 2.3 km/s, so (using same formulas) the upper
end of you cable is at 5* r ~ 200 Mm = 200 E6m.
Critical question is the tension in your cable, given by weight of the
upper portion (pulling up) and weight of the lower portion (pulling
down) over the crossection of the cable (A).
The mass of the upper portion is A* D * 4 *r, but what is it's weight?
It depends on geometry of the cable. Here is one practical estimate
for 100 Mm upper end:
"..Ultimately, the craft carrying the end of the ribbon will be parked
in an orbit 100,000 km from Earth. Edwards calculates that this
ribbon, several micrometers thick and 20 to 40 centimeters wide, could
support a load of 1,800 kilograms..".
http://www.sciencenews.org/articles/20021005/bob9.asp
The dependence of weight is less then proportional to r, however as a
rough estimate, we can double the A to obtain the sufficient
crossection,
for r=200 Mm. required crossection are here;
http://www.zadar.net/space-elevator/#estimate
hedgie |
Request for Answer Clarification by
wulfhawk-ga
on
27 Jun 2004 22:08 PDT
Not exactly what I was looking for. I may be able to find the
equations in the links you provided, but let me try to refine a
little.
A lunar 'skyhook' is not practical because of the moon's slow spin;
centrifugal force/orbital velocity doesn't balance gravity until the
cable is hundreds of thousands of miles long. Instead, I was struck
by the idea that the moon's lack of atmosphere and low gravity might
allow a different use of centrifugal force.
A very long counter-weighted cable spinning horizontal to the lunar
surface could launch cargo, literally whipping it into orbit. With
the tip-speed at orbital velocity plus, the tip of the cable would
rise and pass miles above the lunar surface, dragging the rest of the
cable with it. The hub base would be located on a mountain or
plateau. So, to understand the scale of the hub base, I need to know
how big the counter-weight is, which means I need to know how much the
cable will mass. I chose 100 miles as a reasonable length. Knowing
the diameter or cross-section of the cable lets me feel the scale of
the cargo launch 'pods'.
I get the intuitive feeling that the varying cross-section required by
cable tens of thousands of miles long will not be important in this
case. My math skills and engineering knowledge are not up to this
calculation in any case.
|
Clarification of Answer by
hedgie-ga
on
28 Jun 2004 18:33 PDT
Wulf, hi
Thanks for the clarification. It always helps to know if client is
looking for set of formulas or some already calculated rules of thumbs.
Now we will look for the later, something like:
The skyhooks' size, and very feasibility, is extremely
sensitive to the strength of the material one proposes to use. A
synchronous cable made of steel has to be $10^{50}$ times bigger in
the middle than at the ends and weigh $10^{52}$ times what it can
support.
http://www.frc.ri.cmu.edu/~hpm/project.archive/general.articles/1987/skyhook.ltx
When we go from steel, to kevlar, to nanotubes, the critical material
parameter M=the density/(tensile strength)
is improving (getting smaller) and so the
P = performance =(payload mass/mass of the cable)
is getting better (getting larger).
So, we can look for this P(M) dependence. Dr. Moravec, above says:
"..Per unit weight, Kevlar is 5 times stronger than steel. A
synchronous Earth skyhook made of it needs a taper of $10^{10}$ (ten
billion), and weighs $10^{13}$ times what it can lift. Better than
steel, but still ridiculous. Single crystal graphite whiskers with 50
times the strength to weight of steel have been grown in
laboratories. Bulk material as strong would permit a synchronous cable
with a taper of only 10, and a mass ratio of 400. Half this strength
is perfectly adequate for an Earth synchronous skyhook.."
This is for a GEO cable, and the profile of the cable A(l) -- the crosssection
as a function of the length is optimised (and so changing ).
It is not clear from your description where the cable is. Is it tied
to the surface of the moon, of the earth, or is the center of gravity
at one of
the lagrangian points L1 to L5 ?
We can either tie one end to a body (moon, earth .. ) or we can have a cable
as it may not be an ellipse. Would probably require a computer model.
It is not complex (but more then $20) if not already done. The cable
not tied to the Earth is complicated, but it would be an interesting
combination of the
sub-orbital fligt idea and of the skyhook idea.
The other interesting idea is be cable with center of gravity at one of
the stationary points of the Earth-Moon system. Are you thinking about a
system where one end would be close to Earth surface end other to the Moon
surface, close but not fixed (tied to it?), close fromtime to time?
A generalised rule may consider a center of gravity around some (possibly
non-Keplerian orbit) and subject to tensile stress, as both ends would
prefer some other orbit. The stress is form of tidal forces, and those
would depend on local gradient of gravitational field at that orbit.
Meaning: at the average point of the orbit of the center of the gravity,
one would go 1km in direction of one end and see hoe much g has changed.
Here g(r) is intensity of graviational field due to the Earth and Moon.
For the order of magnitude calculation, one can assume that anywhere in
the Moon-Earth system this gradient would be less then the GEO orbit,
and so calculation of Moravec (above) can be used as a worst case estimate.
Gravity gradient around the L points, compared to gradient at GEO may be an
interesting, but different search or calculation.
Hedgie
|
Clarification of Answer by
hedgie-ga
on
28 Jun 2004 18:47 PDT
Here is a presentation of L points and tides
http://astro.pas.rochester.edu/~aquillen/ast111/Lectures/Lecture3a.pdf
Roche lobe (explained there) can be seen as another rule of thumb,
as 'tidal distruption' -- dependent on density and tensile strength and
size of an object inorbit -- is the same mechanism, which also breaks
the cable of the space elevator. Interesting and not obvious.
Hedgie
|
Request for Answer Clarification by
wulfhawk-ga
on
28 Jun 2004 20:14 PDT
This fictional lunar based launch system would be located on the lunar
surface, on a mountain peak or plateau. A massive solar-powered hub
would spin a single very long cable parallel to the lunar surface,
with a tip velocity of lunar orbit speed or even lunar escape speed.
Payloads would be launched by 'hooking' to the cable at the relatively
slow-moving hub, then allowed it to whip out the length of the cable.
Like the proverbial bucket on the string swinging around your head,
the cable would travel within a circle horizontal to the lunar
surface; gravity would draw it down, centrifugal forces would draw it
out, and the part of the cable travelling at orbital velocity plus
would pull it up--keeping the entire cable spinning at a safe altitude
miles above the lunar surface.
I appreciate your efforts, hedgie, and sorry for not being more clear.
Specifically, I need a) the estimated mass of the above cable--100
miles long, constructed of carbon nanotube in a bonding matrix,
attached to the hub at one end, travelling at 2.3 kilometers a second
at the other end; b) the estimated cross section of the cable,
considering the loads mentioned above and the launch of worthwhile
payloads from the lunar surface into lunar or even earth orbit.
I'm afraid none of the links you've come up with has supplied these answers.
|
Clarification of Answer by
hedgie-ga
on
29 Jun 2004 01:37 PDT
Wulf,
We are making progress but I am still mystified.
".. the cable at the relatively slow-moving hub, ..".
I understand the hub is on the moon. One end, botom end of the cable
, is tied to the hub ,permanently, right? Hub is slow moving with
respect to what?
Is the hub on some rails which go around the moon? Or are you working
in the frame of reference tied to the Earth-Moon system. It would help
if you can make a drawing. I can point you to some server with free upload
if you do not have one handy. What is moving 'parallel to the moon surface',
that is horizontaly (in the moon coordinate system)? The other (top) end
of the cable?
We can build 'space elevator on the moon'. I am sure those calculations
were made and we can find it. But there still are some constraints. Like
on the Earth, the Center of Gravity of the cable is in synchronous orbit
- the moon-synchronous - in this case and things (gliding on the cable)
will fall back to the moon when UNDER that point (CoG) and 'fall up' to
the sky ABOVE that. It seems here our pictures agrees on this point.
But then I get wondering again when you say:
"keeping the entire cable spinning at a safe altitude".
I thought bottom is tied to the hub. Did you untie it and placed the cable
to the low moon orbit?
It is not clear waht you mean by 'spinnig' - around its own axis
(which one?) or orbiting around the moon.
Also, we would need to resolve the issue of the optimal profile A(l).
Normally, that profile is optimised - to get best possible payload/mass ratio.
If you 'based on your intuition' relax this conditin, that there is no
unique answer. You would have to specify what the profile is, e.g. uniform,
but it wouls make the (mass of cable)/payload much larger, for sure.
To conclude,
SEARCH TERM is
lunar space elevator
do not use word moon, since engines ignore 'to' in 'to the moon'
and give you too many articles on the usual space elevator.
I suggest you enroll the google group:
http://groups.yahoo.com/group/space-elevator/messages/2201?viscount=100
where those issues are actively debated.
hedgie
miles above the lunar surface
|
Request for Answer Clarification by
wulfhawk-ga
on
29 Jun 2004 16:12 PDT
Imagine the hub of the launcher is a helicopter sitting on the surface
of the moon, and the cable/counterweight is the spinning blade. The
intent is to use the spinning momentum to fling payloads off the
surface of the moon. Electricity from solar cells would power the
motors that build the momentum. To save money, it uses only one of
the expensive supercables; the counterweight would be steel and
aluminum, including a reel to recover and relaunch the cable as
needed.
|
Clarification of Answer by
hedgie-ga
on
30 Jun 2004 08:03 PDT
Well liner and wulf
Perhaps your are right that I am a bit behind in the sci-fi
but may be you guys are not reading enough real plans and claculations
made for real (well not real yeat, but more likely to be build) launchers.
To take materials from moon to build L5 colony was not just proposed
but also worked out by O'Neil, long time ago
In my (not so humble opinion) the regular launcher (mass driver or gun)
O'Neil proposed is more sensible that your monster helicopter blade.
I realize I am out of order here, as you did not asks if that is a good system.
I will come to answering your question in just a minute. But first I
will suggest you look at
http://www.amazon.com/gp/browse.html/104-9988693-7497529?node=1036592
And (of course) Clarke described a launche 35 years ago.
http://www.islandone.org/APC/Catapults/04.html
I now see that you are thinking about sling-launcher, as described e.g. here
http://yarchive.net/space/exotic/sling_launcher.html
Your answer is here (quote from the above):
"..Total tension divided by area is tensile stress. The best tether material
on the market today is Spectra 2000, with a tensile strength of 3.25 GPa
and a density of 0.97 g/cc. Putting those values into the equation, we
get:
Vtip = 2.59 km/s.
So, the maximum velocity we can obtain is 2.59 km/s. Any higher tip
velocity and the tether will tear itself to pieces. Notice that this
result is independent of the tether length or angular velocity.
An interesting possibility is using buckytube cable, with a (theoretical)
tensile strength of 150 GPa and a density of 1.3 g/cc. That would give a
tip velocity of 15 km/s! "
Your mass is rho* R * A - with values and definitions above if A is constant.
This article describes problem with horizontal launches
http://groups.msn.com/DaveDietzler/massdriversetc.msnw
Tapered sling is discussed here:
http://www.islandone.org/LEOBiblio/SPBI1SL.HTM
Hedgie
|
,
3 Comments
)