Space Forum / Space Flight / March 2006

Perhaps a better question than debating about this is, what purpose
would this serve? One can use a standard space elevator to launch to
anywhere, especially if it's close. I think going to Jupiter would
still require a rocket, or at the very least, an ion drive, but still,
the need for these things is greatly reduced. So, what is the point of
a link between the moon and earth?

There is no spot anywhere on the surface of the earth where the moon is
constantly visible.

The earth rotates, and no matter where you are on the earth's surface, at
some time of day (or some time of year in polar regions) the moon will not
be "up".

http://www.nasa.gov/centers/goddard/images/content/97114main_lunar_elevator.
jpg

I don't know what that diagram shows, some context might be useful - except
it shows that the Earth attachment point is at the end of the "tramway", not
at the pole.

And the other end of the cable is not attached to the moon.

JRS:  In article <1143552165.258456.259260@t31g2000cwb.googlegroups.com>
, dated Tue, 28 Mar 2006 05:22:45 remote, seen in news:sci.space.tech,
Cray74@gmail.com <Cray74@gmail.com> posted :

Since the ecliptic is over 23 deg from the equator, and the Moon's orbit
is over 5 deg from the ecliptic, a cable attached at the North Pole
would at times drag over *at least* the upper 28 degrees of Earth, even
neglecting droop. ((( Note : this Spring, the Moon is at times almost
circumpolar from the northernmost UK. )))

That would upset the Inuit, many Russians, Lapps, Finns, Norwegians, and
Swedes, who at times would need to jump over the cable each day.

Better to build a rather tall South Pole and attach the cable to the
top.  Something over 900 km high should suffice.

No, you didn't understand the objection: the plane
of the moon's orbit around the earth isn't close to
the earth's equatorial plane.

That moon orbit plane is only five degrees away from
the plane of the earth's orbit around the sun, and
the earth is tipped 23.5 degrees to that same orbit
around the sun, so earth's equatorial plane is
tipped 18.5 degrees to 28.5 degrees to the direction
of the moon, depending on the moon's orbital
position. As a result, a big shoulder of the earth
will be in the way of your polar-based elevator, a
big fraction of the time, just not all of the time.

[Aren't L1 and L2 dynamically unstable anyway? How
does that work?]

You missed "tramway". That elevator is attached to a
moving earth-end base that apparently circles the
planet along a circumferential line of longitude,
once per month, to keep the base roughly facing the
other end at L1 or L2.

That's pretty unlikely to happen for lots of
reasons: it can't be built entirely over land, would
cross long distances over some _deep_ ocean, ocean
rife with state-sized broken off ice shelves to
knock down any tramway bridge supports. That is not
the kind of thing current human technology knows how
to build or defend, and its tramcar would have to be
moving over it at several tens of miles per hour,
while bearing an enormous weight (and off vertical
strain?).

Compare and contrast to that monster tractor that
hauls a Kennedy Space Center rocket from the Vehicle
Assembly Building to launch.  Was that moving at 3
miles per hour, and much, much less loaded than a
tramcar for the elevator would be?

Even in your picture the problem is evident: the
catenary from L2 toward the north pole (not the one
they intend, which is instead toward the current
tram location in that picture) intersects a _huge_
portion of the solid earth, maybe a thousand mile
roughly secant piece of it, guessing just from
looking. Too lazy; hmm: along the curve of the
earth, that's 28.5 * 60 = 1710 nautical miles, to
18.5 * 60 = 1110 nautical miles, depending on moon
orbital position, but the secants would of course be
a fraction shorter.

Being fixed to the pole means the line from the base
to the moon (which in your picture IIUC would sit
somewhere off screen at the right nearly-center, not
where L1 or L2 is shown) would intersect the solid
earth for a time of half of the moon's orbit. The
catenary you mention would droop lower yet toward
the plane of the moon's orbit, and so increase that
fraction to _over_ half of the moon's orbit, *not*
improve the situation as you suggest.

Or so it seems to me.

To make the elevator miss the moon, the "pad" would
instead be a free-standing cantilever tower able to
lift the attachment point hundreds of miles straight
up, and capable of taking the whole stress of the
elevator attachment pretty much straight sideways.

The result would look much like a strung bow with a
very overlong string, being strongly pulled from a
distance, top half only.

I'm guessing we still haven't reached the needed
strength of material even with nanotubes.

HTH

xanthian.

Alternative suggestion: make a big, jump-rope shaped
loop from earth pole to earth pole, another one from
moon pole to moon pole, and attach the main elevator
between them on sliding attachments at each end.

That still wouldn't work, I'd bet, the jump-ropes
would lean toward the attachment points and still
intersect their base spheres, but it would be much
_closer_ to being possible.

Hmm, could a more "W" or "H" shaped jump-rope on each
end work, with the bottoms of the "W" being the part
attached to the poles, and the loose ends of the "W"
bearing counterweights to warp the middle jump-rope
part high away from poles of the sphere below?

More like this (see in fixed pitch font):

  O<-weight           ^         weight -> O
   \                  |                  /
    \          elevator to moon         /  --
     \ tether         |          tether/  / (nearly
      \         +-----o-----+         /  / straight
(pulls \     __/    slides   \__     /  /   really,
jump-   \  _/   ^    here   ^   \_  /  /       or a
rope away\/      \         /      \/  /      kinked
from     / <------jump-rope------> \ <    catenary,
surface /         catenary          \ \    not this
at     /                             \ \   sharp an
poles)/    +---------------------+    \ \    angle)
     / ___/     cigar-shaped      \___ \ --
pole/_/             earth             \_\ pole
      \___  (to save drawing space)___/
          \                       /
           +---------------------+

What on earth kind of "orbits" would those weights
follow, being under strain but trying to follow an
elevated great circle if let loose? I get seasick
contemplating my first guesses, 3D curves like the
edges of a Pringle's chip parallel to the equator.
I'm _so_ glad I don't have to defend or do the math.

xanthian.