-Z- (Z@Gundam.Com)
Sat, 03 Jun 2000 11:48:56 -0700

At 19:52 6/1/2000, you wrote:
>At 19:26 05/31/2000 -0700, -Z- wrote:
>>The position of the Lagrange points is determined by the positions of the
>>Earth and Moon,
> [snipped]
>
> Thanks for the info.. but that still doesn't quite answer my
>question (either that or I'm too dense to see it =). AFAIK, L1 amd L2 are
>also fixed points in relation to the Earth and the Moon, so won't these
>points move at the same speed as the Earth-Moon couple?

It's more correct to say that all five Lagrange points maintain the same
distance and vector (and, thus, a constant rate of change in position)
relative to the Earth, Moon and one another, which has the effect of making
their positions fixed relative to one another.

The Moon, L3, L4 and L5 are all moving at the same speed, because they all
transit the same orbit in the same period. L1 is in a lower orbit, with a
correspondingly shorter circumference, but has the same period, so it's
moving correspondingly slower; L2 is in a higher orbit, with a larger
circumference, but again the same period, so it's moving that much faster.

The Lagrange points don't exist as physical bodies, but rather as points of
equilibrium, like your own center of gravity. You can't tattoo a spot on
your spine and call it the center of gravity, but you can feel that point
-- and its shifts in position -- as if it were a physical part of you. You
can also predict where it will move as you shift your mass around. Move
that center of gravity and you shift the mass, as every gymnast and martial
artist knows.

> (Of course, when you consider the gravitational effects of other
>planets and the sun, L1 and L2 is probably going to change a bit.)

Yes, the Sun throws a few perturbations into the orbits, but its effect on
the Lagrange points of the Earth/Moon system is negligible. There is, in
fact, a set of Lagrange points associated with each of the planets, with
the Sun acting in place of the Earth and the planet in place of the Moon,
which is how Lagrange's theory was confirmed. Asteroids were observed in
the Lagrange points leading and trailing Jupiter. The individual bodies
were named for the heroes of the Illiad, with the Trojans in one group and
the Greeks in the other, but eventually they all became known as the Trojan
asteroids. (There's one Greek among the Trojans and one Trojan among the
Greeks -- spies for the other side!)

>>Not only a sufficient distance, but a fixed distance. O'Neill originally
>>specified that the cylinders would be ballistically coupled. This would
>>set the cylinders in each pair about 70 to 80 km (45 to 50 miles) apart.
>
> I understand that the distance must be fixed, I just didn't make
>that clear in my post. Pardon my ignorance, but why the great distance? 70
>to 80 km apart from each other seems a trifle excessive.

Well, the mirrors stick out quite a ways and effectively "widen" the colony
by a factor of ten. (^_^) This near-invisible "width" is reinforced by
the farming ring.

The space colony isn't just the main cylinder. There's also a ring of 72
"hatbox" cylinders with parabolic mirrors to concentrate sunlight within
them. Each acts as a greenhouse -- this is where the colonists grow their
food. The idea is that all industry and agriculture (and their
accompanying pollution) takes place outside the main cylinder, which is
exclusively residential.

Each agricultural blocks or farming satellite (farmsat) is about 645 km
(2,110 feet) across, but is encompassed by a parabolic mirror about 1.3 km
(4,265 feet) across, so the ring would have a minimum circumference of 95
km (60 miles) and thus a diameter of about 30 km (18.6 miles). Per the
original O'Neill design, the ring would have a diameter of exactly 32 km
(20 miles) and serve as the forward attachment points for the mirrors, but
this introduces an apparently unforeseen problem if one is, indeed, using
the satellites as farms -- they're going to be pulling considerable
Coriolis forces if they are actually rotating in synch with the main
cylinder, which they'd have to do if (but only if) the mirrors were
attached to them.

(In Gundam, the ring is shown at the back end more often than not,
presumably because it looked better. In High Frontier, however, it's
always on the front end and always looks conjoined with the mirror tips.)

In any case, the cylinders would have to be at least 30 km apart -- the
combined radii of the two rings -- to keep the farmsats from
colliding. (Unsupported mirror tips could be offset and even interleaved.

But it's a bit trickier than that. (Ain't it always?) The ballistic
paired cylinders are actually orbiting around a common center of gravity
midway between the two, although the period is so long that it's not
noticeable to the human eye. The minimum separation is equivalent to a
colony-width (or, as we've just seen, nearly a colony-length) in each
direction. That sums up to 90 km (55 miles) -- 10% more than my ballpark
figure.

>>I think you misunderstood me. The rotation of the colonies is fixed by
>>Newtonian law -- with a radius of 3.2 km (2 miles), you need exactly
>>one-half RPM.
>
> Oh, I *definitely* misunderstood you. (^_^) But my question stands:
>won't the collision of the two cylinders, anime physics being in force,
>cause the two cylinders to spin at a rate higher than 0.5 RPM? Granted, the
>collision may not cause such a rapid rotation, but it is possible.

The collision shouldn't have any discernable effect on rotation -- these
are massive objects with considerable inertia and the brief friction of
contact won't be sufficient to add or subtract to the rate of rotation.

What WILL have an effect are the loss of mass of the mirrors, which would
indeed cause the cylinders to rotate marginally faster. What would have
the greatest effect would be the COLLAPSE of all three mirrors and the loss
of the farmsat ring. This would be a significant loss of mass, but that's
not what would cause the colonies to speed up.

Imagine that the colony is an ice skater doing a spin with her arms
extended. Taking off the mirrors and ring would have the same effect as
the ice skater pulling her arms in. She doesn't lose any mass, just
redistributes it, but the effect is dramatic.

The same thing would also happen if the mirrors and ring separated during a
drop.

> In such a case, how would the structure of the cylinder hold up? I
>don't think the cylinder will take some damage due to the rapid spin, so
>it's probable that the Delz faction would stop the spin, or at least slow it
>back to 0.5 RPM so that the stress on the cylinder would be lessened.

Again, a change in spin is only going to be stressful if it's uneven --
it's not the change in rotation per se but the uneven distribution of
change in rotation that causes the torque. You get torque during re-entry
because the leading end of the colony is subjected to more atmospheric
resistance than the trailing end.

-Z-

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