Re: Gravitational field visualization

On Feb 1, 2:58 am, Tim Little <t...@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
On 2009-02-01, IsaacKuo <mech...@xxxxxxxxx> wrote:

Roughly, the gravitational field will be equal to Erik's graph,

Erik's graph was for a solid disk, not an annulus. Particularly near
the inner rim that makes a huge difference.

No, it doesn't make much of a difference if the inner hole
is small.

Also, "roughly" doesn't
cut it here. The potential across the disk needs to be constant to
less than about 10^5 J/kg (10 km altitude).  That's out of 10^12 J/kg
or so of overall potential.

An equipotential surface is simply a surface which is perpendicular
to the gravitational force (including spin pseudogravity).

superimposed with a couple scaled graphs of the ringworld
gravitational field.  The ringworld graphs will be centered on the
thickened areas.

To get an equipotential surface you need to solve a functional
equation for area density.  The exact equation has only the trivial
zero solution if required to be satisfied over the whole width of the
annulus.

You're not taking into consideration spin pseudogravity.

Furthermore, the variation in field strength over the surface also
grows. It does no good to have an approximately constant potential if
near the rim the field is 20 gees.

Within a large portion of the disc, the field strength at the surface
will be roughly constant. Remember, to take into account spin
pseudogravity.

I suspect you're looking at the graphs and thinking you could make it
approximately flat with a few well-placed rings of greater density.
Yes, you can - but not to the one part in ten million required.

I really don't care about the outer 1/3 of the disc. My proposal is
to use a Saturn Ring system instead of a thicker torus in that
region. The mathematics for replacing it with a thicker torus is
more complex, but either way any habitability in that region is
merely a bonus.

It's the inner 2/3 of the disc where the radial component of
gravity is linear that can be counteracted with spin pseudogravity
which is of interest for habitability.

On the inner edge, the problem with a hole is that there's a lot of
pressure trying to push the material toward the center.

On the inner edge the gravity is pulling *outward* (and rather
strongly).

If the hole is small, gravity is pulling outward with an small
amount of force. It's completely overwhelmed by the
pressure of the material around it pushing inward.

 We're not talking about black-hole intensity fields, so
gravity is essentially linear. Hence you could intuitively picture
this result by subtracting the field of the removed smaller disk from
the field of a larger one.

Yes, and this will be a small effect if the "smaller disk" is small.

For an obvious extreme example, consider digging a deep well.
Yes, the removed material will result in some insignificant
gravitational force pulling outward. But this is nothing compared
to the inward pushing force of the well's walls.

Isaac Kuo
.

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