>I made a couple of assumptions about the sphere, and you're right,
>Zeta Cygni WOULD exert a pull. The sphere, however, WOULD also exert
>a pull. Maybe an extremely weak one, like Phobos or Deimos, such that
>without gravity generators you could jump too hard (assuming your
>muscles are used to 1g) or throw something and leave the surface and
>therefore fall under the effects of Zeta Cygni, but there is a stupid
>amount of mass in the sphere, and it would have some influence.

Take a point on the inner surface of the sphere, A. Draw a line L through the center of Zeta Cygni, Z, and A, extending to infinity. Place a plane P normal to L at the point it intersects A. This will divide the sphere into two portions, S and S'. If S is the section of the sphere between A and infinity on L, S' is the larger portion of the sphere.

With me so far? Here's why the sphere exerts no pull on anything inside it: S has a 'positive' gravitational pull G (postitive, in this case, being defined as pulling the pseudo-continent away from the star). But S' _also_ has a gravitational pull, G', and since it's on the opposite side of A, it's pull is negative. S' is, on average, much farther away, but it is _also_ far, far more massive.

The result? G + G' = 0. G' is exactly the same size as G, but its sign is reversed, resulting in no net gravitational pull.

>
>If you want to think about why this is so, look in our own back yard,
>or rather in our orbit. The moon is tidally locked with the earth, so
>that one face of it always faces earth. If what you said was true,
>anything on the earthward side of the moon would automatically fall
>towards earth, since earth has a stronger gravitational pull than the
>moon. Since this doesn't, in fact, happen, then the mass of the moon
>must be able to keep stuff firmly attached to the surface.
>

The difference is that the moon is not a shell around the Earth.

>You were right, all mass centers on ZC. From the OUTSIDE. Beyond
>that is where conventional thought gets a little screwy, both because
>of the sheer scale and because conventional thought was never intended
>to study or explain a dyson sphere.

No, it has nothing to do with scale. The exact same thing would apply to a marble surrounded by a tennis ball. The key thing is the mass of the rest of the shell; you can't ignore it. If you could, what you are saying would be the case.