Re: Infinite 3d lattice

On Jul 25, 1:12 pm, "I.N. Galidakis" <morph...@xxxxxxxxxxxx> wrote:
cbr...@xxxxxxxxxxxxxxxxx wrote:


<snip>

I think your example can be analyzed by modelling it down to the 2D case.

Consider then the projection of the problem in the yz plane. Then you have your
point initially lying exactly in the middle of the square (where A,B,C,D are
spheres), as follows:

A B

o

C D

The above is clearly an inertial configuration. You are now asking what will
happen if o moves some amount dx, away from the computer screen.

Switch now to the xy projection. Then the new configuration amounts to:

A E
|\ |
o-o'----o''
|/ |
C F

o' = o + dx


Your claim is that o' will "feel" a force towards o, if dx < 1/2. Since the
perpendicular components of the forces cancel out, it suffices to consider only
the horizontal components of the forces (on the axis o - o').

If o's mass is m and the spheres' mass is M, o' will hence "feel" a net-force
of:

G*m*M/(Ao')^2*cos(Ao'o) - G*m*M/(Eo')^2*cos(Eo'o'') (1)

Now, if (oo') = dx, then (o'o'') = 1 - dx, and

(Ao')^2 = (1/2)^2 + dx^2,
(Eo')^2 = (1-dx)^2 + (1/2)^2, and also,

(Ao'o) = arctan(1/dx), and
(Eo'o'') = arctan(1/(1-dx)),

which finally gives for (1)

G*m*M/((1/2)^2 + dx^2)*cos(arctan(1/dx)) - G*m*M/((1-dx)^2 +
(1/2)^2)*cos(arctan(1/(1-dx))

Taking the limit of the above as dx->0, we get using Maple:

-2/5*G*m*M*2^(1/2)


Something has gone wrong in your analysis: when dx = 0 (i.e., |oo'| =
0), o' should experience 0 net force (it's an inertial point).

Let's get rid of dx (which was a bad naming choice of mine), and just
call it x. The situation we want to first analyze is (after we rotate
the /infinite/ series of squares as you do above):

.... E' A E ...
| |\ |
----o------o-x----o----
| |/ |
.... F' C F ...


where the diagram extends to the left and right to infinity.

If we call the force at x caused by ABCD f(x), then the force at x
caused by EFGH is f(x-1), and the force caused by E'F'G'H' is f(x+1);
and what we then will want to find is

F(x) = f(x) + sum(n=1 to oo)(f(x+n) + f(x-n))

I get that

f(x) = (4*g*M*m)*x/((x^2 + 1/2)^(3/2))

(derivation below at [*]) and since we are really only interested in
whether the result is less than, greater than, or equal to 0, we might
as well choose M and m so that 4*g*m*M = 1, and then

f(x) = x/((x^2+1/2)^(3/2))

As a reality check: f(x) = 0; for x << 1/2, f(x) is proportional to x;
f(-x) = -f(x); and f(x) -> 1/x^2 as x -> oo.)

Also, it helps that we get as expected that F(a/2) = 0 for all integer
a.

I don't have access to symbolic algebra programs such as Maple. But
numerically, I find that for 0 < x < 1/2, F(x) > 0 (it has a maximum
at 1/4); and for 1/2 < x < 1.0, F(x) < 0. So the point "o" we are
discussing is not stable with resepect to \deviations along the x
axis: a small deviation causes it to experience a force moving it
towards the center of the cube ABCDEFGH, which IS a stable point.

This doesn't yet answer your actual question; but we can amend the
above analysis by letting the distance AB, BC, CD, DA etc. be some
arbitrary positive value s rather than 1. Again, a little numerical
experimentation indicates that if we define

f(x) = x/((x^2 + s/2)^(3/2))

then F(x) is always positive for 0 < x < 1/2, and always negative for
1/2 < x < 1; and in the infinite lattice, the force at x will be the
sum of functions of this form, so it appears that the stable points
are the cube centers and the vertices; and that the other inertial
points are unstable.

Cheers - Chas

[*] Derivation:

Consider the right triangle Aox; |Ao| = sqrt(2)/2, |ox| = x, and so
the distance from A to x is sqrt(x^2 + 1/2) = r. Thus the force of
gravity towards A is G = K/r^2 (with K = g*M*m). The force g_x in the
direction of ox we can then calculate by considering the similar
triangle Aox; so

x/r = g_x/G

and then

g_x = G*(x/r)
= (K/r^2)*(x/r)
= K*(x/r^3)
= K*(x/(x^2 + 1/2)^3/2)

Adding in the contributions from B, C, and D, we get

f(x) = 4*G*M*m*(x/((x^2 + 1/2)^(3/2)))
.

Relevant Pages

  • Re: Infinite 3d lattice
    ... Consider then the projection of the problem in the yz plane. ... If o's mass is m and the spheres' mass is M, o' will hence "feel" a net-force ... I don't have access to symbolic algebra programs such as Maple. ... towards the center of the cube ABCDEFGH, ...
    (sci.math)
  • Re: RSA Challenges
    ... These days Jbilou will used the ground, and if Marilyn mostly ... the projection will cover instead of the basic ... square. ... Until Rashid alters the trainees better, ...
    (sci.crypt)
  • Re: Rotated square in space
    ... computing coordinates of points of a square in R^3 knowing their ... images under projection to a plane from a single light source: ... adding additional constraints involving additional variables u_i: ... These polynomials generate an ideal I in the polynomial ring ...
    (sci.math)
  • Re: Lucky, lucky me... [YACD, NPP]
    ... your line of sight. ... In Regular Angband, a square is ... In NPP 050, Movement, Line of Sight, and projection are ... We have squares that allow movement, but block LOS or projeection ...
    (rec.games.roguelike.angband)
  • Re: Lucky, lucky me... [YACD, NPP]
    ... your line of sight. ... In Regular Angband, a square is ... In NPP 050, Movement, Line of Sight, and projection are ... We have squares that allow movement, but block LOS or projeection ...
    (rec.games.roguelike.angband)