Re: Growth algorithm making more possible

-------- Original-Nachricht --------
Datum: Sun, 3 Dec 2006 11:32:11 +0100

> oh i see.. so i've only covered your worst case!  :)

Only the worst case of tetreders and only one choice.

> change one length, and let the body settle again.

ok until here you got everything right
 =

> the changed interval will be under some stress, and there will be some
> stresses elsewhere to compensate.

Not necessarily there exist cases where no stress is created for example:

a\=3D=3D=3D=3D/b  If you enlarge the =3D=3D=3D line will there be stres? No.
  \  /    the a and b point will change distance to each other and the ab
   \/     remains a stable distance to both a and b
   ab
a1/\=3D=3D=3D=3D/\b1  The same game,enlarge =3D=3D=3D result:No stress, a p=
oints + ab    =

 /  \  /  \   point are stable to each other, b points + ab point are
/____\/____\  stable to each other.(Why is that? cause ab is a rotation =

a2   ab    b2 axis).

c1 /\=3D=3D=3D=3D/\ c2 Enlarge =3D=3D=3D result STRESS, all points are stab=
le to each
  /  \  /  \   othr=E9r
 /____\/____\
 \    /\    /c7
c3\  /c4\  /
   \/____\/ =

   c5    c6

> okay, a lengthening run and a shortening run.  but what happens once
> these tests have been done?

Well you have now knowlege of two things: Stress or no Stress and lists
of stable points to each other.(When my thoughts are right there are maxima=
l two partys of points stable to each other, two lists)


> the way you describe it, i don't need anything more than to observe
> the new stresses, right?

No if there appears no stress you have to know which pointts are stable
to each other.
 =


> you're saying
> stress -> one interval
> no stress -> two intervals

> are you talking about bisecting an interval and introducing a new joint
> halfway?

> i'm really sorry, but i can't understand this.

You always bisect the line you have chosen,(in the original)
a midpoint is created, now if there was Stress you can just add ONE new int=
ervall between the midpoint and another point(if he is not already connecte=
d to it of course).
If you had no Stress your first step is the same, however you have to remem=
ber from which list the other point was, now you can either choose another =
point of the same list and connect it with the midpoint or you take
a random point from one list and connect it to a random point of the other =
list(Of course those points should not be connected already).
The random connection between one point of the first list with one of the s=
econd list is what I call choice one.
The connection of the midpoint with a point, stable to the point we connect=
ed the midpoint with before is choice two.

Example lets look at a tetraeder after bisection:
     m(midpoint) =

    / \                The lists are {a,ab1,ab2}  {b,ab1,ab2}
   /   \     First we choose a list and connect m with some               =

  /_ab1_\    point from that list he does not already share an intervall   =

 a\  |  /b   with: only ab1 and ab2 fullfill that condition. =

   \ | /     Lets say we choose the first list, now we connect ab2 with m. =

    \|/      If we proceed with choice one, we take a random point from
    ab2      the first list {a,ab1,ab2} and a randompoint from the second
             list {b,ab1,ab2}.Unfortunately ab1 and ab2 already
connected with all points of the second list so we would need to choose ano=
ther point from the first list or proceed with choice two.
Now if we choose point a, he may be connected with ab1 and ab2 but he is no=
t connected with b (that was or old connection before the bisection)
so we connect a and b.
Now if we proceed with choice two instead, we take another point
from the first list(remember we got ab2 from it) and connect it with m,
however a already has a connection with m so we either have to search anoth=
er point or proceed with choice one. Fortunately ab1 has no connection
with the midpoint and so we connect ab1 with the midpoint.

So the result of choice one for a "lonely" tetraeder are two tetraeders
and for choice two an octaeder. Note however the list of point we can choos=
e from becomes larger with larger objects :), so the results can become mor=
e complexand interesting.
-- =

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