Re: [ATM] wire spiders

["Mel Bartels" <mbartels@bbastrodesigns.com>]

> > Personally I think you've made the situation more complicated then need

> amount of pre-tensioning does not make any difference for the static case.


I was about to add similar.

I do feel that this tack on the solution is too complicated.  It's a good
exercise to work through but does not lend confidence.

I endeavor to solve problems from first principles.  First up, what's the
problem?  We want the optical train to remain undisturbed.  But to what
precision?  That is a calculation worth doing.

The diagonal suspended in the optical path is the real gotcha here.

Symmetry says that we need to solve the problem by preserving 3
translational and 3 rotational symmetries in the dynamic situation.  The
static situation is an identity transformation from the dynamic situation.

Symmetry as manifested in the physical universe further declares that we
will do this using forces.  These could be electromagnetic forces that
suspend the diagonal.

Symmetry also says that forces tend to be transmitted along the edges or
perimeters or skins of structures.  So we can start with shapes that
preserve as much symmetry as possible and whittle them away, knowing that it
is the outside edge that counts.  We move through vanes to thinner vanes to
very thin vanes that are wires.  It's a calculation from first principles to
determine the minimum number of wires needed to preserve in the dynamic
situation the 3 translation + 3 rotational symmetries.

*Now* that you've arrived at an optimal solution, you can calculate tension
for the static case as a verification and order of magnitude sizing.

Mel Bartels


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