Re: [ATM] wire spiders

["Arjan te Marvelde" <arjan.te.marvelde@hetnet.nl>]

> Think of it this way. Wire is very strong when the force is applied along
> the long axis (in tension) and has little resistance to forces applied 90
> degrees to that (limp noodle). So.....the closer you are to taking the
loads
> in tension the stronger the spider will be.



What do you mean when you say "strong"?

It all has to do with the amount of stretching caused by tension in the
wires, which is a material property.

Let's analyse what  happens:
Imagine a pair of wires on one side of the spider. At all times the sum of
forces is zero, and this must apply separately in axial direction as well as
in radial direction. The force in axial direction, e.g. caused by the
spiders' weight, has to be countered by a differential in wire tension. This
means that the tension in the upper wire decreases by some amount, while the
tension in the lower wire increases with a greater amount. The change in
axial force and in wire tension are proportinal according to sin(A), where A
is the angle with the radial direction, or half the angle of the crossing
wires. Clearly, smaller angles will increase the change in  tension for a
certain change in axial force.

Now imagine the telescope pointing to zenith versus pointing to horizon. In
the first case, the full weight of the spider applies, in the second case
the axial force caused by spider weight is zero. The (axial) displacement of
the spider is determined by Youngs' modulus (E) of the wire material, which
for a certain wire diameter and length determines the amount of force needed
per unit of strain. (see for example
http://www.engineersedge.com/strength_of_materials.htm).
In SI units, for steel: E = 2.10^11 N/m^2

An excercise:
So how much force (F) is needed to move a spider axially over a certain
distance (dX)?
- wire length: L
- wire cross section area: A
- wire angle wrt. radial direction: P

One wire will stretch with dL, while the other will shrink with a slightly
smaller amount dL'. The difference of dL and dL', the differential strain,
causes a differential tension which will precisely counter the axial force
F.

Changes dL and dL' can be solved from the equations:
  [L+dL]^2 = [L*cos(P)]^2 + [L*sin(P) + dX]^2
  [L-dL']^2 = [L*cos(P)]^2 + [L*sin(P) - dX]^2

The differential tension is given by:
  T=E*A*(dL-dL')/L

resulting in an axial force per wire pair:
  F'=T*sin(P)

Assuming steel wires, 0.5mm by 250mm, in 4 pairs, mutual angle 30deg:
 F=0.6N for a dX of 1mm, which corresponds to approximately 62 grams. (What
does a spider weigh?)

Conclusions:
-1- It is against gut feeling that the amount of force calculated in te
example is independent of the amount of pre-tension in the wires. But I
don't know what I did wrong ... anyone?
-2- The force needed is relatively small, at least for the first mm. This
could have consequences for collimation for use between zenith and horizon.
-3- The dynamic behaviour is yet to be derived, as well as radial shift and
torque effects.

I have made a spreadsheet for those who want to calculate their own
situation.

Cheers,
   Arjan te Marvelde


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