Re: [ATM] wire spiders

["tony gondola" <acgna@comcast.net>]

Everyone except engineers know what I mean Arjan......:-)
Personally I think you've made the situation more complicated then need be. 
If each "vane" is made up of two legs leading away from the diagonal mount 
at some angle, the greater this angle is the less tension you'll need to 
apply to the wires to carry the load. This is true of everything from 
sailboat rigging to radio towers and it applies here too. This is 
complicated somewhat by the fact that there's a large range for the gravity 
vector but I still think it's the key for a good design. It's also been 
shown in actual use that offsetting the "vanes" from the optical centerline 
helps to stabilize vibration.

Tony

----- Original Message ----- 
From: "Arjan te Marvelde" <arjan.te.marvelde@hetnet.nl>
To: <atm@atmlist.net>
Sent: Friday, June 30, 2006 11:28 AM
Subject: Re: [ATM] wire spiders


>> 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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