Suspended mirrors installation

Hi there,

I am currently working on an art installation made of different suspended mirrors and would like to create a more accurate model of the needed materials.

I start with defining my overall form and then I place my mirror. But I cannot suspend the mirrors with just straight ropes perpendicular to the ground. There is always some kind of substructure, made of larger steel ropes, that the ropes from the mirrors should be attached to.

So far, I’ve been making an approximation of the length of the suspension ropes, and then I’ve been hanging the mirrors by eyeballing their position.

It’s an ok method but I would like to have greater precision - my aim is to be able to suspend a given mirror at a given height and at a given angle with the shortest possible ropes to the substructure.

Could anyone give me a hint on how to approach this case? The “curve closest point” is not working because I need to take into consideration the mass of the objects…

Why would you use 4 ropes instead of 3? wouldn’t 3 make it way more controlable?

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It’s a good remark.
Theoretically, 3 ropes are enough if we have the precise points of suspension.
In real life, as I am only eyeballing the correct position and have to work with the substructure, sometimes a 4th point is necessary to achieve the correct position of the mirror.

Why would you add angles in the ropes? Why wouldn’t you use just three ropes, using two the same length (red ones) for positioning and the one for adjusting the angle? Maybe I’m compeltely off here, but just having them as far outside as possible should do the trick together with straight hanging ropes.

in this case a projection would work.

I hope I didn’t missunderstand the complexity of the problem.

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Hey Benedict, thanks for sharing your thoughts.

Yeah, in the perfect world, where I could attach the ropes perpendicularly to the ceiling, that would be the best solution. And 3 points would be just enough.

But having a given substructure (sometimes it’s a metal truss for example) most of the time I cannot attach the ropes in a perpendicular way. I have to reach for the closest attachment point on the truss which gives this angle I’m talking about. In other words the points on the mirror are not exactly under the truss.

Maybe these images will help to clarify my problem:


Ok, thanks, I understand the problem better now. I think what you need to do is to see the ropes as vectors that can be split into directional force and gravity:

1 is just for reference.
2 shows the force vectors divided into gravity and direction force. (excuse the lack of correct terms in english)
3. having the forces without gravity, you should get a sum of 0 for it to be stable. however, I am unsure if the deviation from the mass center has an additional effect that I am not aware of or if you can just attach the ropes regularly around the mass center.

the advantage of using 3 ropes would be that you can fix two about how you want and then calculate the third’s vector and negate it to get the force and direction required to obtain the position.
With 4 it might become more complex to get clear values.

I must say that I am absoutely not a expert in physics, and physics class was a very long time ago, so I might have said something completely wrong.

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How would you measure the angles on site.
wouldn’t it be easier to use the length(position of the point) of the on the corresponding HSS.

Thanks again, I think that is the beginning of the solution. I might need to read more about the masses and the vectors (barycenters if I remember well from school).

Intuitively I would guess that these force vectors can be extended to find one point of confluence/intersection that might have some relation to the center of mass.

Hah, good point Baris. Measuring the angles on site is impossible. That’s why I am always subjectively eyeballing the final form and effect.

Much easier is to find the correct attachment points on the grid (the truss/steel ropes) and let the mirror hang naturally in the desired position. And this is the problem to solve…

I might misunderstand, but could you not just measure this distance:

Interesting problem.
But how this relates to grasshopper?

Anyway…


M = center of mass of the mirror
A = near M and slightly higher, 3 wires, forming a tetrahedron, rigid in space.
A is taking most of the load.
B & C = far away from M and lower, little load.

… Sorry for the bad picture, I’m from my phone…

Hey Riccardo,

I’m trying to automatize the whole process. I thought maybe grasshopper had a function or a plugin to calculate easily the position of such objects…

Thanks for the drawing!

Correct me if I’m wrong but doesn’t this solution limit the rotation axis possibilities? Because in this situation only 3 points can be the vertices of the mirror (the ones that are on the ends of extended bisectors of each angle). In other words, we limit the possible rotation axis to such (same in B, and C if at the top).

And also we generate 5 ropes instead of 3…

Oh, then, let’s get some help from gravity.

Try this:


(The mis-alignment is due to the weak forces of the wires… increase it to see perfect alignment.
Weak force is needed only to play around with the grab force… )
hanging mirrors.gh (26.4 KB)

3 wires per mirror, but the 3 wires lie on the same plane, because this solution needs a “bracket” placed behind the mirror that have the main hole exactly behind the mass center and another hole higher and along the line that describe the inclination (on the “uphill” line).

You can find the mass center easily-enough by placing the mirror over a tube and balancing it, mark the extremes and do the same with another angle with the tube.
The height of the main hole in the bracket (distance of the hole from the mirror plane) will correct any error in the mass center (the higher it is).


You can even mount the bracket slightly uphill (#) to have the main hole exactly in the vertical of the mass center, with only two main wires.
The weight of the bracket itself would be the only force to topple the mirror, so a minimal force, and so you could use the second hole with a short wire that connect it to the nearest of the two main wires…

(#) by Tan(a)*D
D = hole center distance to mirror plane
a = angle of mirror from horizontal plane

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I’m not the most expert person in the field, but have now suspended quite a number of very weirdly-shaped art pieces with not-very-easy-to-determine center of mass location

in my experience 3 suspension wires evenly oriented around the center of mass are always the best solutions
but if those 3 wires are perfectly vertical then expect the suspended piece to ondulate/swivel/rotate due to air flows caused by air conditioning/heating/crowds of people walking around: vertical wires = poltergeist

ideally I would always prefere 3 wires, slightly on an angle*, as evenly distributed around the center of mass as possible (120 degrees being the perfect theoretical scenario)

slightly on an angle* = if your wires go diagonal then you introduce horizontal-forces components that you really need to take into account, imagine pulling a rope almost horizontally on two ends to lift a weight


the greater the alpha angle, the greater the Ahoriz and Bhoriz forces comes into play, and they just want to pull the suspended piece apart
image
when you go 3+ wires which lie on different planes, then you need to go 3D, and @benedict vector system is just on spot

I would advice for something like Russell Hibbeler - Engineering Mechanics
this book is huge, but its chapter 3 on the equilibrium of a particle offers a very nice and well understandeable review of a wide range of cases on the topic “how much force is on each of those wires that are keeping my particle suspended?” :+1: literally hands-on problems, easy to follow by someone like me who is not an engineer

regarding on-site installation, rule of thumb is always to expect the unexpected :slight_smile: you probably know better than me :slight_smile:
I always rely on linear measurements that generates angles, like illustrated by @Baris, I’d have suspension cables already cut at length (with some abundance), or at least already marked to the correct length, just to be trimmed last minute when things are matching properly

ideally I would come to installation day with a table that for each mirror indicates:
→ lengths A, B, C, D
→ on which beam/rope they need to be measured along
→ which wire La, Lb, Lc, Ld has to be connected and where
→ but there would just be As Bs Cs, just 3 wires :slight_smile:

but all this also implies to have a very reliable 3D model of both location and art pieces (expecially its center of mass) which is not always easy or even possible at all…

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Hi Riccardo,

I am trying to understand your grasshopper: it predicts where a mirror will hang for given attachment points on the substructure and wire lengths. Is that right?

Regards
Jeremy

… uh… I don’t know if I should answer yes or no… :rofl:

Each mirror have an angle, find the “uphill” direction in the mirror, the you can define a plane where everything lies: the mass center, the wires, the uphill direction.
The links to the top structure must be on that same plane too.
The definition just project the mass center to the nearest point in the structure, and make the wires.

The kangaroo part is just to test the generated structure with wires, to grab it and see it always swing back to the correct position.
You can delete all the kangaroo and goals part.

The idea is that, if you find the mass center precisely-enough you can use just 2 anchor points in the mirror, and 2 in the top structure.

The mirror will reach the correct position because, thanks to that structure, it is the only stable position. 2 triangles + gravity.
They are free to swing, sure.


If there is some air movement, then, imho, the only possible solution is using the tetrahedron.
And, as I was corrected by @Bed , an additional 6th wire between my blue and green wire… creating 2 triangles from B-C-structure.
Quadrilaterals will always swing in a way or another.

The help from gravity is amazing :slight_smile:

The solution with the bracket helps to simplify the problem, although I can strictly use only the preexisting attachment points.

And thanks for the explication of the script, I understand it better now and I’ll try some experiments with it.

Hey Inno,

thanks a lot for the explication (I love how of all the objects you use the Bible to explain gravity :rofl:) .

And thanks for the book reference, it’s great to understand more the equilibrium problem.

The scheme of installation is exactly what I’m trying to achieve. Having the exact points of attachments to the given beam I would be able to position precisely each link on the beam and to precut the exact portion of the steel cable

Thanks for the explanation, Riccardo. That makes the gh clear.

But I think I have to take issue with this part. The middle wire, having no strength in compression, means you no longer have two triangles but a quadrilateral with a redundant diagonal in circumstances when the forces on the structure tend to shorten that diagonal, as when you tighten the “top” wire. Whether that would be enough to distort the installation significantly I don’t know. Could the kangaroo part test the “tension only” aspect of support cables?

the middle wire is not compressed, but the opposite.
The most of the weight force is sustained by the 2 wires that attach near the mass center, both are not compressed but stretched.
The third wire have just a small stretch force.
All 3 wires are stretched.
This is the theory.

But the practice, kangaroo, confirms it all.

Sure, the mass center, projected to the top frame, needs to falls inside the “base” of the main triangle (made by the 2 main wires.