kangaroo settings

hi
all, could somebody help me whit kangaroo settings

i have
this structure made of interactive connected beams,
wat i
want is to preserve the connectivity and structure of this beams( the lines
have to stay straight) after moving the top points, the problem is i didn´t
find the right settings for it( i hope thats possible),
the
best what i get was straight lines in top-wiew but in perspective or site-wiew
was the lines whit kinks( the lines have to be straight in all dimensions).have somebody
any idea for the settings or other way to reach that ?

Thanks a lot for a help

kangaroo settings?

i have this structure made of interactive connected beams, wat i want is to preserve the connectivity and structure of this beams( the lines have to stay
straight) after moving the top points, the problem is i didn´t find the right
settings for it( i hope thats possible), the best what i get was straight lines
in top-wiew but in perspective or site-wiew was the lines whit kinks( the lines
have to be straight in all dimensions). have somebody any idea for the settings
or other way to reach it ?

Attachments:

beams.ghx, 1.1 MB
move.JPG, 96 KB
after.JPG, 102 KB

Replies to This Discussion

Permalink Reply by Daniel Piker on September 4, 2012 at 6:12pm

Hi Eduardo,

Have a look at my reply here:

http://www.grasshopper3d.com/xn/detail/2985220:Comment:648007

I posted some definitions there for reciprocal grid relaxation that I think you might find helpful.

Permalink Reply by eduardo matamala on September 4, 2012 at 6:31pm

Hi Daniel,
Indeed, I have fully reviewed your post and it has served me too for my study and achieve the definition, but really your definition while working and make reciprocal mesh, what I'm looking for is to create a freestanding deck and arrivals on the ground are of similar geometry to a tree, similar to the definition of Ma Mi. That's why I'm trying to implement to achieve your definition of reciprocal mesh this geometry.
Greetings and thanks for the attention.

Permalink Reply by Michael Hickey on May 12, 2011 at 10:36am

Hey Guys, I just stumbled actoss this conversation and realised it was the closest solution to a question i have. Im doing a project at the moment and using kangaroo to form find. We are making a frei otto chain moddel for a roof. I have the kangaroo model working which is great. Our Uni has GSA but hasn't got the form finding lisence with it.

The main problem i have is that say for simplicity, that i only have one sting and hang it. I can change the load on it and the stiffness of it in grasshopper to make the catenery shallow or deep, but i don't know what the optimum arch should be.

I was wondering if you guys would have any idea of how we could determin this, or even have any idea of how loads and stiffness in kangaroo are related to real life.

It'd be great if you had any idea at all.

Regard,

Mike

Permalink Reply by Daniel Piker on April 14, 2012 at 7:09pm

Hi Mike,

A hanging chain type model is a way of minimizing the bending forces in a structure.

However there can still be multiple solutions, depending on what else is constrained or minimized, and to get a single solution, you will have to define precisely what you mean by "the optimum arch".

Take the case of a single chain between 2 fixed points - as we vary the length of the chain we get a family of different arches - all catenaries, and all valid solutions to the problem of finding a pure tension structure which can be inverted to give a pure compression one.

I guess an obvious thing would be to also try and minimize the magnitude of the axial forces at the supports. I'll try and come up with an example of how to do this.

If we are modelling the chain with springs of equal length, with equal loads at each vertex, then we ideally want these springs to have infinite stiffness, so they do not change length at all. Of course this means that if we started with a straight line between 2 points, it would not change shape. So we need to either start from a non-straight initial shape, or give it some slack (set the rest length to be greater than the initial length). The example I posted here may help clarify things:

http://www.grasshopper3d.com/xn/detail/2985220:Comment:571599

Permalink Reply by Michael Hickey on May 15, 2012 at 8:57am

Hi Daniel.

Sorry about the delay. I was finishing up my exams so didn’t have time to respond. With the project we had neglected reaction forces and defined optimum as, I suppose the most economic structure (ie. One which had he least materials but could support the load applied).

Say we take the simple hanging chain you mentioned. If we had the least material then the chain would be quite flat and when inverted would act more in bending and less in axial (im assuming). Equally if we have a lot of material we get a very long chain which forms a more “pointier” shaped curve which would take a lot of axial (but equally if too long / pointy it would deform easier when a horizontal load is applied). Therefore there must me a point in between these two extremes which is the most efficient solution.

I am aware that GSA solves this by applying the load, measuring the vertical displacement of the nodes, and then resetting the model with the opposite magnitude of the displacements of each node applied to the model and then re-applying the loads. This is an iterative process. We were wondering if there is a way that this can be done in Grasshopper, or is there a rule that we could use to help us find this “optimum” solution.

Best regards,

Mike.