I’m making a shell structure using Kangaroo. Although the structure has been successfully created in Rhino, I am getting to grips with the concepts of the overall functionality of values.
For load, while applying the unit vector to multiple points as 0.44, why do different points raise at different height despite the fact that same Z-value is applied to all the points?
SESSION -6 INTRO TO FORM FINDING USING KANGAROO- IMPERIAL COLLEGE LONDON PRACTICE.gh (18.8 KB)
The same load in the Z direction is applied to each point - but that’s not a height - it is the direction and strength the point is being pulled with.
For a simpler example think of a single chain, fixed at either end, hanging under gravity to form a catenary curve. The mass of the chain (and therefore the load) is evenly distributed along its length, but of course the middle part ends up lower.
The strength of the load here just affects how much the springs will stretch.
If you want to model something like a Gaudi style hanging chain model, where the stretch is very small, make the springs very stiff (but longer than their flat length, otherwise it won’t be able to move).
As you tend towards inextensible cables, the actual weight used becomes less and less important - just as a steel cable or a nylon cable of the same length will both form approximately the same catenary shape when hanging under their own weight (assuming they are long enough that bending resistance doesn’t play a large part, and the strength to weight ratio of both materials is high).
Okay. I will give it more thought. Thanks.
An awesome and compact explanation, thank you @DanielPiker !
Makes a lot of sense for simple funicular form finding, but how would one correctly create the creases as can be beautifully seen in the sketches by Heinz Isler?
Sure, one could reduce the rest length of these curves and achieve those results. But what does “reducing the rest length” actually mean?
According to Hooke, a material that has deformed less under the same load must have a higher Youngs modulus / cross sectional area. Does that mean that reducing the rest length is equivalent to increasing the stiffness (in case of Kangaroo the strength parameter)?
Maybe a bit of a crazy question, but what would it mean in terms of forces for the form-found shell if the edges where stretched in the process?
Thanks a lot!!!
Cheers, Rudi
