Can I use Kangaroo to Solve This - Kite Bridle Load Balancing

kangaroo
#1

Is it possible to use Kangaroo (or something else in GH) to solve this…

The diagram represents a bridle on an inflatable kitesurf kite. The vector arrows represent aerodynamic lift from the kite. The lines are a bridle that distributes the load from the kite down the flying line at the bottom of the image.

Assume Points 0,1,2,3,4,5 are fixed.
There’s a line that runs from A0 through a pulley to Point 0
I want to optimise the positions of A0, A1 and A2 to equalise the loads on the lines 5>>A2, 4>>A2, 3>>A1, 2>>A1, 1>>A0

Can assume straight lines between 0,1,2,3,4,5

This is 2D but I want to be able to do this on an arc that curves in the YZ plane too so a 3D solution would be good.

Bridle1.gh (19.8 KB)

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(Daniel Piker) #2

This looks like it would probably be possible with the constant tension goal, or some variation of it.

If all tensions are equal, for networks where every node has 3 connections you get Steiner trees - and the angles between edges connected to the same node are all 120°

Here’s an example of a simple branching structure, with different tensions set for the different levels:

constanttension_demo.gh (11.6 KB)

Notice that the angles formed between the top branches are the same, and are symmetrical about the direction of the branch below

I wonder though, if it matters that the cables attached to the kite balance the lift vectors on the kite, then it might need a slightly different goal.
I did have a ‘Projected Force’ component in Kangaroo 1, that I never brought over to Kangaroo 2, because it was rather obscure and I don’t think anyone ever used it. It might be just what is needed here though. It was similar to the constant tension goal (which can be thought of as a spring that iteratively modifies its stiffness in order to keep the same tension force), except it tried to keep the projection of its force onto a given direction constant…

This is a direction for Kangaroo I find interesting, because I think it can be used for some of the kinds of problems that people often consider only possible with the force density method.

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#3

Thanks for the reply… I’m on Rhino5… none of the kangaroo2 components loaded… was that a R6 def or can I upgrade kangaroo2 for R5 to get that def to work?

I have K1 so I’ll look at the projected force in that

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(Daniel Piker) #4

I saved that definition in Rhino5
Yes, you’ll probably need the latest Kangaroo2 from Food4Rhino

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(Kiteboardshaper) #5

Yes, I use Kangaroo 1, make all the tow points fixed in 3D space and model the bridal as springs. Found best results with zero rest lengths and varying the stiffness per cascade to adjust the balance. I do include the flying lines in the model.

Cheers

DK

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(Kiteboardshaper) #6

https://rhino.github.io/components/kangaroo2/constantTension.html

I’d not seen this constant tension goal before, had been thinking about trying to do something similar using hoopsnake.

Way to busy this week to look at my kite design stuff, but will update my bridle code to this as soon as I get a chance.

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(Kiteboardshaper) #7

Had a quick go at using constant tension goal and K2 last night but couldn’t get anything close to a usable bridle. Need to work on it more. Springs and K1 might get your there quicker Martyn.

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(Kiteboardshaper) #8

Also I think your load goal is misplaced, the load at each tow point with an arch style bridle cannot be equal.

Assuming each cascade divides the load by half (exactly the case with a Cabrinha IDS style pulley in each connection point) then each tow point sees about 50% less load than the one lower.

Surf plan allows you to dial the tension per tow point somewhat, so must be using maths similar to the tension goal. Like I said need to work on it more, but K1 and springs has got me flying kites straight out of the bag.

Cheers

DK

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(Daniel Piker) #9

I agree that springs are generally easier to work with, as the changing stiffness of the constant tension goal can sometimes make it harder to control (the energy doesn’t always have a clear ‘basin’ to settle in like it does with the squared distance based energy of springs). If you want to exactly specify tensions though, it can do things that would be very tricky with just springs.
However, there’s no need to switch back to K1, the length goal in K2 does the same thing.
Perhaps what might help here is just some better tools for seeing the tension in each cable for the actual structure as you manually adjust the rest lengths in the form finding.

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#10

I’m reminded of the Smart Geometry 2016 and LaceWall modelling we did, particularly the concept of treating the lengths and/or starting positions of a cable network as a search space (where in this case, the desired tension might serve as the fitness criteria). Anywho, some examples from that that may inspire ideas:



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(Kiteboardshaper) #11

Hi Daniel,

At the moment I set the rest length to zero which means the whole bridle tends to a ‘min energy state’ - problem is that the tension would calculate to infinite on each cascade as there is a divide by zero in the maths?

It would be useful to get some sort of load indication at each tow point - even just a colour spectrum like your example above would work.

Cheers

DK

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(Daniel Piker) #12

It’s important to distinguish between the form-finding properties and the actual built material properties.

Zero rest length springs are very useful for finding configurations of cables which are all in tension, but as you point out, trying to calculate forces in the normal way, you get a division by zero when you do EA/L.
That’s why you need an additional calculation to convert the form finding properties into actual material properties. I made some examples for this for a recent AAG workshop. Here’s one for zero length springs:
PreTension_zerolength.gh (15.4 KB)

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(Daniel Piker) #13

…and for comparison, something similar for constant tension elements:
PreTension_ConstantForce.gh (14.1 KB)

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#14

The end goal here is one for debate really… I’ve seen this done by one manufacturer by equalising the loads on the bridle but I guess this was equalising the first legs, not A0>>A1 and A1>>A2.

The idea is that the aerodynamic lift is not trying to pull the kite out of shape but having said that, as soon as you turn the kite there is a big twist applied along the length of the kite so all the loads change anyway. The pulley helps deal with this but during a turn the wingtips need more support and when flying straight the centre needs more support.

Thanks for all the info above, I will read through it all and have a go with K2

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(Kiteboardshaper) #15

How’s your project coming along Martyn, anything flying yet?

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#16

It’s parked at the edge of the work window at the moment while we try and finish a house!

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