Let´s say we have a simple catenary cable with a start point and an end point. The catenary has sag. My questions is how to find the optimal sag in regards to forces to be anchored. I think it is geometrical problem at the end.
Mean to you? The more sag you have the heavier your piece of cable will be. Optimal could be the shortest? In which case you have a straight line
Actually the reverse is true. The force needed to tension a straigh cable without any sag in the physical world aproaches the infinite. Go and see the cables suspended for the trains or the electricty lines. It will be very very cost effective if you were right though…
Do you want to allow for the cable mass? How is a force applied to the cable? Uniform distributed load per metre span? Or per metre along the cable? Or at discrete points? Is optimum == minimum tension in the cable?
The weight of the cable is 0.65 kg/m. Uniform distributed load per metre span. And yes optimim is minimum tension in cable.
This is exactly what I am pondering about. I want to optimize a simple catenary under these conditions.
So… put in a slider to define the sag between,say, 1% and 100% of the span and calculate the tension?
Or this sounds like the kind of thing that kangaroo is designed for, but i’ve Never used it…
Actually the limit is a lot smaller. Let´s say we have a certain distance from the cable´s low point too the ground as a limit. it will be something a lot smaller than from 1% to 100%. But yes you got the idea very well. Come to think of it Kangaroo might be the best bet. I also have used it very little and from tutorials.
@DanielPiker Hi Daniel. Can somethig like this be done in Kangaroo?
I thought this kind of calculation was pretty standard:
It’s definitely an appropriate approach I’d say. You might also have a look at its extension K2Engineering, if you want to work with calibrated mechanical properties and forces.Though for simple cases of one sagging cable, perhaps using a Kangaroo/K2Engineering model is a bit overkill (edit: as @akilli pointed out).
I actually think using Kangaroo would be overkill here.
This is also something that doesn’t depend on the density or stiffness, just assuming the cable is not stretching any significant amount.
It’s something that should be fairly straightforward to compute analytically, but if you don’t want to do the math, the quickest brute force approach is probably something like this with the standard catenary component and galapagos:
catenary_min.gh (16.3 KB)
Seems to suggest the sweet spot is around 56° from horizontal
I threw this together quickly though, so double check
Exactly! This is more or less what I needed here Thank you guys!