Tag Archives: cable-chain

Research Tutorial

Dynamic Arches...

Tinkering about with SystemModeler a little further, I've managed to finally build a sprung arch, complete with dampers on the revolute joints.  I'm intending on using this principle in my research to create folded structures, so it's interesting to see what effect the spring stiffness will have on the behaviour of the arch during the unpacking process - specifically looking at the accelerations on the masses at key points.

The thing that I was struggling with was creating a structure that had a set of equations that could be solved, the key concept I was initially missing was the closing of the structure with the special type of revolute joint to complete the chain.  Without this special revolute chain the equations are essentially unsolvable, so it's important that one of these joints sits in the system somewhere.

Sprung Arch

Another concept is that the structure in the video has 3 straight segments, each 1m long; but the supports are only 2m apart.... forcing the arch to pop into a stable shape that balances the weights at each of the joints.  This is essentially what makes the arch wobble when solving the initial set of equations.  Next step is applying external forces and measurement points along the structure for displacement etc...

General Research Tutorial

Cable-Chains...

Part of the work that I've been undertaking on Mathematica is to create a series of sheets that will calculate the geometry of a cable-chain arch with a set of given parameters.  In part I’m interested in how the cable-chain arch can behave as a deployable structure and looking to build on the work of (Li, Vu, & Richard, 2011) to see how practical a cable-chain structure can be made with regards economy, efficiency, and robustness.  Essentially a cable-chain structure is a curved arch that is sub-divided into straight sections, with cables spanning across the base of two struts as can be seen in the figure below.  Simple versions of these types of structures are widely used for the likes of temporary and deployable aircraft hangers to create large open spans.

Pantographs

Now that I’ve got the makings of a simple Mathematica sheet up and running and I’ve taught myself some rudimentary programming and graphics manipulation skills I’ve managed to start to knock together what I feel are some high quality illustrations for my thesis.

I’ve done this with a mixture of Mathematica to create the base diagram, which I save as a PDF.  I then import the PDF into OmniGraffle to annotate the diagram and then export to a PNG file to maintain the transparent background, this figure I can then host for linking into blogs etc.  Below is a sample figure which shows how the number of segments (nSeg) affects the internal area available for habitation within a typical parabolic arch.

Cable Arches

Given that both of my brothers are colour blind and I’ve never done the test, I’m not convinced on my choice of colour schemes, but the good news is that it won’t take long to change if it turns out I’ve made my figures look like something off the set of Austin Powers.

So far I’m finding OmniGraffle quite limited compared to Visio that I’ve been using for my diagrams for perhaps 20 years or so.  I decided to use OmniGraffle though as most of my writing work is done on a Mac, although I also have a PC so I can always create the more complex diagrams on Visio if need be, especially as I’ve managed to get a legitimate copy from work for £12.

I'd love to hear how other engineers and academics approach creating technical figures and sketches on their Macs though, I've a feeling that I'm really missing out on something and there's got to be a much slicker workflow out there.

References:

Li, Y., Vu, K. K., & Richard, J. Y. (2011). Deployable Cable-Chain Structures: Morphology, Structural Response And Robustness Study. Journal for the International Association for Shell and Spatial Structures, 52(168), 83-96.

 

Research Teaching Tutorial

Kriss Kross...

I've been really busy with various things associated with my new house lately and not really had much time to tinker and meddle with Mathematica or my PhD.  One of the key elements that I need to investigate is to calculate the internal area of a polygon where cables cross...  This will vary, depending on how many internal cables that I have and so I need a quick way of finding out the internal area for lots of different permutations.

Internal Area

This is a simple polygon problem which can be easily calculated once you know the cartesian co-ordinates for each of the key points using the following equation which is presented for the general condition.  It's not a hard calculation, as really it's just triangles, but it is a tedious calculation, which means it's perfect for Mathematica or Excel.

Area = \left| {\frac{{\left( {{x_1}{y_2} - {y_1}{x_2}} \right) + \left( {{x_2}{y_3} - {y_2}{x_3}} \right) \ldots + \left( {{x_n}{y_1} - {y_n}{x_1}} \right)}}{2}} \right|

However, the tricky part comes from determining where the cables actually cross over as that is where the cartesian co-ordinates will come from that are needed to determine the areas.  The calculation of the co-ordinates can be done a few different ways, but I've decided to solve the problem through using some vector based geometry, consider the simple condition below which is just for two lines that cross, how do we find the point where the lines intersect?

Pantographs

You could describe the two lines in the general term of y=mx+c, but this isn't what I want to do within Mathematica as I want to automate this for a wide range of values and lines and so iterative geometry using vectors is likely to be far more beneficial.  Representing these two lines in vector format using cartesian co-ordinates gives:

\begin{array}{l} Lin{e_1} = \left( {1 - \lambda } \right){P_1} + \lambda {P_2}\\ Lin{e_2} = \left( {1 - \mu } \right){P_3} + \mu {P_4} \end{array}

Now we know the equations in vector format it becomes possible to solve both of these equations to determine λ and μ which when back substituted into either equation will give us the co-ordinate for the intersection point p5 in (x,y) format.

This isn't that difficult to do by hand, but requires a bit of vector manipulation to expand the brackets and given my low attention span I would inevitably get it wrong after a few run throughs, but in Mathematica the problem gets much more simple, so for the example below we'll assume that the four co-ordinates are at 4m away from the origin (0,0) that is point p1.

{p1, p2, p3, p4} := {{0, 0}, {4, 4}, {0, 4}, {4, 0}};

Solve[{(1 - λ) p1 + λ*p2 == (1 - μ) p3 + μ*p4}, {λ, μ}, Reals];

p5 = Flatten[(1 - λ) p1 + λ*p2 /. %]

Executing this codes gives the intersection as being at point (2,2) in a fraction of a second, which is where you would expect it to be.  Now to explain the three lines of code and what they do, I'm sure a more experienced user could make this code much neater but doing it this way makes sense to my tiny brain.

The first line simply sets the variables p1-p4 with their corresponding cartesian co-ordinate position, giving the point two in the top right corner 4m up and 4m across, p3 is along the bottom 4m away, p4 is straight up in the top left corner 4m above point p1, as shown in the figure below.

Example

The second line then solves the two vector equations of the lines for λ and μ and makes sure that they are real numbers, rather than imaginary.  The real part probably isn't necessary, but it's just a force of habit that I put this in when using the solver as a just incase.

The final line substitutes the λ value back into the equation and then stores this as variable p5 so that I can use this in up and coming calculations, this was the fiddly part for me and makes use of the strange /. % input that allows the equation to effectively go back and substitute the solutions from the Solve command into the equation specified and recalculate and because my equation has the λ within it Mathematica knows this is what I want substituting in even without me being explicit.   I spent ages trying to work out this final stage, but now I've cracked it as a principle I can start to roll it out on the other sheets that I've been working on and this was the reason for writing this post incase it too helps someone else.  Now I can directly write the solution outputs from Solve, FindRoot, and NSolve and then carry on using these within my calculations rather than having to manually type them in.

This process lets me know that the intersection point is at (2,2) and now I know that the process works I can start to extrapolate this algebraically for the general conditions and start to calculate the areas of my polygons.

All of the figures in this post have been written using OmniGraffle, which I'm slowly getting to grips with, I can't make it dance as well as I can Visio, but I think it will help create some nice looking diagrams for my thesis once I get more proficient with it.

 

General

Geometry...

I've posted a few times that I feel my mathematics skills have slowly atrophied having been out in industry and that this I feel will be my achilles heel when it comes through to my research.  One of the areas that I've been investigating is the relationship between cable-chain structures and their behaviour.
Screen Shot 2012-06-03 at 16.40.54
Depending on how many segments you divide the arc into will determine the geometry and efficiency of the cable-chain arch, now if it's a semi-circle that's not too hard to calculate the geometry using simple triangles.  In degrees the internal angle of each triangle will be given by the following formula, where nseg is the number of segments you're dividing the arc by:-

Internal{\rm{{ \,\,}}}Angle = \frac{{180}}{{{n_{seg}}}}

This then can allow you to calculate the length of the segment along the arc as set out below using the Cosine rule which I haven't used for probably about 20 years, this though is only half the problem as you still have to determine the length of the cables.
Screen Shot 2012-06-03 at 18.42.40

Length = \sqrt {2{R^2} - 2{R^2}Cos\theta }

Length = \sqrt {2{R^2} - 2{R^2}Cos\left( {\frac{{180}}{{{n_{seg}}}}} \right)}

Next to determine the length of the cable, you will need to combine two of the triangles previously calculated and which gives double the internal angle and repeat the process for the cable that is shown highlighted in red below.
Screen Shot 2012-06-03 at 18.42.56

Length = \sqrt {2{R^2} - 2{R^2}Cos\left( {2\theta } \right)}

Length = \sqrt {2{R^2} - 2{R^2}Cos\left[ {2\left( {\frac{{180}}{{{n_{seg}}}}} \right)} \right]}

This all seems fine and well for simple circles, but how would you know that a semi-circle gives the most efficient arrangement for a cable-chain?  What about if it's a parabola?  Well fortunately for the mathematically out of practice determining the equation for a parabolic curve is pretty easy for us Mathematica users...  Let's say I know I have a draped cable that spans 4100m horizontally between supports and has a 500m drape at the middle.  We can define three points on this cable easy enough as can be seen below:

Screen Shot 2012-06-03 at 19.22.03
This is where Mathematica starts to flex its muscles for me and why I feel I get good value from the student version that I've purchased.  I don't know what the Mathematica code for determining a line through the points is, but if I press the = key twice I get access to the Wolfram Alpha Servers and try and describe using every day English what it is that I'm trying to do...  Following a blog post from Mathematica on determining arc lengths using calculus I tried the following term which doesn't use any mathematical code at all, simply describes what it is that I'm trying to do:-

Parabola through the points (0,500), (2050,0), (4100,500)

Which gives the me the following output, which is pretty helpful to the maths novice who is trying to explore and understand the relationship between the data, the expression, and the graph...

Screen Shot 2012-06-03 at 19.29.49

Which makes my life a lot easier as I can start to tinker with geometry very quickly now using this approach.  One of the cool things about working this way is that you can then start to dissect the code and determine what the correct format should be when creating Mathematica sheets and it makes learning Mathematica much much easier than say MatLab or Maple.

Whilst creating code like this would not be difficult for someone with a little rudimentary programming experience, it isn't that intuitive for a newb like myself, but now I can start to adapt this syntax to suit what I need...  The next step is to try and work out how to subdivide this into nseg number of equal length links... but that's a job for another day.  Maybe I'll need to start using polar geometry to make my life easier? Who knows... but at least I'm not being held to ransom by my elderly fuzzled brain anymore and I can explore and play with the maths rather than being pummelled by them.

I'd love to hear from other Mathematica users, especially to hear what their experiences are and particularly to hear if I'm missing a trick as I'm going through and learning the syntax...