Tag Archives: Triangle

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...

General

Patterns...

One of the challenges when researching in an engineering field is to determine what patterns your data may present you with.  This can be an unfamiliar skill for those of us that have come from industry as we're used to dealing with certainty when designing buildings, not uncertainty.  One of things I've been messing about with lately is patterns and series of numbers as I've been learning Mathematica, using Roozbeh Hazrat's book to help.

Bokeh Basket...

This has let me calculate the palindromic prime numbers less than 10,000 and other long winded sums in a single line of code.  It has had me thinking about other series of numbers such as the Fibonacci series which is present in nature and Pascal's triangle... (below)  I've been seeking these types of patterns and puzzles out to try and sharpen my powers of observation and help with rebuilding my maths skills which have atrophied over the years.

\begin{array}{c} {\rm{1}}\\ {\rm{1\, 1}}\\ {\rm{1\, 2\, 1}}\\ {\rm{1\, 3\, 3\, 1}}\\ {\rm{1\, 4\, 6\, 4\, 1}} \end{array}

Pascal's triangle is quite a simple pattern to determine, you simply work through the line above and it helps you create the next line below in the series.  Take the first number, it's always 1, then add the first two numbers together on the line above to get your next number, rinse repeat to see what you get...

\begin{array}{c} {\rm{1}}\\ {\rm{1 1}}\\ {\rm{1 }}\underbrace {{\rm{(1 + 1)}}}_2{\rm{ 1}} \end{array}

In the example above when you add together the two numbers on the second row, you get the number 2.... I've highlighted this with a bracket on the example above to illustrate how you get the number 2 on the third line...  try working through this process to see if you can get the pattern above to repeat and see if you can follow the logic.

Now this is all fine and well, but Pascal's triangle is well known and it isn't going to win you a bet down the pub... the Aha! moment that goes with puzzles is the key to a successful puzzle (Badger, Sangwin, Ventura-Medina, & Thomas, 2012)  the simpler the explanation, the more readily the solution will be accepted (Michalewicz & Michalewicz, 2008).

So following the pattern theme of this post, let's see what we can make of the following pattern and see if it can help sharpen up the old grey stuff... and if you crack the puzzle, maybe you can try it down your local and see if it earns you a free pint...

\begin{array}{c} 1\\ 1{\rm{ 1}}\\ {\rm{2 1}}\\ {\rm{1 2 1 1}}\\ {\rm{1 1 1 2 2 1}} \end{array}

What is the next line for this pyramid?  If you're struggling and would like a hint, then try saying the series aloud.

The solution is 3 1 2 2 1 1 because there are THREE ONES, TWO TWOs, and ONE ONE on the line above...

If you're struggling to solve this puzzle, then you can highlight the text from this line to the line above to reveal the answer.

References:

Badger, M., Sangwin, C. J., Ventura-Medina, E., & Thomas, C. R. (2012). A guide to puzzle-based learning in STEM subjects. Birmingham: University of Birmingham.

Michalewicz, Z., & Michalewicz, M. (2008). Puzzle-based learning: An introduction to critical thinking, mathematics, and problem solving. Melbourne: Hybrid Publishers.