# Monthly Archives: June 2012

General

## Time…

Now that I’m through the massive pile of end of term marking and I’m starting to think about getting some research done through the summer, this goes hand in hand with time management skills.  I spend a great deal of my time in unstructured activities supporting students during term time, so during semester 3 I become quite hard nosed about the time slots with students.   Once I’ve made my list of things that I’ve got to do, then I’ll typically I’ll go for a quick win and tick off some of the easier things to give myself a sense of initial progress. With the remaining tasks then I create a Gantt chart to programme out the works through the summer period to see if I’m being overly ambitious with my intentions.

Once I’ve protected my time and planned out my summer work, then the next step is simply doing it…

I have to confess to being one of the great procrastinators and I have a couple of techniques that work well in getting me up and started on tasks and I encourage my students to make use of these techniques to help them keep on top of their dissertation writing, the two techniques I use are: the pomodoro technique & breaking the chain.

Once you’ve completed 4 sessions, give yourself a longer break, half an hour, an hour… that part’s up to you, but what you’re likely to find is that you’ve been super productive for the two hours that you’ve running the pomodoro technique and actually taking an hour off at this point might be needed to let your brain cool down… it just depends how hardcore the tasks are that you’ve been doing.  One tip here though is to set a timer for your breaks, otherwise they will over run.

I use a mixture of timers when working with the pomodoro technique, either a cheap mechanical timer from Sainsbury’s which cost £3, a pomodoro mac app that puts completed pomodoros into iCal that I can’t link too here as it’s no longer available in the UK for some reason, or pomodroido on my phone.  If you’re the sharing type, you’re likely to want to broadcast your pomodoro sessions over Twitter or your social media network of choice… please be aware though, this is really annoying for those of us not working on your tasks so please exercise some discretion.

Firstly let’s say that every day you spend a minimum of 10 minutes cleaning, 30 minutes writing, and 20 minutes doing exercise… that’s an hour you need to squeeze in every day.  But when you do these three tasks you are allowed to put a big fat black cross on your calendar, preferably in a fat marker pen to be highly visible.  Every day you do these tasks, you earn a big fat black cross on your calendar… the key is to make the longest chain of crosses possible…  Try and keep your daily tasks below 60 minutes to allow them to be achievable, but simply do them… 10 minutes of tidying every day will strangely make a large difference to the tidiness of your house and 30 minutes of writing, whilst in itself isn’t a lot… but it helps to keep your hand in on writing and stops you going a whole week without having done any!  ( A common curse for PhD students).  20 minutes of exercise every day could be something as simple as heading out for a walk that day, you can vary the level of intensity to suit.

I’m not pretending that this would be all you’d need to do to keep the writing up, but by doing 30 minutes every day that’s 3.5 hours of writing per week, that could be a few thousand words you never would have had being there in your thesis ready to edit and sculpt into some sort of tangible form when you’ve time to edit.  You’re still going to need to make writing a priority during your working week, but at least your brain and fingers will be well trained in making a start after your core sessions each day.

I’m hoping this helps someone a little out there, either through sharing the apps that I use, or giving some food for thought on how a little and often can be a real boost to your productivity.  If you’ve found anything helpful, please drop me a line or leave me a comment.. if you’ve another technique that works for you or have an immunisation for procrastination I’d REALLY like to hear from you…

General

## Summer loving…

Well, having completed my first complete academic year, I’m now entering into semester 3 and I need to start planning what it is that I want to achieve over the summer whilst most of the students are away.  There’s a common misconception it would appear, that academics all bugger off to Tuscany for the summer running care free and naked through the hills with the wind blowing through their hair, finally returning in September to teach again.

The truth of the matter is that this is perhaps the only time in the year when we can actually get stuff done that we want to, we’ve access to the laboratory and we’re able to crank through all of those things that we’ve just not had opportunity to get round to through the academic year due to other pressures.  The key to getting things done though is to create a list of what it is that you want to achieve… so here’s my list (in no particular order yet as I need to put timescales next to them).

• Review COMSOL and attend a training session.
• Develop a series of Mathematica sheets for calculating geometry.
• Learn ANSYS.
• Learn Inventor.
• Update lecture materials for the MSc and some of the first year modules.
• Write 2 chapters for my PhD.
• Complete my Lynda.com training session for Blackboard 9.1.
• Completely re-populate all of my Blackboard area, embedding Mathematica.
• Complete WriteTEL2 programme with Napier.
• Write resit papers and mark any resubmissions from repeating students.
• Obtain ethics approval for my innovative structures teaching programme.
• Finish two journal papers.
• Become registered as a STEM ambassador.
• Complete a ‘Meet the Engineer’ session at MOSI.
• Arrange next year’s Technical Lecture Programme for the IStructE
• Complete my examination marking for the IStructE
• Complete the Advanced Structural Behaviour Exam paper for the IStructE.
• Submit draft textbook outline to McGraw Hill.
• Define a solid workflow for creating technical illustrations, MS Visio or Adobe Illustrator.
• Explore Wolfram SystemModeller.
• Find and create a pool of study buddies…

In amongst this I’ve also a significant build on with our new house, although we’ve a good builder on board so hopefully it shouldn’t drain too much of my time.  I also intend to take some time out for a holiday, as well as doing some hill walking with a little landscape photography thrown in too…. but here’s the rub… when I look at the list above I find myself thinking that’s quite a lot of stuff to fit in through the summer months, but the difference being that actually I find the activities listed above good fun and I’ve actively taken on board these activities through my own volition… they’re not crap that has been dumped on me by an inconsiderate boss.

This ability to be highly self-indulgent is perhaps the thing I like the most having joined academia, although I still feel guilty come every Friday when I’ve not completed a timesheet… but just to be able to make time in my own diary to head off and read about something I find interesting is incredibly liberating…

## Arc…

Following on from my last post where I was investigating how to determine the parabolic equation when only provided with three co-ordinates, I wanted to extend that knowledge and create a Mathematica sheet to allow me to do this without having to rely on Wolfram Alpha.  This is for two reasons, firstly if I don’t have an internet connection, then I’m screwed as Wolfram Alpha isn’t going to work for me… secondly I need to encourage a deeper understanding of the procedure and process if I’m going to get the most from Mathematica.

The first part is trying to determine the parabolic equation when I know three points on the curve, typically these are going to be the start, midpoint, and the end support point.  To give an appreciation of the geometry, it’s important to firstly understand that parabolic curves are defined by the common equation:

$f(x) = a{x^2} + bx + c$

Determining the length of a parabola with cartesian co-ordinates is done through applying the equation below over the limits being considered.

$Length = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx}$

As three co-ordinates are known along the curve, the equations can be expressed as follows, with (x1,y1) (x2,y2) (x3,y3) being taken as the co-ordinate points… this gives a nest of simultaneous equations that can be solved using traditional matrix methods or in this case elimination could be used given that one of the input variables is 0.

$\begin{array}{l} ax_1^2 + b{x_1} + c = {y_1}\\ ax_2^2 + b{x_2} + c = {y_2}\\ ax_3^2 + b{x_3} + c = {y_3} \end{array}$

Taking the co-ordinates above as (0,500) (2050,0) (4100,500) for the input variables, this gives the simultaneous equations below to solve:-

${a{{\left( 0 \right)}^2} + b\left( 0 \right) + c = 500}$

${a\left( {2050} \right)_{}^2 + b\left( {2050} \right) + c = 0}$

${a\left( {4100} \right)_{}^2 + b\left( {4100} \right) + c = 500}$

However, Mathematica is well equipped for solving simultaneous equations and simply requires that the expressions are stacked up neatly and logically…. there can be a few problems with mixing bracket types, but this is just down to practice.  To solve the simultaneous equations above, the following code will solve them, the double equals signs are used just to push up the accuracy of the expressions and Reals is used as an additional parameter just incase any of the solutions are imaginary numbers.

Solve[{a x1^2 + b x1 + c == y1, a x2^2 + b x2 + c == y2, a x3^2 + b x3 + c == y3}, {a, b, c}, Reals]

Which will return the following solution:-

$a = \frac{1}{{8405}};b = – \frac{{20}}{{41}};c = 500$

Substituting this back into the general expression determines the parabolic equation for this particular curve to be…

$f(x) = \frac{{{x^2}}}{{8405}} – \frac{{20x}}{{41}} + 500$

The last thing to do is to work out the length of the curve between the two support points, this is a simple process determined by:-

$Length = \int\limits_0^{4100} {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx}$

For those of us with too little time on our hands, this calculation can easily be automated and defined within Mathematica using the following code, where f'[x] is used to differentiate the term within the brackets.

f[x_]:=500 – (20 x)/41 + x^2/8405;

NIntegrate[Sqrt[1 + (f'[x])^2], {x, 0, 4100}]

Which gives the length of the cable to be  4,257.24m  which matches Wolfram Alpha’s answer perfectly… the next step is defining equal length links to segment this curve to start creating the geometry for parabolic cable-chain structures… that’s definitely a post for another day.

The intention with this post is to help anyone who is trying to define parabolas and arc lengths using Mathematica… or traditional methods, if they’ve got time on their hands…

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.

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.

$Length = \sqrt {2{R^2} – 2{R^2}Cos\theta }$

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

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.

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

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

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:

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…

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.

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.