Monday, 31 January 2011

Chemistry 1: The Periodic Table

Okay, I'm temporarily setting aside my first chemical rant to provide something a little more useful and objective: a neat way to remember the first twenty or so elements of the periodic table in order.  This mnemonic was given to me by my uncle (thanks Graham!) and I had to extend it in order to complete the first row of the transition elements, which came in useful from my A-levels onwards.

The mnemonic goes like this:

Ha, Here Lies Benjamin Bones; Cry Not Old Friend Needlessly, Nature Magnifies All Simple People, Sometimes Clowns Are Kings.  Callous Scoundrels Tickle Viciously; Crumbling Magnolias Fear Cold Nights, Cute Zones Gather Geraniums.

As you can see, it's a little uneven, but it works.  One important note:  the words are expansions of the chemical symbols for the elements, and are not always helpful in remembering the element names where the symbol and name aren't connected in English.  For example, "nature" is an expansion of Na, which is sodium; similarly, "kings" is the keyword for K, the symbol for potassium.

I'd suggest consulting a periodic table, like this one, to help remember where to put them all (and where the line breaks go :-)

The only other mnemonic for learning the periodic table in sequence that I've found while skimming the web is this one, which makes some sense but doesn't cover much territory:

Harry He Likes Beer But Can Not Obtain Food

Others seem far more complicated, more interested in electronic configurations (which are far too complicated and easier to just sit down and work out, then learn by sight).

So, long live Benjamin Bones... except he's died already!

Saturday, 29 January 2011

Mathematical Problems 4B: Close-packed spheres

In a previous post, I looked at close-packed circles - arranging circles hexagonally and calculating how much of the available area they fill.  That was fairly straightforward, and gave me an idea on how to calculate the volume occupancy of close-packed spheres.  Put it another way - how many Maltesers could I fit in a box if I could fill it to capacity (minimum spare volume left over)?

Let's start by putting them in a square arrangement - so that they're all in columns and rows.  How much space will they fill?  The easiest way to solve this is to think of one sphere inside its cube-shaped box.

The volume of a sphere S = 4/3 π r3
And the volume of the cube it sits in is 2r x 2r x 2r (since the cube has to be twice the radius of the sphere in height, width and depth)  C = 8 r3

Therefore, the ratio of the two volumes is S/C which is 4 π / 24 (notice that the r cancels - it doesn't matter how big the sphere is) and this is equal to 52.36% - only half of the available volume.

Next, let's look at hexagonal packing - arranging the circles so that they form hexagon patterns, instead of squares.

If we consider just one of the spheres, enclosed in a regular hexagonal prism, then we have this arrangement:

The volume of a sphere is...

And the volume of the hexagonal prism it occupies is found by multiplying the area of the base by the height.  The height is 2r (it's twice the radius of the sphere in height) and we can look at the base as being made up of six equilateral triangles...

So the volume of the hexagon, H, is 

And now that we have the volume of the hexagonal prism, H, and the volume of the sphere, S, we can work out how much of the prism is being filled by the sphere.

So, if we want to maximise the number of Maltesers in a box of chocolates, it makes sense to arrange them hexagonally, and not cubically.  80% volume coverage, compared to just 52% for the cubic arrangement, is definitely worth having!

Hexagonally arranged Maltesers (a Christmas present!)

On a more theoretical note, science textbooks regularly quote that close-packed spheres fill 80% of their volume, but I've never noticed any of them prove it.  So, this blog post counts as closure from a figure that's been drifting around since my A-level chemistry days, and which has recurred frequently since then.  None of the books seemed bothered enough to spend time on it - perhaps it's too much like maths and not enough like chemistry!

Next time - something different.  It might be projectiles, or escape velocity (conveniently related to each other) or it might be something about chemistry - in which case it'll be a rant (consider this advance notice!).

Tuesday, 25 January 2011

Physics Experiment: Determine g with a pendulum

Having done some work on determining pi by mathematical methods, I'm now going to use it in conjunction with some experimental work to determine the value of g, which is acceleration due to gravity. Any reference book will tell you the value of g is approximately 9.81 ms-2, but I'm going to do an experiment to show what it is. It's not a difficult experiment, and it doesn't require any specialised scientific equipment. To give you an idea, I did this using a toddler fireguard for my vertical surface, a piece of sewing cotton for my pendulum, and in the absence of any respectable small mass, used a small pine cone tied to the end of it. I also used a standard stopwatch on a digital watch (it's accurate to 1/100th of a second, although I'm not).

The relationship between a pendulum and g is described in the following limerick:

If a pendulum's swinging quite free
Then it's always a marvel to me
That each tick plus each tock
Of the grandfather clock
Is 2 pi root L over g

In order to improve the accuracy of my results, I counted the time taken for ten complete 'swings' or periods, and quoted this. I also repeated each ten-swing measurement three times, so that I could take an average and identify any anomalies. And somehow, saying that, I feel like I'm writing up a GCSE science coursework piece!

Here are my results...

length (l, in metres) 10 swings
(10T, seconds)
Run 2Run 3
0.162 8.458.318.35
0.237 10.099.9010.04
0.321 11.6311.6711.66
0.344 12.0911.9012.17
0.410 12.9812.9513.00
0.475 14.2314.1314.03

I calculated the average value of 10T, and hence T and then T2, which I can use to determine g, with the following rearrangement:

An alternative, if I'd wanted to plot a graph of my data, is to determine g by finding the slope of the appropriate plot.  Using the following rearrangement, it's possible to plot T2  against l and have a slope of 4pi2/g

However, I'm going at it in number-crunching form, using the formula above.  My results for g are as follows:

length (l, in metres) g in ms-2
0.162 9.136
0.237 9.338
0.321 9.332
0.344 9.3436
0.410 9.612
0.475 9.395

So, not perfect, but given the nature of the experiment - me with a fireguard and a pine cone - it's not too bad at all, and I feel quite pleased at having worked out something so massively significant with such basic equipment, and I feel it proves that science isn't just for big-budget departments!

Next time, determining the distance to the moon using the same principle as for geostationary satellites (except that this one is a bit bigger, a bit further away and not geostationary!).

Sunday, 16 January 2011

Maths Problems 5: Geostationary Satellites

Now that I've shown how to estimate (or calculate) pi with a high degree of accuracy and precision, it's time to start using it!

Modern communications around the world rely on being able to bounce a signal of a satellite in orbit around the earth, so that it can be relayed beyond the horizon of the person sending it (or better still, over a specific area of the earth). The ability to bounce the signal off the satellite is very important, and relies on one important factor - knowing where the satellite is, and better still, having it stay directly above the same point on the earth, so that its apparent position in the sky is fixed.

In order to do this, the satellite needs to occupy a geostationary orbit, meaning that it goes around the earth at the same rate as the earth rotates on its axis - once every 24 hours. A geostationary satellite appears to stay in the same point in the sky because it's rotating at the same rate as the earth.

The question is - how can this be done? The answer is to put the satellite in orbit at a specific height above the earth, and above the earth's equator. The rest is physics. Well, it's not rocket science, is it?

Firstly, there are two key formulae to use - the first is Newton's universal law of gravity, because gravity is what holds a satellite in orbit, and the second is the laws of motion for an object moving in a circle (we'll be assuming that the satellite follows a circular orbit).

The first formula is Newton's universal law of gravitation, describing the force of attraction between two bodies.  It is:

G is the universal gravitational constant, 6.67300 × 10-11 m3 kg-1 s-2
M is the mass of the heavier body (in this case, the Earth)
m is the mass of the lighter body (in this case, the satellite)
and r is the distance between the centres of gravity of the two bodies.

The second formula is the equation which describes the force required to keep an object moving in a circular path.

In this formula, F is the force which points towards the centre of the circle, m is the mass of the object (in this case, the satellite), v is the angular velocity (how quickly it's moving in a circle), and r is the radius of the circular path being described by the body.

The first thing I'm going to do is unpack v in the second equation.  v, the angular velocity, is defined in the same way as all speeds as the distance travelled divided by the time taken.  For a circular path, the distance is the circumference of the circle, and the time is the time taken to complete one circle (or orbit), T.
We can plug v2 into the formula for circular motion, and we can also equate the two formulae, since the force that will hold the satellite in orbit is the force of gravity.

By cancelling and rearranging, we have one expression for the radius of a geostationary satellite held in orbit by the Earth's gravity.

One thing to notice here is that the mass of the satellite, m, does not appear in our final expression.  The period of rotation of a satellite depends only on the height of its orbit.  The time taken to complete an orbit is only affected by the height of the orbit - the mass of the satellite is not important.

Anyway, we know the time taken for one orbit has to be 24 hours, and all the other components of the formula are constants, so we can plug them in and calculate r.

G is the universal gravitational constant, 6.67300 × 10-11 m3 kg-1 s-2
M is the mass of  the Earth, 5.9742 × 1024 kg
T is the time to complete one orbit, 24 hrs = 86 400 s

Therefore, r = 42,243 km.

However, this is not the altitude of the satellite from the Earth's surface.  Remember that when I first gave Newton's law of gravity, the r was the distance between the centres of gravity of the two bodies.  The size of the satellite is small enough to be ignored, but we must subtract the radius of the Earth from this value, to give the orbital height.  Radius of Earth = 6378.1 km, so orbital radius = 35,865 km.

Later edits of this post will include some nice diagrams, but for now I'm happy to have posted the result!

NASA gives the height as approx 35,790 km using a more exact value of the length of one day, while Wikipedia has a similar figure and more information on geostationary orbits.

In my next post... I'm not sure.  Possibly more on orbits generally, including calculating a Moon - Earth distance, or a Sun - Earth distance, or a practical experiment to determine g (acceleration due to gravity on Earth).

Sunday, 9 January 2011

Mathematical Problems, 3D - Pi from infinite polygon

In this final post on π (for the time being at least), I'm going to look at another way of calculating π, based on the principle I first used for calculating a minimum value for it.  Back then, I used a square inside a circle to give a minimum value, but it occurred to me later that it's possible to use a hexagon inside a circle (and we can show that the perimeter of a hexagon inside a circle is 6r) and that the figure would become more accurate if I could use a polygon with more sides.

What about a polygon with 8 sides, or 12, or 20, or n sides?  Consider the following diagram, where the line EF  is a side of a regular polygon ABCDEF which has all its corners on the circumference of a circle of radius r.  In this case, the diagram shows a regular hexagon, but the theory applies to any polygon which has n sides.

Since this is a regular n-sided polygon, the angle EOF is 360/n and the angle EOG is 180/n. Additionally, EGO is a right-angled triangle, so we can use trigonometry to solve this triangle. If we call the length of one side l, this is the line EF, and EG is l/2. Using trigonometry, we can see that sin 180/n = l/2 /r

This rearranges to give l, the length of one side, as l = 2r sin 180/n

And the total perimeter, P, of the polygon which has n sides of length l, is P = n 2r sin 180/n

Now, 2r =d, the diameter of the circle, so P = n d sin 180/n

And for a circle, π is the ratio P/d and for this polygon, P/d = n sin 180/n
The advantage of this is that we can immediately plug in a large value of n to give an approximation of π. Here are some values of n and π based on this formula (and one day I'll work out how to put a table into this blog).

n - π
100 - 3.141076
200 - 3.141463
300 - 3.141535
1000 - 3.141587
2000 - 3.141591
3000 - 3.1415920
10,000 - 3.14159260191
100,000 - 3.14159265307

I must say I like this method; it's simple trigonometry (not calculus, and not sampling either) and I was very surprised at how easy it was to obtain a reasonable value of π from a polygon with just 100 sides.

One of my regular readers has asked me to calculate the value of π for a polygon on the outside of a circle; I'll leave that as an exercise for the reader, and point you toArchimedes' method for calculating π - it's got a nice flash display for the calculation of the internal and external polygon. Another benefit of this method over the previous sampling method is that this is a one-off calculation - we can calculate π from a large number of sides without having to take a large number of measurements. No chance of crashing the spreadsheet then!

Next time, something different - geostationary satellites - what they are, and why they have to orbit at a specific height - and what that height is!

Friday, 7 January 2011

Maths Problems 3C: Estimating pi by sampling

In response to my first post about pi, my friend Chris Timbey pointed out that he's previously written a computer program which will estimate pi by determining if a random point in a square is also in a quadrant drawn within that square.  The diagram below shows how this would work.

Not only is pi the ratio of a circle's circumference to its diameter (C = pi x d), but the area of a circle is pi r2, so we have another way of approximating the value of pi.
The cartesian equation (using x and y) for a circle is x2 + y2 = 1. By taking random values of x and y and determining if the value of x2 + y2 is greater than or less than 1, we can calculate the ratio of the area of the quadrant to the area of the square.

This is where a computer comes in very handy.  The basic program works thus:
1.  Obtain random values of x and y between 0 and 1.
2.  Square x and y, and sum the two values.
3.  If the sum of x2 + y2 is less than 1, then count this as 'within the quadrant', otherwise count it as outside the quadrant.
4.  Repeat the first three steps a large number of times.
5.  Calculate the proportion of counts 'within the quadrant' to the total number of counts (both inside and outside the quadrant).
6.  Multiply this proportion by 4, since there are four quadrants in a circle.

For example:
Random value of x = 0.252
Random value of y=0.881
x2 + y= 0.840 so this is within the quadrant.

This takes a large number of iterations to produce an accurate result.

Here are my results; iterations on the left, value of pi on the right:
100  -  3.08
200  -  3.12
1000 - 3.212
5500 - 3.167272 (recurring)
10,000 - 3.13040, also 3.1736, 3.1312  (repeating the experiment with 10,000 new random pairs).
15,000 - 3.13040 (11739/15000)
Beyond this, my spreadsheet starts grinding to a slow and painful halt, but at least by 10,000 iterations it's mostly hitting pi to one decimal place.  This will always be an approximation, since it's based on a fraction (no matter how precise), and pi, being an irrational number, doesn't much like being expressed as a fraction!

Next time - approximation of pi based on a regular polygon with thousands of sides.

Thursday, 6 January 2011

Mathematical Problems 4: Close packed circles

While I was at university, I remember studying close-packed spheres and how they form in crystalline structures.  We were given the figure that close-packed spheres occupy the largest proportion of space available - more than they'd fill if they were arranged in squares, for example.

To explain what close-packed circles look like, here's a diagram.

Sometimes, they're described as hexagonally close-packed; as you can see from the diagram, the circles form hexagonal arrangements.  This makes it easier to calculate how much of the area the circles are filling... especially if we break the hexagon down into equilateral triangles...

Now, the triangle marked in bold contains three sixths of a circle - half the circle, in other words, which has an area of A = 0.5 x pi r2

The triangle is an equilateral triangle (all sides are equal, all angles are equal at 60 degrees).  I don't know any shortcuts for working out the area of an equilateral triangle, so I'll do it long hand:

Area = base x height x 0.5

The base = the diameter of a circle = 2 r
The height is found through trigonometry:  tan 60 = h / r   therefore h = r tan 60

Area = 2 r x r tan 60 x 0.5
Area of triangle = r2 x tan 60

Area of half a circle (contained within the triangle) = 0.5 x pi x r2

The proportion of the triangle's area which is covered by the semi circle = area semicircle / area triangle

Proportion = 0.5 x pi x r2  / r2 x tan 60  and the r2 cancel, so proportion = 0.5 pi / tan 60

pi / 2 =1.5707
tan 60 = 1.732

Proportion = 1.5707 / 1.732 = 0.9069 which is 91%.

Now I accept I haven't worked it out for packing by spheres in space, but I thought I'd start simple and work from there...  maybe next time!

Tuesday, 4 January 2011

Mathematical Problems, 3B - A lower value of Pi

Returning to pi, and this time using geometry to calculate a minimum value.  Instead of using a square around the outside of a circle, this method will use a square within a circle.  Last time we looked at square ABCD, this time, it's WXYZ.

Now the circumference of the circle is greater than the perimeter of WXYZ.  We know this for sure because the shortest distance between two corners of a square is the straight line that connects them, and the circle is a curved line and therefore must be longer.

If we call the centre of the circle O, then we can see that WOX is a right-angled triangle, and WO = OX = radius of the circle.

Using Pythagoras, we can determine the length WX:

WX 2 = WO2 + OX2

And since WO = OX = radius of the circle, r, then WX2 = 2 r2
And WX = sqrt (2 r2)  = sqrt 2 x r

Now, the full perimeter of the square is four times WX (since the square has four sides),
4 WX = 4 sqrt (2 r2)  = 4 x sqrt 2 x r

We want to describe pi in terms of the diameter of the circle, not the radius, so substitute d = 2r and this gives

Perimeter of square (4 WX) = 4 x sqrt 2 x d/2
Perimeter of square = 2 x sqrt 2 x d

We know the circumference of the circle is pi x d
The perimeter of the square is 2 x sqrt 2 x d
Therefore, pi is greater than 2 x sqrt 2  (approximately 2.82)

Combining this with the result from the previous post gives us the approximation

2 sqrt 2< pi < 4

In my next post, I'll work out how to use superscripts and square root signs in HTML (I hope) and I'll show a way of approximating pi using statistics rather than geometry.  In a future post, I'll also look at close-packed circles, and calculate how much of the available area they can fill.