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 2	Run 3
0.162	8.45	8.31	8.35
0.237	10.09	9.90	10.04
0.321	11.63	11.67	11.66
0.344	12.09	11.90	12.17
0.410	12.98	12.95	13.00
0.475	14.23	14.13	14.03

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!).

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 to Archimedes' 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!

My Series on Determining The Value Of Pi

1. An upper limit on the value of pi simple but a place to start!
2. A lower limit on the value of pi also simple, but leading to better approximations.
3. Approximating pi by sampling
4. Approximating pi using an infinite polygon

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 + y² = 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.

My Series on Determining The Value Of Pi

1. An upper limit on the value of pi
2. A lower limit on the value of pi
3. Approximating pi by sampling
4. Approximating pi using an infinite polygon

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, then extend this to close-packed spheres.

My Series on Determining The Value Of Pi

1. An upper limit on the value of pi
2. A lower limit on the value of pi
3. Approximating pi by sampling
4. Approximating pi using an infinite polygon

Mathematical Problems, 3A - The Value of Pi

Back when I was doing my A-levels, I remember learning about how it's possible to evaluate pi in various ways, one of which was through calculus.  I can only remember the basics, and I'm sure I can't recall how to do it now - at least not without some help!

The value of pi is the ratio of the circumference of a circle (the distance all the way around the outside of the circle) divided by the diameter (the straight-line distance from one side to the other, through the centre).  It's been known historically to be about 3, but I'm going to make some approximations from first principles.

Firstly, drawing a square around a circle, so that the square touches the circle.  See the diagram above, with the square ABCD around the circle.

If the diameter of the circle is d, then the perimeter of the square is 4d.  We can see by inspection that the square's perimeter is longer than the circle's circumference, and we've called the circumference pi d.  Therefore, we know that pi is less than 4.

It's not a dramatic result, I know, but it's a start!

Next, we need to look at setting a minimum value for pi, and we'll look at this next time.

My Series on Determining The Value Of Pi

1. An upper limit on the value of pi
2. A lower limit on the value of pi
3. Approximating pi by sampling
4. Approximating pi using an infinite polygon