Saturday, June 27, 2009

The Mathematics of Measurement

We are surrounded by measurement devices. The modern world is abound with instruments providing values for temperature, humidity, weight, time, speed, force, pH, radioactivity, power, height, voltage, current and even our attractiveness to the opposite sex ! It is a naive person indeed who accepts a measurement on face value. Accurate and reliable measurement of any quantity is difficult and errors, whether they be random or systematic, are normal.

For example, if you are told that the temperature of your house is 26.56632 C, a thinking person would ask:

How do you know the value to such accuracy ?

At what time did you take this value and does it vary with time ?

Is the value an average of many values taken from many positions in the house or is taken from a set position in the house ?

If taken from one position, is this value representative of the "house" as a whole ?

If it is an "average" value, how precisely is this average calculated ?

Are there any corrections made for the way the thermometers are distributed in the house ?

For example, if there are ten thermometers in the basement and only one in the front lounge, wouldn't a straight averaging of these values give a distorted figure ?

How much variation is there in the values "averaged" ?

Does this variation in the case of multiple values relate to the position of the measurement or is apparently random ?

Have the thermometers been calibrated against a standard ?

These questions all converge onto two main points: What does the measurement tell us about the system we are studying and how accurate is the measurement ?

Mathematics is highly useful in evaluating many of these issues. For example, statistics can be used to evaluate variation in measurements and calculus can be used to "average" values and quantify variation. Above all, mathematics can ease the hand waving and provide quantifiable answers to these questions.

For example, imagine you are calculating the distance travelled by a trolley moving at constant velocity, using the very simple formulae:

distance (m) = velocity (m) x time (m) or D = V x t

The velocity has been measured as 22.35 m/s and the time has been measured as 10.00 s. What is the error associated with this calculation ?

Given there crude measurement, we can assume that there the random error of the measurement is one half of the last graduation of the device. Simply put, if you are using a mm graduated ruler, we can assume that the error associated with the rule is +/- 0.5 mm. This may not be correct, for example, if my sight is poor the error may become larger or if the graduations on the ruler have been badly printed, this assumption may also be too low. Another possibility is that I incorrectly placed the ruler and introduced a large systematic error (as opposed to the "random" errors I have been discussing) However, without any more information the "half the smallest graduation" principle is a reasonable starting point for our deliberations.

Therefore, in this calculation, we can assume the that the velocity is 22.35 +/- 0.005 m/s and the time was 10.00 +/- 0.005 s.

Method 1.

We know from the fundamental derivation of calculus that dD/dt is approximately equal to (small change in D/small change in t) or more simply put the gradient of the curve at a particular point is approximately equal to the ratio of a small change in the resultant variable to a small change in the independent variable. This principle can be used to approximate the error using the following formulae:

error in D = dD/dt x error in t = V x error in t = 22.35 x 0.005 = 0.11175 m.

Therefore, we have the result of 223.5 +/- 0.1 m. This approach ignores any error in the V value, as it treats the problem as being D = f(t). This would be fine if V was truly constant or the error associated with V was very small compared to that associated with t. This approach is particularly useful when the function is complex (e.g. D = Vcos (t^2)) and other methods are difficult to use.

Method 2.

We estimate the error by calculating the answer using the most pessimistic values and take this answer away from the value calculated without considering the error. In this case:

(22.355 x 10.005) - (22.35-10.00) = 223.66175 - 223.5 = 0.16175m

Therefore, the answer is 223.5 +/- 0.16 m.

Method 3.

It can be shown by a simple proof, that when the errors associated with measurements are relatively small, that when two values are multiplied together, the relative error (absolute error/value) of the new value is the sum of the relative errors of the original values. In our example, this results in:

error in D = D ((error in V/V)+(error in t/t)) = 223.5 ((0.005/22.35)+(0.005/10)) = 0.16175 m

Therefore, the answer is 223.5 +/- 0.16 m.

Clearly, the first method underestimated the error and the results from the final two techniques should be used in this case. This simple example illustrates some of the complexity in determining what a measurement really means and how mathematical approaches are useful and dealing with the complex issues associated with measurement.

1 comment:

1. Sometimes I'm thinking that university degrees are products that need to be sold. Universities get a lot and a lot of money from students but the degrees do not guarantee them jobs.

In fact, there are many graduates of all universities which include the Group of 8 graduates who are jobless.

Guess what? the government will even increase the number of CSP places (previously know as HECS) in 2010 (it will clear off all the full fee places).

I don't know where this country is going but don't you feel guilty, sometimes, Geoff that you're telling your students that they will become this and that and finally many of them will be jobless (or get a job as a sales person, door to door, or a call center person).

I know the academics get plenty of money from teaching (and researching) at universities (yet they are relatively easy jobs compared to the ones in the industries).

Stanley Kosasih, B.Eng (Biomedical) Swinburne Grad.Cert.Ed (Dev.Studies) Monash