described. A new modelling example will be given in the next section to illustrate how the modelling flow chart works.

3.4 The methodology in practice

We now need an example that demonstrates the methodology, showing how a structured approach can benefit the modelling process. For this purpose, we shall take a well-known situation of common experience.

Background to the problem

It is about to rain; you have to walk a short distance of about 1 km between home and college. As there is some hurry, you do not bother to take a raincoat or umbrella but decide to ‘chance it’. Suppose that it now starts to rain heavly and you do not turn back; how wet will you get?

This seems a simple matter of getting out of the rain as soon as possible but, if variation in the direction of the rainfall is taken into account, it may not follow that the best strategy is to run as fast as possible over the distance. We shall now attempt to model this problem with the help of the flow chart structure.

Box 1: Identify the real problem

Given particular rainfall conditions, can a strategy be devised so that the amount of rain falling on you is minimised? The model will be ‘deterministic’ since it will depend entirely on the input factors such as the following.

1.How fast is it raining?

2.What is the wind direction?

3.How far is the journey and how fast can you run?

We shall need to develop a formula for the amount of rain collected which is dependent on these factors. Suppose that the data available are as follows:

These data are typical for average behaviour but could be altered to cover more extreme cases. Box 2: Formulate a mathematical model

The first objective is to set up the simplest model possible.

Suppose that you run the whole kilometre journey at 6 m s −1 . Therefore,

Now ignore the rain direction, and merely consider rain collection from the given data of 2 cm h −1 , i.e. 2/3600 cm s −1 . Thus, over the whole journey of 167 s,

It is now necessary to give some data about the surface area of the body which is being rained on. Suppose for simplicity that the human frame is represented as a rectangular block 1.5 m high, 0.5 m across and 0.2 m deep. Then

i.e.

On the assumption that all these surfaces collect rain, then

(3.1)

(So it is like having about two bottles of wine poured over you!)

The rules have been broken somewhat here by rushing through a quick result for illustration purposes, and achieving an answer that is reasonably plausible as well. It will be useful now to go back to the flow chart and to proceed through in a more detailed fashion incorporating this time all the relevant features, thus developing a second and more general model.

Assumptions

It has already been decided that the human frame can be represented by a rectangular block. A diagram helps in explaining the situation to be modelled and this is shown in Figure 3.2. Other assumptions which we shall make are that the rain speed is constant throughout and also that you move through the rain at a constant speed.

Some of these quantities shown in Table 3.1 are not variables at all but have numerical values from the data provided. It is nevertheless convenient to retain

Figure 3.2

Table 3.1 List of the factors

symbols while the model is being constructed. In fact, r, θ, v, t and C are variables while the others are ‘parameters’ in the sense that they do not vary for this particular situation. There is a need to distinguish between the velocity of rain and the collection of rain. If rain was a continuous flow of water (like a river), then the velocity of the rain would give us the collection rate over a certain area. However, this is clearly incorrect since rain is a stream of droplets which gives rise to the idea of rain intensity.

A rain intensity factor I is introduced to deal with this situation. From the data given above, the rain speed is 4 m s −1 and the rain collection is taken as 2 cm h −1 . However,

(3.2)

compared with the collection rate of 2 cm h −1 , i.e. the ratio is 7.2 × 10 5 . This discrepancy is allowed for by introducing I as the measure of rain intensity. For these data, I = 1/(7.2 × 10 5 ). Thus generally, if I = 0, it has stopped raining, while increasing I gives heavier rain. Ultimately, I = 1.0 would correspond to continuous flow like a river.

Figure 3.3

We are now ready to draw up the equations relating the variables listed above. There is no question of the laws of motion, probability effects, etc., to worry about, merely an evaluation of the rain collection capacity. With speed taken as constant, then

(3.3)

Also the key factor now to be considered in assessing how wet you get is the relative direction of the rain with respect to your direction of travel. These relative effects are conveniently shown in Figure 3.3.

Now, since the rain is coming down at an angle, we can see that in any situation only your top and front will be getting wet. This is in accord with experience.

Box 3: Obtain the mathematical solution

The amount of rain falling on you will now be calculated in the two cases. First, consider your top surface area, for which

and

Therefore,

In time D / v,

(3.4)

Next consider your front surface area, for which

and

Therefore,

In time D / v,

(3.5)

By addition of equations (3.4) and (3.5), the total amount C of rain collected is

(3.6)

It is now a question of extracting from equation (3.6) the information that we want on how wet the journey will be. First, values for some of the quantities used can be substituted. From the earlier data, we have h = 1.5, w = 0.5 and d = 0.2; also r = 4 and D = 1000.0. Also substituting the value for I from equation (3.2), then

(3.7)

The variables retained are v and θ, since you can choose v and θ is the rain direction, which we shall want to vary in investigating the mathematical solution. Thus, given θ, what v is chosen so that C is minimised?

Box 4: Interpret the mathematical solution

Equations (3.6) and (3.7) are now examined. First, note that, if the rain intensity I is zero, then C = 0 which means that you stay dry. Second, the value of θ will determine whether the rain is facing you or blowing in from behind. We shall evaluate equation (3.7) to show what happens in particular cases.

Case 1: θ = 0°

In this case, as θ = 0°, the rain is falling straight down. From equation (3.7),

This expression is smallest for the largest possible value of v, i.e. v = 6 in this case. Substituting v = 6 gives

(3.8)

Case 2 θ = 30°

In this case, as θ = 30°, the rain is driving in towards you. From equation (3.7),

Again this is at its smallest when v = 6, i.e.

(3.9)

Case 3 Negative θ

Now suppose that the rain is coming from the rear so that θ is negative. Taking θ = −α, say, then from equation (3.7), we obtain

This expression can become negative for α sufficiently large, which means that the model must be examined more carefully since it is not possible for C to be negative! It is best to return to equation (3.5) to analyse the situation. With θ = −α, there are two cases to consider according to how fast you move through the rain.

In the first case, for v < r sin α, it rains on your back and so

This gives the total collection formula now as

Putting in the data again, we get

(3.10)

If you now increase your speed to 4 sin α, this expression is reduced to

which corresponds to the top rain amount only. Thus, if the rain is at an angle of 30° from the rear, you should walk at 4 sin 30° = 2 m s −1 , in which case

(3.11)

Effectively this means that you are just keeping up with the rain. If the speed falls below 2 m s −1 , then the rain collected increases due to that falling on your back.

In the second case, should you be unable to resist the temptation to run at a speed of more than 2 m s −1 , then you will be catching up with the rain. This is the case where v > r sin α and the contribution from equation (3.5) is now

So the total amount of rain now collected is

When v = 6 and α = 30,

(3.12)

Box 5: Compare with reality