FIG. 12.1
Definition sketch for the rainfall problem.
Analysis of the Problem
The basic dimensions and units of our problem are the following: mass of water (kilograms), distance (meters), and time (seconds). The main variables are a person's velocity Up (m/s), raindrop velocity Ur (m/s), and rainfall rate r (m/s or mm/hr). To characterize the person's dimensions we select an esthetically unattractive but mathematically simple rectangular prism with height H, width B, and thickness D. The person moves in a direction parallel to the D dimension, that is, perpendicular to the BH plane.
A definition sketch is presented in figure 12.1. We note that the combination of velocities Up and Ur establishes a resultant velocity , inclined at an angle θ to the BH plane.
The volume of rain crossing plain ab per second is
This is the amount of water collected on the person's head and shoulders (i.e., the top of the prism). Similarly, the volume of rain crossing plane bc per second is
and this amount of water is retained on the person's front (i.e., the front of the prism).
Next we need an expression to describe the volume of water q3 collecting on the person's shoes and socks (i.e., the bottom of the prism) due to pavement splashing. This is a more difficult problem. A complete examination of this feature would involve three aspects: (a) dimensional analysis, (b) mathematical modeling, and (c) experimental investigation.
Dimensional analysis is a very useful technique that starts, in our problem, with the qualitative statement that the volume of splashing water may depend on a number of parameters. That is, we suppose that
This expression indicates that the magnitude of q3 is probably affected by the depth of water on the pavement (d), by the bottom dimensions (B and D), by the person's velocity (Up), and by the rainwater's mass density (ρ), dynamic viscosity (μ), surface tension (σ), and gravitational force (g).
A dimensional analysis of the variables listed in equation (12.3) yields the relationship
where Re, We, and Fr are, respectively, the so-called Reynolds, Weber, and Froude numbers. These quantities are very important in the field of fluid mechanics. Barenblatt (1987) is an easy-to-understand book that deals with all of these topics.
Augmenting the techniques of dimensional analysis are straightforward mathematical analysis and experimental studies. With regard to our rainfall problem, reference is made to the experiments of Mutchler and Larson (1971) concerning the mechanics of raindrop splash erosion of soil. Utilizing their results, as well as those of dimensional and mathematical analyses, we establish that the volume of water splashing onto the bottom plane BD per second is
where c is a dimensionless constant, d is the water depth on the pavement, and is the kinematic viscosity of water.
Utilizing the relationships and we combine equations (12.1), (12.2), and (12.5) to obtain
which expresses the total volume of water collected on the person per second. The total number of seconds during the journey of length L is Using , the total volume of water collected on the trip is
This equation can be put into the following dimensionless form:
in which is the wetness factor, S = Up/Ur is the speed ratio, n = D/H is the aspect ratio, and finally f = cdUp2/vr is the splash number.
A word about this last parameter, the splash number. This is simply an index or scale to describe how wet the pavement or ground surface is. In this sense, it is like the Dow-Jones index for the stock market, the Richter scale for earthquakes, or the Beaufort number for a sea-wind condition. A short list of splash numbers is presented in table 12.1.
The main result of our analysis is equation (12.8). This expression indicates that the amount of wetness, Z, depends on the person's speed as a fraction of raindrop velocity. Several plots of (12.8) are shown in figure 12.2 for various values of splash number f, with a constant value of aspect ratio, n = 0.15.
TABLE 12.1
Splash number f for various surfaces
surface description value of f
Dry pavement or dry grass 0
Damp grass 5
Wet pavement 10
Puddly pavement 15
Gutter or very shallow stream 20
FIG. 12.2
Wetness factor Z as a function of the speed ratio S for various values of the splash number f.
In the figure we note several things. First, if f = 0, that is, the pavement or grass is dry, there is no optimum point on the curve. In this case, a person should run as fast as possible to minimize the amount of rain collected. However, for values of f > 0 there are optimum speed points that do provide minimum wetness. Furthermore, as f increases, the optimum speed decreases. In this case, a person should adjust his or her speed in order to minimize wetness.
Now one could determine the “best” speed ratio directly from figure 12.2. However, if we utilize the methods of differential calculus we can be more precise. This involves a step known as “taking the first derivative” of equation (12.8). Basically, this procedure allows us to compute the “slope” at any point on one of the curves of figure 2.12. We note that when the slope is zero (i.e., a line parallel to the S-axis), we have identified the value of S corresponding to the minimum value of Z. Of course, this is what we are after.
If we take the first derivative of equation (12.8) and set it equal to zero, we obtain
where Sm is the value of the speed ratio corresponding to the minimum value of the wetness factor, Zm. Now in a great many problems involving determination of minimum values it is possible to solve an equation analogous to (12.9) to obtain the desired quantity. In our problem, however, we cannot obtain an explicit solution for Sm. Nevertheless, for specified values of f and n, it is not difficult to compute Sm from equation (12.9) by trial-and-error calculation. Table 12.2 lists the minimum points for the curves displayed in figure 12.2. The dashed line in the figure is the locus of these points.
There is considerable information about the velocity of raindrops. A typical study is that of Dingle and Lee (1972), who show that the terminal velocity of a drop depends primarily on its diameter. A list of raindrop velocities at sea level, computed from their equation, is shown in table 12.3.
TABLE 12.2
Optimum velocity coordinates
TABLE 12.3
Raindrop velocities at sea level
Diameter, mm Ur, m/s
1.0 3.9
2.0 6.6
3.0 8.1
4.0 8.8
5.0 9.1
Of course, in a particular rainstorm there is a wide distribution of raindrop diameters and hence raindrop velocities. If the rainfall rate r = 25 mm/hr, the mean diameter is about 1.9 mm, and if r = 100 mm/hr, the mean drop size is approximately 2.5 mm. Accordingly, a typical raindrop velocity is around 6 to 7 m/s.
An Example
A young lady leaves her office at the end of the day to face her customary (L = 500 m) trip to the train station. Unfortunately, there is a pouring-down rain. Indeed, it has been raining all afternoon; the soggy route to the station can best be characterized as “puddly pavement” (f = 15).
The young lady's dimensions, brutally simplified, are B = 55 cm, D = 24 cm, and H = 160 cm; accordingly, n = D/H = 0.15. Before she steps out into the rain to begin her trek, she quickly ascertains that the rainfall rate is r = 25 mm/hr and the raindrop velocity is Ur = 6.5 m/s.
Then she makes the following rapid calculations. From table 12.2 she determines that Sm = 0.278 and Zm = 2.110. Since Sm = Up,m/Ur and Ur = 6.5 m/s, she easily computes that her jogging speed to get least wet should be Up,m = 1.81 m/s = 5.93 ft/s. At this speed she will be able to reach the train station in T = 500/1.81 = 276 seconds = 4.6 minutes.
As she jogs to the station, she remembers the first of the relationships following equation (12.8):
Making certain that all quantities are in consistent (MKS) units, she substitutes the numbers into (12.10):
to determine that of water has collected or will collect on her person during
her jog. The weight of this water is 991 grams or about 2.19 pounds. Finally, from the relative values of the terms of equation (12.8) she determines that 224 grams (24.6% of the total) of water collects on her head and shoulders, 453 grams (45.7%) on her jacket and slacks, and 294 grams (29.7%) on her shoes and socks.
This point concerning distribution of collected water is illustrated in figure 12.3 for the case f = 15 and n = 0.15. We note from the figure that the Z1 and Z3 curves descend and ascend, respectively, with increasing speed ratio S. However, the Z2/Z (frontal wetness) curve possesses a maximum point. To determine this point we simply construct the ratio of the second term of equation (12.8) to the sum of the three terms. Then, as before, we take the first derivative of the resulting expression and equate it to zero. The answer is
FIG. 12.3
Percentage distribution of collected rainfall. Top plane, frontal plane, Z2, bottom plane, Z3. Splash number f = 15; aspect ratio n = 0.15.
in which S* is the speed ratio corresponding to the maximum value of Z2/Z. In our example, S* = 0.247, (Z2/Z)* = 0.459, and Z* = 2.116. We note that the critical point (S*, Z*) is very close to the more important critical point (Sm, Zm).
Back to our clever young lady. Clearly, if she is wearing a new hat or has a new hairdo she should, according to the Z1/Z curve of figure 12.3, run as fast as possible. Alternatively, if she has a new pair of shoes then, from the Z3/Z curve, she should walk quite slowly. Finally, if she has a new hat, a new pair of shoes, and a new jacket and slacks, she should take a taxi to the station.
13
Great Turtle Races: Pursuit Curves
We have some small turtles that are very intelligent and well trained; they are what we call smart turtles. They can be ordered to stay put, to walk or trot along a straight line or some curved path, and to follow other specific instructions. They refuse to swim and they cannot fly; hence their movements are restricted to zero, one, or two dimensions, that is, to a point, a straight line, or a plane, respectively.
With that as a prologue, we examine what is termed the curve of pursuit problem. This is quite an ancient problem in mathematics. Evidently it was first studied by Leonardo da Vinci in the early sixteenth century, and it has attracted the attention of mathematicians ever since.
With reference to the definition sketch of figure 13.1, we have two curves in the x-y plane. Curve A is the path of the so-called pursuer and curve B that of the pursued. That is, at any particular moment the pursuing turtle (or frigate, cheetah, bad guy on horseback) is at point P(x, y) and moves along path A at a velocity u. At the same moment, the pursued turtle (or pirate ship, gazelle, stagecoach) is at point Q(m, n) and moves along path B at a velocity v.
The curve of pursuit problem follows: if the pursuer P always heads directly toward the pursued Q, and if the path B of the pursued is specified, then what is the path A of the pursuer?
FIG. 13.1
Definition sketch for the curve of pursuit problem.
The key phrase is “…always heads directly toward…” This means that the slope of the pursuit curve is
where dy/dx is the slope of curve A. The first derivative of this equation yields
In addition, we have the relationship u = kv, where u and v are the velocities of the pursuer and pursued; k is a constant which may be either larger or smaller than one. Since u = ds/dt and v = dσ/dt, where ds and dσ are incremental distances along the two curves, we obtain
which can be written in the form
Finally, we assume that the path of the pursued, m = g(n), is a known relationship. This information, along with equations (13.2) and (13.4), gives us what we want: the path of the pursuer, y = f(x).
Sounds easy? Not always. Even with some substantial simplifications already (e.g., a plane surface and constant velocities) this can be a difficult problem. For example, suppose we have a duck paddling at constant speed v around the periphery of a large circular pond. A small alligator, at the center of the pond at time t = 0, sights the duck and swims with velocity u directly toward the tasty meal. Under what conditions will the alligator catch the duck, and what will be the alligator's path?
It turns out that this is not a simple problem. In fact, as Davis (1962) and Melzak (1976) show, an exact answer is impossible though, of course, numerical and graphical solutions are available. As we might expect, when k > 1 (the alligator swims faster than the duck) capture occurs. However, when k < 1 the duck is safe; the frustrated alligator simply swims around in a circle, a so-called “limit cycle,” of radius kR, where R is the radius of the pond.
We go to an easier problem. As shown in figure 13.2, two turtles are situated on a horizontal plane. Initially turtle A, the pursuer, is located at the origin (0, 0) and is facing east. Turtle B, the pursued, is at (a, 0) and is facing north. At t = 0, the turtles commence the race. Pursued turtle B has instructions to proceed at constant velocity v along the straight line m = a. Accordingly, dm/dx = 0 and (13.2) becomes
and so from (13.4) we obtain
This differential equation is solved to provide the answer we want, y = f(x).
FIG. 13.2
Curve of pursuit with u = v (k = 1).
For the first race the two turtles are ordered to move at the same velocity, that is, u = v, and so k = 1. In this instance, the solution to equation (13.6) is
This equation describes the path of pursuing turtle A and is displayed in figure 13.2 with a = 100 cm and u = v = 1.0 cm/s. Not surprisingly, turtle A never catches up with turtle B. However, we note that at the start of the race there was a distance a = 100 cm between them. Because turtle A took the indicated shortcut, he eventually closed the gap to 50 centimeters—but no more.
For all ensuing races, the turtles are instructed to move at speeds such that k ≠ 1. In this case, we obtain the following solution to equation (13.6):
If k < 1, turtle A will never capture turtle B. However, if k > 1, capture will always result; this will occur when x = a. So from (13.8) we immediately have
where yc is the capture point. For example, if k = 1.5 then yc = 6a/5; if k = 2, yc = 2a/3; and if k = 3, yc = 3a/8.
Reversing the question, with what speed must turtle A move to capture turtle B at a specified yc? Solving equation (13.9) for k gives
Suppose we specify that yc = a. Then from this equation we get k = (1/2)(1 + √5) = 1.61803.
What a nice surprise. Once again, we run across one of history's truly famous numbers. You will recall we studied it in an earlier chapter: the golden number or golden ratio, ø = (1/2) (1 + √5) = 1.61803.
One thing can be said with certainty about ø: it shows up in the strangest places. Its appearance here as the speed ratio, k = ø, in our turtle pursuit problem indeed demonstrates the amazing characteristics of ø to show up unexpectedly. Numerous other remarkable appearances and features of this fascinating number are given by Huntley (1970) and by Kappraff (1990).
FIG. 13.3
Curve of pursuit with k = 1/2(1 + √5)
A plot of this particular pursuit curve is shown in figure 13.3. The capture point is yc = a = 100 cm and turtle B's speed is v = 1.0 cm/s. Turtle A, with speed u = kv = 1.618 cm/s, covers a distance of S = 161.8 cm in a path that is almost circular. The precise path, of course, is determined by substituting k = (1/2) (1 + √5) into equation (13.8).
In the preceding pursuit curve analysis we stipulated constant velocities of the pursuer and the pursued. A closely related problem is one that stipulates constant distance between the two.
As before, equation (13.1) applies. The pursuer always heads directly toward the pursued. However, this time we impose the condition of constant distance:
FIG. 13.4
The tractrix curve with separation distance a = 100 cm.
where, as indicated in figure 13.1, L is the required distance between P(x, y) and Q(m, n). Again, the path of the pursued, m = g(n), is specified. This information, along with equations (13.1) and (13.11), determines the path of the pursuer, y = f(x).
We consid
er the same relatively easy problem as before: m = a. In this case,
Solving for (n – y), substituting into (13.1), and integrating gives the solution
This equation is called the tractrix; its shape is shown in figure 13.4. Recall that we ordered turtle A, starting at the origin, always to maintain a constant distance a from turtle B as the latter moves along. However, the same problem would be presented had we simply instructed turtle B to attach a string to turtle A and drag him along.
The tractrix has numerous interesting properties and these are discussed by Lockwood (1961) and by Steinhaus (1969). In equation (13.13) we make the substitution r = a – x, to obtain
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 10