In the coordinate system of our example, with Singapore at λ = 105°E, the equation of the flight curve between λ = 105°E and λ = 15°E is given by equation (24.14) of the previous chapter. With ø0 = 0 and q = 1, we have
in which λ = λ* – 15°, where λ* is the longitude east of the Greenwich meridian of a point on the curve.
For the portion of the flight between λ = 15°E and 75°W, we employ equation (24.16). With ø0 = 0 and ρ = arctan√2, we get
where, now, λ = λ* +15° and λ* is the longitude west of Greenwich of a point on the path.
FIG. 25.3
The ø0 = 0 route westward from Singapore.
From equation (24.18), with ø0 = 0, the maximum latitude occurs at λm = arctand 1/√2); this longitude corresponds to λ* = 20.26°W. The maximum latitude is øm = ρ = 54.74°N and S.
The coordinates of the flight path are listed in table 25.1.
The path of this fantastic journey is shown in figure 25.3. Initially, the trip takes us across the Indian Ocean, the island of Madagascar, the tip of southern Africa, and the south Atlantic. Then we fly over Argentina, northern Chile, and Peru and reach the midpoint of the journey at a point about 400 kilometers east of Quito, Ecuador.
Homeward bound, we cross Colombia, the Caribbean, and Haiti and then follow the Gulf Stream over most of the north Atlantic. After skirting the southern tip of Newfoundland, we reach our maximum northern latitude approximately 800 kilometers directly south of Iceland. This is followed by the grand tour of Europe, starting with Ireland and England and ending with Greece. We then fly over northeastern Egypt, Saudi Arabia, and Yemen, and finally have another long flight over the Indian Ocean before arriving back in Singapore.
TABLE 25.1
Coordinates of the path of the ø0 = 0 route westward from Singapore
λ* ø
105°E 0.0
90°E 1.4
75°E 5.6
60°E 12.7
45°E 22.3
30°E 33.7
15 E 45.0
0 52.0
15°W 54.6
30°W 54.1
45°W 50.3
60°W 41.3
75°W 0.0
From our earlier analysis, we determined that S/R = 13.68 for the ø0 = 0 curve. Since our trip is only half the total and with R = 6,370 km, the length of our journey, S, is 43,570 kilometers. The total area enclosed by the route is one-fourth the area of the world, or 127.5 million square kilometers.
The map showing our journey in figure 25.3 is the Robinson projection of the world. If you want to know more about this projection—and a great many more—you should consult the excellent album of map projections by Snyder and Voxland (1989).
Before we leave the subject, a word of caution. Check out Bob's Air Tours before you go on this long and expensive trip. For example, they advertise “We Take You All Around Africa?” Do they ever! Look at the map.
Finally, why not take an hour or two sometime and make a plot of Bob's “ø0 = 0 Route” eastward from Singapore. Where does it go? Where in the world are the points of maximum latitude? This assignment is easy; you have all the numbers in table 25.1.
This is enough! We have carried out extensive analyses of topics relating to baseball seams and tennis ball grooves. Topics involving basketball seams do not present much of a problem. We shall save the investigation of golf ball dimples for another day.
26
Lengths, Areas, and Volumes of All Kinds of Shapes
An intriguing book you might want to examine sometime is Engineering in History, by Kirby et al. (1990). One thing becomes quite apparent as you read this informative text: for thousands of years, humans have been making calculations about things they build.
The Great Pyramid of Cheops in Egypt is an excellent example. Constructed over 4,500 years ago, this structure had a height H of 480 feet and an average side length L of 750 feet. So its volume V = HL2/3 was approximately 90 million cubic feet. This required 2.1 million blocks of limestone, each block measuring 3.5 feet on a side and weighing over 6,000 pounds.
Of course, the ancient Egyptians did not use our units of pounds and feet; they had their own units of measurements. Never mind. The point is that in order to build the enormous pyramids, they had to compute numerous quantities. With these observations, we begin our own calculations about some interesting shapes.
The Washington Monument
Although it may not be the most beautiful structure in the nation's capital, the Washington Monument is certainly the most familiar, the tallest, and probably the most impressive. It is 555 feet 5 inches in height and is made entirely of marble and granite. As its name implies, this magnificent structure is a memorial to America's first president.
Following many years of deliberation and planning, the cornerstone of the monument was laid on July 4, 1848 during the term of office of President James Polk. The original design called for a very elaborate Parthenon-type structure as the base for a 500-foot square shaft with a star on the top. The plan called for the shaft to be built first; the base was to be constructed later on.
Things went along fairly well for the next five years. Unfortunately, after the structure had risen to a height of 152 feet, funds ran out in 1854 and for the ensuing twenty-five years there was no further progress on construction. Eventually, in 1880, Congress appropriated the necessary funds and work on the monument was resumed. The earlier plan for the fancy base was discarded and the design of the shaft was somewhat modified. The monument was finally completed in December 1884. A photograph of the imposing structure is shown in Figure 26.1.
At the top of the 500-foot tapered column is a 55-foot pyramid-ion and on top of that is a nine-inch aluminum capstone. An elevator operates to the 500-foot level but there are 897 steps for people who prefer to walk up and down.
What Is the Surface Area of the Washington Monument?
The architectural shape of the monument is an obelisk; notable examples of such structures are seen among the monuments of ancient Egypt. A drawing of the structure is shown in figure 26.2, in which the proportions are substantially distorted so that dimensions can be displayed more clearly.
We now carry out calculations to determine the surface area.
The Sides of the Main Column. There is a slight taper to the main column. From the dimensions given in figure 26.2, we easily determine that the slant length of the sides is ft. So the area of each of the four trapezoidal sides is given by (1/2)(55.0 + 34.4)(500.5) = 22,372 ft2. Consequently, the total surface area of the main column is A1 = 89,488 ft2.
FIG. 26.1
The Washington Monument. (Photograph provided by National Park Service.)
The Pyramidion at the Top. The slant length of the sides is Therefore the area of each of the four triangular sides is (1/2)(34.4)(57.6) = 991 ft2. Hence, the total area of the pyramidion is A2 = 3,964 ft2.
FIG. 26.2
The Washington Monument. Not to scale. Dimensions in feet.
Total Surface Area. If we now add A1 and A2, we obtain A = 93,452 ft2. This is the total surface area of the monument.
What Is the Total Volume of the Washington Monument?
To simplify our computations for the volume, we (a) temporarily remove the pyramidion at the top and (b) extend the tapered sides of the main column until they meet high in the sky. The geometry of this situation is shown in figure 26.3. Next, utilizing the geometrical property of similar triangles, we get the equation
We easily solve this equation to obtain H = 1,336 ft. This height is more than double the actual height of the monument, and indeed exceeds the height of the Empire State Building in New York.
FIG. 26.3
A sharp-pointed Washington Monument. Not to scale. Dimensions in feet.
Back to our computations: The volume of a pyramid is given by the expression V = (1/3)BH, in which B is the area of the base and H is the height.
The Main Column. First, we calculate the volume of the entire—but mythical—1,336-foot pyramid. Acc
ordingly,
V1 = (1/3)(55.0)2(1,336) = 1,347,133 ft3.
Next, we compute the volume of the top—but also mythical—pyramid of height 1,336 – 500.4 = 835.6 ft. We have
V2 = (1/3)(34.4)2(835.6) = 329,605 ft3.
So the volume of the main column is V1 – V2 = 1,017,528 ft3.
The Pyramidion at the Top. The volume of the 55-foot pyramid at the top of the monument is
V3 = (l/3)(34.4)2(55.0) = 21,695 ft3.
Total Volume. From the preceding results, we easily obtain the total volume of the monument: V = 1,039,223 ft3.
How Many Golf Balls Would Fit Inside the Washington Monument?
This is something you have always wanted to know, right? We hasten to add that we are not being frivolous or disrespectful when we ask this question. As we shall see in a moment, we are simply utilizing this physical setting to pose an interesting problem in solid geometry.
It is assumed that the entire structure is empty (i.e., no stairs, no elevator) and that the walls are of zero thickness (in fact, they are 15 feet thick at the base). So the total volume is V = 1,039,223 ft3. The diameter of a golf ball is D = 1.65 in and so its radius is R = 0.825 in. Consequently, the volume of a golf ball is υ = (4/3)πR3 = 2.352 in3 = 0.001361 ft3.
Here comes the problem. No matter how you arrange a collection of uniform-diameter spheres in a given space, only a certain fraction of the space will be occupied by the spheres (i.e., golf balls); the remaining fraction—the so-called pore or void space—will be occupied by air.
So we have the following problem in solid geometry. What arrangement of uniform-diameter spheres, in contact with other spheres, will provide (a) the least dense and (b) the most dense packing?
Well, it turns out that the least dense is a cubic packing. It has a porosity (ratio of pore space to total space) of n = 0.4764. On the other hand, the most dense is termed the rhombohedral packing with porosity n = 0.2595.
This kind of information concerning the porosity of what are called “packed beds” is very important to chemical and petroleum engineers in the design of columns and tanks for the processing of petrochemical products. For these purposes, engineers usually select porosity values in the range n = 0.35 to 0.38.
Back to our problem: Using a total volume V = 1,039,223 ft3 and selecting a porosity n = 0.37 means that the volume occupied by the pores (i.e., air) is Vp = (0.37)(1,039,223) = 384,513 ft3. Consequently, the volume occupied by the golf balls (i.e., spheres) is Vs = 654,710 ft3.
Since the volume of a single golf ball is υ = 0.001361 ft3, the total number of balls that would fit inside the Washington Monument is
This is quite a few golf balls.
A RESEARCH ASSIGNMENT. Perhaps you would like to examine more closely the solid geometry problem concerning the least dense and the most dense packing arrangements of uniform-diameter spheres. Here's a suggestion for some simple experiments: glue or tape some ping-pong balls together to construct a few layers of spheres in various patterns. Then see how the layers fit together to provide loose and dense packings.
Can you express these geometrical patterns in terms of the mathematics needed to compute the porosity? Suggested references are Steinhaus (1969) and Greenkorn (1983).
Blimps
We are now going to calculate the volumes and surface areas of a few things with which we are all familiar. We start with blimps. These are the big airships we frequently see sailing over football stadiums, along crowded beaches and numerous other places.
The largest airship—blimp—ever constructed was the U.S. Navy's ZPG 3-W. Four of these were built for the navy by Goodyear in the late 1950s. We shall calculate some things about these enormous blimps after we develop the mathematical relationships needed for our calculations.
FIG. 26.4
Definition sketch for an ellipse. We assume that a blimp has an elliptical shape.
Ellipses, Ellipsoids, and Blimps
The main tools we require for our analysis are obtained from analytic geometry and elementary calculus. However, if you have not studied these subjects or if you have forgotten them, do not be discouraged. The equations and formulas we derive are easy to understand and apply.
An ellipse with semimajor axis of length a and semiminor axis of length b is shown in figure 26.4. From analytic geometry we know that the equation of an ellipse is
where, as shown in the figure, x and y are the coordinates of any point P on the curve.
Now suppose we rotate the ellipse about its major axis (i.e., the x-axis). We create a body of revolution called a prolate ellipsoid (or prolate spheroid). It looks like a watermelon. Now the general equation of an ellipsoid is
in which the z-axis passes through the origin perpendicular to the x-y plane; c is the half-length in the z-direction. In the case of the prolate ellipsoid, c = b. If we let r2 = y2 + z2, then (26.2) becomes
FIG. 26.5
Definition sketch for a prolate ellipsoid.
which is the equation of the prolate ellipsoid.
Alternatively, suppose we rotate the ellipse about its minor axis (i.e., the y-axis). This time we generate a so-called oblate ellipsoid (or oblate spheroid). This has somewhat the shape of a pumpkin. Indeed, the earth itself is an oblate ellipsoid because the length of its polar diameter, 2b is slightly less than the length of its equatorial diameter, 2a. This time, c = a. If we let r2 = x2 + z2, then (26.2) becomes
This is the equation of the oblate ellipsoid. In this chapter, we are interested in only the prolate ellipsoid, so we set (equation 26.4) aside.
How to Calculate the Volume of a Blimp
First we calculate the volume of a prolate ellipsoid by using integral calculus. With reference to figure 26.5, the volume of the thin disk is dV = πr2dx. Clearly, the summation of the volumes of all the thin disks, from x = –a to x = +a, gives the volume of the entire ellipsoid. Now, from (26.3) we have r2 = b2(1 – x2/a2). So we set up the following problem in integral calculus and obtain a relatively easy solution:
We assume that the blimp has the shape of a prolate ellipsoid. Then, if L = 2a is the length of the blimp and D = 2b is its diameter, (equation 26.5) becomes simply
With this formula we easily compute the volume of our blimp.
How to Calculate the Surface Area of a Blimp
This problem is a bit more complicated. Again, with reference to figure 26.5, the surface area of the narrow strip along the outer edge of the thin disk is dA = 2πrds. It should be clear that ds Accordingly, the summation of the areas of all the narrow strips, from x = – a to x = + a, gives the total surface area of the ellipsoid. So we have
From (equation 26.3) we obtain expressions for r and for dr/dx. Substituting these into (26.7) and carrying out the integration yields the result
Again using L = 2a and D = 2b as the blimp's length and diameter, respectively, we obtain the final answer:
This formula gives us the surface area of the blimp.
The Navy's ZPG 3-W Marine Patrol Airship
This was the largest airship ever constructed. Its length was L = 403 ft and its diameter D = 85 ft. It is reported that its volume was about 1.5 million cubic feet.
If we substitute these numerical values of L and D into (equation 26.6), we obtain the computed volume V = 1,525,000 ft3, which is pretty close to the reported value. Substituting the numbers into (26.9) gives the calculated surface area A = 86,200 ft2. This is an area of almost exactly 2.0 acres.
Lifting Capacity of Blimps
The ability of a blimp to lift not only its own weight but also the weight of an appreciable payload is due to the fact that a large fraction of its total volume contains a gas that is lighter than air. In the old-time dirigibles, especially those of the Germans, hydrogen was used as the buoyant gas. Hydrogen, of course, is by far the lightest known gas but it is also extremely dangerous to use. The explosion of its hydrogen tanks is what caused the famous Hindenburg to crash and burn at Lakehurst, New Jersey in May 1937.
Because hydrogen is so hazar
dous, many of the early dirigibles and all of the present-day blimps use helium as the buoyant gas.
QUESTION FOR YOUR COFFEE BREAK. If the specific weight of air is W1 = 0.077 lb/ft3 and the specific weight of helium is w2 = 0.011 lb/ft3, and if helium occupied the entire volume of the ZPG 3-W blimp, what was the blimp's total lifting capacity?
Answer. 100,650 pounds.
Footballs
There are rigid specifications concerning the proper size and shape of footballs. For example, the National Football League stipulates that the football be
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