As indicated in table 24.1, the radius of a baseball is R = 3.68 cm. So we finally determine that the length of the seam on a baseball is 40.46 centimeters or 15.93 inches.
Please check these results by direct measurements on your plastic ball and your baseball. Since there are 216 stitches on a baseball, how many stitches per inch do you determine (a) by counting and (b) by computing?
QUESTION FOR YOUR COFFEE BREAK You own a smart ant that always runs at a speed of 1.0 centimeters per second. How long will it take the ant to cover the length of a tennis ball groove assuming no stops en route?
Answer. Almost exactly 30 seconds.
WHAT IS THE ANGLE BETWEEN THE TWO PLANES? On the way back from coffee break, you should stop at the market and buy either a nice round ball of Dutch cheese or a medium-size cantaloupe. On one or the other, please reproduce the art work you have created on your plastic ball.
Next, with your knife or small saw, carefully cut the cheese (or cantaloupe) through the planes defined by the circles centered at C′ and C″. Then place two pieces of stiff cardboard against the flat surfaces; the edges of the cardboard will meet along a line passing through point T. You will observe a certain angle between the cardboard planes. Measure this angle with your scale or protractor.
If you have used the value ø0 = 25° throughout and if you have carried out the slicing and measuring operations with reasonable care, you will determine that the angle between the planes is about 85°.
To solve the problem mathematically gets us into the topic of solid analytic geometry. The problem is pretty easy. It starts with recognizing that the angle between the two planes is the same as the angle between the two lines perpendicular to the planes. Call this angle σ. It can be computed from the relationship
TABLE 24.2
where l, m, and n are the so-called direction cosines of the two lines, one passing through C′ (subscript 1) and the other through C″ (subscript 2).
In our problem, it is easy to compute the direction cosines because C′ is located along λ = –90 and C″ along ø = 0. Without going into details, we obtain the answer
If ø0 = 25 (baseball), we get σ= 84.85; if ø0 = 40 (tennis ball), then σ = 89.60, and if ø0 = 45 then σ=90. If ø0 = 90 then σ=180 or σ = 0, if you prefer. In this limiting case, the two planes coincide; this is rather obvious.
Summary of Results
Before we go to a final topic or two, we summarize the main results obtained so far; this information is listed in table 24.2.
FIG. 24.5
Circular-arc curves for various values of ø0.
As a visual aid, some painted ping-pong balls are shown in figure 24.5. The six balls show the circular arc curves (i.e., seams or grooves) for six values of ø0. A wonderment: Why did they select ø0 = 25° for a baseball and 40° for a tennis ball?
If I Know the Longitude then What Is the Latitude?
The question is rephrased. We have a large zeppelin we are ordered to fly from point A′, λ = –90, ø = ø0, to point A″, λ = 90 – ø0, ø = 0. For reasons we are not permitted to know, we are instructed to follow a baseball-seam-type curve in our journey from A′ to A″.
We have state-of-the-art navigational equipment aboard and so we know our position (λ, ø) at all times. Here is the question: Knowing our longitude λ what should be our latitude ø in order to be on course. In other words, what is the equation of the curve ø = ƒ(λ) between A′ and A″?
It is necessary to solve the problem in two parts: (a) for λ < 0 (i.e., in the western hemisphere) and (b) for λ > 0 (the eastern hemisphere). The Greenwich meridian is λ = 0. We are not going to go through all the details. However, you might want to do this as a homework problem.
Region λ < 0
On your plastic ball, with the nice curve corresponding to ø0 = 25, select any point on the curve midway between A′ and T. Label this point P′ and draw, through P′, the arc of a great circle from the North Pole to the equator.
Using the spherical triangle C′NP′ and the law of cosines, equation (24.1), we write
Utilizing equations (24.4) several times, employing (24.5) to express ρ in terms of ø0, and going through some algebra—including the solution of a quadratic equation—we obtain the answer:
in which
This equation is a little messy but you can easily handle it on your calculator. It is important to remember that (24.14) refers to the region west of λ = 0.
Region λ > 0
On your plastic ball, select another point, P″, on the curve somewhere between T and A″. This time, using spherical triangle C″ NP″ and the law of cosines, we get
where, as in equation (24.5),
Maybe you can find a neater way to express this answer. Again, remember that (24.16) is for the region east of λ = 0.
Well, with equations (24.14) and (24.16), we have what we want: the equation of the curve ø = ƒ(λ). You might feel like computing this curve with ø0 = 25 and comparing the results with the curve you have already plotted on your plastic ball.
Here's another topic: You will observe on your plastic ball that in the λ > 0 region, the curve reaches a maximum latitude øm. This is a very interesting result. Using differential calculus, you can show, from (24.16), that this maximum point has the coordinates
Remember that this result is obtained from equation (24.16), which is valid only for λ > 0. If ø0> 45, there are no maxima in the eastern hemisphere (λ > 0) but there are in the western (λ < 0). We can determine those maxima by inspection: λm = –90, øm = ø0. Take a look at the ping-pong balls in figure 24.5.
One more thing needs to be established. We would expect that at λ = 0, the value of ø should be the same whether it is computed from (24.14) (for λ < 0) or (24.16) (for λ > 0). This is indeed the case, as you might want to confirm. Regardless of the value of ø0 the curve always passes through point T: λ = 0, ø = 45.
Furthermore, at point T we want the slope of the curve, dø/dλ, computed from (24.14) to be the same as the slope computed from (24.16). In other words (back to our zeppelin flight), we do not want our ancient airship to undergo any sudden drastic course change as it sails over λ = 0, ø = 45—a point 50 kilometers east-northeast of Bordeaux, France.
As we know by now, computing the slope of a curve means taking its first derivative. You have to be rather brave to differentiate equation (24.16) (for λ > 0) and almost foolhardy to try to take on equation (24.14) (for λ < 0).
This is clear enough. However, we do not need to know the slope at every point along the curve—only the slope near λ = 0. So what we shall do is “linearize” equations (24.14) and (24.16). This means that for small λ we can use the approximations sin λ = λ and cos λ = 1.
Doing this to (24.14) yields the simple expression
where, as before, Taking the first derivative of this equation and then substituting λ = 0 gives
Linearizing equation (24.16) provides the relationship
where Differentiating this equation and substituting λ = 0 again yields (24.20), as we would hope. Therefore the curve ø = ƒ(λ) is continuous at λ = 0, that is, there is no abrupt angle change.
TABLE 24.3
To be precise, on a sphere the slope of a curve is given by the equation
where θ is the angle between the curve and the meridian. This expression was developed in chapter 22, in which we computed the curve called the rhumb line or loxodrome.
Well, since equation (24.20) gives us the value of dø/dλ at point T and since ø = 45 at point T, we can easily calculate the meridian angle θ at T for any value of the basic parameter ø0. Table 24.3 lists these values of θ along with the maximum point coordinates.
PROBLEM We return to our zeppelin and its proposed flight from λ = –90, ø = ø0 to λ = 90 – ø0, ø = 0. Select any value of ø0 you like except ø0 = 90 (basketball)—there are no facilities for handling zeppelins at the North Pole. Calculate the zeppelin's path, ø = ƒ(λ), and plot it on your plastic ball. Determine the length of it
s journey; the radius of the earth is R = 6,370 km.
From your atlas, find out where the journey of the zeppelin begins and ends. If its velocity V = 100 km/ hr, how many days does the journey take? Where in the world is the point of maximum latitude? As the zeppelin crosses the point λ = 0, ø = 45 (near Bordeaux, France), what is its heading from true north?
In this chapter, we have carried out analyses on a number of fairly complicated problems involving spherical coordinates. The excellent book by Gellert et al. (1977) covers the subjects of spherical geometry and spherical trigonometry in a manner that is easy to understand.
Although we are not yet through with the problem of the baseball seam, this is a good place to pause. In the next chapter, we continue with the same general subject but with applications to topics that have nothing whatsoever to do with baseballs.
Sincerest appreciation is expressed to the Wilson Sporting Goods Company for providing very helpful information concerning the geometry of baseball templates.
25
Baseball Seams, Pipe Connections, and World Travels
We have established that the length of the seam on a baseball is 10.99 times the radius of the ball and that the length of the groove on a tennis ball is 9.39 times its radius. Very interesting. What else can we get out of our lengthy analysis?
This is a good question. We are now going to look at two problems that have nothing to do with a baseball but everything to do with the mathematics we developed in analyzing a baseball. The first problem is quite a practical one in civil engineering; the second problem deals with some aspects of cartography and geography.
A Nice Way to Connect Pipes
Here's the engineering problem. We have a deep well (or indeed an array of wells) penetrating a ground water aquifer or a petroleum reservoir. In the case of the aquifer, we plan to pump water into an injection well for ground water recharging, or perhaps to build a barrier to eliminate salinity intrusion from the ocean.
In the case of the petroleum reservoir, we intend to inject water into an oil-bearing layer to force the oil to move to nearby wells, where it is pumped out. When the reservoir was pumped the first time, it was not possible to remove all the oil, so this “secondary oil recovery” technique of injecting a “water flood” is used.
Some distance from the top of the injection well, we have a horizontal pipe that will deliver water to the well head. In the case of ground water recharging, the water in the delivery pipe is probably the effluent from a waste treatment plant. In the case of secondary oil recovery, the water is being delivered from a process tank in which chemicals are added to reduce the surface tension between the sand particles and the oil, so that more oil can be extracted.
In either case, the engineers have intentionally left a suitable distance between the end of the elevated horizontal water delivery pipe and the well head located at ground level. There are good reasons for leaving this gap. The engineers want to install a curved connection pipe between the two points (a) to allow for significant expansion and contraction of the steel pipe due to temperature changes and (b) to provide for a smooth transition of the flow of water from the horizontal delivery pipe to the vertical injection wall. If the connection were simply a 90° elbow at the well head, friction losses and hence pumping costs would be substantially increased.
With that introduction we get to the problem. Our approach closely follows that given by Kosko (1987) in his analysis of “three-dimensional pipe joining.” A definition sketch for our problem is shown in figure 25.1.
The surveyors in the field measure and report the following two quantities to the engineering design office: (a) the height h of the horizontal delivery pipe and (b) the horizontal distance L between the end of the delivery pipe and the well head. With this information concerning the numerical values of h and L, the design engineers then compute the minimum latitude ø0 and radius R of an equivalent but entirely fictitious sphere. The remainder of the problem is easy. Using the same symbols, the equations we derived in the previous chapter are employed to compute
FIG. 25.1
Definition sketch for the pipe connection problem. A' : water delivery pipe (elevation h), A”: well head (elevation 0).
The length of the two circular arcs of the connection pipe between A' and A'' : S
The bend radius: r
The bend angle: α
The angle between the planes of the two arcs: σ
The location of the maximum height: hm
We now back up. As noted, the only information obtained from the field is the vertical height h and the horizontal distance L. How can ø0 and R be determined with only this information?
To answer this, we first employ the plane law of cosines,
which, with regard to figure 25.1, takes on the form
Solving for cos γ gives
From the figure, it is observed that γ = 180 – δ. We also recall, from our baseball analysis, that δ = ø0; we emphasized this very important feature in the previous chapter. Accordingly, γ = 180 – ø0 and cos(180 – ø0) = –cos ø0. Substituting this result into (25.3) and carrying out a bit of algebra gives
We also know that h/R = sin ø0. Consequently, substituting R = h/sin ø0 into (25.4), remembering that sin2ø0 = 1 – cos2 ø0, and doing a little more algebra, we obtain the formulas we want:
Therefore, with h and L known, we can calculate the values of ø0 and R of the fictitious sphere.
Let us use some numbers. From the field survey we obtain h = 15.55 ft and L = 68.49 ft. So from equations (25.5) we determine that ø0 = 25.0° and R = 36.80 ft. This is a great big baseball.
From table 24.2, we get ρ = 47.57°, r = R sin ρ = 27.16 ft, α = 106.67°, σ = 84.85°, and S =1/8(10.99)R = 50.55 ft. The distance S = 50.55 ft is the arc length of each of the two connection pipes. Remember that only one-fourth of the entire baseball seam is involved in our pipe connection problem. At the junction between the two pipes (λ = 0°, ø = 45°), the welded flanges must be drilled to allow an angle change, σ = 84.85°, between the planes of the two circular arc pipes.
The photographs presented in figure 25.2 illustrate what we have done. The photo on the left shows half of the sphere and the ø0 = 25° seam. The one on the right is our connection pipe with the equivalent fictitious sphere removed.
Now it is your turn to solve a problem: design and build your own pipe connection. Start with some dimensions for h and L, and then compute ø0 and R. Forget about the sphere if you like but, by all means, visualize it.
FIG. 25.2
The pipe connection problem for the case ø0 = 25°, with (left) and without (right) the equivalent sphere.
Suggested materials of construction for your pipe connection are (a) some fairly heavy duty electrical wire from the hardware store (it is easy to bend and shape) or (b) a hula hoop you can probably get at the toy store (the material is rigid and the radius is fixed but you can easily set the correct angle change σ at the junction point and clamp it with a dowel).
If you can acquire two hula hoops, you will have enough to build the entire curve for the case ø0 = 45°. Alternatively, you may prefer to construct half of the ø0 = 0° curve. We take a close look at this ø0 = 0° case in the following section. For both cases, ø0 = 0° and ø0 = 45°, the completed space curves are interesting structures. You might even classify them as attractive examples of modern art.
Halfway Around the World Along the ø = 0 Curve
Here, we revisit maps and cartography. Have you ever wondered where a curve shaped like a baseball seam would go if it were projected onto a globe of the earth? Have you ever imagined how a tennis ball groove would look if it were plotted on a Mercator projection of the world? Have you ever speculated about what major city a basketball-seam-type curve would traverse if one of the seams coincided with the Greenwich meridian on a map? Well, if you are like most people you will say, in response to these questions, “Yes indeed.”
We conclude our study of the mathematics of baseball seams wit
h the following example. It features the world geography aspects of this interesting problem of spherical trigonometry.
That well-known travel agency, Bob's Air Tours, has been promoting a fantastic journey, beginning and ending in Singapore (λ = 105°E, ø = 0°), that flies the ø0 = 0 route. The advertising agency has recommended to Bob's that they use something a lot more exciting than the phrase “The ø0 = 0 Route” to promote the tour.
In any event, the trip starts and stops in Singapore, which has great hotels and shopping. It is also on the equator, which makes our example easier. As you have gathered, this journey follows the ø0 = 0 curve on a sphere. As shown in figure 25.3, outbound it goes to the southwest and inbound returns from the northwest.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 19
