Special Analysis: Practical Math for the New American Century

As a college teacher of mathematics, as well as economics, finance and other business subjects, I have always believed that it is important to provide students with real world, practical applications of the skills they are learning in the abstract. Just about every math teacher has heard the refrain, "Where would I use that in the real world?" That's usually a subtle declaration by the student that he or she has a life that is in some way more "real" than that of the teacher. Patience is always demanded in such a situation. So, too, is a ready application that uses the skill under current investigation in some setting that will resonate with the learner. The more clearly the example portrays a situation the student might encounter, the more likely the student is to drop his or her defenses and be willing to proceed with the admittedly often difficult task of mastering the subject matter and the mechanics of the quantitative analysis.

Sometimes, a teacher must be a bit disingenuous. The real "real world" application might require mathematics far too robust for the student at his or her current level. That's where some professors are likely to fall down: their mastery is at such a high and sophisticated level that simplified but clear examples of a technique will elude them. The professor who honestly aspires to be a great teacher will, on the other hand, seize an opportunity, reduce or eliminate the complexities that obscure underlying principles, and proceed with a realistic, if somewhat not-quite-real-world, presentation. In the process, the professor might garner a side benefit: cutting through the layers of far more sophisticated analysis can sometimes—not always, but sometimes—offer a personal clarity of understanding that has been lost to the years of advanced studies, gruelingly boring seminars, and mind-numbingly dry field literature.

This article offers a simplified—but still vividly realistic—portrayal of one of those real-world problems that life in the 21^st Century has thrown in the path of people not just here in the United States, but all over the world. Admittedly, it does have some particular currency with the peoples of America and Europe, but it truly is a problem in which math students everywhere can find value and a reason to care about mastering a few basic mathematical skills.

Without further motivation, then, we'll use the mathematics of empirical formulæ to compute the killing power of a nuclear bomb, with more extensive and exhaustive information available from the Federation of American Scientists.

First, we need to break down the killing power of a nuclear explosion into the three classical categories:

• pressure (the blast wave, which is like a wind);
• heat (which is more technically referred to as thermal radiation);
• nuclear radiation (the "nuke" stuff, as in ionized and sub-atomic particles).

All three of these effects become bigger as the yield of the device increases. In mathematics, we say each of these three factors varies directly with yield.

Also, and somewhat less obviously, each of these three effects varies against the others as the yield of the bomb is varied. In other words, these three effects don't increase at the same rate as the yield of the bomb is increased. In fact, these three effects also vary against each other based on several other factors besides plain old bomb yield. For example, a device detonated up in the sky, as opposed to being detonated at ground level, might spread a crushing blast wave downward over a larger area, but some of the fireball of thermal energy is wasted in so doing. For the purposes of what we're doing here, though, we'll assume that the bomb is detonated at the distance above ground level that maximizes the effectiveness of all three killing effects. We call this the bomb's optimal height.

Moving along, the standard basis for a first-round, rather rough analysis—at least within a wide range of yields—is a two-and-a-half kiloton device. The reason this is a good basis is that the three kill effects are about "equal" at that yield in terms of having about the same kill radius: right around one kilometer, assuming the following thresholds for mortal injury:

• blast: 4.6 pounds per square inch of overpressure;
• thermal radiation: 8 Calories per cm² (creating 3rd degree burns);
• nuclear radiation: 500 rem.

Now, if we do bomb yields as multiples of the 2.5 kiloton bad boy, the equation governing the kill radius for each of the three effects comes out as follows:

• kill radius of blast = (yield)^0.41
• kill radius of thermal radiation = (yield)^0.33
• kill radius of nuclear radiation = (yield)^0.19

Let's take as our first example a nuclear bomb with a two kiloton yield. First things first: we need to convert that yield into a multiple of the base, 2½ kiloton device, and then we'll be ready to crank.

2.0÷2.5=0.80

So, in other words, we're looking for the kill radii of the three effects for a bomb that has 80% of the yield of the base device. Now, we can do the calculations.

• kill radius of the blast: (0.80)^0.41 = 0.91 kilometers = 2986 feet
• kill radius of the heat: (0.80)^0.33 = 0.93 kilometers = 3051 feet
• kill radius of the radiation: (0.80)^0.19 = 0.96 kilometers = 3150 feet

In this example, the nuclear radiation sets the overall kill radius of the weapon: 3150 feet.

Now, let's look at a four kiloton bomb.

Again, we need to convert the yield of the device under analysis into a multiple of the base, 2½ kiloton device.

4.0÷2.5=1.60

So, in this example, we're looking for the kill radii of the three effects for a bomb that has 160% of the yield of the base device. Here come the calculations.

• kill radius of the blast: (1.60)^0.41 = 1.22 kilometers = 3978 feet
• kill radius of the heat: (1.60)^0.33 = 1.17 kilometers = 3831 feet
• kill radius of the radiation: (1.60)^0.19 = 1.09 kilometers = 3587 feet

Well, isn't that interesting: now, it's the overpressure that sets the overall kill radius. As a matter of fact, you've probably just figured out that, as the yield of the device increases, the overall kill radius becomes more and more defined by the blast contour, not by the fireball cooking people or by the radiation microwaving them.

And finally, we can do one last algebra trick to get some more information from the equations. We'll do the calculations backward to see what yield would be required to achieve a given kill radius for each of the effects. For example, suppose we wanted to know what yield would be required to create a kill radius of a half-mile. That would be a diameter of one full mile, of course.

The first thing we'll need to do is to convert 0.5 miles into kilometers, since the equations require metric units. Using the standard conversion tables, we get

0.5 mile × 1.6093 km/mile = 0.8047 km

Next, we write the three equations, with yield, of course, being the unknown for which we must solve.

• blast: (yield)^0.41 = 0.8047 km
• thermal radiation: (yield)^0.33 = 0.8047 km
• nuclear radiation: (yield)^0.19 = 0.8047 km

Solving any one of the three equations above is just a matter of using the algebra rule of taking the inverse power of the exponent on the variable. Here's how it works in the abstract: if we have the problem

• x⁴ = 81; we'll power up both sides by the reciprocal of the 4 (recalling that, as long as you do exactly the same thing to both sides of an equation, it's okay).
• (x⁴)^¼ = (81)^¼
• x^4×¼ = 81^0.25
• x = 3, where we usually use a calculator to take a number to a decimal power, unless it's a pretty obvious one we've memorized.

Okay, now that we've made it through that little detour, we can do the same trick to solve for the yields in the preceding set of equations. We'll just need to take the reciprocal power from the yield side and apply it to the number on the right side for each of the equations. If we run the reciprocals of the powers through the calculator we find that, for 0.41, the reciprocal is about 2.4390; for 0.33, the reciprocal is about 3.0303; and for 0.19, the reciprocal is around 5.2632.

Now, we can do the calculations, but we have to remember that these equations are doing yields based upon multiples of a 2.5 kiloton device; that means at the very end, we'll have to take each number we get for a yield and multiply it by 2.5 to get an actual kiloton value.

• blast: required yield = (0.8047)^2.4390 = 0.59 = 1.47 kiloton
• thermal radiation: required yield = (0.8047)^3.0303 = 0.52 = 1.29 kiloton
• nuclear radiation: required yield = (0.8047)^5.2632 = 0.32 = 0.80 kiloton

Well, there you go: all it takes is a device that isn't even yielding a full kiloton, because with the low yield devices, it's the radiation that does the maximum damage, not the blast or the firestorm.

And there you have it: math really is useful for everyday kinds of problems.



The Dark Wraith has spoken.