Practical Math for the New American Century
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 sometimesnot always, but sometimesoffer 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 simplifiedbut still vividly realisticportrayal of one of those real-world problems that life in the 21st 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).
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 analysisat least within a wide range of yieldsis 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.
- kill radius of blast = (yield)0.41
- kill radius of thermal radiation = (yield)0.33
- kill radius of nuclear radiation = (yield)0.19
- 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
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.
- 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
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
- blast: (yield)0.41 = 0.8047 km
- thermal radiation: (yield)0.33 = 0.8047 km
- nuclear radiation: (yield)0.19 = 0.8047 km
- 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
- x4×¼ = 810.25
- 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.
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
And there you have it: math really is useful for everyday kinds of problems.
The Dark Wraith has spoken.