How does one calculate the effectiveness of a nuclear strike? Although people tend to view nukes as absolute weapons, they are not guaranteed to kill their target. This is especially true if the target is hardened against attack. So how does one go about determining the likelihood of a nuclear weapon achieving a kill? The first step is to determine if the weapon will be airbursted (detonated hundreds of meters in the air) or groundbursted (detonated near or on the ground). Groundbursts are more effective against hardened targets while airbursts are better at destroying soft targets and minimizing fallout. In this post, I will be discussing only groundbursts.
After determining where to detonate the nuclear device, one must calculate its lethal radius (LR). This is the maximum distance from the detonation at which the target can still be destroyed. For ground bursts, the LR=2.62*Y1/3/H1/3, where Y is the warhead yield in megatons and H is the target hardness in psi. So, if the U.S. were to target a Russian silo hardened to 2,500 psi with a 350 kt W78 thermonuclear warhead, the LR would be equivalent to 2.62*.351/3/25001/3, which equals .136 nautical miles. In other words, anything with a hardness of 2,500 psi or below that lies within .136 nautical miles of ground zero would be destroyed.
But this is not enough for us to calculate the probability of a nuclear strike being effective. To do that, one must also know the circular error probable (CEP). This is a circle that bounds the area around the target in which the warhead will land 50% of the time. To visualize this, think of a dartboard. The center of the dartboard is the target, and the outer circumference of the board represents the CEP. That means that 50% of the darts will land somewhere on the dartboard while the other 50% land somewhere on the wall. Once we know the CEP, we can plug it into the equation for the single strike kill probability (SSKP): 1-.5(LR/CEP)2. The CEP of a Minuteman III ICBM, which delivers the W78 warhead, is 120 meters (.065 nautical miles). The SSKP is, therefore, 1-.5(.136/.065)2, or .9519. This means that the strike will achieve a kill 95.19 percent of the time.
But one more calculation needs to be made that incorporates the weapon’s reliability. After all, there is a chance that the rocket motor misfires or guidance systems fail, resulting in the warhead never reaching its target. A conservative estimate puts U.S. ICBM reliability at 85%. By multiplying the SSKP by .85, one yields a total kill probability (TKP) of 80.9%.
Is this sufficiently high to execute a strike? It depends on the risk aversion of the decisionmaker. But regardless of one’s risk tolerance, what should be clear is that single nuclear strikes are not always sufficient to ensure a “kill.” This has implications for deterrence and nuclear strategy, and it also reveals interesting things about American motives (namely that the U.S. clearly seeks to maintain a counterforce arsenal). But most importantly, I think, it removes some of the mystique around nukes. They are indubitably powerful weapons, but they are also not omnipotent. This is something that we would do well to remember.