Convexity of intercept time in short-range re-entry vehicle defence: how to optimally intercept a fast re-entry vehicle travelling in a lofted trajectory

Ballistic trajectories are determined by the gravitational field as well as the initial conditions of the vehicle: the launch location and the launch velocity. At short ranges (relative to the Earth's radius), we can consider the Earth as flat and the gravitational force as a constant. We will focus on short-range problems in this article. Cases of intermediate and long-range ballistic missile defence are more complex and can be found, for example, in (Buontempo, 2015; Carnegie Endowment for International Peace, 2021; Weitz, 2013). This is also a high-level analysis that examines the physics of short-range ballistic missile defence – environmental conditions such as atmospheric parameters are not considered, nor are tactics and manoeuvres. Detailed models have been the subject of intensive research and examples may be found in Ann et al. (2020), Carpentier (2014) and Lui et al. (2023).

There are two possible trajectories for getting from the origin location A to the final location B: a depressed trajectory with a (low) launch angle less than $45o$ and a lofted trajectory with a (high) launch angle of more than $45o$⁠, as shown below.

Figure 1 shows several trajectories of re-entry vehicles (RVs). The RV assumes an initial speed equal to $MACH5$⁠. There are two types of trajectories: the dashed lines correspond to the low-angle trajectories, while the solid lines correspond to the high-angle trajectories. We refer to the low-angle (high-angle) trajectories as the depressed (lofted) trajectories. The depressed trajectories have a maximal $z≤20km$ and the lofted trajectories have a maximal $z≥120km$⁠, where $z$ is the altitude of the RV. Here, the RV is launched from the ground $(z=0)$⁠, travels from right to left and aims for the defense location at the origin $(0,0)$⁠.

Lofted trajectories are advantageous for RVs as they force the defence’s radars to scan a large volume of space, which reduces search capabilities (Glover and Hagan, 1971). In addition, a lofted trajectory reduces the risk of civilian casualties in the event of a mid-air interception, since such a trajectory is located high above the ground. For example, the North Korean Hwasong-17 and Hwasong-18 intercontinental RVs employ lofted trajectories (Lendon and Bae, 2023). In addition, the Iranian Emad missile, employed in the 2024 attacks on Israel, is known to exit and re-enter the atmosphere (Pleitgen, 2024). This implies a lofted trajectory. The presence of a lofted trajectory is confirmed based on footage obtained from the BBC (Spender, 2024). We re-sketch the original figure from the BBC and estimate an angle of $48o$⁠, which necessitates a lofted trajectory (Figure 2).

In general, the mathematics of ballistic trajectories is complicated, but they can be determined using astrodynamics through Kepler and Gauss algorithms (Bate et al., 1971; Vallado, 1997). However, short-range ballistic trajectories are considerably simplified since they are quadratic functions of time. Yet, the intercept time formula of an RV by a ballistic interceptor is not simple due to the non-linearity of the gravitational force, even at short ranges. Adding to the complexity is the non-zero detonation range: an interceptor explodes when it reaches a certain non-zero range to the RV.

In this article, we derive the formula of the intercept time that accounts for the RV initial location, the RV initial velocity, the interceptor initial location, the interceptor launch delay time, the interceptor speed and the detonation range and gravity. We show through the convexity of the intercept time as a function of the interceptor launch delay time that if the RV's speed is greater than the interceptor's speed and if the RV's trajectory is lofted, then there is an optimal delay that minimizes the intercept time. We refer to this phenomenon as an anomaly in intercept time (Nguyen et al., 2022). Minimizing the intercept time affects intercept location, engagement opportunities and the probability of raid negation (Armstrong, 2014; Bourn, 2012; Cranford, 2004; Menq et al., 2007; Mury and Nguyen, 2007; Nguyen, 2014; Nguyen et al., 1997; Soland, 1987; Wilkening, 1999).

The article is organized as follows. First, we describe the scenarios. Second, we derive an analytical formula for the intercept time. Third, we derive the feasible volume of approach. Fourth, we show that there is an optimal launch delay to minimize the intercept time. Fifth, we show that there is only one minimal intercept time. Sixth, we interpret the results through examples. Finally, we provide a discussion and a conclusion.

We assume that the RV is like the Scud missiles deployed during the Gulf War by Iraq in 1990–1991 and the interceptor is like the Patriot missiles. Tables 1 and 2 show their characteristics (Wikipedia, 2022a, b). We also assume that the target location of the RV and the launch location of the interceptor are the same to illustrate the anomaly.

The intercept occurs when the interceptor reaches the detonation range from the RV. We can write the intercept equations as:

Where

$T=tint$ is measured from the time when the interceptor is launched; $(xi0yi0zi0)$ is the location of the launch site of the interceptor; $(xm0ym0zm0)$ is the location of the RV at the interceptor launch time (this is not necessarily the location of the launch site of the RV); $(xαyαzα)$ is the location of detonation on the sphere of radius $α0$ (detonation range) measured from the location of the RV; $u⇀$ is the velocity of the RV at time $t$ (it is not necessarily the velocity of the RV at its launch time) and it is known while $v$ is the speed of the interceptor. Even though gravity terms cancel in Equation (1), the intercept location $R⇀int$ must include gravity since the vehicles are both subject to gravitational acceleration as seen below:

We aim to determine the direction of $vˆ=(cos⁡θ·sin⁡φsin⁡θ·sin⁡φcos⁡φ)$ so that the intercept time is as short as possible. It is practically impossible to solve for the direction of $vˆ$ analytically from Equations (1) and even with the help of symbolic software such as Maple, Mathcad or Mathematica. That is, to minimize the intercept time as a function of $(θ,φ)$ would involve solving a sixth-order polynomial with a few hundred terms once the derivative is radicalized (Nguyen and Nguyen, 1996). Therefore, this is a non-trivial mathematical problem.

Below, we derive a closed-form solution for the shortest intercept time that satisfies Equations (1) and (2) using geometry instead of calculus. Using Equation (1), we get:

Substituting Equation (4) into Equation (2), we get:

Since we know the speed of the interceptor:

We observe that Equations (5) and (6) are now equations for spheres in the variables $(vxvyvz)$⁠. Each sphere has a different radius and a different centre. Equation (5) dictates that the first sphere has a radius equal to $α0/T$ and its centre is at $(ux+xm0−xi0Tuy+ym0−yi0Tuz+zm0−zi0T)$⁠. Equation (6) dictates that the second sphere has a radius equal to $v$ and its centre is at the origin $(0,0,0)$⁠.

To have a common solution in $(vxvyvz)$⁠, the two spheres must intersect one another. If the spheres do not intersect, the $(vxvyvz)$ solution of one sphere will not correspond to the $(vxvyvz)$ solution of the other sphere. This means that the distance between the two spheres' centres is greater than or equal to the difference of their radii and less than or equal to the sum of their radii as shown in Figure 3. This can be expressed as:

We note that $t$ is the only unknown in Equation (7). Nguyen and Nguyen (1996) determine the shortest intercept time $tint$ satisfying Equation (8):

Where $β=uv$⁠, $α=α0Δr0$⁠, $cos(θ0)=u⇀·Δr⇀0u·Δr0$ , $Δr⇀0=(xi0−xm0yi0−ym0zi0−zm0)$ and $F=α+β·cos(θ0)$⁠.

$tint$ is measured from the launch time $t$ of the interceptor. Hence, the total intercept time is given by $T←t+T=t+tint(t)$ measured from the time that the RV is launched. Based on this formula, we plot the total intercept time $T$ as a function of $χ=t/TF$ (where $TF$ is the time of flight of the RV and $0≤χ≤1$⁠) in Figure 4. Four ranges are considered: the RV is initially $50,100,150and200km$ away from the interceptor location. There are four scenarios (clockwise from top left):

1. Slow interceptor (interceptor's speed is less than RV's speed) and depressed RV's trajectory;

2. Slow interceptor and lofted RV's trajectory;

3. Fast interceptor (interceptor's speed is greater than RV's speed) and lofted RV's trajectory and

4. Fast interceptor and depressed RV's trajectory.

We can see clearly the anomaly in Figure 4: there is a minimum in the total intercept time as a function of launch time (Figure 4, top right panel). This corresponds to the scenario where the interceptor's speed is less than the RV's speed and the RV's trajectory is lofted. In all the remaining scenarios, the total intercept time is monotonically increasing as a function of launch time.

In the following sections, we will examine carefully when and why the anomaly occurs.

In the formula for $tint$⁠, there is a radical (square root). If the argument of the radical is negative, then $tint$ is complex, which means that it is not physically possible to intercept the RV. Assuming that the RV is initially beyond the detonation range $(α<1)$ and knowing that $F2≥0$⁠, this means that $tint$ can only be complex when $β=u(t)v>1$⁠. Therefore, if the interceptor's speed is less than the RV's speed then it can happen that intercept is impossible.

Analysing the argument of the radical, we obtain a feasible volume of approach for the interceptor. That is, if the interceptor is slower than the RV, then to intercept the RV, the interceptor must initially be in the feasible volume of approach, as depicted below. In general, the interceptor's initial location must be in front of the RV's direction. The feasible volume of approach becomes larger with a non-zero detonation range. This feasible volume is a cone subtended by an angle $θf$ (Figure 5):

Knowing the feasible volume of approach tells us that if the RV's initial launch angle is too steep, then it is not possible for the interceptor to intercept the RV. However, if the defence waits for a certain time, the velocity of the RV will change as it turns towards its target and eventually the interceptor's location will be within the feasible volume of approach. This is shown in Figure 6 where the interceptor location is the black, hollow circle on the left-hand side and the RV's launch location is the red dot on the right-hand side of the horizontal line. Initially, the interceptor's location is outside of the angle subtended by the red lines (red feasible volume of approach); hence intercept is not possible. But at a later time, the RV has moved to the green dot, the interceptor's location is now inside of the angle subtended by the green lines (green feasible volume of approach); hence intercept is possible. The feasible volume of approach changes with time because the velocity vector and the location of the RV change with time due to gravity.

The simple argument of evolving a feasible volume of approach suggests why the interceptor should delay its launch time so that the intercept is optimal.

The intercept time formula can be rewritten as follows:

$S$⁠, $D$ and $N$ are defined as follows:

Where $TF$ is the time of flight of the RV and $R$ is the horizontal range of the RV. It is convenient to analyze the intercept time in the dimensionless parameter $χ,$ which is normalized by the time of flight. If the results are true in $χ$ then they are true for all times of flight.

Note that Equation (10) contains a removable singularity when $D=0$ (Nguyen and Nguyen, 1996):

So, if $D=0$⁠, we get:

We will assume that $D≠0$⁠. The special case where $D=0$ will be explained later. We will show that the derivative of the total intercept time, $T′$ (where $/TF=χ+tint(χ)$ ), is negative for a value of $χ$ on the left-hand side and is positive for a value on the right-hand side (Figure 4, top right panel). For conciseness, we refer to $T/TF$ as $T$ since $TF$ is a constant parameter in the problem. Since $T′$ is a continuous function, there will be a $χ$ such that $T′(χ)=0$ which corresponds

to a minimum.

$T$ is decreasing on the left-hand side.

Let $Z=S2+D·N$⁠, then:

Note that $N≥0$ as it is proportional to the distance from the interceptor to the RV minus the detonation range and we assume that the RV is initially beyond the detonation range. Also, $S2≥0$ as $S$ is real. So, if $Z→0+$⁠, this makes $D<0$ since $Z=S2+D·N$⁠. This also means that the interceptor's speed is less than the RV speed at time $χ$⁠. We claim that

That is, the total intercept time $T$ is decreasing on the left-hand side if $limZ→0+Z′>0$⁠. We first assume that $α=0$ and solve for $χ$ when $Z=0$⁠. There is a double root at $χ=1$ as well as two non-trivial roots:

It can be checked that: $0<χ+<1$ and $χ−<0$⁠. So $χ+$ corresponds to the left-hand side of the feasible launch time. In addition,

Since $χ+$ is the only root strictly between zero and one, this means that $Z$ must cross the zero value in an increasing way. Hence, $Z′(χ+)>0$⁠.

When $α>0$⁠, we get:

Where $Zα$ is the value of $Z$ when the interceptor reaches the detonation range from the RV. That is, $N=0$⁠. An example of $Z$ as a function of $χ$ is shown in Figure 7. The red curve corresponds to $α>0$ (⁠$γ=7.187$⁠, $tan⁡θ=−11.921$ and $α=0.3$⁠) while the blue curve corresponds to $α=0$ (⁠$γ=1.682$ and $tan⁡θ=−2.619$⁠). Note $χ+$ moves to the left when $α>0$⁠. This means that the interceptor can be launched earlier and still intercept the RV if given a non-zero detonation range.

The proof is like Case 1.

$T$ is increasing on the right-hand side.

We now show that $T′>0$ as $χ→1−$⁠. For simplicity, we write $χ→1−$ to refer to the case where the defense waits for the RV to reach the detonation range of its aimed destination. Technically, for small $α$⁠, $χ∼1+α·cos⁡θ$ where $(cos⁡θ<0)$⁠. Note that $N=0$ and $tint=0$ at $χ→1−$⁠, as the RV has reached the weapon range. Therefore, differentiating Equation (11) and set $N,tint=0$ yields:

$N′$ and S can be expressed as:

Thus, $N′=2·α·γ−2·S$⁠. Hence, as the interceptor reaches its detonation range to the RV $(χ→1)$⁠:

Thus, total intercept time $T$ is increasing at $χ→1$⁠. This proves the existence of the anomaly.

It is very simple to repeat the analysis when $D=0$⁠. The results on $T′$ still hold when $D=0$⁠. For example, since $tint=12·NS$ , $N≥0$ and $Z=S$ then

This requires $T$ to be decreasing on the left-hand side.

Given that the RV's speed is initially greater than the interceptor's speed: $γ2<1+tan2(θ)$⁠, we know that $D(0)<0$⁠. There are two cases to consider:

1. $D$ can be zero $(γ≥1)$

2. $D<0$ $(γ<1)$

In Case 1, the RV's speed is initially greater than the interceptor's speed, but later, due to gravity, the RV's speed can be equal to the interceptor's speed. In Case 2, the RV's speed is always greater than the interceptor's speed. We will provide the proof for Case 1 below and infer the proof for Case 2 from that of Case 1.

In order to provide tactics to ballistic missile defence operators, it is important to know exactly where the minimal intercept time occurs and whether there is more than one minimum. We will show that the intercept time is a convex function in $χ$⁠. This means that there is only one minimum. Before providing the proof, we will sketch out the essential ideas. We recall that $T=χ+tint(χ)$ and so $T″=tint″≥0$⁠. We will show that $tint″≥0$ by a new technique. We will illustrate the technique with an example.

But we have shown that $(S2+D·N)′=S′+D′·N2·S$ with $D′·N2·S≠0$⁠. Therefore, this is a contradiction.

when $D=0$⁠, $FSign=0$ is zero. The derivative $FSign$ is also zero when $D=0$⁠:

However, the second derivative of $FSign$ is not zero at $D=0$⁠. This holds because

$FSign″]D=0=A/B]D=0$ where $B$ is a power of $S]D=0$ which is not zero (Observation 1) and $A$ is a polynomial in $χ$ that does not divide $D(χ)$ evenly. By the fundamental theorem of algebra, $A]D=0≠0$ as $A$ and $D$ do not have overlapping roots. We have also confirmed this graphically. There are two roots in $χ$ when $D=0$⁠. This means that each of these roots are double roots and so $tint′·D2$ is either non-negative or non-positive. What is more, these are all the roots as

$FSign=FSign′=0$ only when $D=0$ (for $D$ real & $χ$ real and in the feasible range). To do that, we determine $tint′·D2$ at a feasible $χ$ (at $χ→1−$⁠) and show that $tint′·D2<0$ for that $χ$⁠. That is sufficient to show that $tint′·D2<0$ for all feasible $χ$⁠. We plot $FSign=tint′·D2$ as a function of $χ$ in Figure 8 to illustrate this is the case.

From Equation (17), we get:

as $limχ→1−S>α·γ$⁠.

Since $tint′·D2≤0$ and $D2≥0$⁠, then $tint′≤0$⁠. In the same way, we can show that $tint″·D2≥0$⁠. Therefore, $tint″≥0$⁠. Since $T=χ+tint$⁠, we know $T″=tint″≥0$⁠. This completes the proof of convexity. Knowing there is exactly one minimum lying between zero and one, we determine the value of $χ=χmin$ using bisection (Wikipedia, n.d.a, n.d.b, accessed 2024). This guarantees a solution to the desired accuracy.

We will show that

This means that the right-hand side of Equation (19) has at least a double root at $D=0$⁠. Simple calculus yields:

Differentiating Equation (11) and letting $D=0$⁠, we get:

Where $tint]D=0=12·NS$⁠. Using Equations (12), (21) and (22), we establish Equation (20). Furthermore, we can verify that

Using the same argument as the one used in Observation 2. Since the right-hand side of Equation (19) is the product of Equation (20) and $D$⁠, this implies that Equation (19) has a double root at $D=0$⁠. In addition, $D$ is a quadratic function in $χ$ so each root of $D$ is a double root as seen in Figure 9. We plot Equation (20) and $D$ as a function of $χ$ in Figure 10. The blue curve corresponds to $D$ and the red curve corresponds to Equation (20). They clearly have the same roots.

We now show that $limχ→1−tint″>0$⁠. Note that we can obtain $limχ→1−tint″$ by differentiating Equation (11) twice and setting:

Also, when $χ→1−$⁠:

This yields:

In essence, we show in Case 1 that $FSign2$ has the same roots as $D(χ)$⁠. So, if $D(χ)<0$ then $D$ has no real roots and so $FSign2$ also has no real roots (in the feasible range). This means that $FSign2$ is either positive or negative but not both. In addition, Equation (18) tells us that $T″$ has the same sign as $FSign2$⁠, and we show above that $T″>0$ for $χ→1−$⁠. This concludes that $T″>0$⁠, in the feasible range. An illustration of $FSign2$ is shown in Figure 11 where $γ=0.8<1$⁠.

To validate the results derived in the previous sections, we compare them to those of MANA (Map-Aware Non-uniform Automata), which is an agent-based model developed by the Defence Technology Agency of New Zealand (Lauren and Stephen, 2002). Agent-based simulations and characteristics abound and their descriptions can be found, for example, in Castle and Crooks (2006) and Diallo et al. (2015). Our results (dashed line) and those from MANA (bold line) are consistent, as shown in Figure 12.

To understand the results, we analyse the intercept time with a range of $200km$ (solid yellow curve) shown in Figure 12. In Case 1, if the defence waits for $50s$ after the RV is launched, the intercept occurs at approximately $300s$⁠. In Case 2, if the defence waits for $100s$ after the RV is launched, the intercept occurs at approximately $250s$⁠. This includes the $100s$ delay. That is, the RV is launched at time zero, the interceptor is launched at time $100s$ and the intercept occurs at time $250s$⁠. This means that the time of flight of the interceptor is $150s250−100=150s$⁠. At this interceptor launch delay time, we note that the RV intercept time is also a minimum. With an interceptor launch delay time equal to $50s100s$ the RV intercept time is equal to $300s250s$⁠. The time line for Case 1 and Case 2 are shown in Figure 13.

There are many advantages to intercepting the RV at optimal delay. First, the time of flight of the interceptor is shorter, which increases the reliability of the operation of the interceptor. Second, an early intercept generally means the RV is intercepted further from the target, which reduces the secondary effects of debris. Third, having more time is always a benefit. In this example, a saving of $100secs$ out of $350secs$ is $28percent,$ which is substantial in time-critical intercepts.

Figure 14 shows the straight-line solutions: the RV trajectory is affected by gravity until the launch of the interceptor. After the launch of the interceptor, both vehicles move in straight lines in the solution space since the gravitational acceleration terms cancel. Figure 15 shows the actual physical solutions: both vehicles are subjected to gravity even after the launch of the interceptor. That is, gravitational acceleration is added back to both trajectories even though the gravitational terms cancel mathematically after the launch of the interceptor. In the solution space, Figure 14 shows that the intercept occurs at a higher altitude than that of Figure 15. This holds as gravity pulls both vehicles towards the ground. The intercept time is, however, unchanged.

The intercept time formula allows examination of the detonation range. The detonation range of Patriots is in the order of meters, which is much less than the range of RV (in the order of kilometres). In the future, it is possible that interceptors will have greater detonation ranges. Figure 16 shows the non-linearity of the intercept time when the speed ratio $β$ decreases (slow interceptors) and the detonation range increases. This happens because the detonation range becomes more important when the interceptor is slower. As $β$ increases, the intercept time becomes more linear, as shown by the straight contours. Eventually, when $β>1$ there is no anomaly.

In this article, we have shown that there is an anomaly in intercept time: the wisdom of launching the interceptor as early as possible does not apply here. There is an optimal launch delay which results in a saving of $28percent$ in time in the scenario considered. We show exactly when and why the anomaly occurs. This happens when the interceptor's speed is less than the RV's speed and the RV travels in a lofted trajectory. As described, there are actual RVs, such as North Korean Hwasong 17 and Hwasong 18 missiles, which travel in lofted trajectories, and they are harder to detect. Hence, it is an advantage to the offense. With hypersonic RVs, it will also not be surprising that the defence needs to intercept a faster RV. A non-zero detonation range becomes a key parameter in such a scenario as it raises the capabilities of the interceptor, especially against a fast RV. In fact, there have been cases in Ukraine where the Patriot defence system has intercepted a hypersonic RV (the hypersonic speed was greater than the Patriot's speed).

From a methodological viewpoint, we have developed simple tools to analyse a complex and non-linear problem. Without these tools, it would not have been possible to derive the intercept time formula. The formula provides insights into the convexity of the problem, yielding a unique optimal launch time. It tells us when and why the anomaly occurs. The formula also yields the concept of a feasible volume of approach. That is, based on the initial conditions, such as locations and velocities, it may not be feasible at all to intercept an RV. The defence would need to wait for the RV to change its velocity vector before an intercept is possible. Hence, there is an optimal launch delay which can be determined with the desired accuracy in an efficient way. In a two-dimensional maritime context, the feasible volume of approach is known as the limiting lines of approach. The time intercept formula quantifies the effects on the detonation range, both on the intercept time and the feasible volume of approach. That is, the intercept occurs earlier, and the feasible volume of approach gets larger with a non-zero detonation range. If we had simplified the trajectories as straight lines, then the anomaly would not have been discovered and the intercept time would have had the wrong pattern: a monotonically increasing function as opposed to a convex function in launch delay time. In general, we believe that convexity can be exploited in many other defence and optimization problems.