📂Dynamics

Lanchester's laws

Laws

First Law

In melee or close combat, combat power is proportional to the size of the force.

Second Law

In modern or long-range combat, combat power is proportional to the square of the size of the force.

Description

Lanchester’s Laws describe the number of casualties in combat between two groups, categorized into the First Law (linear law) and the Second Law (square law).

• Linear Law: War, before the distribution of firearms, was a melee where spears and shields clashed. There were front and rear lines; while the front line was engaged in combat, many in the rear line could not participate due to being out of range. Hence, even if one side had a numerical superiority, if they couldn’t engage in combat, their effective combat power at that moment could be considered equal. Therefore, it’s advantageous for the side with fewer troops to induce the course of the war to follow the linear law.

• Square Law: Long-range combat frees up spatial constraints on attacks. To take this to the extreme, it could be assumed that all soldiers on both sides are engaging in combat simultaneously. As such, having more troops is beneficial for making the war’s course follow the square law.

It is important to note that Lanchester’s Laws are not meant to precisely describe historical battles. Rather, they indicate that if the command cannot strategize to leverage these laws in their favor, they will end up playing into the hands of the enemy’s preferred law.

Most maneuver tactics and ancient formations aimed to convert numerical superiority into actual combat power. Across ages and cultures, the superior force always desires to encircle and annihilate the enemy. Assuming equal combat power for each soldier, it is far more advantageous to create a many-on-one situation, even in the front lines. The more troops that partake in a battle, the more the situation aligns with Lanchester’s square law.

Conversely, when defending, it’s necessary to utilize everything from terrain to fortifications to follow Lanchester’s linear law, especially when at a numerical disadvantage. If the defenders can hold their position with minimal losses, the attackers, due to logistics, will be forced to act hastily.

Let’s now discuss these explanations in terms of mathematics and population dynamics.

Model

First Law

$$ \begin{align*} A ' =& - \beta \\ B ' =& - \alpha \end{align*} $$

Second Law¹

$$ \begin{align*} A ' =& - \alpha B \\ B ' =& - \beta A \end{align*} $$

Variables

• $A(t)$: Represents the number of individuals in group $A$ at the time of $t$.
• $B(t)$: Represents the number of individuals in group $B$ at the time of $t$.

Parameters

• $\alpha>0$: The attack coefficient of group $A$ against $B$.
• $\beta>0$: The attack coefficient of group $B$ against $A$.

Derivation

Without loss of generality, let’s assume that the size of the forces at the start of the battle (the initial value) starts at $A_{0} > B_{0} > 0$ for convenience, denoted as $\alpha = \beta = c = 1$. If every individual soldier’s combat power is equal, the differential equation would be $A_{0} > B_{0}$, making it impossible for $A$ to lose. Assuming both groups fight to complete annihilation without surrender, when $t \to \infty$, then $B(t) = 0$. Now, by calculating $\displaystyle a := \lim_{t \to \infty} A(t)$, we aim to derive Lanchester’s Laws.

Derivation of the First Law

Assuming the number of participants in the battle remains constant, casualties would thus remain constant as well, allowing us to model this as follows: $$ \begin{align*} { { d A } \over { d t } } =& - \alpha = -1 \\ { { d B } \over { d t } } =& - \beta = -1 \end{align*} $$ Organizing this in relation to $- d t$ gives: $$ - d t = d A = d B $$ Integrating from the start of the battle $t = 0$ to the end $t = \infty$ yields: $$ \int_{0}^{\infty} -1 dt = \int_{A_{0}}^{a} dA = \int_{B_{0}}^{0} dB $$ Here, the left side is irrelevant for finding out about $a$, so we need not focus on it. Only calculating the definite integral of the middle term and the right side: $$ a - A_{0} = 0 - B_{0} $$ Organizing in relation to $A$ gives: $$ a = A_{0} - B_{0} $$ In other words, combat power is simply proportional to the size of the force.

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Derivation of the Second Law

Assuming all troops participate in the combat, the attack power appears as the size times the attack coefficient, modeled as follows: $$ \begin{align*} { { d A } \over { d t } } =& - \alpha B = -B \\ { { d B } \over { d t } } =& - \beta A = -A \end{align*} $$ Organizing this in relation to $A(t),B(t)$ gives: $$ { { 1 } \over { - B(t) } } dA(t) = { { 1 } \over { - A(t) } } dB(t) $$ Multiplying both sides by $-AB$ gives: $$ A(t) dA(t) = B(t) dB(t) $$ Integrating from the start of the battle $t = 0$ to the end $t = \infty$ gives: $$ \int_{A_{0}}^{a} A dA = \int_{B_{0}}^{0} B dB $$ Calculating the definite integral gives: $$ {{ a^{2} - A_{0}^{2} } \over { 2 }} = {{ 0 - B_{0}^{2} } \over { 2 }} $$ Organizing in relation to $a^{2}$ gives: $$ a^{2} = A_{0}^{2} - B_{0}^{2} $$ In other words, combat power is proportional to the square of the size of the force.

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Example

The scene from a match between Lee Young-ho (Random Zerg) and Kim Taek-yong (Protoss) shows that, starting at 7 minutes and 59 seconds (as the start time is linked, it starts playing immediately), only 3 zealots fought against 12 zerglings, resulting in the deaths of 2 zealots and 8 zerglings. Normally, a zealot is stronger than a zergling, making it an asymmetric power situation as per $\alpha>\beta$; resource-wise, both cost minerals equal to $300$. This is not crucial to the battle itself, so let’s disregard it.

In the video, the zealots are positioned behind minerals, taking geographical advantage. Despite their large numbers, the zerglings are unable to engage in actual combat effectively, circling around due to lack of space. In contrast, the zealots are not facing more than 4 zerglings at a time, attacking without losses. This situation avoids a direct confrontation in open space, following the pattern of Lanchester’s linear law.