AP CALCULUS BCCalculus at the Battle of Trafalgar

This article applies calculus to the naval strategy used at the Battle of Trafalgar.

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Authored by

  • Dan Teague
    The North Carolina School of Science and Mathematics
    Durham, North Carolina

Calculating a Strategy

The summer of 2015 marked the 210th anniversary of the British naval victory over a combined French and Spanish fleet in the waters off Cape Trafalgar. During the Napoleonic wars, naval warfare followed certain rules that seem rather formal to us today. The ships in each fleet lined up in a row, sailing parallel to their opponents and firing as they sailed past each other (Figure 1). They repeated this maneuver until one fleet disabled or sank the other. This is known as the directed fire model or conventional combat model.

The diagram shows two sets of ships parallel to each other.  The first row has four white ships, and the second row has five black ships.  Arrows from the white ships point down toward the black ships, and arrows from the black ship point up toward the white ships.

Figure: The White Fleet takes a beating. Image courtesy of the author.

In such an engagement, the fleet with superior firepower will inevitably win. To represent this battle, we begin with a system of differential equations that models the interaction of two fleets in combat. Suppose we have two opposing forces, fleet A with  ships and fleet B with  ships initially, and A left parenthesis t right parenthesis and B left parenthesis t right parenthesis ships at t units of time after the battle is engaged. Given the style of combat at the time of Trafalgar, the losses for each fleet will be proportional to the effective firepower of the opposing fleet. That is, fraction numerator d A over denominator d t end fraction equals negative b B space a n d space fraction numerator d B over denominator d t end fraction equals negative a A and  where a and b are positive constants that measure the effectiveness of the ship's cannonry and personnel, and A and B are both functions of time. These equations indicate that the rate at which one navy loses ships depends on only two things: the number of ships in the opposing fleet and the effectiveness of the opposition fire. Assume that the effectiveness does not change throughout the battle so that the rate at which a navy loses ships is proportional to the number of ships in the opposing fleet.

The Questions for Battle

  1. Assume that a equals b equals k (the two fleets are equal in battle), and show that the total number of ships still fighting is decreasing exponentially by considering fraction numerator d over denominator d t end fraction left parenthesis A plus B right parenthesis.

  2. Assume that a equals b equals k, and show that the difference in the size of the two fleets is increasing exponentially by considering fraction numerator d over denominator d t end fraction left parenthesis A minus B right parenthesis.

  3. Use the results from questions (1) and (2) to solve for A and B as functions of time.

  4. If a is not equal to b, you can't use the techniques in questions (1) and (2) to determine the number of ships in each fleet as a function of time. If fraction numerator d A over denominator d t end fraction equals negative b B and  , find  , substitute, and solve for  . You will have two solutions to the second-order differential equation. The general solution will be the sum of these solutions. Use this solution to find  . Ensure that the solutions to question (3) are special cases of this more general solution.

  5. If  and fraction numerator d B over denominator d t end fraction equals negative a A, find fraction numerator d A over denominator d B end fraction and solve for A in terms of B. This equation will give you the expected number of ships remaining in fleet A when B=0.

  6. The commander of the British fleet was Admiral Nelson. In the now famous Battle of Trafalgar, he exhibited cunning military strategy. In one account of the battle, Nelson expected to have 27 ships in the British fleet (B) and predicted that the French/Spanish Armada (A) would have 34 ships. In planning his strategy, Nelson believed that the British fleet was better prepared (and better led) than the French/Spanish Armada. Suppose that a equals 0.75 b.

    If Nelson's 27 ships fought a conventional battle against the 34 ships in the French/Spanish Armada with a equals 0.75 b, would he win? How many ships would remain in the winning fleet?

  7. Instead of sailing parallel to the French/Spanish fleet, Nelson planned to sail through the middle of the fleet, cutting it in half and fighting two separate conventional battles. In one battle, he would have numerical superiority and consequently win that battle. In the other, he would have fewer ships and lose. But, with the ships that remained in the battle that he had won, would he be able to defeat the ships remaining in the French/Spanish fleet in the battle that they won? In a third and decisive battle, the British fleet would be victorious. Note that Nelson assigned himself the task of leading the portion of his fleet expected to lose its battle.

    Show, using the results from question (5), how Nelson could arrange his 27 ships to defeat a larger fleet of 34 ships using a three-battle plan as described above with a equals 0.75 b. According to our model, how many ships would we expect to survive the final battle?

Solutions

Answer the questions in full before viewing the Solutions and Commentary.

  1. Assuming the combat effectiveness of the combatants is equal, when we add the two equations together, we get fraction numerator d A over denominator d t end fraction plus fraction numerator d B over denominator d t end fraction equals negative k B minus k A. Recall that the sum of derivatives is the derivative of a sum. So we actually have fraction numerator d left parenthesis A plus B right parenthesis over denominator d t end fraction equals negative k left parenthesis A plus B right parenthesis.

    This equation is a variation of the basic differential equation for exponential functions fraction numerator d P over denominator d t end fraction equals negative k P, where P equals A plus B. The solution to fraction numerator d P over denominator d t end fraction equals negative k P is P left parenthesis t right parenthesis equals P subscript 0 e to the power of negative k t end exponent, and so the solution to our equation is A left parenthesis t right parenthesis plus B left parenthesis t right parenthesis equals left parenthesis A subscript 0 plus B subscript 0 right parenthesis e to the power of negative k t end exponent

    This means that the total number of ships in the battle decreases exponentially over the time of the battle.

  2. In a similar manner, we can consider the difference in the two defining differential equations. fraction numerator d A over denominator d T end fraction minus fraction numerator d B over denominator d t end fraction equals negative k B plus k A equals k left parenthesis A minus B right parenthesis is equivalent to fraction numerator d left parenthesis A minus B right parenthesis over denominator d t end fraction equals k left parenthesis A minus B right parenthesis space and so

    In the course of the battle, the difference in the number of ships that remain in the winning fleet A and those remaining in the losing fleet B increases exponentially.

  3. We now have two equations in two unknowns, A and B. Solve for A by adding the two equations from questions (1) and (2) together to eliminate B. This yields 2 A left parenthesis t right parenthesis equals left parenthesis A subscript 0 minus B subscript 0 right parenthesis e to the power of negative k t end exponent plus left parenthesis A subscript 0 minus B subscript 0 right parenthesis e to the power of k t end exponent and A left parenthesis t right parenthesis equals fraction numerator left parenthesis A subscript 0 minus B subscript 0 right parenthesis e to the power of negative k t end exponent plus left parenthesis A subscript 0 minus B subscript 0 right parenthesis e to the power of k t end exponent over denominator 2 end fraction. Solving for B, we find that B left parenthesis t right parenthesis equals fraction numerator left parenthesis A subscript 0 minus B subscript 0 right parenthesis e to the power of negative k t end exponent plus left parenthesis A subscript 0 minus B subscript 0 right parenthesis e to the power of k t end exponent over denominator 2 end fraction.

  4. If  and  , then  . This shows that the second derivative of A is proportional to A. What function is this? Only the exponential function has this property. However, there are two solutions, since the exponent could be either positive or negative. Both A subscript 1 left parenthesis t right parenthesis equals C subscript 1 left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis and A subscript 2 left parenthesis t right parenthesis equals C subscript 2 left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesissatisfy the differential equation  . The general solution, then, is the sum of the two. A left parenthesis t right parenthesis equals C subscript 1 left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis plus C subscript 2 left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis is the general solution to the second-order differential equation  .

    Now,  . This means that fraction numerator d B over denominator d t end fraction equals negative a left parenthesis C subscript 1 space left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis plus C subscript 2 subscript 1 left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis right parenthesis. By integration, we find that B left parenthesis t right parenthesis equals negative C subscript 1 square root of a over b end root left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis plus C subscript 2 subscript 1 square root of a over b end root left parenthesis e to the power of square root of 2 delta t end root end exponent right parenthesis. Since we have A left parenthesis 0 right parenthesis equals A subscript 0 and B left parenthesis 0 right parenthesis equals B subscript 0, we can solve for C subscript 1 equals fraction numerator A subscript 0 minus square root of begin display style delta over 2 end style end root B subscript 0 over denominator 2 end fraction and C subscript 2 equals fraction numerator A subscript 0 minus square root of begin display style delta over 2 end style end root B subscript 0 over denominator 2 end fraction. Notice that if a = b = k, we have the same solutions as in question (3).

  5. We know that  and  . So, from the chain rule and inverse function rule, we have fraction numerator d A over denominator d B end fraction equals fraction numerator d A over denominator d t end fraction times fraction numerator d t over denominator d B end fraction equals fraction numerator begin display style fraction numerator d A over denominator d t end fraction end style over denominator begin display style fraction numerator d B over denominator d t end fraction end style end fraction equals fraction numerator negative b B over denominator negative a A end fraction and integral a A d A equals integral b B d B. This is a separable equation, so integral a A d A equals integral b B d B, and it follows that  .

    The initial conditions give us c equals a A subscript 0 squared minus b B subscript 0 squared, so A equals square root of b over a B squared plus A subscript 0 squared minus b over a B subscript 0 squared end root. The end of the battle occurs when B = 0. Substituting into the equation above, we find that the number of ships in fleet A at the end of the battle is A equals square root of A subscript 0 squared minus b over a B subscript 0 squared end root.

    If fleet B wins, then the expected number of surviving vessels is B equals square root of B subscript 0 squared minus a over b A subscript 0 squared end root. By using this equation, we can determine the expected number of ships remaining in the winning fleet at the end of the battle. For example, two equal forces ((a = b)) with A subscript 0 = 20 and B subscript 0 = 15 would result in a victory for A with 13 ships remaining after the battle. This surprisingly large number is a result of the two exponential functions above. The total number is decreasing exponentially, but the difference in the size of the fleets is increasing exponentially.

  6. If Nelson's 27 ships fought a conventional battle against the 34 ships in the French/Spanish Armada with a = 0.75b, then B subscript 0 squared minus a over b A subscript 0 squared equals 27 squared minus 3 over 4 left parenthesis 34 squared right parenthesis less than 0, and Nelson would lose. The expected number of ships remaining in the winning French/Spanish fleet would be square root of 34 squared minus 4 over 3 left parenthesis 27 squared right parenthesis almost equal to end root 13.56 or 14 ships. (We will consider a "fractional ship" as still capable of fighting if the fraction is greater than one-half.) This would not be a good day for Admiral Nelson and the British.

  7. The number in the Survive column is the number of ships remaining after the battle. For example, looking at the first row, the British have 27 ships and the Armada has 34. The British split the battle into a fight of 4 British ships against 17 Armada. They lose, with the Armada having 16 ships left. The remainder of the British fleet (23 ships) fights a battle against the remaining 17 in the Armada. The British win with 18 ships surviving. Those 18 winning British ships from the second battle fight the 16 winning Armada ships from the first battle. The British win with 11 ships remaining at the end.

    The table shows that Nelson has many options. However, he will want the first battle (the one he loses) to last as long as possible, so that the second battle will have the best chance of concluding before the survivors of the first battle can rejoin the fight. Having 7 ships in the first battle and 20 in the second allows Nelson to defeat the larger fleet with an expected 5 ships surviving.

    The actual battle had more ships and did not follow this outline, but Nelson's original, creative strategy changed naval history.

References

Bartkovich, Kevin, et al. Contemporary Calculus Through Applications. Providence, Rhode Island: Janson Publications, 1995.

Bonder, Seth. "Mathematical Modeling of Military Conflict Situations." In Proceedings of Symposia in Applied Mathematics, edited by Saul Gass. Vol. 25, Operations Research Mathematics and Models. Providence, Rhode Island: American Mathematical Society, 1981.

Coleman, Courtney S. "Combat Models." In Modules in Applied Mathematics, edited by William Lucas. Vol. 1, Differential Equation Models, edited by Martin Braun, Courtney Coleman, and Donald Drew. New York: Springer-Verlag, 1983.