📚 The Franco-Prussian War 1870–1: A Mathematical Case Study | 普法战争 1870–1：数学案例研究
The Franco-Prussian War of 1870–1 is usually studied in history, but it also provides a rich dataset for Edexcel A-Level Mathematics. By treating mobilisation figures, casualties, artillery ranges, and battle outcomes as numerical variables, you can revise ratios, percentages, exponential growth, probability, statistical tests, projectile motion, and linear models.
1870–1 年普法战争通常出现在历史课程中，但它也为 Edexcel A-Level 数学提供了丰富的数据集。如果把动员人数、伤亡、火炮射程和战斗结果当作数值变量，就可以复习比率、百分比、指数增长、概率、统计检验、抛体运动和线性模型。
1. Historical Data as a Mathematical Dataset | 作为数学数据集的历史数据
Mathematics turns historical descriptions into measurable comparisons. The Franco-Prussian War lasted from 19 July 1870 to 10 May 1871 and involved two main sides: France and the Prussian-led North German Confederation.
数学将历史描述转化为可测量的比较。普法战争从 1870 年 7 月 19 日持续到 1871 年 5 月 10 日，主要交战双方是法国和普鲁士领导的北德意志邦联。
Several key engagements can be summarised in a data table. Using approximate historical estimates makes it easier to apply ratio, percentage, and probability methods later.
几次关键战役可以汇总成数据表。使用近似历史估计值可以更方便地应用后续的比率、百分比和概率方法。
| Event | Date | French strength | German strength |
|---|---|---|---|
| Battle of Sedan | 1 Sept 1870 | 120,000 | 200,000 |
| Siege of Metz surrender | 27 Oct 1870 | 180,000 | — |
| Siege of Paris | 19 Sept 1870 – 28 Jan 1871 | ≈ 400,000 defenders | ≈ 200,000 besiegers |
These figures are approximate, but they are suitable for modelling because the order of magnitude is consistent with the historical record.
这些数字是近似值，但它们适合建模，因为其数量级与历史记录一致。
2. Army Mobilisation Ratios | 军队动员比率
At the start of the war France had an active army of around 400,000 men, while the German states mobilised roughly 1,200,000 men. The ratio of German to French mobilised strength was therefore about 3 : 1.
战争开始时，法国现役军队约为 40 万人，而德意志各邦动员了约 120 万人。因此德法动员兵力之比约为 3 : 1。
Ratio = 1,200,000 ÷ 400,000 = 3 : 1
You can simplify ratios like algebraic fractions. If one side has strength a and the other has strength b, the ratio a : b is reduced by dividing both terms by their greatest common divisor. This helps to compare unequal forces quickly.
你可以像化简代数分数一样化简比率。如果一方兵力为 a，另一方兵力为 b，那么 a : b 可以通过将两项同时除以最大公约数来化简。这有助于快速比较兵力不等的双方。
At Sedan the ratio of German to French troops was 200,000 : 120,000, which simplifies to 5 : 3. A ratio greater than 1 shows a German numerical advantage.
在色当战役中，德法兵力之比为 200,000 : 120,000，化简为 5 : 3。比值大于 1 表明德军在人数上占优势。
3. Casualty Percentages and Index Numbers | 伤亡百分比与指数
At the Battle of Sedan, French losses included about 3,000 killed, 14,000 wounded, and 83,000 captured. Adding these gives 100,000 French personnel lost out of approximately 120,000 engaged.
在色当战役中，法军伤亡包括约 3,000 人死亡、14,000 人受伤、83,000 人被俘。这些相加为 10 万人损失，而参战人数约为 12 万人。
French loss percentage = (100,000 ÷ 120,000) × 100 ≈ 83.3%
German losses at Sedan were approximately 9,000 killed and wounded out of about 200,000 engaged. This gives a much lower loss percentage.
德军在色当的伤亡约为 9,000 人，而参战人数约为 20 万人。这给出了一个低得多的损失百分比。
German loss percentage = (9,000 ÷ 200,000) × 100 = 4.5%
comes (83.3 ÷ 4.5) × 100 ≈ 1851, meaning the French loss rate was about 18.5 times higher.
指数可以将这两个百分比进行比较，把德军损失率设为 100。法军损失率则为 (83.3 ÷ 4.5) × 100 ≈ 1851，这意味着法军损失率约为德军的 18.5 倍。
4. Exponential Mobilisation and Doubling Time | 动员的指数增长与倍增时间
The speed of mobilisation can be modelled with exponential growth. Suppose the Prussian-led force grew from 300,000 to 1,200,000 in about
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