📚 The Tour De France | 环法自行车赛中的数学
The Tour de France is one of the most famous sporting events in the world. Every July, cyclists race about 3,500 kilometres across France in 21 stages. Behind the drama of the race lies a rich collection of mathematics: speed calculations, percentages, gradients, unit conversions and statistics. In this article, we will explore how IGCSE mathematics helps us understand the greatest bike race on Earth.
环法自行车赛是世界上最著名的体育赛事之一。每年七月，自行车手们在21个赛段中骑行约3500公里，穿越法国各地。在这场比赛的戏剧性背后，蕴藏着丰富的数学知识：速度计算、百分比、坡度、单位换算和统计。在本文中，我们将探索IGCSE数学如何帮助我们理解这场地球上最伟大的自行车赛事。
1. Speed, Distance and Time | 速度、距离和时间
The most fundamental formula in cycling mathematics is the relationship between speed, distance and time. The formula is:
骑行数学中最基本的公式是速度、距离和时间之间的关系。公式为：
speed = distance ÷ time
From this single equation, we can rearrange it to find distance (distance = speed × time) or time (time = distance ÷ speed). For example, if a rider completes a 180 km flat stage in 4 hours and 30 minutes, their average speed is 180 ÷ 4.5 = 40 km/h.
从这个单一方程出发，我们可以重新排列来求距离（距离 = 速度 × 时间）或时间（时间 = 距离 ÷ 速度）。例如，如果一名车手以4小时30分钟完成180公里的平路赛段，他的平均速度是180 ÷ 4.5 = 40 km/h。
In the Tour de France, time is measured to the nearest second. A rider who finishes a 30 km time trial in 38 minutes and 45 seconds has a finishing time of 38.75 minutes, and their average speed is 30 ÷ (38.75 ÷ 60) ≈ 46.45 km/h.
在环法自行车赛中，时间精确到秒。一名车手用38分45秒完成30公里的个人计时赛，其完赛时间为38.75分钟，平均速度为30 ÷ (38.75 ÷ 60) ≈ 46.45 km/h。
2. Calculating Average Speed | 计算平均速度
The average speed of a stage is not simply the mean of the speeds at different moments. It is defined as the total distance divided by the total time. This is known as the harmonic mean in disguise.
一个赛段的平均速度并不是不同时刻速度的简单平均值。它定义为总距离除以总时间。这实际上是调和平均数的一种表现形式。
Consider a rider who completes a two-part stage. In the first 100 km, they ride at 40 km/h. In the next 50 km, they ride at 30 km/h. The total time is (100 ÷ 40) + (50 ÷ 30) = 2.5 + 1.667 = 4.167 hours. The average speed is 150 ÷ 4.167 ≈ 36 km/h. Notice that this is not (40 + 30) ÷ 2 = 35 km/h, because the rider spent more time at the slower speed.
考虑一名车手完成一个分为两段的路程。前100公里以40 km/h骑行，后50公里以30 km/h骑行。总时间为(100 ÷ 40) + (50 ÷ 30) = 2.5 + 1.667 = 4.167小时。平均速度为150 ÷ 4.167 ≈ 36 km/h。注意，这并不是(40 + 30) ÷ 2 = 35 km/h，因为车手以较慢速度行驶的时间更长。
This idea is tested in IGCSE questions where students must calculate average speed over a journey with multiple segments. Always use total distance ÷ total time, never the arithmetic mean of the speeds.
这一概念在IGCSE考题中经常出现，学生必须计算包含多个路段行程的平均速度。始终使用总距离 ÷ 总时间，绝不要使用速度的算术平均值。
3. Gradients and Road Slopes | 坡度与路面斜率
Mountain stages are where the Tour de France is often won. Climbing roads are described by their gradient, which is the percentage rise divided by the horizontal distance. A 10% gradient means the road rises 10 metres for every 100 metres of horizontal distance.
山地赛段往往是环法自行车赛的决胜之地。爬坡路段用坡度来描述，即垂直上升高度除以水平距离的百分比。10%的坡度意味着道路每水平前进100米就上升10米。
In mathematics, the gradient of a line is calculated as:
在数学中，直线的斜率计算如下：
gradient = rise ÷ run = change in vertical height ÷ change in horizontal distance
For example, the famous Alpe d’Huez climb has an average gradient of about 8.1%. Over a horizontal distance of 13.8 km, the total vertical rise is 13.8 × 1000 × 0.081 ≈ 1118 metres. In reality, the climb rises 1120 metres, which matches closely.
例如，著名的阿尔普迪埃爬坡平均坡度约为8.1%。在13.8公里的水平距离上，总垂直上升高度为13.8 × 1000 × 0.081 ≈ 1118米。实际上，该爬坡上升1120米，两者非常接近。
Sometimes the question gives the vertical rise and the actual road distance. Since the road is the hypotenuse of a right-angled triangle, we use Pythagoras’ theorem to find the horizontal distance. If a climb rises 800 m over a road length of 10 km, the horizontal distance is √(10000² − 800²) ≈ 9968 m, giving a gradient of 800 ÷ 9968 ≈ 8.03%.
有时题目给出垂直上升高度和实际道路距离。由于道路是直角三角形的斜边，我们使用勾股定理来求水平距离。如果一段爬坡在10公里道路长度上上升800米，水平距离为√(10000² − 800²) ≈ 9968米，坡度为800 ÷ 9968 ≈ 8.03%。
4. Unit Conversions in Cycling | 骑行中的单位换算
Cyclists and commentators use different units for speed depending on the country. In France, speeds in the Tour are given in kilometres per hour (km/h), but some students and scientists prefer metres per second (m/s). Converting between these units is a core IGCSE skill.
自行车手和评论员使用的速度单位因国家而异。在法国，环法赛的车速以公里每小时（km/h）为单位，但一些学生和科学家更喜欢米每秒（m/s）。在这些单位之间进行换算是IGCSE的核心技能。
The conversion factor is simple. Since 1 km = 1000 m and 1 hour = 3600 seconds, we divide km/h by 3.6 to get m/s. Therefore, a rider travelling at 54 km/h is moving at 54 ÷ 3.6 = 15 m/s.
换算系数很简单。由于1公里 = 1000米，1小时 = 3600秒，将km/h除以3.6即可得到m/s。因此，以54 km/h行驶的车手速度为54 ÷ 3.6 = 15 m/s。
Conversely, to convert m/s to km/h, multiply by 3.6. A sprinting cyclist reaching 20 m/s is travelling at 20 × 3.6 = 72 km/h. These conversions appear frequently in exam questions about real-life speed.
反过来，将m/s转换为km/h时，乘以3.6即可。一名冲刺车手达到20 m/s时，其速度为20 × 3.6 = 72 km/h。这些换算频繁出现在关于现实生活速度的考题中。
5. Team Time Trial Strategies | 团队计时赛策略
In a team time trial, riders start together and the finishing time is taken from the fifth rider to cross the line. This creates a fascinating mathematical problem: how fast can a group travel compared with an individual rider?
在团队计时赛中，车手们一起出发，完赛时间以第五名过线车手的时间为准。这产生了一个有趣的数学问题：一个团队相比个人车手能骑多快？
Research shows that riding in a group reduces aerodynamic drag by up to 30%. If the power output of each rider is roughly the same, the group can maintain a higher speed. Suppose each rider individually could sustain 45 km/h. With a 15% speed benefit from drafting, the group speed becomes 45 × 1.15 = 51.75 km/h.
研究表明，团队骑行可使空气阻力降低多达30%。如果每位车手的功率输出大致相同，团队可以维持更高的速度。假设每位车手个人能维持45 km/h，得益于跟车带来的15%速度增益，团队速度变为45 × 1.15 = 51.75 km/h。
This is a direct application of percentage increase. If a 30 km team time trial is completed at 51.75 km/h, the time taken is 30 ÷ 51.75 ≈ 0.5797 hours, or about 34.8 minutes. An individual riding at 45 km/h would take 40 minutes — a saving of over 5 minutes.
这是百分比增加的直接应用。如果30公里的团队计时赛以51.75 km/h完成，所需时间为30 ÷ 51.75 ≈ 0.5797小时，约为34.8分钟。个人以45 km/h骑行则需要40分钟——节省了5分多钟。
6. Points and Classifications | 积分与排名系统
The Tour de France has several different competitions, each with its own scoring system. The green jersey is awarded to the rider with the most sprint points, while the polka-dot jersey goes to the best climber. The mathematical skills of addition and data interpretation are essential here.
环法自行车赛有多个不同的竞赛，每个都有自己的积分系统。绿衫颁发给冲刺积分最高的车手，而圆点衫则授予最佳爬坡车手。加法和数据解读的数学技能在此至关重要。
| Stage Finish Position | 1st | 2nd | 3rd | 4th | 5th |
| Points Awarded | 50 | 30 | 20 | 15 | 10 |
If a rider finishes 1st, 3rd and 5th in three flat stages, their total points are 50 + 20 + 10 = 80. If another rider finishes 2nd in all three stages, their total is 30 + 30 + 30 = 90. Despite never winning a stage, the second rider leads the points classification.
如果一名车手在三个平路赛段中分别获得第1、第3和第5名，他的总积分为50 + 20 + 10 = 80。如果另一名车手在三个赛段中均获得第2名，他的总分为30 + 30 + 30 = 90。尽管从未赢得赛段冠军，第二名车手仍领跑积分榜。
This illustrates why consistency matters in the points competition. Solving these problems requires careful addition and comparison of totals, a skill directly tested in IGCSE data handling questions.
这说明了在积分竞争中稳定性的重要性。解决这些问题需要仔细的加法和总数比较，这是IGCSE数据处理题中直接考察的技能。
7. Statistics of the Yellow Jersey | 黄衫的统计分析
The yellow jersey is worn by the rider with the lowest total time. Over 21 stages, the general classification is a real-world exercise in adding and comparing decimals. Each rider’s total time is the sum of their stage times, and the leader has the minimum total.
黄衫由总时间最少的车手穿着。在21个赛段中，总排名是加法和比较小数的真实练习。每位车手的总时间是其各赛段时间之和，领先者拥有最小的总时间。
Suppose after three stages, Riders A and B have the following times:
假设三个赛段后，车手A和B的时间如下：
| Stage | Rider A | Rider B |
| 1 | 4h 12m 30s | 4h 11m 45s |
| 2 | 5h 03m 18s | 5h 04m 52s |
| 3 | 3h 48m 56s | 3h 49m 10s |
Converting everything to seconds, Rider A’s total is (4 × 3600 + 12 × 60 + 30) + (5 × 3600 + 3 × 60 + 18) + (3 × 3600 + 48 × 60 + 56) = 15150 + 18198 + 13736 = 47084 seconds. Rider B’s total is 15105 + 18292 + 13750 = 47147 seconds. Rider A leads by 47147 − 47084 = 63 seconds, or 1 minute 3 seconds.
将所有时间转换为秒，车手A的总时间为(4 × 3600 + 12 × 60 + 30) + (5 × 3600 + 3 × 60 + 18) + (3 × 3600 + 48 × 60 + 56) = 15150 + 18198 + 13736 = 47084秒。车手B的总时间为15105 + 18292 + 13750 = 47147秒。车手A以47147 − 47084 = 63秒，即1分3秒的优势领先。
Students should practise working with mixed units of time and converting them to a single unit such as seconds before adding. This prevents errors in carrying between minutes and seconds.
学生应练习处理混合时间单位，并在相加之前将其转换为秒等单一单位。这可以避免在分钟和秒之间的进位错误。
8. Energy and Proportional Reasoning | 能量与比例推理
A Tour de France rider burns approximately 6000 calories per stage. This energy comes from food, and the relationship between distance, power and energy involves proportional reasoning. If a rider burns 6000 calories over a 180 km stage, the calories per kilometre are 6000 ÷ 180 ≈ 33.3 calories per km.
环法车手每个赛段大约消耗6000卡路里。这些能量来自食物，而距离、功率和能量之间的关系涉及比例推理。如果一名车手在180公里的赛段中消耗6000卡路里，每公里的卡路里消耗为6000 ÷ 180 ≈ 33.3卡路里/公里。
Proportional reasoning also applies to food intake. If a rider needs to consume 90 grams of carbohydrates per hour, over a 5-hour stage they need 90 × 5 = 450 grams of carbohydrates. A banana contains about 27 grams of carbohydrates, so the rider would need approximately 450 ÷ 27 ≈ 16.7 bananas, or about 17 bananas.
比例推理同样适用于食物摄入。如果一名车手每小时需要摄入90克碳水化合物，在5小时的赛段中他需要90 × 5 = 450克碳水化合物。一根香蕉约含27克碳水化合物，因此该车手大约需要450 ÷ 27 ≈ 16.7根香蕉，即约17根。
These calculations use direct proportion: if one quantity doubles, the other doubles. Recognising when two quantities are in direct proportion is a key IGCSE skill that appears in many real-life contexts.
这些计算使用正比例关系：如果一个量翻倍，另一个量也翻倍。识别两个量何时成正比是IGCSE的一项关键技能，出现在许多现实生活情境中。
9. Describing the Route with Bearings | 用方位角描述路线
Some Tour de France stages involve transfers between the start and finish towns. On a map, we can describe the straight-line direction of a transfer using three-figure bearings. A bearing is an angle measured clockwise from north, written as three digits.
环法赛的一些赛段涉及起点和终点城镇之间的转场。在地图上，我们可以使用三位方位角来描述转场的直线方向。方位角是从正北方向顺时针测量的角度，写成三位数。
Suppose a team bus travels 40 km east and then 30 km north. To find the direct distance from the starting point, we use Pythagoras’ theorem: √(40² + 30²) = √2500 = 50 km. To find the bearing of the final position from the start, we use trigonometry: tan θ = 30 ÷ 40 = 0.75, so θ = tan⁻¹(0.75) ≈ 36.9°. The bearing is 036.9°, or 037° to the nearest degree.
假设一辆队车先向东行驶40公里，再向北行驶30公里。要求从起点到终点的直线距离，我们使用勾股定理：√(40² + 30²) = √2500 = 50公里。要求终点相对于起点的方位角，我们使用三角函数：tan θ = 30 ÷ 40 = 0.75，因此θ = tan⁻¹(0.75) ≈ 36.9°。方位角为036.9°，四舍五入为037°。
This combination of Pythagoras’ theorem and trigonometry is a classic IGCSE examination topic. It shows how right-angled triangle mathematics applies to navigation in the real world.
这种勾股定理与三角学的结合是IGCSE考试中的经典主题。它展示了直角三角形数学如何应用于现实世界中的导航。
10. Analysis of Race Data | 比赛数据分析
The Tour de France produces enormous amounts of data. Analysing this data uses the statistical measures studied in IGCSE: mean, median, mode and range. Consider the finishing times of a group of 7 riders in a mountain time trial: 38m 20s, 39m 05s, 39m 05s, 40m 12s, 41m 48s, 42m 30s, 44m 15s.
环法自行车赛产生海量数据。分析这些数据需要使用IGCSE中学习的统计量：平均数、中位数、众数和极差。考虑一组7名车手在山地计时赛中的完赛时间：38分20秒、39分05秒、39分05秒、40分12秒、41分48秒、42分30秒、44分15秒。
The mean is found by adding all the times and dividing by 7. In seconds, the sum is 2300 + 2345 + 2345 + 2412 + 2508 + 2550 + 2655 = 17115 seconds. The mean is 17115 ÷ 7 ≈ 2445 seconds, or about 40m 45s. The median is the middle value: 40m 12s. The mode is 39m 05s because it appears twice. The range is 44m 15s − 38m 20s = 5m 55s.
平均数通过将所有时间相加并除以7得到。以秒计算，总和为2300 + 2345 + 2345 + 2412 + 2508 + 2550 + 2655 = 17115秒。平均数为17115 ÷ 7 ≈ 2445秒，约为40分45秒。中位数是中间值：40分12秒。众数是39分05秒，因为它出现了两次。极差为44分15秒 − 38分20秒 = 5分55秒。
These measures give different perspectives on the data. The mean is affected by the very slow rider at 44m 15s, while the median and mode are more resistant to extreme values. Understanding which measure is most appropriate is a higher-level IGCSE skill.
这些统计量提供了看待数据的不同视角。平均数受到44分15秒这位较慢车手的影响，而中位数和众数对极端值更具抵抗性。理解哪种统计量最合适是IGCSE的高阶技能。
11. Probability and Race Outcomes | 概率与比赛结果
Probability can be used to model race outcomes. Suppose a bookmaker estimates that a sprinter has a 40% chance of winning Stage 1, a 30% chance of winning Stage 2, and the two stage results are independent. The probability of winning both stages is 0.40 × 0.30 = 0.12, or 12%.
概率可用于模拟比赛结果。假设博彩公司估计一名冲刺车手有40%的概率赢下第1赛段，30%的概率赢下第2赛段，且两个赛段的结果相互独立。赢下两个赛段的概率为0.40 × 0.30 = 0.12，即12%。
The probability of winning at least one of the two stages is found using the addition rule. P(win at least one) = P(win S1) + P(win S2) − P(win both) = 0.40 + 0.30 − 0.12 = 0.58, or 58%.
至少赢下两个赛段中一个的概率使用加法法则计算。P(至少赢一个) = P(赢S1) + P(赢S2) − P(赢两个) = 0.40 + 0.30 − 0.12 = 0.58，即58%。
This is a direct application of the multiplication rule for independent events and the addition rule for mutually non-exclusive events. These are standard topics in the IGCSE probability syllabus and are brought to life by the uncertainty of sport.
这是独立事件乘法法则和非互斥事件加法法则的直接应用。这些是IGCSE概率大纲中的标准主题，而体育比赛的不确定性使它们变得生动。
12. Summary and Exam Tips | 总结与考试技巧
The Tour de France offers a rich source of mathematical problems. The key skills we have explored are the speed-distance-time formula, calculating gradients as percentages, unit conversions, statistical measures, proportional reasoning, bearings and probability. All of these appear regularly in IGCSE Edexcel Mathematics examinations.
环法自行车赛为数学问题提供了丰富的素材。我们探讨的关键技能包括速度-距离-时间公式、以百分比计算坡度、单位换算、统计量、比例推理、方位角和概率。所有这些都经常出现在IGCSE Edexcel数学考试中。
To succeed in exam questions based on real-life contexts, always read the question carefully, identify the formula or concept needed, and check your units before calculating. Convert all times to the same unit, use the correct conversion factor for speed, and remember that average speed is total distance divided by total time.
要在基于现实情境的考题中取得成功，务必仔细审题，确定所需的公式或概念，并在计算前检查单位。将所有时间转换为相同单位，使用正确的速度换算系数，并记住平均速度等于总距离除以总时间。
Finally, practise drawing the right-angled triangle for gradient problems and applying Pythagoras’ theorem where the road is the hypotenuse. With consistent practice, you will be ready to tackle any cycling-themed mathematics question with confidence — just like a champion climber attacking the mountains.
最后，练习为坡度问题绘制直角三角形，并在道路为斜边时应用勾股定理。通过持续练习，你将能够自信地解决任何以骑行为主题的数学题——就像冠军爬坡手进攻高山一样。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导