📚 Year 9 OCR PE Formula & Theorem Quick Reference Handbook | 九年级OCR体育公式定理速查手册
This handbook provides clear explanations, key equations, and practical examples for every formula and theorem covered in the Year 9 OCR Physical Education syllabus. Use it for quick revision, homework, and deepening your understanding of how science underpins sports performance.
本手册为九年级OCR体育课程涵盖的每一个公式和定理提供清晰的解释、关键方程与实际示例。适合用于快速复习、课后作业以及深入理解科学如何支撑运动表现。
1. Maximum Heart Rate (MHR) | 最大心率 (MHR)
Maximum Heart Rate is the highest number of beats per minute your heart can safely achieve during maximal exercise. The simplest and most widely used estimation in Year 9 PE is: MHR = 220 − age. For example, a 13‑year‑old student has an estimated MHR of 207 bpm. This value forms the baseline for setting training zones.
最大心率是你的心脏在极限运动中每分钟所能安全达到的最高心跳次数。九年级体育中最简单、最广泛使用的估算方法是：最大心率 = 220 − 年龄。例如，一名13岁学生的估算最大心率为207次/分。这个数值是设定训练区间的基础。
Athletes use MHR to ensure they work at the right intensity: aerobic training typically targets 60–80% MHR, while anaerobic intervals may push above 85% MHR.
运动员利用最大心率来确保训练强度适宜：有氧训练通常将目标设定在最大心率的60%–80%，而无氧间歇训练则可能要求超出最大心率的85%。
2. Karvonen Formula (Heart Rate Reserve) | 卡沃南公式（心率储备）
The Karvonen formula is a more personalised method to calculate target heart rate because it accounts for resting heart rate (RHR). The heart rate reserve is MHR − RHR. The target heart rate at a given intensity is then calculated as:
Target HR = ((MHR − RHR) × %Intensity) + RHR
卡沃南公式是一种更个性化的目标心率计算方法，因为它考虑了安静心率（RHR）。心率储备 = 最大心率 − 安静心率。给定强度下的目标心率则按下式计算：
目标心率 = ((最大心率 − 安静心率) × 强度百分比) + 安静心率
If a 14‑year‑old has an RHR of 70 bpm and wants to train at 70% 65 bpm. This method respects individual fitness differences and is more precise than using %MHR alone.
如果一名14岁学生的安静心率为70次/分，并希望以70%的强度进行训练，首先求得最大心率 = 206次/分。心率储备 = 206 − 70 = 136次/分。然后目标心率 = (136 × 0.70) + 70 = 95 + 70 = 165次/分。这种方法尊重个体体能差异，比单纯使用最大心率百分比更精确。
3. Training Intensity Zones (% of MHR) | 训练强度区间（最大心率百分比）
Training intensity is often prescribed as a percentage of maximum heart rate. There are five broad zones relevant to the OCR PE course:
- Zone 1 – Very Light (50–60% MHR): warm‑up and recovery
- Zone 2 – Light (60–70% MHR): aerobic base building, fat burning
- Zone 3 – Moderate (70–80% MHR): aerobic endurance, improvement of cardiovascular fitness
- Zone 4 – Hard (80–90% MHR): anaerobic threshold, lactate tolerance
- Zone 5 – Maximum (90–100% MHR): speed and power, short bursts
训练强度通常以最大心率的百分比来规定。与OCR体育课程相关的有五大区间：
- 1区 – 非常轻松（50%–60% 最大心率）：热身与恢复
- 2区 – 轻松（60%–70% 最大心率）：有氧基础建设、燃脂
- 3区 – 中等（70%–80% 最大心率）：有氧耐力、提升心血管适能
- 4区 – 高强度（80%–90% 最大心率）：无氧阈、乳酸耐受力
- 5区 – 极限（90%–100% 最大心率）：速度与爆发力、短时冲刺
A cross‑country runner might spend most of the session in Zone 3, whereas a sprinter performing repeated 60‑metre dashes would work in Zone 5. Understanding these zones allows performers to design sport‑specific training programmes.
一名越野跑选手可能会将大部分训练时间花在3区，而一名进行重复60米冲刺的短跑选手则会在5区运动。理解这些区间有助于运动者设计针对专项的训练计划。
4. Body Mass Index (BMI) | 身体质量指数 (BMI)
Body Mass Index is a simple screening tool that compares weight to height in order to classify individuals into categories such as underweight, normal weight, overweight and obese. The formula is:
BMI = weight (kg) ÷ height² (m)
身体质量指数是一种简单的筛查工具，通过比较体重与身高，将人体分为偏瘦、正常体重、超重和肥胖等类别。其公式为：
BMI = 体重（千克）÷ 身高²（米）
a height of 1.65 m has a BMI of 55 ÷ (1.65²) ≈ 20.2, falling within the normal range (18.5– aches use skinfold callipers or bioelectrical impedance for a more accurate body composition assessment.
例如，一名体重55千克、身高1.65米的学生，其BMI = 55 ÷ (1.65²) ≈ 20.2，属于正常范围（18.5–24.9）。在运动中，仅凭BMI无法区分肌肉与脂肪；一名肌肉量高的橄榄球运动员可能具有高BMI值却体脂很低。因此，教练会使用皮脂卡钳或生物电阻抗法来进行更精确的身体成分评估。
5. Speed, Distance, Time | 速度、距离、时间
The fundamental relationship between speed, distance and time is essential for analysing performance in almost every event. The core equation is:
Speed = Distance ÷ Time
速度、距离和时间之间的基本关系对于分析几乎所有运动项目的表现都至关重要。核心等式为：
速度 = 距离 ÷ 时间
To find distance: Distance = Speed × Time. To find time: Time = Distance ÷ Speed. In a 100‑ late how far they sprint during a match if their average speed and playing time are known.
求距离时：距离 = 速度 × 时间。求时间时：时间 = 距离 ÷ 速度。在一项12.5秒完成的100米短跑中，平均速度为 100 ÷ 12.5 = 8 米/秒。利用该等式，足球运动员也可以在已知平均速度与比赛时间的情况下，计算出整场比赛中冲刺的距离。
Always ensure units are compatible: if distance is in metres and time in seconds, speed will be in metres per second (m/s). For kilometres per hour, convert metres to kilometres and seconds to hours.
务必保持单位一致：若距离以米、时间以秒为单位，则速度为米/秒（m/s）。若需以千米/时表示，则应将米转换为千米、秒转换为小时。
6. Acceleration | 加速度
Acceleration is the rate at which velocity changes. It is calculated using the change in velocity over the time taken for that change:
Acceleration (m/s²) = (Final velocity − Initial velocity) ÷ Time
加速度是速度变化的快慢程度。它通过速度变化量除以发生变化所用的时间来计算：
加速度（米/秒²）=（末速度 − 初速度）÷ 时间
leration, often called deceleration, occurs when a performer slows down. In sprint starts, high acceleration is crucial; elite athletes aim to maximise force production in the first few strides to generate a large change in velocity.
如果一名自行车手在4秒内从2米/秒加速到10米/秒，其加速度为 (10 − 2) ÷ 4 = 2 米/秒²。负加速度，通常称为减速度，发生在运动者减速时。在短跑起跑中，高加速度至关重要；精英运动员力求在前几步产生最大力量，以实现速度的急剧变化。
Newton' ion) links directly to this concept: the greater the force applied, the greater the acceleration for a given mass. An explosive lower‑body push generates a large acceleration from the blocks.
牛顿第二定律（力 = 质量 × 加速度）与此概念直接相关：在质量不变的情况下，施加的力越大，加速度越大。爆发式双腿蹬伸能从起跑器产生巨大的加速度。
7. Work Done | 做功
In physical terms, work is done when a force moves an object through a distance in the direction of the force. The equation is:
Work (J) = Force (N) × Distance moved in direction of force (m)
在物理学中，当一个力使物体沿力的方向移动一段距离时，即做了功。等式为：
功（焦耳）= 力（牛顿）× 沿力方向移动的距离（米）
rbell 0.5 m, the work done is 200 × 0.5 = 100 J. In sport, work done provides insight into energy expenditure; cycling up a hill requires more work than cycling on a flat road because the vertical displacement is greater.
向上举起一根20千克的杠铃：所需力即为重力 = 质量 × 引力场强度 ≈ 20 × 10 = 200 牛。若举重运动员将杠铃上提0.5米，所做的功为 200 × 0.5 = 100 焦。在运动中，做功直观反映能量消耗；骑自行车上坡比在平路上骑行需要做更多的功，因为垂直位移更大。
8. Power Output | 功率输出
Power measures how quickly work is done or how rapidly energy is transferred. It is a crucial factor in explosive sports such as jumping, throwing and sprinting. The fundamental equation is:
Power (W) = Work done (J) ÷ Time taken (s)
功率衡量做功的快慢或能量转移的速率。它是跳跃、投掷和短跑等爆发性运动中的关键因素。基本等式为：
功率（瓦）= 做功（焦）÷ 所用时间（秒）
e work = force × distance, and distance ÷ time = velocity, power can also be expressed as:
Power (W) = Force (N) × Velocity (m/s)
因为 功 = 力 × 距离，而 距离 ÷ 时间 = 速度，所以功率也可以表示为：
功率（瓦）= 力（牛）× 速度（米/秒）
If a rower exerts 400 N of force on the oar and moves it at 3 m/s, the mechanical power output is 1200 W. Instantaneous power generation is a better predictor of athletic performance than strength alone, as it accounts for speed of movement.
如果一名赛艇运动员对桨施加400牛的力，并以3米/秒的速度拉动桨，则其机械功率输出为1200瓦。瞬间功率生成能力比单纯的力量更能预测运动表现，因为它还考虑了动作速度。
9. Moment of Force (Torque) | 力矩
A moment is the turning effect produced by a force around a pivot. In the human body, joints act as pivots, and muscles create moments to produce movement or maintain posture. The formula is:
Moment (Nm) = Force (N) × Perpendicular distance from pivot to line of force (m)
力矩是力绕支点产生的转动效应。在人体中，关节充当支点，肌肉产生力矩以实现运动或维持姿势。公式为：
力矩（牛·米）= 力（牛）× 支点到力作用线的垂直距离（米）
To hold a dumbbell at arm' e moment of the weight is 30 × 0.6 = 18 Nm. The muscle creates an equal anticlockwise moment to keep the arm horizontal. Athletes use the principle of moments when adjusting technique to reduce injury risk, for example by keeping a weight closer to the body during a lift to decrease the moment arm.
要保持哑铃在手臂伸直时的水平，三角肌必须产生一个与哑铃重量力矩大小相等、方向相反的力矩。若哑铃重30牛，距肩关节0.6米，其力矩为 30 × 0.6 = 18 牛·米。肌肉则产生18牛·米的逆时针力矩以维持手臂水平。运动员在调整技术时运用力矩原理来降低受伤风险，例如在举重过程中让负荷更靠近身体，以缩短力臂。
10. Mechanical Advantage and Levers | 机械效益与杠杆
Levers allow a small effort to move a large load, or to increase the speed and range of movement. Every lever has three parts: fulcrum (pivot), load (resistance) and effort (muscle force). Mechanical advantage describes the ratio of load to effort:
Mechanical Advantage = Load ÷ Effort
杠杆能使较小的施力移动较大的负荷，或增大运动的速度与幅度。每个杠杆由三部分组成：支点、负荷（阻力）和施力（肌肉力）。机械效益表示负荷与施力的比值：
机械效益 = 负荷 ÷ 施力
The OCR course distinguishes three classes of levers found in the human body:
| Class | 类别 | Arrangement | 排列 | Sport example | 运动示例 | Mechanical Advantage | 机械效益 |
|---|---|---|---|
| First | 一类 | Fulcrum between load and effort | 支点在负荷与施力之间 | Neck joint heading a football | 颈部关节顶足球 | Can be >1 or <1 | 可大于或小于1 |
| Second | 二类 | Load between fulcrum and effort | 负荷在支点与施力之间 | Calf raise standing on toes | 站立提踵 | Always >1 (large load moved) | 始终大于1（可移动较大负荷） |
| Third | 三类 | Effort between fulcrum and load | 施力在支点与负荷之间 | Bicep curl | 肱二头肌弯举 | Always <1 (favours speed & range) | 始终小于1（有利于速度与幅度） |
Most limbs operate as third‑class levers, sacrificing mechanical advantage for a wide range of fast motion. Sprinters exploit this design to achieve high limb velocity.
人体大多数肢体以第三类杠杆运作，牺牲机械效益以换取大范围的快速运动。短跑运动员利用这一构造来实现较高的肢体速度。
11. Calorie Expenditure (MET) | 卡路里消耗（代谢当量）
used while sitting quietly (≈ 3.5 ml O₂/kg/min). The formula for total calories burned is:
Calories = MET value × body weight (kg) × duration (hours)
运动过程中的能量消耗可使用代谢当量（MET）进行估算。1 MET代表安静坐姿时的能量消耗（约为3.5毫升氧气/千克/分钟）。总消耗热量的公式为：
卡路里 = MET值 × 体重（千克） × 运动时长（小时）
. For example, a 55 kg student who plays football for 1.5 hours burns approximately 7 × 55 × 1.5 = 577.5 kcal. This equation helps nutritionists design appropriate dietary plans for athletes and makes energy balance tangible for GCSE PE analysis.
不同活动具有不同的MET值：以8千米/时速度跑步约为8 METs，踢足球约为7 METs，中等强度的骑车约为5.5 METs。例如，一名55千克的学生踢足球1.5小时，大约消耗 7 × 55 × 1.5 = 577.5 千卡。该等式帮助营养师为运动员设计合理的膳食计划，并使能量平衡在GCSE体育分析中变得具体化。
d labels energy is often given in kilojoules, so conversions may be required: multiply kcal by 4.184 to obtain kJ.
请记住 1 千卡 = 4.184 千焦。食品标签上的能量常以千焦表示，因此可能需要换算：将千卡乘以4.184即可得到千焦值。
12. Predicted VO₂ Max from 20‑Meter Shuttle Run | 20米折返跑预测最大摄氧量
The 20‑metre shuttle run (beep test) is a common method to estimate maximum oxygen uptake (VO₂ max) without laboratory equipment. VO₂ max, measured in ml/kg/min, indicates the body's maximal capacity to transport and utilise oxygen during intense exercise. A widely recognised Léger equation for children and adolescents is:
VO₂ max = 31.025 + (3.238 × speed) − (3.248 × age) + (0.1536 × age × speed)
20米折返跑（哔哔测试）是无需实验室设备即可估算最大摄氧量（VO₂ max）的常用方法。VO₂ max 以毫升/千克/分钟为单位，表示身体在剧烈运动中运输和利用氧气的最大能力。一个广泛认可的儿童与青少年Léger方程为：
VO₂ max = 31.025 + (3.238 × 速度) − (3.248 × 年龄) + (0.1536 × 年龄 × 速度)
In this equation, 'speed' refers to the running speed in km/h at the final completed stage. For a 14‑year‑ l/kg/min. Higher VO₂ max values are associated with stronger aerobic endurance, benefiting distance runners, swimmers and team sport athletes who cover large areas.
在该方程中，'速度'指最后完成的阶段所对应的奔跑速度，单位为千米/时。一名14岁学生完成第8阶段（速度11.5千米/时）： 8 × 11.5) − (3.248 × 14) + (0.1536 × 14 × 11.5) ≈ 31.025 + 37.237 − 45.472 + 24.73 ≈ 47.52 毫升/千克/分钟。较高的VO₂ max 值与较强的有氧耐力相关，有益于长跑运动员、游泳选手以及需要大范围跑动的集体项目运动员。
Keep in mind that this prediction is an estimate; factors such as motivation, pacing and environmental conditions can influence the bleep test score. Nevertheless, the formula is invaluable for tracking fitness changes across a school year.
需要注意的是，该预测仅为估算值；动机、节奏和环境条件等因素都会影响哔哔测试成绩。尽管如此，该公式在整个学年中追踪体能变化时仍极具价值。
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