📚 IGCSE WJEC PE: Formula and Theorem Quick Reference Handbook | IGCSE WJEC 体育：公式定理速查手册
This quick reference handbook summarises the essential formulas, equations and biomechanical principles required for IGCSE WJEC Physical Education. It is designed to help you revise exercise physiology, movement analysis, training calculations and body composition measures.
本速查手册汇总 IGCSE WJEC 体育所必需的关键公式、方程和生物力学原理，帮助你复习运动生理学、动作分析、训练计算和身体成分测量等内容。
1. Maximum Heart Rate and Heart Rate Training Zones | 最大心率与心率训练区间
The simplest estimate of maximum heart rate (HRmax) is HRmax = 220 − age. For example, a 15-year-old student has an estimated HRmax of 205 beats per minute (bpm). This value is used to set aerobic and anaerobic training intensities.
最大心率（HRmax）最常用的估算公式是 HRmax = 220 − 年龄。例如，15 岁学生的估算最大心率为 205 次/分（bpm）。该数值用于设定有氧和无氧训练强度。
HRmax = 220 − age
Training zones are expressed as a percentage of HRmax. Coaches use these zones to target different energy systems and training adaptations.
训练区间通常以最大心率的百分比表示。教练利用这些区间来针对不同的供能系统和训练适应。
| Training zone / 训练区间 | %HRmax | Typical use / 典型用途 |
|---|---|---|
| Light intensity / 低强度 | 50–60% | Warm-up, recovery / 热身、恢复 |
| Aerobic zone / 有氧区间 | 60–70% | Endurance base / 耐力基础 |
| Aerobic threshold / 有氧阈值 | 70–80% | Cardiovascular fitness / 心血管适能 |
| Anaerobic threshold / 无氧阈值 | 80–90% | Lactate tolerance / 乳酸耐受 |
| Maximal zone / 最大强度区 | 90–100% | Speed and power / 速度与爆发力 |
2. Karvonen Formula and Heart Rate Reserve | 卡沃宁公式与储备心率
The Karvonen formula uses heart rate reserve (HRR) to produce a more individualised target heart rate. First calculate HRR = HRmax − resting HR, then apply the target heart rate equation below.
卡沃宁公式使用储备心率（HRR）来计算更具个体化的目标心率。先计算 HRR = HRmax − 静息心率，再代入以下目标心率公式。
Target HR = ((HRmax − resting HR) × %intensity) + resting HR
Example: a 16-year-old has HRmax = 204 bpm and resting HR = 60 bpm. At 70% intensity, target HR = ((204 − 60) × 0.70) + 60 = (144 × 0.70) + 60 = 160.8 bpm.
示例：16 岁学生 HRmax = 204 bpm，静息心率 = 60 bpm。在 70% 强度下，目标心率 = ((204 − 60) × 0.70) + 60 = (144 × 0.70) + 60 = 160.8 bpm。
The Karvonen method is often more accurate than using %HRmax alone because it accounts for differences in resting heart rate.
卡沃宁法通常比单独使用最大心率百分比更准确，因为它考虑了静息心率的个体差异。
3. Cardiac Output, Stroke Volume and Heart Rate | 心输出量、每搏输出量与心率
Cardiac output (Q) is the volume of blood pumped by the heart per minute. It is calculated by multiplying stroke volume (SV) by heart rate (HR).
心输出量（Q）是心脏每分钟泵出的血液量，由每搏输出量（SV）乘以心率（HR）计算得出。
Q = SV × HR
Stroke volume should be converted to litres before substitution if Q is needed in L/min. For example, if SV = 70 mL/beat and HR = 72 bpm, then Q = 0.070 L × 72 = 5.04 L/min.
如果心输出量以 L/min 表示，每搏输出量需先换算为升。例如，SV = 70 mL/次，HR = 72 bpm，则 Q = 0.070 L × 72 = 5.04 L/min。
During exercise, both stroke volume and heart rate increase, so cardiac output rises significantly to deliver more oxygen to working muscles.
运动时每搏输出量和心率都会增加，因此心输出量显著上升，从而向工作肌肉输送更多氧气。
4. Minute Ventilation and Respiratory Rate | 每分通气量与呼吸频率
Minute ventilation (VE) is the total volume of air moved into or out of the lungs per minute. It is the product of tidal volume (TV) and breathing frequency (f).
每分通气量（VE）是每分钟进出肺部的空气总量，等于潮气量（TV）乘以呼吸频率（f）。
VE = TV × f
At rest, tidal volume is about 0.5 L and breathing frequency is about 12 breaths/min, so VE ≈ 6 L/min. During maximal exercise, tidal volume may rise to 3 L and frequency to 40 breaths/min, giving VE ≈ 120 L/min.
静息时潮气量约为 0.5 L，呼吸频率约为 12 次/分，因此每分通气量约为 6 L/min。最大运动时潮气量可上升至 3 L，呼吸频率可达 40 次/分，每分通气量约为 120 L/min。
Minute ventilation increases during exercise mainly because both tidal volume and breathing frequency increase to supply more oxygen and remove carbon dioxide.
运动时每分通气量增加，主要是因为潮气量和呼吸频率同时上升，以提供更多氧气并排出二氧化碳。
5. Body Mass Index and Body Composition Classification | 身体质量指数与身体成分分类
Body Mass Index (BMI) is a simple estimate of body composition based on mass and height. It is widely used in health and fitness screening.
身体质量指数（BMI）是根据体重和身高对身体成分进行简单估算的指标，广泛用于健康和体适能筛查。
BMI = body mass (kg) ÷ (height (m))²
Example: a student with mass 60 kg and height 1.65 m has BMI = 60 ÷ (1.65 × 1.65) = 60 ÷ 2.7225 = 22.0 kg/m².
示例：一名学生体重 60 kg、身高 1.65 m，则 BMI = 60 ÷ (1.65 × 1.65) = 60 ÷ 2.7225 = 22.0 kg/m²。
| BMI classification / BMI 分类 | BMI range (kg/m²) / BMI 范围 |
|---|---|
| Underweight / 体重过轻 | < 18.5 |
| Normal weight / 正常体重 | 18.5 – 24.9 |
| Overweight / 超重 | 25.0 – 29.9 |
| Obese / 肥胖 | ≥ 30.0 |
For adolescents, BMI should be interpreted using age- and sex-specific centile charts rather than adult cut-offs alone.
对于青少年，BMI 应结合年龄和性别特定的百分位曲线图来解释，而不能仅使用成人界值。
6. Basal Metabolic Rate and Energy Balance | 基础代谢率与能量平衡
Basal metabolic rate (BMR) is the energy used by the body at complete rest to maintain basic functions such as breathing, circulation and cell repair. A simple IGCSE estimate is shown below.
基础代谢率（BMR）是身体在完全静息状态下维持呼吸、循环和细胞修复等基本功能所需消耗的能量。以下是一个适用于 IGCSE 的简单估算公式。
BMR ≈ body mass (kg) × 24 kcal/day
This simplified formula assumes an average metabolic rate of about 1 kcal per kg of body mass per hour. A 65 kg student would have an estimated BMR of approximately 65 × 24 = 1560 kcal/day.
该简化公式假设每公斤体重每小时约消耗 1 kcal。65 kg 学生的估算 BMR 约为 65 × 24 = 1560 kcal/day。
Energy balance compares energy intake from food with energy expenditure from BMR and physical activity. If intake exceeds expenditure, body mass tends to increase.
能量平衡比较食物摄入的能量与来自 BMR 和身体活动的能量消耗。如果摄入大于消耗，体重往往增加。
Energy balance = energy intake − energy expenditure
A positive energy balance leads to weight gain, while a negative energy balance leads to weight loss. For weight management, the balance must be considered over days and weeks.
正能量平衡导致体重增加，负能量平衡导致体重下降。体重管理需要从数天到数周的时间尺度上考虑能量平衡。
7. Work, Power and Mechanical Efficiency | 功、功率与机械效率
In biomechanics, work is done when a force moves an object through a distance. Work is measured in joules (J).
在生物力学中，当力使物体移动一段距离时，就做了功。功的单位是焦耳（J）。
Work (J) = force (N) × distance moved (m)
Power is the rate of doing work, measured in watts (W). It can also be calculated as force multiplied by velocity.
功率是做功的速率，单位是瓦特（W）。功率也可用力量乘以速度计算。
Power (W) = work (J) ÷ time (s) = force (N) × velocity (m/s)
Example: a shot putter applies a force of 50 N to accelerate a shot over 2 m in 0.5 s. Work = 50 × 2 = 100 J and power = 100 ÷ 0.5 = 200 W.
示例：一名铅球运动员在 2 m 距离内对铅球施加 50 N 的力，用时 0.5 s。功 = 50 × 2 = 100 J，功率 = 100 ÷ 0.5 = 200 W。
Mechanical efficiency compares useful work output with total energy input. The human body is typically less than 30% efficient during exercise.
机械效率比较有用功输出与总能量输入。人体运动时的机械效率通常低于 30%。
Efficiency (%) = (useful work output ÷ total energy input) × 100
8. Speed, Acceleration, Force and Momentum | 速度、加速度、力与动量
Speed is the distance travelled per unit time. Acceleration is the rate of change of velocity. These are essential for analysing sprinting, jumping and throwing movements.
速度是单位时间内经过的距离，加速度是速度的变化率。这些概念对于分析短跑、跳跃和投掷动作至关重要。
Speed (m/s) = distance (m) ÷ time (s)
Acceleration (m/s²) = (final velocity − initial velocity) ÷ time (s)
Newton’s second law states that the force acting on an object is equal to the object’s mass multiplied by its acceleration.
牛顿第二定律指出，作用在物体上的力等于物体的质量乘以加速度。
Force (N) = mass (kg) × acceleration (m/s²)
Momentum is the product of mass and velocity. It is conserved in collisions and is useful when analysing contact situations in sport.
动量是质量与速度的乘积。动量在碰撞中守恒，在分析运动中的接触情况时很有用。
Momentum (kg m/s) = mass (kg) × velocity (m/s)
A larger momentum makes an athlete harder to stop, which is why fast, heavy rugby players can be difficult to tackle.
动量越大，运动员越难被截停，这就是为什么快速且体重大或质量大的橄榄球运动员很难被擒抱的原因。
9. Moment of Force, Levers and Mechanical Advantage | 力矩、杠杆与机械效益
A moment is the turning effect of a force about a pivot. It is calculated as force multiplied by the perpendicular distance from the pivot to the line of action of the force.
力矩是力绕支点产生的转动效应，等于力乘以从支点到力作用线的垂直距离。
Moment (N m) = force (N) × perpendicular distance (m)
In a lever system, mechanical advantage (MA) compares the effort arm and the resistance arm. A mechanical advantage greater than 1 means the lever can move a larger load with a smaller effort, but usually over a shorter distance.
在杠杆系统中，机械效益（MA）比较力臂和阻力臂。机械效益大于 1 表示杠杆能用较小的力移动较大的负荷，但通常移动距离较短。
Mechanical advantage = effort arm ÷ resistance arm
| Lever class / 杠杆类别 | 更多咨询请联系16621398022（同微信）
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