📚 Edexcel PE Formula & Theorem Quick Reference | 爱德思体育公式定理速查手册
Success in Year 13 Edexcel Physical Education requires a confident grasp of the quantitative and theoretical building blocks that underpin every unit – from physiology and biomechanics to sport psychology. This quick-reference handbook brings together the essential formulas, laws and models you will be expected to apply in written papers and data-response questions. Each entry is explained with clear working examples, variable definitions and links to relevant sporting contexts, helping you move from memorisation to fluent application.
在爱德思 A-level 体育的 Year 13 考试中，无论是生理学、生物力学还是运动心理学模块，都需要熟练运用一系列公式与定理。这本速查手册将核心的定量公式、运动定律和经典理论模型浓缩在一起，每条均配有变量解释、计算示例以及与运动情境的关联解读，帮助你从机械记忆过渡到灵活解题，在笔试和数据分析题中精准得分。
1. Maximum Heart Rate & Karvonen Formula | 最大心率与卡沃宁公式
Maximum heart rate (HRmax) provides a simple benchmark for exercise intensity zones. The classic population-level estimate subtracts age from 220, though individual variation is significant. In exam answers, always state that the formula gives an estimate, not a precise measure.
最大心率（HRmax）是划分运动强度区间的基本参照。常用的人口估算公式为 220 减去年龄，但个体差异较大。答题时务必说明该公式仅提供估算值，而非精确测量值。
HRmax = 220 – age (years)
The Karvonen formula refines target heart rate by incorporating resting heart rate (HRrest) and a chosen intensity percentage. It uses heart rate reserve (HRR), giving more personalised training zones.
卡沃宁公式引入安静心率（HRrest）和训练强度百分比，计算心率储备（HRR），从而得出更个性化的目标心率区间。
Target HR = [(HRmax – HRrest) × % intensity] + HRrest
Example: an 18-year-old athlete with HRrest 60 bpm aiming for 70% intensity would have HRmax = 202, HRR = 142, Target HR = (142 × 0.70) + 60 = 159.4 bpm ≈ 159 bpm.
示例：一名 18 岁运动员安静心率 60 bpm，目标强度 70%，则 HRmax = 202，HRR = 142，目标心率 = (142 × 0.70) + 60 = 159.4 bpm，约 159 bpm。
2. Cardiac Output & Stroke Volume | 心输出量与每搏输出量
Cardiac output (Q) is the volume of blood pumped by the heart per minute. It is a central variable when discussing cardiovascular drift, steady-state exercise or maximal performance.
心输出量（Q）是心脏每分钟泵出的血量，是分析心血管漂移、稳态运动和最大运动表现时的核心变量。
Q = stroke volume (SV) × heart rate (HR)
Stroke volume is influenced by venous return, myocardial contractility and afterload. At rest, typical values are SV ≈ 70 mL, HR ≈ 72 bpm, giving Q ≈ 5 L/min. During maximal exercise, a trained athlete may reach SV ≈ 150 mL, HR ≈ 200 bpm, Q ≈ 30 L/min.
每搏输出量受静脉回流量、心肌收缩力和后负荷的影响。安静时 SV 约 70 mL，HR 约 72 bpm，Q ≈ 5 L/min；最大运动时，训练有素的运动员可达 SV ≈ 150 mL，HR ≈ 200 bpm，Q ≈ 30 L/min。
Changes in Q during prolonged exercise can be explained by cardiovascular drift: HR rises gradually to compensate for a declining SV caused by dehydration and redistribution of blood to the skin.
长时间运动中 Q 的变化可由心血管漂移解释：因脱水和血液向皮肤再分配，SV 逐渐下降，HR 代偿性升高以维持 Q。
3. Oxygen Consumption (VO₂) & METs | 摄氧量与代谢当量
Absolute VO₂ describes the volume of oxygen consumed per minute. Relative VO₂ normalises this to body mass, allowing fair comparisons between individuals of different sizes.
绝对摄氧量描述每分钟消耗的氧气体积；相对摄氧量将绝对摄氧量除以体重，便于不同体型个体间的公平比较。
Relative VO₂ (mL/kg/min) = absolute VO₂ (mL/min) ÷ body mass (kg)
Metabolic equivalents (METs) express energy cost as multiples of resting metabolic rate. 1 MET ≈ 3.5 mL O₂/kg/min. Moderate activities (3-6 METs) and vigorous activities (>6 METs) can be classified using this scale.
代谢当量（MET）以安静代谢率的倍数表示能量消耗。1 MET ≈ 3.5 mL O₂/kg/min。中等强度运动约 3–6 METs，高强度运动 >6 METs。
In exam data-response questions, students may be asked to calculate VO₂ from MET values or estimate energy expenditure: kcal/min = METs × body mass (kg) × 0.0175 (for general conversion).
在数据分析题中，考生可能需要从 MET 值计算 VO₂，或估算能量消耗：kcal/min = METs × 体重 (kg) × 0.0175（通用转换公式）。
4. Respiratory Quotient (RQ) | 呼吸商
Respiratory quotient is the ratio of carbon dioxide produced to oxygen consumed at the cellular level (or at the mouth, when termed respiratory exchange ratio, RER). It indicates which fuel is being predominantly metabolised.
呼吸商（RQ）是细胞层面生成的二氧化碳量与消耗的氧气量之比（若在口鼻处测量则称为呼吸交换率 RER），可指示身体主要利用哪种燃料供能。
RQ = CO₂ produced ÷ O₂ consumed
- Carbohydrate: RQ ≈ 1.0
- Fat: RQ ≈ 0.7
- Protein: RQ ≈ 0.8
- 混合饮食：RQ 通常约 0.82–0.85
Interpretation in sport: during high-intensity exercise, RQ rises towards 1.0 as carbohydrate oxidation dominates; at low intensities, fat utilisation yields RQ closer to 0.7.
运动中的解读：高强度运动时 RQ 趋向 1.0，反映糖类氧化占优势；低强度时脂肪利用增多，RQ 接近 0.7。
5. Speed, Velocity, Acceleration & Momentum | 速度、速率、加速度与动量
Biomechanics relies on precise motion descriptors. Speed is a scalar, velocity is a vector. Acceleration is the rate of change of velocity, and momentum links mass with velocity.
生物力学依赖精确的运动描述量。速率为标量，速度为矢量；加速度是速度的变化率；动量将质量与速度联系起来。
Speed = distance ÷ time
Velocity = displacement ÷ time
Acceleration = Δvelocity ÷ time
Momentum = mass × velocity (p = m × v)
Example: a 75 kg rugby player sprinting at 8 m/s has momentum 600 kg·m/s. Tackling requires absorbing or changing this momentum, which links directly to impulse.
示例：一名 75 kg 的橄榄球运动员以 8 m/s 冲刺时动量 p = 600 kg·m/s，擒抱时需要吸收或改变这一动量，这直接与冲量相关。
6. Impulse & Newton’s Laws | 冲量与牛顿运动定律
Impulse equals the change in momentum and is the product of force and the time over which it acts. Athletes manipulate impulse time to maximise or minimise force effects.
冲量等于动量的变化量，也是力与其作用时间的乘积。运动员通过改变冲量时间来实现力的最大化或缓冲效果。
Impulse = force × time (J = F × Δt) = Δ(mv)
Newton’s three laws underpin all linear motion: Law of Inertia (first), Law of Acceleration (second, F = ma), and Law of Action-Reaction (third). For A-Level PE, applying F = ma in sporting examples is essential.
牛顿三定律奠定所有直线运动的基础：惯性定律（第一定律）、加速度定律（第二定律，F = ma）和作用与反作用定律（第三定律）。在 A-Level 体育中，能将 F = ma 应用于运动情境至关重要。
Second law: F = m × a, or a = F/m
For a given force, a lighter object accelerates more. This explains why reducing body mass (while maintaining force output) improves sprint acceleration.
在给定作用力下，质量越小的物体加速度越大。这就解释了为何在保持发力能力的同时降低体重能提升冲刺加速度。
7. Projectile Motion & Magnus Effect | 抛体运动与马格努斯效应
Projectile motion is influenced by release height, release velocity, release angle and air resistance/surface properties. Factors affecting the horizontal and vertical components are often separated for calculations.
抛体运动受出手高度、出手速度、出手角度和空气阻力/物体表面特性影响。通常将水平和竖直分量分开计算。
Horizontal displacement: s_h = v × cosθ × t
Vertical displacement: s_v = v × sinθ × t – ½ g t²
The Magnus effect explains the curved flight path of a spinning ball. A spinning object drags air, creating a pressure difference: higher pressure on one side, lower on the other, producing a lateral force and deviation.
马格努斯效应解释了旋转球的弧线轨迹。旋转球拖动周围空气，形成一侧高压、一侧低压的压力差，产生侧向力使球偏离直线路径。
Topspin causes air to move faster above the ball (low pressure) and slower below (high pressure), creating a downward force – useful in tennis serves and volleyball spikes.
上旋球使球上方气流加快（低压）、下方减慢（高压），产生向下的力，在网球发球和排球扣球中极为有用。
8. Levers & Torque | 杠杆与力矩
Levers in the human body are classified by the relative positions of fulcrum, effort and load. Torque (moment of force) describes rotational effectiveness.
人体内的杠杆根据支点、动力点和阻力点的相对位置分为三类。力矩（力偶矩）描述力的转动效果。
Torque (T) = force × perpendicular distance from fulcrum
| Lever class | Arrangement | Example |
|---|---|---|
| First class | Fulcrum between effort and load | Neck extension |
| Second class | Load between fulcrum and effort | Ankle plantarflexion (calf raise) |
| Third class | Effort between fulcrum and load | Biceps curl |
Most human levers are third-class, favouring speed and range of movement over force production. Torque calculations help explain why a muscle must generate far greater force than the external load when its insertion is close to the joint.
人体内多数杠杆为第三类，以牺牲力输出换取速度和活动幅度。力矩计算可解释为何肌肉止点靠近关节时，肌肉产生的力必须远大于外部负荷。
9. Friction & Air Resistance | 摩擦力与空气阻力
Friction allows athletes to start, stop and change direction. It is determined by the coefficient of friction (μ) and the normal reaction force.
摩擦力使运动员能够启动、停止和变向。其大小取决于摩擦系数（μ）和法向反作用力。
Friction (F) ≤ μ × normal reaction force (R)
Air resistance (drag) is a fluid friction force that opposes motion. It increases with the square of velocity, cross-sectional area and the drag coefficient. Minimising frontal area (e.g., crouching in cycling) reduces resistive drag.
空气阻力（曳力）是一种流体摩擦力，与速度的平方、迎风面积和阻力系数成正比。减小迎风面积（如自行车运动员伏低身体）可降低阻力。
Drag force = ½ × ρ × Cd × A × v²
(ρ = air density, Cd = drag coefficient, A = frontal area, v = velocity)
Streamlining, skin suits and body position are all strategies to lower Cd or A, improving efficiency in sports such as cycling, skiing and sprinting.
流线型设计、紧身服装和身体姿势均为降低 Cd 或 A 的策略，可提升自行车、滑雪、短跑等项目的效率。
10. Inverted-U Theory & Drive Theory | 倒U理论与驱力理论
The Inverted-U hypothesis predicts that performance improves as arousal increases up to an optimal point, beyond which further arousal causes performance decline. The optimum varies with skill level, personality and task complexity.
倒U假设认为，唤醒水平由低升高时运动表现随之提高，达到最佳唤醒点后若继续升高则表现下降。最佳唤醒点因技能水平、个性和任务复杂度而异。
Performance = f (arousal, with an optimal point)
Drive Theory proposes a linear relationship between arousal and performance, especially for well-learned tasks or dominant responses. In formulaic terms: Performance = Habit × Drive (arousal).
驱力理论提出，尤其是对于熟练的任务或主导反应，唤醒与表现呈线性关系。用公式表达：表现 = 习惯 × 驱力（唤醒）。
For simple, gross-motor skills, higher arousal can improve performance (drive theory); for complex, fine-motor skills, the Inverted-U model predicts that over-arousal harms performance. Examiners often ask students to critique these two models.
对简单的大肌肉群技能，高唤醒能提升表现（驱力理论）；对复杂的精细技能，倒U模型预测过度唤醒会损害表现。考官常要求考生比较、评述这两种模型。
11. Hick’s Law & Fitts’ Law | 希克定律与菲茨定律
Hick’s Law relates reaction time to the number of stimulus-response choices. It is a fundamental principle in skill acquisition and interface design in sport.
希克定律将反应时间与刺激-反应选项的数量联系起来，是技能习得和运动界面设计中的基本原理。
Reaction time (RT) = k × log₂(N + 1), where N = number of choices
Fitts’ Law describes the speed-accuracy trade-off in rapid aimed movements. Movement time increases with distance to the target and decreases with target width.
菲茨定律描述快速瞄准运动中速度与准确性的权衡。移动时间随目标距离增大而增加，随目标宽度增大而减小。
Movement time (MT) = a + b × log₂(2D/W), where D = distance, W = target width
In sport, offering a goalkeeper fewer cues (late ball release) reduces N and can increase response speed. In tennis, wider racket sweet spots (increased W) reduce MT demands.
在运动中，减少守门员需处理的线索数量（如晚出手球）可降低 N、加快反应；在网球中，增大球拍甜区（增大 W）可降低对移动时间的需求。
12. Energy Systems & ATP Yield | 能量系统与ATP生成
The three energy systems – ATP-PC, lactic acid (glycolytic) and aerobic – differ in rate and capacity of ATP resynthesis. Quantitative values are often required in exam responses to justify training adaptations.
人体三大供能系统——ATP-CP 系统、乳酸能（糖酵解）系统和有氧系统——在 ATP 再合成速率和总量上各有不同。考试常要求提供具体数值以支持训练适应性的论点。
| System | ATP yield per mole of fuel | Peak power (duration) | Fuel source |
|---|---|---|---|
| ATP-PC | 1 ATP per PC molecule | Very high (up to 10 s) | Phosphocreatine |
| Lactic acid | 2 ATP per glucose | High (~30–60 s) | Muscle glycogen/ glucose |
| Aerobic | 38 ATP per glucose (approx.), >100 ATP per fatty acid | Low–moderate (hours) | Glycogen, fats, protein |
During maximal sprinting, the ATP-PC system dominates, providing rapid ATP but limited to about 10 seconds. As exercise continues, the glycolytic system becomes the primary contributor, producing lactate and H⁺ ions that contribute to fatigue.
最大强度冲刺时以 ATP-CP 系统为主，ATP 生成快但仅维持约 10 秒；持续运动时糖酵解系统成为主要供能者，产生乳酸和 H⁺ 离子，导致疲劳。
The crossover concept illustrates how the proportion of fat and carbohydrate oxidation shifts with exercise intensity – the “crossover point” is the intensity at which carbohydrate becomes the dominant fuel.
交叉点概念说明脂肪与糖氧化比例随运动强度的变化——”交叉点”指糖类成为主导燃料的运动强度临界值。
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