📚 A-Level OCR PE: Formula & Theorem Quick Reference Handbook | A-Level OCR 体育：公式定理速查手册
This handbook provides a concise but comprehensive reference for all key formulas, principles, and theorems you need to master for the OCR A-Level Physical Education specification. It covers biomechanics, exercise physiology, and relevant physical laws, with each entry explained in plain English and Chinese, so you can revise efficiently. Only unicode symbols are used throughout – no LaTeX – making it easy to copy and study on any device.
本手册为 OCR A-Level 体育课程中所有关键公式、原理和定理提供了简明而全面的速查参考，涵盖生物力学、运动生理学和相关的物理定律。每个条目均用通俗易懂的英文和中文解释，方便你高效复习。全文仅使用 Unicode 符号，无 LaTeX，可在任何设备上轻松查阅和学习。
1. Speed, Velocity & Acceleration | 速度、速率与加速度
Speed (m/s) = Distance (m) ÷ Time (s), v = d/t. It is a scalar quantity and only tells you how fast an object moves, ignoring direction. In many OCR PE questions, you calculate average speed over a sprint or a phase of play.
速度 (m/s) = 距离 (m) ÷ 时间 (s)，v = d/t。速度是标量，只反映物体运动的快慢，不考虑方向。在 OCR 体育考题中，你常需计算短跑或比赛某个阶段的平均速度。
Velocity (m/s) = Displacement (m) ÷ Time (s), v = Δs / Δt. Displacement is the straight-line distance in a specific direction, so velocity is a vector. Acceleration (m/s²) = Change in velocity (m/s) ÷ Time (s), a = (v − u)/t. Negative acceleration means deceleration. Remember to use final velocity (v) and initial velocity (u).
速率 (m/s) = 位移 (m) ÷ 时间 (s)，v = Δs / Δt。位移是特定方向上的直线距离，因此速率是矢量。加速度 (m/s²) = 速度变化量 (m/s) ÷ 时间 (s)，a = (v − u)/t。负加速度表示减速。记住使用末速度 (v) 和初速度 (u)。
2. Momentum, Impulse & Newton’s Laws | 动量、冲量与牛顿定律
Momentum (kg·m/s) = Mass (kg) × Velocity (m/s), p = m × v. Momentum is always conserved in a closed system, which is vital for collision analysis in sport (e.g., rugby tackle). The total momentum before impact equals total momentum after impact: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
动量 (kg·m/s) = 质量 (kg) × 速度 (m/s)，p = m × v。在封闭系统中动量守恒，这对分析运动中的碰撞至关重要（如橄榄球擒抱）。碰撞前的总动量等于碰撞后的总动量：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
Impulse (N·s) = Force (N) × Time (s), Impulse = F × t. Impulse equals change in momentum: Ft = Δp = m(v − u). To reduce injury, increase impact time (e.g., crash mats in gymnastics) so that force decreases for the same momentum change.
冲量 (N·s) = 力 (N) × 时间 (s)，冲量 = F × t。冲量等于动量的变化量：Ft = Δp = m(v − u)。为减少受伤，应延长碰撞时间（如体操用的防摔垫），这样在动量变化相同时可以减小受力。
Newton’s First Law (Inertia): an object remains at rest or in uniform motion unless acted upon by an external force. Second Law: F = m × a (Force = mass × acceleration). Third Law: for every action there is an equal and opposite reaction. These laws underpin all movement analysis in OCR PE.
牛顿第一定律（惯性定律）：物体将保持静止或匀速直线运动状态，除非有外力迫使它改变。第二定律：F = m × a（力 = 质量 × 加速度）。第三定律：作用力与反作用力大小相等、方向相反。这三条定律是所有 OCR 体育运动分析的基础。
3. Forces, Weight & Friction | 力、重力与摩擦力
Weight (N) = Mass (kg) × Gravitational field strength (g = 9.81 m/s² on Earth), W = m × g. Always remember weight is a force, not mass. Normal reaction force acts perpendicular to the surface and balances weight on flat ground when no other vertical forces apply.
重力 (N) = 质量 (kg) × 重力场强度（地球 g = 9.81 m/s²），W = m × g。务必记住重力是一种力，不是质量。法向反作用力垂直于接触面，在平地上且无其他竖直外力时与重力平衡。
Friction (F) = Coefficient of friction (μ) × Normal reaction force (R), F = μR. Static friction (before movement) is higher than sliding friction. Factors affecting μ include surface roughness, temperature, and presence of lubricants. In sport, friction is essential for grip (studded boots) but can be a disadvantage (slippery artificial turf without proper footwear).
摩擦力 (F) = 摩擦系数 (μ) × 法向反作用力 (R)，F = μR。静摩擦力（运动开始前）大于滑动摩擦力。影响 μ 的因素有表面粗糙度、温度和润滑剂的存在。运动中，摩擦力对抓地力至关重要（如钉鞋），但不当的摩擦力也会成为劣势（如没有合适鞋具的光滑人造草皮）。
4. Centre of Mass & Stability | 质心与稳定性
The centre of mass (CoM) is the point where all body mass is considered concentrated. In a uniform gravitational field, it is the same as the centre of gravity. For a symmetrical human in anatomical position, the CoM lies around the level of S2 (second sacral vertebra), but it moves with limb position. Stability increases when the CoM is lower, the base of support is larger, and the line of gravity falls centrally within the base.
质心 (CoM) 是假设身体全部质量集中于一点的位置。在均匀引力场中，质心与重心重合。对于解剖姿势下对称的人体，质心大约在第二骶椎 (S2) 水平，但会随肢体位置移动。质心越低、支撑面越大、重力线越接近支撑面中心，稳定性就越高。
The equation for moment (torque, Nm) = Force (N) × Perpendicular distance from pivot (m), M = F × d. In OCR PE, this is used to explain balance and turning effects: a gymnast on a beam increases stability by lowering CoM; a rugby player spreads feet and bends knees to resist a tackle.
力矩 (扭矩，Nm) = 力 (N) × 到支点的垂直距离 (m)，M = F × d。OCR 体育用它解释平衡和转动效应：体操运动员在平衡木上通过降低质心增强稳定性；橄榄球运动员通过分腿加屈膝来抵抗擒抱。
5. Levers & Mechanical Advantage | 杠杆与机械利益
A lever consists of a rigid bar, a pivot (fulcrum), an effort force, and a load. The three classes of lever are defined by the relative positions of the fulcrum (F), effort (E), and load (L). Class 1: F between E and L (e.g., neck extension – effort by posterior neck muscles, load head weight, fulcrum at atlanto-occipital joint). Class 2: L between F and E (e.g., calf raise – ball of foot fulcrum, load body weight through tibia, effort by gastrocnemius). Class 3: E between F and L (most common in the body, e.g., biceps curl – elbow fulcrum, effort biceps on radius, load in hand).
杠杆由刚性杆、支点、动力和阻力组成。根据支点 (F)、动力 (E) 和阻力 (L) 的相对位置分为三类。第一类：支点在动力和阻力之间（如头部后伸——动力来自颈后肌群，阻力是头部重力，支点在寰枕关节）。第二类：阻力在支点和动力之间（如提踵——跖球部为支点，体重通过胫骨成为阻力，动力由腓肠肌提供）。第三类：动力在支点和阻力之间（身体中最常见，如肱二头肌弯举——肘关节为支点，动力是肱二头肌在桡骨上的附着点，阻力在手上）。
Mechanical advantage (MA) = Effort arm length ÷ Load arm length. Class 2 levers have MA > 1 (force advantage, e.g., calf raise produces large force to lift body). Class 3 levers have MA < 1 (speed and range of motion advantage, e.g., throwing). OCR uses this to analyse sporting movements – sprinters rely on class 3 levers for rapid leg movement.
机械利益 (MA) = 动力臂长 ÷ 阻力臂长。第二类杠杆 MA > 1（省力，如提踵可产生较大力量抬升身体）。第三类杠杆 MA < 1（速度和活动幅度优势，如投掷）。OCR 借此分析运动动作——短跑运动员依赖第三类杠杆实现快速腿部动作。
6. Projectile Motion & Equations of Uniform Acceleration | 抛体运动与匀加速方程
For projectiles released at an angle θ from the ground with initial velocity u (m/s), horizontal component uₓ = u cosθ, vertical component uᵧ = u sinθ. Range = (u² sin 2θ) / g. Maximum height = (u² sin²θ) / (2g). Time of flight = (2u sinθ) / g. The optimal angle for maximum range in a vacuum is 45°, but air resistance and release height above landing surface alter this in real sport.
对于以初始速度 u (m/s) 从地面以角度 θ 抛出的抛体，水平分量 uₓ = u cosθ，竖直分量 uᵧ = u sinθ。射程 = (u² sin 2θ) / g。最大高度 = (u² sin²θ) / (2g)。飞行时间 = (2u sinθ) / g。真空中最大射程的最佳角度是 45°，但实际运动中空气阻力和出手点与落地点的高度差会改变最佳角度。
The four SUVAT equations are essential for linear motion with constant acceleration:
v = u + a t
s = u t + ½ a t²
v² = u² + 2 a s
s = ½ (u + v) t
Here, s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time. Apply these to biomechanics problems such as vertical jump height or sprinter’s acceleration phase.
四个 SUVAT 方程是匀加速直线运动的核心：s = 位移，u = 初速度，v = 末速度，a = 加速度，t = 时间。将这些方程应用于生物力学问题，如垂直纵跳高度或短跑运动员的加速阶段。
7. Angular Motion, Torque & Moment of Inertia | 角运动、力矩与转动惯量
Angular velocity (rad/s) = Angular displacement (rad) ÷ Time (s), ω = θ/t. Angular acceleration (rad/s²) = Change in angular velocity (rad/s) ÷ Time, α = (ω − ω₀)/t. Torque (Nm) = Moment of inertia (kg·m²) × Angular acceleration (rad/s²), T = I × α.
角速度 (rad/s) = 角位移 (rad) ÷ 时间 (s)，ω = θ/t。角加速度 (rad/s²) = 角速度变化量 (rad/s) ÷ 时间，α = (ω − ω₀)/t。扭矩 (Nm) = 转动惯量 (kg·m²) × 角加速度 (rad/s²)，T = I × α。
Angular momentum = I × ω, and is conserved when no external torque acts (e.g., an ice skater spins faster by pulling arms in, reducing I, so ω increases). Moment of inertia depends on mass distribution relative to the axis of rotation: I = Σ m r². The further the mass from the axis, the greater I and the harder to rotate. In diving and gymnastics, athletes manipulate body shape to control rotation speed.
角动量 = I × ω，当无外力矩作用时角动量守恒（如滑冰运动员收拢手臂可减小 I，使 ω 增大而转得更快）。转动惯量取决于质量相对于转轴的分布：I = Σ m r²。质量离轴越远，I 越大，越难转动。在跳水和体操中，运动员通过改变身体形态来控制旋转速度。
8. Bernoulli’s Principle & Magnus Effect | 伯努利原理与马格努斯效应
Bernoulli’s principle: where fluid velocity is high, pressure is low; where fluid velocity is low, pressure is high. Mathematically, P + ½ ρ v² + ρ g h = constant (incompressible, inviscid flow). The simplified relationship is often used: faster airflow → lower pressure. This explains lift on an aerofoil and the flight of a discus or javelin.
伯努利原理：流体流速高的地方压强低，流速低的地方压强大。数学表达为 P + ½ ρ v² + ρ g h = 常数（适用于不可压缩、无黏性流体）。简化关系常被使用：流速越快 → 压强越低。这解释了翼型的升力以及铁饼或标枪的飞行。
The Magnus effect arises when a spinning ball experiences a pressure differential due to the relative air speeds. For a backspinning ball, the top surface moves against the airflow (reducing relative velocity, leading to higher pressure), while the bottom surface moves with the airflow (increasing relative velocity, lowering pressure). The resultant force lifts the ball. This is crucial in tennis topspin, football free kicks, and cricket swing. OCR expects you to sketch flow lines and identify pressure zones.
马格努斯效应源于旋转的球体因相对气流速度不同而产生的压力差。对于后旋球，球体上表面与气流方向相反（相对速度降低，压强较高），而下表面与气流方向相同（相对速度升高，压强较低）。合力使球上升。这在上旋网球、足球任意球和板球飘移中至关重要。OCR 要求你能画出流线并标出压力区域。
9. Energy, Work & Power | 能量、功与功率
Work done (J) = Force (N) × Distance moved in the direction of force (m), W = F d cosθ (θ is angle between force and displacement). In many OCR PE scenarios, the force and movement are in line (θ = 0°, cosθ = 1). Kinetic energy (KE) = ½ m v². Gravitational potential energy (GPE) = m g h. Conservation of energy: total mechanical energy remains constant in the absence of external work (e.g., a pole vaulter’s kinetic energy converts to strain energy in the pole and then to GPE).
功 (J) = 力 (N) × 在力的方向上移动的距离 (m)，W = F d cosθ（θ 为力与位移的夹角）。在许多 OCR 体育情景中，力和运动方向一致（θ = 0°，cosθ = 1）。动能 (KE) = ½ m v²。重力势能 (GPE) = m g h。能量守恒：没有外力做功时，机械能总量保持不变（如撑杆跳高运动员的动能转化为撑杆的弹性势能，再转化为重力势能）。
Power (W) = Work done (J) ÷ Time (s), P = W/t. Also Power = Force × Velocity (when force and velocity are parallel). In sport, power reflects the rate of energy transfer and is crucial for explosive actions like jumping and sprinting. A Wingate test measures anaerobic power; VO₂max relates to aerobic power.
功率 (W) = 做功 (J) ÷ 时间 (s)，P = W/t。功率也 = 力 × 速度（当力与速度方向一致时）。在体育中，功率反映能量转化的速率，对跳跃和冲刺等爆发性动作至关重要。温盖特测试可测量无氧功率；最大摄氧量则与有氧功率相关。
10. Fluid Resistance, Drag & Terminal Velocity | 流体阻力、空气阻力与终极速度
Drag force (FD) = ½ CD ρ A v², where CD is the drag coefficient, ρ is fluid density, A is cross-sectional area, and v is velocity relative to the fluid. Reducing A (e.g., cyclists in tuck position) or CD (skin suits, dimples on golf balls) decreases drag and improves performance. At terminal velocity, weight = drag force, so net force is zero, and velocity remains constant (e.g., a skydiver in free fall).
空气阻力 FD = ½ CD ρ A v²，其中 CD 为阻力系数，ρ 为流体密度，A 为横截面积，v 为相对于流体的速度。减小 A（如自行车运动员的蜷缩姿势）或 CD（紧身衣、高尔夫球的凹坑）可降低阻力并提高成绩。当达到终极速度时，重力 = 阻力，合力为零，速度保持不变（如跳伞者自由落体）。
In swimming, water resistance is much higher than air resistance, so drag reduction techniques are vital. The same formula applies, but the density ρ of water is about 800 times that of air, making drag dramatically larger. Streamlining body position, caps, and technical suits all help reduce CD or A.
在游泳中，水的阻力远大于空气阻力，因此减阻技术至关重要。同一公式适用，但水的密度 ρ 约为空气的 800 倍，使得阻力大得多。流线型身体姿势、泳帽和科技泳衣均有助于降低 CD 或 A。
11. VO₂max & Respiratory Exchange Ratio (RER) | 最大摄氧量与呼吸交换率
VO₂max is the maximum volume of oxygen the body can utilise per minute per kilogram of body mass (ml/kg/min). The Fick equation: VO₂ = Cardiac Output (Q) × a-vO₂ difference (arterio-venous oxygen difference). Cardiac output = Stroke Volume (SV) × Heart Rate (HR). Therefore, VO₂max = SV × HR × a-vO₂ diff. This links cardiovascular and muscular systems – a high VO₂max requires large SV and effective oxygen extraction.
最大摄氧量 (VO₂max) 是身体每分钟每千克体重能利用的最大氧气量 (ml/kg/min)。菲克方程：VO₂ = 心输出量 (Q) × 动静脉氧差 (a-vO₂ diff)。心输出量 = 每搏输出量 (SV) × 心率 (HR)。因此 VO₂max = SV × HR × a-vO₂ diff。这联系了心血管系统与肌肉系统——高 VO₂max 需要较大的 SV 和有效的氧摄取。
Respiratory Exchange Ratio (RER) = VCO₂ produced ÷ VO₂ consumed. At rest, RER ≈ 0.80 (mixed diet). During intense exercise, RER can exceed 1.0 as CO₂ production rises due to buffering of lactic acid. RER values indicate fuel utilisation: 0.7 mainly fats, 0.85 mixed, 1.0 mainly carbohydrates. OCR students must interpret RER data in exercise tests.
呼吸交换率 (RER) = CO₂ 排出量 ÷ O₂ 摄取量。安静时 RER 约为 0.80（混合膳食）。剧烈运动时，乳酸缓冲使 CO₂ 产生增加，RER 可超过 1.0。RER 值指示燃料利用：0.7 主要为脂肪，0.85 混合，1.0 主要为碳水化合物。OCR 考生须学会解读运动测试中的 RER 数据。
12. Levers in the Body & Practical Application | 人体中的杠杆与实际应用
A quick revision list of body lever examples: Class 1 – neck extension (atlanto-occipital joint); triceps extension (less common, E between elbow and hand load but technically E and L on same side of F). Class 2 – calf raise (ankle joint), push-up phase (fulcrum at feet, load CoM, effort arms). Class 3 – biceps curl, hamstrings at knee (flexion), quadriceps in kicking (extension). Remember the mechanical disadvantage of most joint levers is compensated by muscle architecture and speed of contraction.
人体杠杆速记：第一类 – 头部后伸（寰枕关节）；肱三头肌伸肘（少见，动力在肘和手之间但技术上动力和阻力在支点同侧）。第二类 – 提踵（踝关节）、俯卧撑推起阶段（支点脚，阻力质心，动力手臂）。第三类 – 肱二头肌弯举、腘绳肌屈膝、股四头肌踢球伸膝。记住，多数关节杠杆的机械劣势由肌肉结构和收缩速度所弥补。
To calculate force in a lever system: Effort force × Effort arm = Load × Load arm (in equilibrium). Example: during a biceps curl, if load 100 N at 30 cm from elbow, effort arm 5 cm, the required biceps force = (100 N × 30 cm) / 5 cm = 600 N. This illustrates the high muscle forces even for small loads, highlighting the importance of tendons and joint adaptation.
计算杠杆系统中的力：动力 × 动力臂 = 阻力 × 阻力臂（平衡时）。示例：肱二头肌弯举时，若阻力 100 N 距肘关节 30 cm，动力臂 5 cm，则肱二头肌所需发力 = (100 N × 30 cm) / 5 cm = 600 N。这说明即使轻负荷肌肉也需很大力量，凸显了肌腱和关节适应的重要性。
Published by TutorHao | Physical Education 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