📚 Year 13 OCR PE: Formula & Theorem Quick-Reference Handbook | Year 13 OCR 体育：公式定理速查手册
This quick-reference handbook consolidates the essential formulas, laws and theorems required for Year 13 OCR Physical Education. It covers linear and angular mechanics, fluid dynamics, lever systems, cardiorespiratory calculations and body composition indices. Each entry is paired with a sporting example to reinforce application, helping you move confidently between theory and high-mark examination answers.
本速查手册汇总了 Year 13 OCR 体育所需的公式、定律与定理。内容涵盖线性与角运动力学、流体力学、杠杆系统、心肺计算和身体成分指数。每一条目均配有运动实例，强化公式应用，助你在理论与高分答案间游刃有余。
1. Linear Kinematics – SUVAT Equations | 线性运动学 – SUVAT 方程
The four SUVAT equations describe uniformly accelerated motion along a straight line. They link displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). In OCR PE, these are used to analyse sprint phases, long jump take-offs and any skill where acceleration is constant.
四个 SUVAT 方程描述匀加速直线运动，关联位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)。在 OCR 体育中，它们用于分析短跑阶段、跳远起跳等加速度恒定的动作。
v = u + at
Gives final velocity when initial speed and acceleration are known over a time interval. For a 100 m sprinter who accelerates at 3 m/s² for 2 seconds from a stationary start, v = 0 + (3 × 2) = 6 m/s.
已知初速度和加速度，求某段时间后的末速度。若 100 m 短跑运动员从静止以 3 m/s² 加速 2 秒，v = 0 + (3 × 2) = 6 m/s。
v² = u² + 2as
Links velocity and displacement without time. Useful for calculating the speed of a high jumper at take-off if the approach run acceleration and distance are known.
关联速度与位移，不含时间变量。当需要从助跑加速度和距离推算跳高起跳速度时，十分实用。
s = ut + ½ at²
Predicts displacement when initial velocity and acceleration are constant. A long jumper’s horizontal travel during the flight phase can be estimated using this equation, treating the horizontal acceleration as zero after take-off.
当初速度与加速度恒定时可预测位移。跳远腾空阶段的水平位移可用此式估算，起跳后水平加速度可视为零。
s = ½ (u + v) t
Calculates displacement from the average of initial and final velocities. This is handy for assessing the distance covered during a cycling time-trial start when velocity changes linearly.
利用初末速度的平均值求位移，适用于自行车计时赛起跑阶段速度线性变化时的距离估算。
2. Linear Kinetics – Newton’s Laws & Momentum | 线性动力学 – 牛顿定律与动量
Newton’s three laws and the impulse–momentum relationship are central to explaining how forces cause movement in sport. OCR exam questions frequently ask you to apply these laws to a named sporting movement.
牛顿三定律与冲量–动量关系是解释体育运动中力致动作的核心。OCR 试题常要求将定律应用于具体运动动作。
Newton’s First Law (Inertia): A body remains at rest or in uniform motion unless acted upon by an external resultant force. The reluctance of a rugby scrum half to move until pushed illustrates inertia.
牛顿第一定律（惯性）：物体保持静止或匀速直线运动状态，除非有外合力迫其改变。橄榄球传球前九号位队员的静止状态即体现惯性。
Newton’s Second Law: F = ma. The acceleration of a body is directly proportional to the net force and inversely proportional to its mass. A 75 kg hockey player struck by a stick force of 150 N experiences an acceleration of 2 m/s².
牛顿第二定律：F = ma。加速度与合外力成正比，与质量成反比。75 kg 曲棍球运动员受到 150 N 球棍击打力时，加速度为 2 m/s²。
Newton’s Third Law: For every action there is an equal and opposite reaction. A swimmer pushing water backwards generates a forward reaction force; the explosive drive in a sprint start is another clear demonstration.
牛顿第三定律：作用力与反作用力大小相等、方向相反。游泳者向后推水产生向前的反作用力；短跑起跑时的爆发蹬伸也是清晰的例证。
Momentum: p = mv. Momentum (kg·m/s) is the product of mass and velocity. During a tackle in rugby, the total momentum before collision equals the total momentum afterwards (conservation of momentum).
动量：p = mv。动量（kg·m/s）为质量与速度的乘积。橄榄球擒抱碰撞前后总动量守恒（动量守恒定理）。
Impulse–Momentum Theorem: Ft = Δ(mv). Impulse (force × time) equals the change in momentum. A gymnast landing from a vault uses a longer landing time by bending knees to reduce peak force, illustrating impulse management.
冲量–动量定理：Ft = Δ(mv)。冲量（力 × 时间）等于动量的变化。跳马落地时屈膝延长作用时间，减小峰值冲击力，正是冲量管理的体现。
3. Work, Energy and Power | 功、能与功率
Work and energy formulas quantify how athletes transfer chemical energy into mechanical output. Power is the rate of doing work, a key indicator of explosive performance.
功与能量公式量化运动员将化学能转化为机械能的过程。功率是做功的快慢，是爆发性表现的关键指标。
Work = F d cosθ
Work done (J) is force (N) × displacement (m) × cos of the angle between force and displacement. Pushing a sled horizontally (θ = 0°, cos0° = 1) with 200 N over 10 m gives 2000 J of work.
做功（J）= 力（N）× 位移（m）× 力与位移夹角余弦。沿水平方向推 10 m 阻力橇（θ = 0°, cos0° = 1），力 200 N，则做功 2000 J。
Kinetic Energy: KE = ½ mv²
Energy possessed by a moving body. A 0.4 kg football travelling at 20 m/s has KE = ½ × 0.4 × (20)² = 80 J. Sprinters accelerate to maximise KE before take-off in long jump.
物体因运动具有的能量。0.4 kg 足球以 20 m/s 运动时，动能 = ½ × 0.4 × (20)² = 80 J。短跑运动员在跳远起跳前加速以最大化动能。
Gravitational Potential Energy: GPE = mgh
Energy stored due to height (h) in a gravitational field (g ≈ 9.81 m/s²). A 70 kg pole-vaulter lifted to a centre of mass height of 5 m gains GPE = 70 × 9.81 × 5 ≈ 3434 J.
因高度位置储存的能量（g ≈ 9.81 m/s²）。一名 70 kg 撑竿跳高运动员重心升至 5 m 时，势能 ≈ 70 × 9.81 × 5 ≈ 3434 J。
Power: P = W / t = Fv
Power (W) is work done per unit time, or force × velocity at an instant. A rower producing 4000 J of work over 5 seconds outputs 800 W. In sprint cycling, monitoring P = Fv helps gauge mechanical efficiency.
功率（W）= 单位时间做功 / 瞬时力 × 速度。赛艇运动员 5 秒内做功 4000 J，输出功率 800 W。在场地自行车冲刺中，监测 P = Fv 可评价机械效率。
4. Angular Motion Basics | 角运动基础
Angular motion descriptors are essential for analysing rotation in gymnastics, diving, figure skating and club/stick swings. The fundamental relationships mirror linear equations but use angular quantities.
角运动描述符对分析体操、跳水、花样滑冰及球杆挥击中的旋转至关重要。基本关系与线性方程类似，但使用角量。
Angular velocity: ω = θ / t
Rate of change of angular displacement (rad/s). A discus thrower completing a 540° (3π rad) turn in 1.2 s has an average ω = 3π / 1.2 ≈ 7.85 rad/s.
角位移变化率（rad/s）。铁饼运动员在 1.2 秒内完成 540°（3π rad）旋转，平均角速度 ω = 3π / 1.2 ≈ 7.85 rad/s。
Angular acceleration: α = Δω / t
Rate of change of angular velocity (rad/s²). A gymnast accelerating from a spin of 5 rad/s to 12 rad/s in 0.5 s undergoes α = (12 − 5)/0.5 = 14 rad/s².
角速度变化率（rad/s²）。体操运动员 0.5 秒内从 5 rad/s 加速旋转至 12 rad/s，角加速度 α = (12 − 5)/0.5 = 14 rad/s²。
Relationship between linear and angular velocity: v = rω. The linear speed of a point on a rotating segment equals the radius (m) × angular velocity (rad/s). A tennis racket head at a radius of 1.1 m from the shoulder, rotating at 30 rad/s, moves at 33 m/s.
线速度与角速度关系：v = rω。旋转环节上一点的线速度 = 半径(m) × 角速度(rad/s)。距肩关节 1.1 m 的网球拍拍头以 30 rad/s 转动时，线速度为 33 m/s。
Centripetal force: Fc = mω²r = mv²/r
Net force directed towards the centre of rotation, required for circular motion. A hammer thrower spinning a 7.26 kg ball at 12 rad/s on a 1.2 m wire requires Fc = 7.26 × (12)² × 1.2 ≈ 1255 N.
指向旋转中心的合外力，维持圆周运动所必需。链球运动员用 1.2 m 钢丝以 12 rad/s 旋转 7.26 kg 球，需向心力 ≈ 7.26 × (12)² × 1.2 ≈ 1255 N。
5. Angular Kinetics – Moment of Inertia & Angular Momentum | 角动力学 – 转动惯量与角动量
Rotational analogues of mass and momentum explain how athletes manipulate body shape to control spin speed. These principles underpin twists, somersaults and pirouettes.
转动惯量和角动量是质量和动量的旋转对应概念，解释运动员如何通过改变身体姿态控制旋转速度。这是扭身、翻腾和旋转变速的基础。
Moment of inertia: I = Σmr²
Resistance to angular acceleration, dependent on mass distribution about the axis. A diver tucking brings mass closer to the axis, reducing I and increasing spin rate (conservation of angular momentum).
抵抗角加速度的能力，取决于质量相对转轴的分布。跳水运动员抱膝使质量靠近转轴，减小 I，从而增加旋转速度（角动量守恒）。
Angular momentum: L = Iω
Angular momentum (kg·m²/s) remains constant if no external torque acts. In a figure skating spin, the skater pulls arms in (decrease I) and ω rises dramatically.
角动量（kg·m²/s）在无外力矩作用下守恒。花样滑冰旋转中，运动员收拢双臂（减小 I），ω 显著增大。
Torque: T = F d (perpendicular distance)
The rotational effect of a force. T = Iα is the angular equivalent of F = ma. A gymnast on uneven bars generating a torque of 120 N·m by applying force at 0.8 m from the bar produces rotational acceleration.
力的转动效应。T = Iα 是 F = ma 的角运动对应。高低杠运动员在距杠 0.8 m 处施力，产生 120 N·m 力矩，驱动旋转加速。
6. Projectile Motion | 抛体运动
Projectile formulas break motion into independent horizontal and vertical components. Factors such as release velocity, angle and height determine the parabolic flight path of shots, long jumpers, footballs and shuttlecocks.
抛体公式将运动分解为独立的水平与垂直分量。出手速度、角度和高度等因素决定铅球、跳远、足球和羽毛球的抛物线轨迹。
Horizontal component: vₓ = v cosθ; displacement: sₓ = v cosθ × t. The horizontal motion is constant when air resistance is negligible.
水平分量：vₓ = v cosθ；位移：sₓ = v cosθ × t。当空气阻力可忽略时，水平运动为匀速。
Vertical component: vᵧ = v sinθ, ruled by SUVAT with a = −g. Time of flight, maximum height and range can be predicted through: t = 2 v sinθ / g (for symmetric level projectiles).
垂直分量：vᵧ = v sinθ，服从含 a = −g 的 SUVAT 方程。飞行时间、最大高度和射程可预测：t = 2 v sinθ / g（对称水平抛体）。
Optimal projection angle: 45° maximises range if release and landing are at the same height. In shot put, the release height is above the landing area, so the optimal angle is slightly below 42°.
最佳出手角度：度如出手与落地同高，45° 可最大化射程。铅球出手点高于落地，最佳角度略低于 42°。
Parabolic path equation: y = x tanθ − (g x²) / (2 v² cos²θ). This is used to predict whether a basketball shot will clear the defender’s reach.
抛物线轨迹方程：y = x tanθ − (g x²) / (2 v² cos²θ)。此式用于预判篮球投篮能否越过防守者指尖高度。
7. Fluid Mechanics & Drag | 流体力学与阻力
Fluid forces dramatically affect performance in cycling, swimming, skiing and ball sports. Understanding drag and lift helps explain technique modifications and equipment design.
流体力对自行车、游泳、滑雪和球类运动表现影响显著。理解阻力和升力有助于解释技术调整与器材设计。
Drag force: Fd = ½ Cd ρ A v²
Where Cd = drag coefficient, ρ = fluid density, A = cross-sectional area, v = velocity. Doubling velocity quadruples drag. A cyclist in a tucked position reduces A and Cd, lowering aerodynamic resistance.
其中 Cd = 阻力系数，ρ = 流体密度，A = 迎风截面积，v = 速度。速度加倍，阻力增至四倍。自行车运动员低伏姿势减小 A 和 Cd，降低气动阻力。
Lift (Magnus effect): Fₗ = ½ Cₗ ρ A v². A spinning football curves because differential pressure arises from relative air speeds on opposite sides. Topspin creates a downward force, making a tennis ball dip faster.
升力（马格努斯效应）：Fₗ = ½ Cₗ ρ A v²。旋转的足球因其两侧相对气流速度不同产生压差而弧线飞行。上旋产生向下的力，使网球更快下坠。
Bernoulli’s principle: An increase in fluid speed occurs simultaneously with a decrease in pressure. This principle explains discus and javelin aerodynamic lift as air travels faster over the curved upper surface.
伯努利原理：流体流速增加时压强降低。这解释了铁饼和标枪飞行中因上表面气流较快而产生空气动力学升力。
8. Levers & Mechanical Advantage | 杠杆与机械优势
The body’s musculoskeletal system operates as a series of lever systems. Classifying levers and calculating mechanical advantage explains why some movements prioritise speed while others favour force.
人体肌肉骨骼系统如同一系列杠杆。杠杆分类和机械优势计算解释了为何某些动作优先产生速度，而另一些则注重力量。
Lever classes & formula: Mechanical Advantage (MA) = Effort Arm / Resistance Arm. An MA > 1 means the lever amplifies force (force multiplier); MA < 1 amplifies speed and range of motion (speed multiplier). In the body, most levers are third-class (MA < 1) to favour rapid, extensive movement.
杠杆级别与公式：机械优势 (MA) = 力臂 / 阻力臂。MA > 1 为省力杠杆；MA < 1 为增速杠杆，放大速度与活动范围。人体多数杠杆为第三类（MA < 1），以利于快速大范围运动。
Second-class lever example: The ankle joint during plantar flexion (standing on toes). The ball of the foot is the fulcrum, body weight is the resistance, and the calf muscle provides effort. MA > 1, offering force advantage for push-off.
第二类杠杆实例：踝关节跖屈（踮脚尖）。足球为支点，体重为阻力，小腿肌群提供动力。MA > 1，为蹬伸提供省力优势。
Third-class lever example: The elbow during a bicep curl. The elbow is the fulcrum, the weight in the hand is the resistance, and the bicep insertion on the radius provides effort. MA < 1, so a large muscle shortening translates into fast, large hand displacement – essential for throwing and striking.
第三类杠杆实例：肱二头肌弯举。肘为支点，手中哑铃为阻力，肱二头肌在桡骨附着点为力点。MA < 1，大块肌肉小幅度缩短即可产生手部大幅度快速位移——对投掷和击打尤为关键。
9. Cardiovascular & Respiratory Formulae | 心血管与呼吸系统公式
Cardiorespiratory equations underpin the interpretation of laboratory tests and field-based fitness assessments. They link heart function, ventilation and metabolic gas exchange.
心血管与呼吸方程是实验室测试和现场体能评估的基础，联系心脏功能、肺通气与代谢气体交换。
Cardiac output: Q = HR × SV
Cardiac output (L/min) = heart rate (beats/min) × stroke volume (L/beat). An endurance athlete at maximal effort may exhibit HR = 195 bpm, SV = 0.15 L, giving Q = 29.25 L/min.
心输出量（L/min）= 心率（次/分）× 每搏输出量（L/次）。耐力运动员最大用力时 HR = 195 bpm，SV = 0.15 L，Q = 29.25 L/min。
Mean arterial pressure: MAP ≈ (2 × diastolic + systolic) / 3
Estimates the average blood pressure in arteries. Important when discussing vascular resistance during exercise.
估算动脉平均血压，在讨论运动时血管阻力时常用。
Minute ventilation: V̇E = f × VT
V̇E (L/min) = breathing frequency (breaths/min) × tidal volume (L). During intense exercise, f may rise to 50 breaths/min and VT to 2.5 L, giving V̇E = 125 L/min.
每分通气量 V̇E (L/min) = 呼吸频率（次/分）× 潮气量（L）。剧烈运动时 f 可升至 50 次/分，VT 至 2.5 L，V̇E = 125 L/min。
Respiratory exchange ratio: RER = VCO₂ / VO₂
Indicates fuel utilisation. RER ≈ 0.7 for fat oxidation, 1.0 for carbohydrate, and >1.0 during high-intensity anaerobic efforts when CO₂ production exceeds O₂ uptake.
显示底物利用情况。RER ≈ 0.7 为脂肪氧化，1.0 为碳水化合物，高强度无氧运动时 CO₂ 产生量超过摄氧量，RER > 1.0。
VO₂ max (absolute & relative):
Absolute VO₂ max (L/min) is the maximum volume of oxygen the body can use per minute. Relative VO₂ max (ml/kg/min) = absolute VO₂ max / body mass (kg). Used to compare aerobic capacity across athletes of different sizes.
绝对最大摄氧量（L/min）为每分钟身体能利用的最大氧气量。相对最大摄氧量（ml/kg
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