📚 Year 11 SQA Physical Education: Formula & Theorem Quick Reference Handbook | Year 11 SQA 体育：公式定理速查手册
This handbook is designed to be your go-to resource for every essential formula and theorem you need in SQA Physical Education. From calculating training zones with heart‑rate formulas to applying Newton’s laws in movement analysis, these equations bridge the gap between theory and practical performance. Each entry is explained with clear examples so you can use it confidently in written assignments and exam responses.
这本手册是你查询 SQA 体育课程所有核心公式与定理的速查资料。从用心率公式计算训练区间，到运用牛顿定律进行动作分析，这些方程将理论与实际运动表现紧密相连。每个条目都配有清晰示例，让你能够在书面作业和考试答题中自信运用。
1. Maximum Heart Rate Estimation | 最大心率估算
The maximum heart rate (HRmax) is the ceiling your cardiovascular system can reach during all‑out effort. A standard field‑based estimate is:
最大心率（HRmax）是心血管系统在全力以赴时所能达到的峰值。常用的现场估算公式为：
HRmax = 220 − age
For a 16‑year‑ for adolescents the formula HRmax = 208 − 0.7 × age provides greater accuracy, giving a 16‑year‑old an estimated HRmax of about 196 bpm.
对于16岁青少年，HRmax ≈ 204 次/分。研究表明，针对青少年群体，公式 HRmax = 208 − 0.7 × 年龄 更为精确，估算出一位16岁学生的最大心率约196 次/分。
Coaches use HRmax to set training intensities. For example, a moderate‑intensity session may target 65–75% HRmax, while high‑intensity intervals demand 85–95% HRmax.
教练利用最大心率设定训练强度。例如，中等强度课可设定为 65–75% HRmax，而高强度间歇则需要达到 85–95% HRmax。
2. Karvonen Formula for Target Heart Rate | 卡沃南靶心率公式
The Karvonen method refines intensity prescription by incorporating resting heart rate (HRrest). It uses heart rate reserve (HRR) to set individualised target zones:
卡沃南公式通过纳入静息心率（HRrest）来优化强度设定。它利用心率储备（HRR）设定个性化目标区间：
Target HR = HRrest + (HRmax − HRrest) × Intensity
If a 16‑year‑ as HRrest = 62 bpm and HRmax = 204 bpm, the HRR is 142 bpm. For a training intensity of 70% (0.70), the target heart rate becomes:
如果一位16岁学生静息心率为62 bpm，最大心率为204 bpm，则心率储备为142 bpm。若训练强度设定为70%（0.70），靶心率计算如下：
62 + (142 × 0.70) = 161.4 bpm
This formula is widely used to design aerobic training programmes because it adapts automatically as resting heart rate decreases with fitness gains.
该公式被广泛用于设计有氧训练计划，因为随着体能提升静息心率下降，它能自动调整训练区间。
3. Body Mass Index (BMI) | 身体质量指数 (BMI)
BMI is a quick screening tool that relates body weight to height. While it does not measure body fat directly, it helps identify weight categories relevant to health and sport.
BMI 是一项将体重与身高关联的快速筛查指标。虽然它不直接测量体脂，但有助于识别与健康和运动相关的体重等级。
BMI = weight (kg) ÷ height² (m²)
e: A performer weighing 65 kg and 1.72 m tall has a BMI = 65 ÷ (1.72 × 1.72) ≈ 22.0 kg/m².
示例：一位运动员体重65 kg，身高1.72 m，则 BMI = 65 ÷ (1.72 × 1.72) ≈ 22.0 kg/m²。
| BMI Range (kg/m²) | Weight Category | 体重等级 |
|---|---|
| Below 18.5 | Underweight | 偏瘦 |
| 18.5 – 24.9 | Healthy weight | 健康体重 |
| 25.0 – 29.9 | Overweight | 超重 |
| 30.0 and above | Obese | 肥胖 |
In SQA PE, BMI can be used to discuss how body composition affects performance in activities like gymnastics, rock climbing or endurance running.
在 SQA 体育中，BMI 可用于探讨身体成分如何影响体操、攀岩或耐力跑等运动的表现。
4. Basal Metabolic Rate (BMR) Estimation | 基础代谢率估算
BMR estimates the energy your body needs to maintain basic life functions at rest. The Mifflin‑St Jeor equation is one of the most reliable predictions:
BMR 用于估算身体在静息状态下维持基本生命活动所需的能量。Mifflin‑St Jeor 方程是当前最可靠的预测公式之一：
Male BMR = 10 × weight(kg) + 6.25 × height(cm) − 5 × age(y) + 5
Female BMR = 10 × weight(kg) + 6.25 × height(cm) − 5 × age(y) − 161
A 16‑year‑ imated BMR of (10×68) + (6.25×178) − (5×16) + 5 = 680 + 1112.5 − 80 + 5 = 1717.5 kcal/day.
一位16岁男性运动员（68 kg、178 cm）的 BMR 估算值为 (10×68) + (6.25×178) − (5×16) + 5 = 680 + 1112.5 − 80 + 5 = 1717.5 千卡/天。
Total daily energy expenditure (TDEE) is calculated by multiplying BMR by an activity factor. This helps performers plan nutrition for training and recovery.
每日总能量消耗（TDEE）由 BMR 乘以活动系数计算得出，有助于运动员规划训练与恢复期间的营养摄入。
5. Speed, Velocity & Acceleration | 速度与加速度
These are fundamental descriptors of motion. Speed is a scalar quantity; velocity includes direction.
这些是描述运动的基本量。速度是标量，而速度矢量包含方向。
Speed = distance ÷ time
Acceleration = (final velocity − initial velocity) ÷ time
. If her velocity changes from 0 to 9.0 m/s in 2.5 s, acceleration = 9.0 ÷ 2.5 = 3.6 m/s².
如果一名短跑运动员用12.0秒完成100米，平均速度 = 100 ÷ 12.0 = 8.33 m/s。若她的速度在2.5秒内从0增至9.0 m/s，则加速度 = 9.0 ÷ 2.5 = 3.6 m/s²。
In SQA PE, you may calculate speed to analyse a winger’s sprint or use acceleration to explain performance in a 40‑m dash test.
在 SQA 体育中，你可以计算速度来分析边锋的冲刺，或用加速度解释40米冲刺测试中的表现变化。
6. Newton’s Laws of Motion | 牛顿运动定律
Newton’s three laws form the foundation of sports biomechanics. They explain how forces create, alter and stop movement.
牛顿三定律是运动生物力学的基础，它们解释了力如何产生、改变和停止运动。
First Law – Inertia: A body remains at rest or in uniform motion unless acted upon by an external force. This is why a stationary goalkeeper must generate force to dive, and why a sliding player continues until friction stops him.
第一定律——惯性：一切物体总保持匀速直线运动状态或静止状态，直到外力迫使它改变。这解释了为什么静止的守门员必须主动发力才能扑球，以及为什么滑铲的球员会继续滑行直到摩擦力使他停下。
F = m × a
Second Law – Acceleration: The acceleration of an object is directly proportional to net force and inversely proportional to mass. A rugby player with greater mass requires more force to accelerate at the same rate as a lighter player.
第二定律——加速度：物体加速度的大小跟作用力成正比，跟物体的质量成反比。体重更大的橄榄球运动员需要更大的力才能获得与体重较轻者相同的加速度。
Third Law – Action–Reaction: For every action there is an equal and opposite reaction. When a swimmer pushes water backwards, the water pushes the swimmer forwards.
第三定律——作用力与反作用力：两物体间的作用力和反作用力总是大小相等、方向相反。游泳者向后推水，水便向前推动游泳者。
7. Work, Energy & Power | 功、能量与功率
Mechanical work and power quantify the effectiveness of a performer’s effort. They are crucial for evaluating explosive movements like jumps or lifts.
机械功和功率量化了运动员发力的效率，对于评价跳跃或举重等爆发性动作至关重要。
Work = Force × distance moved in direction of force
Power = Work ÷ time
k done = 2000 × 0.8 = 1600 J. If the lift takes 1.6 s, power output = 1600 ÷ 1.6 = 1000 W.
如果举重运动员施加2000 N向上的力，将杠铃竖直移动0.8 m，做功 = 2000 × 0.8 = 1600 J。若该动作耗时1.6秒，则功率输出 = 1600 ÷ 1.6 = 1000 W。
Power can also be expressed as the product of force and velocity (P = F × v), useful for sustained activities like cycling or rowing where velocity can be measured continuously.
功率也可表示为力与速度的乘积（P = F × v），这对于自行车或划船等可持续测量速度的项目特别有用。
8. Angular Motion & Linear Velocity | 角运动与线速度
Many sporting actions involve rotation about an axis – a golf swing, a discus throw or even a gymnast’s giant circle. The relationship between angular and linear motion is captured by:
许多运动动作都包含绕轴的转动——高尔夫挥杆、铁饼投掷甚至体操大回环。角运动与线运动之间的关系可用以下公式表达：
v = r × ω
otation (m), and ω is angular velocity (rad/s). Angular velocity is defined as ω = Δθ / t.
其中 v 表示线速度（m/s），r 为旋转半径（m），ω 为角速度（rad/s）。角速度的定义为 ω = Δθ / t。
When a tennis player increases the length of the lever arm (by straightening the elbow) during a serve, r increases, so linear velocity at the racket head rises even if ω remains constant. This generates a faster serve.
网球运动员在发球时通过伸直肘关节加长力臂，r 增大，即使角速度不变，拍头的线速度也会提升，从而发出更快的球。
9. Impulse and Momentum | 冲量与动量
Impulse governs how forces applied over time change an object’s momentum. This theorem explains how to increase projectile speed or reduce impact forces.
冲量决定了力在时间上的积累如何改变物体的动量。这一定理能解释如何提高抛射体速度或减小冲击力。
Impulse = Force × time = change in momentum = m × Δv
To throw a cricket ball faster, a fielder applies a large force over a longer follow‑through time, increasing the impulse and thus the ball’s final velocity.
为了让板球投得更快，外野手通过更大的力量配合更长的随挥时间，增加了冲量，从而提高了球的最终速度。
In landing mechanics, bending the knees extends the time over which impact force is absorbed, reducing the peak force on the body – a direct application of the impulse‑momentum relationship.
在落地缓冲中，屈膝延长了吸收冲击力的时间，减小了作用于身体的峰值力，这正是冲量–动量关系的直接应用。
10. Lever Systems & Mechanical Advantage | 杠杆系统与机械效益
Levers in the human body are formed by bones, joints and muscles. Mechanical advantage (MA) determines whether a lever favours speed or strength.
人体中的杠杆由骨、关节和肌肉构成。机械效益（MA）决定了一组杠杆是偏向速度还是力量。
Mechanical Advantage = effort arm ÷ resistance arm
If the effort arm is longer than the resistance arm, MA > 1 (force amplifier, e.g. standing on tiptoes – second‑class lever). If MA < 1, the lever amplifies speed and range of motion (e.g. the elbow joint during a throw – third‑class lever).
若力臂长于阻力臂，MA > 1（增长力，例如提踵——第二类杠杆）。若 MA < 1，杠杆放大速度与运动幅度（例如投掷时的肘关节——第三类杠杆）。
Most sporting levers are third‑class because they allow faster limb movement at the expense of force. A footballer generating a powerful kick uses a rapid, third‑class lever action at the knee.
运动中的杠杆大多数是第三类，因为它们能牺牲力量来换取更快的肢体运动速度。足球运动员大力抽射时，膝关节快速伸展正是利用了第三类杠杆的性质。
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