📚 Year 7 CIE Physical Education: Quick-Reference Formula & Theorem Handbook | Year 7 CIE 体育：公式定理速查手册
This handbook collects the essential formulas, equations, and principles that Year 7 CIE Physical Education students need to master. From heart rate calculations to lever mechanics, each section provides a concise explanation in English followed by its Chinese translation, helping bilingual learners build confidence in both languages. Use this as a quick revision tool before tests, lab activities, or fitness assessments. The formulas are presented with clear examples and notation suitable for the CIE curriculum, ensuring accurate application in practical and theoretical contexts.
本手册汇集了 Year 7 CIE 体育课程中学生必须掌握的核心公式、方程和原理。从心率计算到杠杆力学，每一节先用英文解释，再配以中文译文，帮助双语学习者在两种语言中都能建立信心。在考试、实验活动或体能测评前，可将其作为快速复习工具。公式均配有清晰的示例和符合 CIE 课程要求的符号，确保在实践与理论场景中准确应用。
1. Maximum Heart Rate (MHR) Estimation | 最大心率估算
Maximum heart rate is the highest number of beats per minute your heart can safely achieve during all‑out exercise. The simplest and most widely used formula in Year 7 CIE PE is:
最大心率是指在全天候极限运动中，心脏每分钟所能安全达到的最高心跳次数。Year 7 CIE 体育中最简单且应用最广的公式为：
MHR = 220 – Age
For example, a 12‑year‑old student would have an estimated MHR of 220 – ormula provides a starting point for setting training intensities, though actual MHR may vary slightly due to genetics and fitness level. Teachers often use it to design age‑appropriate cardiovascular workouts.
例如，一名 12 岁学生的估算最大心率为 220 – 12 = 208 次/分钟 (bpm)。此公式为设定训练强度提供了基准，尽管实际最大心率会因遗传和体适能水平略有差异。教师常借此设计适合年龄的心血管训练。
2. Target Heart Rate (THR) Zone | 靶心率区间
The target heart rate zone represents the ideal pulse range for improving aerobic fitness. Once MHR is known, the THR zone is calculated by applying percentage ranges directly to MHR:
靶心率区间是提升有氧适能的理想脉搏范围。得出 MHR 后，直接将百分比区间乘以 MHR 即可求得 THR 区间：
Lower THR = MHR × 0.60
Upper THR = MHR × 0.85
A 12‑year‑ his zone ensures the heart is working hard enough to gain endurance benefits without excessive strain. Moderate‑intensity sessions typically use 60–70% MHR, while vigorous sessions use 70–85% MHR.
12 岁学生（MHR ≈ 208 bpm）应当在 208×0.60 ≈ 125 bpm 至 208×0.85 ≈ 177 bpm 的区间内锻炼。在此区间内运动，既能确保心脏得到足够刺激以增强耐力，又不会造成过度负荷。中等强度训练通常使用 60–70% MHR，高强度训练则使用 70–85% MHR。
3. Heart Rate Reserve (HRR) and Karvonen Formula | 心率储备与卡沃宁公式
A more personalised method uses resting heart rate (RHR) to establish training zones. The Karvonen formula accounts for individual fitness levels by calculating the heart rate reserve (HRR):
一种更个性化的方法利用静息心率 (RHR) 来确定训练区间。卡沃宁公式通过计算心率储备 (HRR) 来反映个体差异：
HRR = MHR – RHR
THR = (HRR × Desired Intensity%) + RHR
If a student has an RHR of 70 bpm and wants to train at 70% intensity, the target heart rate becomes [(208 – es that fitter individuals with lower RHR need different training loads to achieve the same relative effort.
若某学生的 RHR 为 70 bpm，欲以 70% 强度训练，其靶心率为 [(208 – 70) × 0.70] + 70 = (138 × 0.70) + 70 = 96.6 + 70 ≈ 167 bpm。该方法认识到，RHR 较低的体能较好者，要达到相同的相对努力程度需采用不同的训练负荷。
4. Body Mass Index (BMI) | 身体质量指数
BMI is a screening tool used to categorise weight status in relation to height. The formula is universal and straightforward:
BMI 是一种根据身高筛查体重状态的工具，公式通用且简单：
BMI = Weight (kg) ÷ Height² (m²)
h Organization classification for adults provides context, but for children and adolescents, BMI‑for‑ ear 7 learners understand the relationship between mass and stature in health assessments.
一名体重 45 kg、身高 1.50 m 的学生，其 BMI 为 45 ÷ (1.50×1.50) = 45 ÷ 2.25 = 20.0 kg/m²。世界卫生组织的成人分类法可作为参照，不过对儿童青少年则使用年龄别 BMI 百分位数。然而，原始的 BMI 数值仍能帮助 Year 7 学生理解健康评估中体重与身高的关系。
5. Speed and Velocity | 速度与速率
In PE, speed is a scalar quantity measuring how fast an object moves, while velocity includes direction. The basic speed equation and an advanced kinematic equation for constant acceleration are both useful:
在体育中，速率是标量，衡量物体运动快慢；速度则包含方向。基本速率公式以及匀加速下的运动学方程都很有用：
v = d / t (Speed)
v² = u² + 2as (Constant acceleration)
- peed (m/s), d = distance (m), t = time (s), a = acceleration (m/s²), s = displacement (m).
- v = 最终速率/速度 (m/s), u = 初速 (m/s), d = 距离 (m), t = 时间 (s), a = 加速度 (m/s²), s = 位移 (m)。
can determine final velocity from a stationary start (u = 0) over a given acceleration distance.
例如，跑步者用 12.5 秒跑完 100 m，其平均速率为 100/12.5 = 8.0 m/s。分析短跑起跑时，利用 v² = u² + 2as 可从静止启动 (u = 0) 和给定的加速距离求出最终速度。
6. Acceleration | 加速度
Acceleration describes the rate at which velocity changes. It is fundamental in analysing sprints, jumps, and throws:
加速度描述速度变化的快慢，是分析短跑、跳跃和投掷的基础：
a = (v – u) / t
If a cyclist increases velocity from 4 m/s to 10 m/s in 3 seconds, acceleration is (10 – tes deceleration. Year 7 students apply this concept when measuring reaction times off starting blocks or calculating how quickly a ball changes speed after being kicked.
若自行车运动员速度在 3 秒内由 4 m/s 增至 10 m/s，加速度为 (10 – 4) / 3 = 2 m/s²。负值表示减速。Year 7 学生可在测量起跑反应时间或计算踢球后球速变化时应用此概念。
7. Work, Power, and Torque | 功、功率与力矩
Work and power connect force to movement. Torque measures the turning effect of a force, essential for understanding levers in the human body:
功和功率将力与运动联系起来。力矩衡量力的转动效应，是理解人体杠杆的基础：
W = F × d (Work, in joules)
P = W / t = F × v (Power, in watts)
τ = F × r (Torque, in N·m, where r = lever arm)
- in direction of force (m), v = velocity (m/s), r = perpendicular distance from pivot (m).
- F = 力 (N), d = 沿力方向的位移 (m), v = 速度 (m/s), r = 力臂（到支点的垂直距离）(m)。
Lifting a 20‑N weight through 0.8 m does 16 J of work. If the lift takes 2 s, power output is 8 W. In the elbow joint, the biceps muscle applies a force at a small lever arm while the weight in the hand acts at a larger lever arm – the torque produced determines whether flexion can occur smoothly.
将一个 20 N 的重物上提 0.8 m，做功 16 J。若举起耗时 2 s，输出功率为 8 W。在肘关节中，肱二头肌在短力臂上施力，手中的重物在长力臂上作用——产生的力矩决定了屈肘是否顺畅。
8. One-Repetition Maximum (1RM) Estimation | 单次最大负荷估算
Since directly testing 1RM can be unsafe for young adolescents, CIE PE often uses estimation formulas. The Epley and Brzycki equations are the most common:
由于直接测试 1RM 对青少年可能不安全，CIE 体育常使用估算公式，其中最常用的是 Epley 和 Brzycki 方程：
Epley: 1RM = Weight Lifted × [1 + (Reps / 30)]
Brzycki: 1RM = Weight Lifted × 36 / (37 – Reps)
Epley estimates 1RM = 15 × (1 + 10/30) = 15 × 1.333 ≈ 20.0 kg; Brzycki gives 15 × 36 / (37 – 0 kg. Both are valid for reps between 1 and 10. These predictions help set safe resistance‑training loads.
若一名学生以 15 kg 做 10 次卧推，Epley 估算 1RM = 15 × (1 + 10/30) = 15 × 1.333 ≈ 20.0 kg；Brzycki 公式为 15 × 36 / (37 – 10) = 540/27 = 20.0 kg。两种方法在 1–10 次重复范围内均有效，这些预测可帮助设定安全的抗阻训练负荷。
9. Basal Metabolic Rate (BMR) Estimation | 基础代谢率估算
BMR is the energy required to sustain vital functions at rest. The Mifflin‑St Jeor equation is internationally recognised and often referenced in CIE health‑related fitness topics:
BMR 是静息状态下维持生命机能所需的能量。Mifflin‑St Jeor 方程被国际认可，常在 CIE 健康体适能主题中引用：
Male BMR = 10 × W + 6.25 × H – 5 × A + 5
Female BMR = 10 × W + 6.25 × H – 5 × A – 161
(W = weight in kg, H = height in cm, A = age in years.) For a 12‑year‑old girl weighing 40 kg and measuring 150 cm, BMR ≈ 10×40 + 6.25×150 – 5×12 – 161 = 400 + 937.5 – 60 – s the baseline for calculating total daily energy expenditure and planning balanced diets.
(W = 体重 kg, H = 身高 cm, A = 年龄 岁。) 一名 12 岁、体重 40 kg、身高 150 cm 的女孩，BMR ≈ 10×40 + 6.25×150 – 5×12 – 161 = 400 + 937.5 – 60 – 161 = 1,116.5 kcal/天。BMR 构成了计算每日总能量消耗和规划均衡饮食的基线。
10. Energy Expenditure During Activity | 活动能量消耗
The metabolic equivalent (MET) system allows students to estimate calories burned during different exercises:
代谢当量 (MET) 系统使学生能够估算不同运动的燃脂量：
Energy (kcal) = MET × Body mass (kg) × Time (hours)
| Activity | MET Value | 活动 | MET 值 |
| Walking (4 km/h) | 3.0 | 步行 (4 公里/小时) | 3.0 |
| Jogging (8 km/h) | 8.0 | 慢跑 (8 公里/小时) | 8.0 |
| Swimming (moderate) | 6.0 | 游泳 (中等强度) | 6.0 |
| Football (recreational) | 7.0 | 足球 (娱乐) | 7.0 |
es (0.5 hours) burns approximately 8.0 × 40 × 0.5 = 160 kcal. MET tables appear in fitness‑testing coursework and help explain energy balance alongside BMR.
一名 40 kg 的学生慢跑 30 分钟 (0.5 小时) 大约消耗 8.0 × 40 × 0.5 = 160 kcal。MET 表出现在体适能测评作业中，与 BMR 一起帮助解释能量平衡。
11. Lever Systems and Mechanical Advantage | 杠杆系统与机械效益
The human skeleton operates as a system of levers. Understanding the mechanical advantage (MA) of a lever helps explain why some movements favour speed while others favour force:
人体骨骼如同一套杠杆系统。理解杠杆的机械效益 (MA) 有助于解释为何某些动作利于速度，而另一些利于力量：
MA = Effort arm / Load arm
- First‑ and load on opposite sides of fulcrum (e.g., neck extension); MA can be >1, =1, or <1.
- Second‑class lever: load between fulcrum and effort (e.g., calf raises); MA >1 – good for force.
- Third‑class lever: effort between fulcrum and load (e.g., biceps curl); MA <1 – good for speed and range of motion.
- 第一类杠杆：用力点与阻力点在支点两侧（如颈后伸），MA 可大于、等于或小于 1。
- 第二类杠杆：阻力点在支点和用力点之间（如提踵），MA >1——省力。
- 第三类杠杆：用力点在支点和阻力点之间（如肱二头肌弯举），MA <1——利于速度和活动范围。
Most limb movements in sport are third‑class levers, sacrificing force for the rapid distance coverage needed in throwing, kicking, and running.
体育运动中绝大多数肢体动作属于第三类杠杆，牺牲力量以换取投掷、踢击和奔跑所需的快速位移。
12. Hydration and Fluid Needs | 水合与液体需求
Maintaining hydration during exercise is critical for performance and safety. A simple daily fluid requirement formula used in CIE health modules is:
运动中保持水合对运动表现和安全至关重要。CIE 健康模块中使用的每日液体需求量简单公式为：
Daily fluid (ml) ≈ Body mass (kg) × 30 to 35
ended daily water intake ranges from 45 × 30 = 1,350 ml to 45 × 35 = 1,575 ml. During high‑ l fluids containing electrolytes and carbohydrates are advised. Coaches also use the sweat‑loss calculation: pre‑exercise weight – post‑exercise weight (kg) = fluid loss (litres), stressing that 1 kg lost ≈ 1 litre of sweat.
对于 45 kg 的 Year 7 学生，每日推荐饮水量在 45 × 30 = 1,350 ml 至 45 × 35 = 1,575 ml 之间。若进行超过一小时的高强度运动，建议额外补充含电解质和碳水化合物的液体。教练还会使用排汗量计算：运动前体重 – 运动后体重 (kg) = 体液流失量 (升)，强调 1 kg 体重减少约等于 1 升汗液。
Published by TutorHao | Physical Education Revision Series | aleveler.com
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