Curriculum/DP Design/B3.3 Mechanical Systems Application and Selection

Mechanical Systems Application and Selection | B3.3

Guiding questionHow can mechanical systems be incorporated into product design?

Overview and teacher commentary will appear here.

Building on A3.3, this topic applies mechanical systems to real design contexts — computing mechanical advantage, velocity ratio and efficiency, and constructing gear, cam and lever systems. Calculations are central to this topic: expect to work through multi-step problems and explain your reasoning in terms of design decisions.

Mechanical Systems Application — B3.3

Students must be able toCalculate mechanical advantage in gear, pulley, belt and lever systems.

Belt drive diagram showing drive pulley, driven pulley, and belt
English

Mechanical advantage (MA) is the ratio of the output load to the input effort in a machine. For a belt drive or pulley system, MA is determined by the ratio of pulley diameters — or equivalently by the inverse ratio of rotational speeds:

MA = Load / Effort = d₂ / d₁ = N₁ / N₂

where d₂ is the driven (output) pulley diameter, d₁ is the drive (input) pulley diameter, N₁ is the input speed and N₂ is the output speed.

Effect on torque and speed:

  • If d₂ > d₁ (MA > 1): output torque increases, output speed decreases — a force multiplier.
  • If d₂ < d₁ (MA < 1): output speed increases, output torque decreases — a speed multiplier.

Characteristics of belt drives: Belts transmit power by friction between belt and pulley. They are quiet, cost-effective, and can span large distances between shafts. Under heavy loads, belts can slip — this protects components from shock overload but reduces efficiency and speed accuracy. V-belts wedge into grooves to increase friction and reduce slip.

中文

机械优势(MA)是机器输出载荷与输入力之比。对于带传动或滑轮系统,MA由带轮直径之比决定,也等于转速之反比:

MA = 载荷 / 力 = d₂ / d₁ = N₁ / N₂

其中 d₂ 为从动(输出)带轮直径,d₁ 为主动(输入)带轮直径,N₁ 为输入转速,N₂ 为输出转速。

对扭矩和转速的影响:

  • 若 d₂ > d₁(MA > 1):输出扭矩增大,输出转速降低——起力的放大作用。
  • 若 d₂ < d₁(MA < 1):输出转速增大,输出扭矩减小——起速度放大作用。

带传动特点:皮带依靠与带轮之间的摩擦力传递动力。传动噪音小、成本低,且能跨越轴间较大距离。在重载下皮带可能打滑,这在保护机构免受冲击的同时会降低效率和速度精度。V形带楔入槽内,可增大摩擦力并减少打滑。

Worked example — belt drive MA

GivenFormulaResult
Drive pulley d₁ = 100 mm, driven d₂ = 300 mmMA = d₂ / d₁MA = 300 / 100 = 3
MA = 3, input speed N₁ = 300 rpmN₂ = N₁ / MAN₂ = 300 / 3 = 100 rpm
The driven pulley produces 3× the torque but rotates at 1/3 the speed.

Students must be able toCalculate velocity ratios for gear-, pulley- and belt-driven systems.

Crossed belt drive and compound belt drive diagrams
English

The velocity ratio (VR) describes how much the rotational speed changes from input to output. VR and MA are reciprocals — a system that multiplies force (MA > 1) reduces speed (VR < 1) by the same factor.

For a simple belt drive or pulley:

VR = N₂ / N₁ = d₁ / d₂

Crossed belt drives reverse the rotation direction of the driven pulley — the belt crosses between the two pulleys. The VR formula is unchanged; only the direction differs.

Compound belt drives connect multiple pulley pairs in series. Two pulleys share a common shaft so that when the first driven pulley rotates, it drives the next stage. The total VR is the product of each stage's VR:

VR_total = VR₁ × VR₂ × VR₃ …

For gear trains, VR is calculated using the number of teeth instead of pulley diameters:

VR = T_driven / T_driver = N_driver / N_driven

中文

速比(VR)描述旋转速度从输入到输出的变化程度。VR与MA互为倒数——放大力的系统(MA > 1)会以同等比例降低转速(VR < 1)。

对于简单带传动或滑轮:

VR = N₂ / N₁ = d₁ / d₂

交叉带传动使从动带轮的旋转方向相反——皮带在两带轮间交叉。速比公式不变,仅方向不同。

复合带传动将多组带轮串联。两个带轮共用同一轴,当第一个从动带轮转动时驱动下一级。总速比为各级速比之积:

VR_total = VR₁ × VR₂ × VR₃ …

对于齿轮系,速比用齿数代替带轮直径计算:

VR = 从动齿轮齿数 / 主动齿轮齿数 = 主动转速 / 从动转速

System typeVR formulaDirection change?
Simple belt / pulleyd₁ / d₂No
Crossed beltd₁ / d₂Yes — reverses
Compound beltVR₁ × VR₂ × …Depends on stages
Gear pairT_driver / T_drivenYes — reverses
Compound gear trainVR₁ × VR₂ × …Depends on stages

Students must be able toCalculate efficiency for gear- and belt-driven systems.

Power flow diagram showing input power, losses and output power
English

Efficiency (η) is the fraction of input power that reaches the output as useful work. No real machine reaches 100% — some energy is always lost to friction and heat.

η = (P_out / P_in) × 100%

Power in gear systems — power is the product of torque and angular velocity:

P = τ × ω    where ω = N × (2π / 60) rad/s

To convert rpm to rad/s, multiply by 2π and divide by 60.

Power in belt drives — power is transmitted by the difference in belt tensions between the tight side and the slack side:

P = F_E × v_belt    where F_E = T₊ − T₋

T₊ is the tight-side tension and T₋ is the slack-side tension. Belt speed v_belt = π × d₁ × N₁ / 60 (m/s, with d₁ in metres).

Why real systems fall short of 100%:

  • Gear tooth friction — as teeth slide against each other, energy converts to heat. This is the primary loss in most gear trains.
  • Churning losses — gears and belts must push through lubricating oil or grease, creating resistance.
  • Bearing friction — shaft bearings absorb some input power.
  • Belt slip — the belt slides slightly on the pulley under heavy load, reducing speed accuracy and transmitting less power.
中文

效率(η)是输出功率占输入功率的比例。任何真实机械都无法达到100%效率——总有部分能量因摩擦转化为热量而损失。

η = (P_out / P_in)× 100%

齿轮系统功率——功率是扭矩与角速度的乘积:

P = τ × ω    其中 ω = N × (2π / 60) rad/s

将rpm转换为rad/s,需乘以2π再除以60。

带传动功率——功率由皮带紧边与松边张力之差传递:

P = F_E × v_belt    其中 F_E = T₊ − T₋

T₊ 为紧边张力,T₋ 为松边张力。带速 v_belt = π × d₁ × N₁ / 60(m/s,d₁ 单位为米)。

实际系统功率损失原因:

  • 齿面摩擦——齿面相互滑动时,能量转化为热量,是大多数齿轮系的主要损耗。
  • 搅油损失——齿轮和皮带在润滑油或润滑脂中运动产生阻力。
  • 轴承摩擦——轴承吸收部分输入功率。
  • 皮带打滑——重载时皮带在带轮上轻微滑动,降低速度精度并减少功率传递。

Worked example — gear power and efficiency

StepWorkingResult
Convert speed to rad/sω = 120 × (2π / 60)12.57 rad/s
Calculate input powerP = τ × ω = 1.8 Nm × 12.5722.6 W
Calculate efficiencyη = (13.5 / 15) × 100%90%

When a torque-based power calculation gives a different value from a given input power, the figures likely represent different operating conditions (e.g., theoretical vs. measured). State your assumptions and show all working.

Students must be able toCalculate gear ratios and belt-driven system ratios, calculate the speed of rotation of a gear system at several points including initial input and final output speed, and construct systems that use gears to increase or decrease speed and motion.

Compound gear train diagram showing multiple gear pairs on shared shafts
English

Gear and belt systems are selected by designers to achieve specific combinations of speed, torque and direction. Choosing the right system requires calculating what will happen at each stage of the mechanism.

Direction of rotation in gear trains: In a simple gear train, adjacent gears rotate in opposite directions. An idler gear placed between two gears reverses direction again — the output then turns the same way as the input without changing the speed ratio. Belts and pulleys do not reverse direction (unless crossed).

Compound gear trains mount two gears on a shared shaft. This allows large speed changes in a compact space. The total velocity ratio is the product of each gear pair's ratio:

VR_total = VR₁ × VR₂ × VR₃ …

Calculating speed at multiple stages: Work stage by stage. For each gear pair: N_out = N_in × (T_driver / T_driven). For each belt stage: N_out = N_in × (d_drive / d_driven).

Overdrive gearboxes have VR < 1, meaning output speed is greater than input speed but output torque is reduced. Used in bicycle derailleur top gears and vehicle transmissions during motorway cruising — where higher speed and less force are needed.

中文

设计师通过选择齿轮和带传动系统来实现特定的转速、扭矩和旋转方向组合。选择合适的系统需要计算机构中每个阶段的变化。

齿轮系中的旋转方向:在简单齿轮系中,相邻齿轮旋转方向相反。在两齿轮之间加入惰轮可再次反转方向,使输出与输入同向旋转,而不改变总速比。皮带和带轮不会改变旋转方向(除非使用交叉带传动)。

复合齿轮系将两个齿轮安装在同一轴上,可在紧凑空间内实现大幅度的速度变化。总速比是每对齿轮速比的乘积:

VR_total = VR₁ × VR₂ × VR₃ …

多级转速计算:逐级计算。对每对齿轮:N_out = N_in × (T_driver / T_driven)。对每级带传动:N_out = N_in × (d_drive / d_driven)。

超速传动的VR < 1,即输出转速大于输入转速,但输出扭矩减小。应用于自行车变速器最高挡和汽车高速公路行驶档——此时需要较高速度而非较大力。

Worked example — compound belt drive (multi-stage)

StagePulley diametersVROutput speed
Stage 1d₁ = 400 mm → d₂ = 200 mmVR₁ = 200/400 = 0.5N₂ = 60 × (400/200) = 120 rpm
Stage 2d₃ = 80 mm → d₄ = 50 mmVR₂ = 50/80 = 0.625N₄ = 120 × (80/50) = 192 rpm
OverallVR_total = 0.5 × 0.625 = 0.3125N₄ = 192 rpm (input N₁ = 60 rpm)

For compound belt drives, you cannot divide the final pulley diameter by the first — you must multiply individual stage VRs. A common exam mistake is treating it like a simple two-pulley system.

Students must be able toAnalyse how cam systems translate rotary motion into reciprocating motion, construct mechanical systems that use cams, and interpret diagrams that represent the use of cams in a system.

Cam and follower diagram showing cam profile with rise, fall and dwell phases
English

A cam is a specially shaped rotating component. A follower is held against its surface and is pushed up and down as the cam rotates — converting continuous rotary motion into reciprocating (back-and-forth) motion. The cam's profile (its outline shape) determines exactly how the follower moves.

Cam motion phases:

  • Rise — the follower moves away from the cam centre as the cam radius increases.
  • Fall — the follower returns toward the centre as the radius decreases.
  • Dwell — the follower remains stationary while the cam radius stays constant.

Follower types: knife-edge (precise but wears quickly), roller (reduces friction, suited to most applications), flat-faced (suited to high-speed cams with gentle profiles).

Historical development: Trip hammers in Han Dynasty China (206 BCE–220 CE) used cams driven by water wheels to produce a hammering action — one of the earliest uses of rotary-to-reciprocating conversion. In the 12th century, Ismail al-Jazari described programmable camshafts in his Book of Knowledge of Ingenious Mechanical Devices. His "Mechanical Servant" automaton used cam lobes on a hidden shaft to produce a sequence of controlled movements — an early form of mechanical programming.

Modern applications:

  • Sewing machine stitch cams — interchangeable cam profiles produce different reciprocating needle patterns, creating different stitches (zigzag, blind hem) without altering any other part of the machine. Different cam shapes encode different stitch programs.
  • Internal combustion engine camshaft — driven at half the crankshaft speed, each cam lobe pushes open an intake or exhaust valve at a precisely timed moment. The cam profile controls valve lift (how far it opens), duration (how long), and timing (when). Once the lobe rotates past, a valve spring closes the valve.
中文

凸轮是一种形状特殊的旋转零件。从动件紧压其表面,随凸轮旋转而上下运动——将连续的旋转运动转换为往复运动(来回运动)。凸轮的轮廓(外形)决定了从动件的运动方式。

凸轮运动阶段:

  • 升程——随凸轮半径增大,从动件远离凸轮中心。
  • 回程——随凸轮半径减小,从动件回向中心。
  • 停歇——凸轮半径恒定时,从动件保持静止。

从动件类型:尖端从动件(精确但磨损快)、滚子从动件(减小摩擦,适用于大多数场合)、平底从动件(适用于轮廓平缓的高速凸轮)。

历史发展:中国汉朝(公元前206年至公元220年)的舂碓(动力锤)利用水车驱动凸轮产生锤击动作——这是旋转运动转换为往复运动最早的应用之一。12世纪,伊斯梅尔·贾扎里在其著作《精妙机械装置知识之书》中描述了可编程凸轮轴。他的"机械侍者"自动机利用隐藏轴上的凸轮叶产生一系列受控动作——这是机械编程的早期形式。

现代应用:

  • 缝纫机线迹凸轮——可更换的凸轮轮廓为针杆产生不同的往复运动,形成不同的线迹(锯齿形、暗缝),无需更换机器其他部件。不同的凸轮形状编码不同的线迹程序。
  • 内燃机凸轮轴——以曲轴一半的转速运转,每个凸轮叶在精确时刻推开进气阀或排气阀。凸轮轮廓控制阀门升程(开启程度)、持续时间(开启时长)和正时(开启时刻)。凸轮叶转过后,阀门弹簧将阀门关闭。

Students must be able toAnalyse the Load (L), Effort (E) and Fulcrum, calculate Load (L) and Effort (E), construct mechanical systems that use levers, and interpret diagrams that represent the use of levers in a system.

Three classes of levers showing positions of fulcrum, effort and load with examples
English

A lever is a rigid beam that rotates about a fixed point called the fulcrum (F). By varying the positions of the effort (E) and load (L) relative to the fulcrum, a lever can multiply force, speed, or distance.

Equilibrium condition — for a lever in balance, the turning moments on each side of the fulcrum must be equal:

E × effort arm = L × load arm    →    MA = load arm / effort arm

Oblique forces — when a force acts at angle θ to the lever rather than perpendicular to it, only the perpendicular component creates a turning moment. Giovanni Batista Benedetti (16th century) recognised this and showed the effective force can be resolved using trigonometry:

(E × sin θ) × effort arm = L × load arm

Biomechanics — Giovanni Alfonso Borelli (1608–1679), the father of biomechanics, proved that most joints in the human body act as third-class levers. Muscle insertion points sit close to the joint (very short effort arm), so large muscular forces produce small loads but move the limb end through a large distance. Human body levers are speed and distance magnifiers, not force multipliers — a design essential for throwing, writing, reaching and all fine motor skills. If the body used second-class levers at the elbow, we would be extraordinarily strong but too slow for daily tasks.

中文

杠杆是绕固定点旋转的刚性杆——固定点即支点(F)。通过改变力(E)和载荷(L)相对于支点的位置,杠杆可以放大力、速度或距离。

平衡条件——杠杆处于平衡时,支点两侧的转矩必须相等:

力 × 力臂 = 载荷 × 阻力臂    →    MA = 阻力臂 / 力臂

斜向力——当力以角度θ作用于杠杆(而非垂直)时,只有其垂直分量产生转矩。乔瓦尼·巴蒂斯塔·贝内代蒂(16世纪)认识到这一点,并证明可用三角法求有效力:

(E × sin θ)× 力臂 = 载荷 × 阻力臂

生物力学——乔瓦尼·阿尔方索·博雷利(1608–1679),生物力学之父,他证明了人体大多数关节作为第三类杠杆运作。肌肉附着点靠近关节(极短力臂),因此大肌肉力产生较小的载荷,但肢体末端移动较大距离。人体杠杆是速度和距离的放大器,而非力的放大器——这对于投掷、书写、伸展和所有精细运动技能至关重要。

ClassPosition of F, E, LMAExample中文示例
First classF between E and LCan be >1 or <1Crowbar, scissors, seesaw撬棍、剪刀、跷跷板
Second classL between F and EAlways >1Wheelbarrow, nutcracker独轮车、胡桃夹
Third classE between F and LAlways <1Bicep curl, tweezers, shovel肱二头肌弯举、镊子、铁锹

Worked example — oblique force (bicep curl)

StepWorkingResult
Load torque50 N × 0.35 m17.5 Nm
Equilibrium with oblique effort (θ = 75°)(E × sin 75°) × 0.04 = 17.5E × 0.966 × 0.04 = 17.5
Solve for EE = 17.5 / (0.966 × 0.04)E ≈ 453 N
Mechanical advantageMA = 50 / 453MA ≈ 0.11

MA = 0.11 confirms this is a third-class lever. The body exerts ~453 N to lift a 50 N load — but the hand moves ~35 cm for every ~3 cm of muscle contraction. Speed and range of motion are gained at the expense of force.

Ten questions covering all six learning objectives. Select one answer per question, then click "Check all answers" to see your score and the explanations.

Q1 · 3.3.1 Mechanical Advantage
A belt drive has a drive pulley diameter of 100 mm and a driven pulley diameter of 300 mm. The mechanical advantage (MA) and the effect on torque and speed are:
MA = d₂/d₁ = 300/100 = 3. A larger driven pulley increases torque but reduces speed by the same factor. Energy is conserved — you cannot gain both force and speed simultaneously.
Q2 · 3.3.2 Velocity Ratios
In a crossed belt drive, the driven pulley rotates:
Crossing the belt between the two pulleys reverses the rotation direction of the driven pulley. The velocity ratio formula (VR = d₁/d₂) is unchanged — only the direction of rotation differs.
Q3 · 3.3.6 Levers
Giovanni Alfonso Borelli (1608–1679) is considered the father of biomechanics because he:
Borelli proved in his 1680 treatise De Motu Animalium that most joints in the human body act as third-class levers, where muscles must exert large forces to produce small loads — but the limb moves through a large distance for a small muscle contraction. Speed, not force, is the primary output.
Q4 · 3.3.4 Gear & Belt Systems
A gear train has three gear pairs with individual velocity ratios of 0.43, 2.125, and 0.44. The overall velocity ratio is approximately:
VR_total = 0.43 × 2.125 × 0.44 ≈ 0.40. Because VR < 1, the output speed is greater than the input speed — this is an overdrive (speed increaser). Output torque is reduced by the same factor.
Q5 · 3.3.3 Efficiency
In a belt drive system, the effective force (F_E) that transmits power is equal to:
Power is transmitted by the difference in tension between the tight side and the slack side: F_E = T₊ − T₋. The slack side still carries tension (to keep the belt on the pulley), but only the net difference does useful work.
Q6 · 3.3.3 Efficiency
A gear has a torque of 1.8 Nm and a rotational speed of 120 rpm. The power output (in Watts) is approximately:
ω = 120 × (2π/60) = 12.57 rad/s. P = τ × ω = 1.8 × 12.57 ≈ 22.6 W. A common error is forgetting to convert rpm to rad/s before using P = τω.
Q7 · 3.3.6 Levers
In the human body, the bicep curl (flexing the forearm at the elbow) is an example of:
The bicep muscle attaches close to the elbow (fulcrum), making this a third-class lever (effort between fulcrum and load). MA is less than 1 — the muscle exerts far more force than the load it lifts, but the hand moves through a much larger distance than the muscle contracts.
Q8 · 3.3.2 Velocity Ratios
The velocity ratio of a belt drive is 0.32. If the drive pulley rotates at 20 rpm, the driven pulley speed is:
VR = N₂/N₁, so N₂ = N₁/VR = 20/0.32 = 62.5 rpm. VR < 1 means the output speed is higher than the input — this is an overdrive configuration.
Q9 · 3.3.5 Cams and Followers
Ismail al-Jazari (12th century) is credited with:
Al-Jazari described programmable camshafts in his Book of Knowledge of Ingenious Mechanical Devices. His "Mechanical Servant" used cam lobes on a hidden shaft — repositioning the cams changed the automaton's sequence of movements, making it an early form of mechanical programming.
Q10 · 3.3.4 Gear & Belt Systems
A compound gear train is preferred over a simple gear train when:
Compound gear trains mount multiple gears on shared shafts, multiplying velocity ratios at each stage. This achieves large overall speed changes (high VR_total) without needing gears of impractical size — making them ideal for compact gearboxes such as vehicle transmissions and electric drill chucks.
Paper 2 requires extended written responses. Write your answer before revealing the example — then compare your approach, not just the content.
Question 1 · 3.3.2 Velocity Ratios · 4 marks
A belt drive has a drive pulley with a diameter of 320 mm rotating at 20 rpm, and a driven pulley with a diameter of 128 mm.

a) Calculate the velocity ratio (VR) of the belt drive.
b) Calculate the rotational speed of the driven pulley.
c) If the belt thickness is 5 mm, recalculate the speed of the driven pulley.
Show example answer

Given: d₁ = 320 mm, N₁ = 20 rpm, d₂ = 128 mm, belt thickness t = 5 mm

a) Velocity ratio:
VR = d₂ / d₁ = 128 / 320 = 0.4

b) Driven pulley speed (ignoring belt thickness):
N₂ = N₁ × (d₁ / d₂) = 20 × (320 / 128) = 20 × 2.5 = 50 rpm

c) Driven pulley speed (including belt thickness):
When belt thickness is considered, the effective diameter becomes (diameter + thickness), as the belt's neutral axis sits at its mid-thickness:
N₂ = N₁ × (d₁ + t) / (d₂ + t) = 20 × (320 + 5) / (128 + 5) = 20 × 325 / 133 = 48.87 rpm

The belt thickness slightly increases both effective diameters, but affects the smaller pulley proportionally more — reducing the final output speed slightly below the no-thickness result.

Question 2 · 3.3.6 Levers · 6 marks
A third-class lever system represents a bicep curl. The load (weight held in the hand) is 50 N at a distance of 0.35 m from the elbow joint (fulcrum). The bicep muscle attaches to the radius bone at a distance of 0.04 m from the fulcrum. The force from the bicep is applied at an angle of 75° to the forearm.

a) Calculate the effective effort force required to hold the load in equilibrium (ignore the weight of the forearm).
b) Calculate the mechanical advantage (MA) of this lever system.
c) Explain why the body uses third-class levers despite their low mechanical advantage.
Show example answer

a) Effective effort force:

Only the component of the bicep force perpendicular to the forearm creates a turning moment. For equilibrium, the moments about the fulcrum must balance:

(E × sin 75°) × effort arm = Load × load arm
(E × 0.966) × 0.04 = 50 × 0.35
E × 0.03864 = 17.5
E = 17.5 / 0.03864 ≈ 452.9 N

b) Mechanical advantage:
MA = Load / Effort = 50 / 452.9 ≈ 0.11

c) Why the body uses third-class levers:

As Borelli proved in De Motu Animalium (1680), human body levers are primarily magnifiers of speed and distance, not force. Although the bicep must exert ~453 N to lift a 50 N load (MA = 0.11), the muscle only contracts approximately 2–3 cm to move the hand through ~35 cm. This large range of motion is essential for throwing, writing, reaching and fine motor skills. If the arm used a second-class lever (which would give MA > 1), movements would require less muscular force but would be much slower — making everyday actions impossible to perform at the speed required.

Question 3 · 3.3.3 Efficiency · 5 marks
A gear system has an input power of 15 W and an output power of 13.5 W.

a) Calculate the efficiency (η) of the gear system.
b) If the input torque is 1.8 Nm and the input rotational speed is 120 rpm, calculate the input power from these values.
c) State two reasons why real gear systems have efficiency less than 100%.
Show example answer

a) Efficiency:
η = (P_out / P_in) × 100% = (13.5 / 15) × 100% = 90%

b) Input power from torque and speed:
ω = N × (2π / 60) = 120 × (2π / 60) = 12.57 rad/s
P = τ × ω = 1.8 × 12.57 = 22.6 W

This value (22.6 W) differs from the given input power of 15 W. This suggests the torque and speed values correspond to a different operating point, or that 15 W is the power after accounting for losses external to the gearbox itself. When discrepancies arise, state your assumption and show all working.

c) Two reasons for efficiency less than 100%:

  1. Friction between meshing gear teeth — as teeth slide and roll against each other under load, kinetic friction converts mechanical energy into heat. This is the primary loss mechanism in most gear trains.
  2. Churning losses — gears in a lubricated gearbox must push through oil or grease. The viscous resistance of the lubricant dissipates energy, particularly at high rotational speeds.
Question 4 · 3.3.4 Gear & Belt Systems · 4 marks
A compound belt drive system has two stages:
Stage 1: drive pulley d₁ = 400 mm, driven pulley d₂ = 200 mm
Stage 2: d₃ = 80 mm (coaxial with d₂), driven pulley d₄ = 50 mm
Drive speed N₁ = 60 rpm

a) Calculate the velocity ratio of each stage.
b) Calculate the overall velocity ratio.
c) Calculate the final output speed N₄.
Show example answer

a) Velocity ratio of each stage:
Stage 1: VR₁ = d₂ / d₁ = 200 / 400 = 0.5
Stage 2: VR₂ = d₄ / d₃ = 50 / 80 = 0.625

b) Overall velocity ratio:
VR_total = VR₁ × VR₂ = 0.5 × 0.625 = 0.3125

Note: You cannot calculate this as d₄/d₁ = 50/400 = 0.125 — that formula only works for a simple two-pulley system. In a compound belt drive, each intermediate pulley pair contributes its own VR and all must be multiplied together.

c) Final output speed:
Stage by stage: N₂ = 60 × (400/200) = 120 rpm. Since d₃ is coaxial with d₂, N₃ = N₂ = 120 rpm.
N₄ = 120 × (80/50) = 192 rpm

Verification: N₄ = N₁ / VR_total = 60 / 0.3125 = 192 rpm ✓

Question 5 · 3.3.5 Cams and Followers · 6 marks
Analyse how cams and followers are used in modern mechanical systems. In your answer, refer to at least one historical development and two modern applications, and explain why cams remain relevant despite advances in electronic control systems.
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A cam and follower system converts rotary motion (the rotating cam) into reciprocating motion (the back-and-forth movement of the follower pressed against it). The cam profile — its shape — encodes a specific motion pattern through its rising, falling and dwell phases.

Historical development: Trip hammers in Han Dynasty China (206 BCE–220 CE) used water-wheel-driven cams to produce a hammering action, one of the earliest mechanical applications of the rotary-to-reciprocating principle. In the 12th century, Ismail al-Jazari described programmable camshafts in his automata — changing the position of pegs and lobes on the shaft changed the sequence of movements. This represents an early form of mechanical programming, where the cam profile acted as the stored instruction.

Modern application 1 — Sewing machine stitch cams: Before computerised sewing machines, interchangeable cam discs produced different stitch patterns. As the cam rotated, a follower translated the motion into the needle bar's side-to-side movement. Changing the cam changed the stitch pattern without modifying any other part of the machine — a mechanically elegant solution to programmable output.

Modern application 2 — Internal combustion engine camshaft: The camshaft rotates at half crankshaft speed. Each cam lobe pushes a follower (tappet or rocker arm) to open an intake or exhaust valve at a precisely timed moment. The cam profile controls valve lift (how far the valve opens), duration (how long) and timing (when relative to piston position). A valve spring closes the valve once the lobe passes. Precision here is critical: incorrect valve timing reduces power, increases emissions and can cause engine damage.

Why cams remain relevant: Electronic control systems and servo motors can replicate cam motion, but mechanical cams offer reliability without sensors, controllers or software. In high-speed, high-temperature environments such as engines, cams require no electrical supply, tolerate oil and heat, and cannot suffer software failure. The internal combustion engine camshaft is one of the highest-volume precision mechanical components in manufacturing. "Camless" valvetrains exist but remain complex and costly by comparison.

History · Mechanical engineering
Ismail al-Jazari — Wikipedia
en.wikipedia.org/wiki/Ismail_al-Jazari
Comprehensive overview of al-Jazari's life and inventions, including his camshaft-driven automata and the "Mechanical Servant." Includes reproductions of manuscript illustrations from the Book of Knowledge of Ingenious Mechanical Devices and a timeline of his most significant contributions to mechanical engineering. Useful for the historical context required in Paper 2 cam questions.
History · Biomechanics
Giovanni Alfonso Borelli — Wikipedia
en.wikipedia.org/wiki/Giovanni_Alfonso_Borelli
Background on the father of biomechanics and his 1680 treatise De Motu Animalium, in which he proved that human body joints operate as third-class levers that magnify speed and distance rather than force. Provides context for the lever calculations and biomechanics questions in this topic.
Video · Belt drives & gear trains
The Engineering Mindset — YouTube channel
youtube.com/@theengineeringmindset
Animated engineering explainer videos covering belt drive calculations (velocity ratio, mechanical advantage, tight and slack side tension) and gear train analysis. Visualisations are particularly useful for understanding how compound systems multiply velocity ratios. Search the channel for "belt drive" and "gear train" to find the most relevant videos for B3.3.
Video · Gear trains & compound systems
Lesics — YouTube channel
youtube.com/@Lesics
3D-animated explanations of mechanical systems including compound gear trains, overdrive gearboxes and camshaft operation. The animations make it easier to visualise how speed and direction change at each stage of a multi-gear system. Particularly useful for understanding why compound gear trains achieve higher velocity ratios than simple pairs of the equivalent size.

Linking Questions

  • How does an understanding of the mechanical systems introduced in A3.3 inform the selection of components in real product design? (A3.3)
  • To what extent does the choice of material affect the efficiency and durability of mechanical systems such as gears and levers? (B3.1)
  • How do production methods and manufacturing tolerances influence the performance of precision mechanical components? (B4.1)
  • In what ways can the energy losses in a mechanical system contribute to the environmental impact of a product over its lifetime? (C2.1)
  • How might a user-centred approach change the selection of mechanical systems in consumer products such as power tools or assistive devices? (B1.1)