The Bicycle Derailleur
A dozen gear ratios, chosen with a lever instead of a gearbox.
Read spotlight →B3.3 is the calculating half of mechanisms, where the ideas from A3.3 turn into design decisions instead of descriptions. Mechanical advantage, velocity ratio and efficiency are the three numbers that let you state what a mechanism is doing for the user. They are the difference between "this gear train makes it easier" and "this gear train trades a factor of four in speed for a factor of four in torque, which is why the user can lift it at all".
Two pieces of advice. Efficiency is the objective students skip most often and the one that matters most in practice, because no real mechanism gives back everything you put into it and the losses are exactly where the design choices live. And do the problems by hand until the relationships feel obvious. These questions are multi-step, the marks sit in the working rather than the final figure, and a habit of showing what you did is worth building now rather than in May.
Students must be able toCalculate mechanical advantage in gear, pulley, belt and lever systems.
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.
Worked example (belt drive MA)
| Given | Formula | Result |
|---|---|---|
| Drive pulley d₁ = 100 mm, driven d₂ = 300 mm | MA = d₂ / d₁ | MA = 300 / 100 = 3 |
| MA = 3, input speed N₁ = 300 rpm | N₂ = N₁ / MA | N₂ = 300 / 3 = 100 rpm |
| The driven pulley produces 3× the torque but rotates at 1/3 the speed. | ||
Interactive
Gear / Velocity-Ratio Calculator
Enter driver and driven sizes (teeth count or diameter, any consistent unit) for one stage, or add stages for a compound drive. Add an input speed and/or torque to see the output.
Stage 1
Stage 2
Stage 3
rpm
N·m
Students must be able toCalculate velocity ratios for gear-, pulley- and belt-driven systems.
The velocity ratio (VR) describes how much the rotational speed changes from input to output. It is always written as driven over driver. In a perfect machine with no friction, VR and MA are equal, so a system that multiplies force by 3 also divides speed by 3. In a real machine some input is lost to friction, so the actual MA is always a little lower than the VR. That gap is what efficiency measures.
For a simple belt drive or pulley:
VR = d₂ / d₁ = N₁ / N₂
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
| System type | VR formula | Direction change? |
|---|---|---|
| Simple belt / pulley | d₂ / d₁ | No |
| Crossed belt | d₂ / d₁ | Yes: reverses |
| Compound belt | VR₁ × VR₂ × … | Depends on stages |
| Gear pair | T_driven / T_driver | Yes: reverses |
| Compound gear train | VR₁ × VR₂ × … | Depends on stages |
Students must be able toCalculate efficiency for gear- and belt-driven systems.
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 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 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.
The choice of material affects every one of these loss mechanisms, from gear tooth wear resistance to belt friction coefficients, as discussed in B3.1 Material Selection.
Worked example (gear power and efficiency)
| Step | Working | Result |
|---|---|---|
| Convert speed to rad/s | ω = 120 × (2π / 60) | 12.57 rad/s |
| Calculate input power | P = τ × ω = 1.2 Nm × 12.57 | 15.1 W |
| Calculate efficiency | η = (13.5 / 15.1) × 100% | 89% |
Work in the order shown: convert rpm to rad/s first, then find power, then compare output with input. Forgetting the rpm conversion is the most common mistake in this calculation.
Discussion
Chasing the last few percent
A well-lubricated gear train can run at 95–98% efficient; a worm gear can sit under 50%. A petrol engine typically wastes 70–80% of its fuel energy as heat before any of it reaches the wheels; an electric drivetrain converts roughly 85–90% of its input into motion. Closing that gap almost always costs something: precision-ground gears cost more than stamped ones, better bearings cost more than bushings, and most loss-reduction measures add mass, complexity or price.
Find a real machine's published efficiency figure. Is the gap between that number and 100% a design failure, or the correct trade-off given what a more efficient version would cost, weigh or take to manufacture? At what point does chasing the last few percent of efficiency stop being worth it?
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.
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.
Calculated gear ratios only hold up if the gears are manufactured to the required tolerances; how production systems achieve this precision is covered in B4.1 Production Systems.
Worked example (compound belt drive, multi-stage)
| Stage | Pulley diameters | VR | Output speed |
|---|---|---|---|
| Stage 1 | d₁ = 400 mm → d₂ = 200 mm | VR₁ = 200/400 = 0.5 | N₂ = 60 × (400/200) = 120 rpm |
| Stage 2 | d₃ = 80 mm → d₄ = 50 mm | VR₂ = 50/80 = 0.625 | N₄ = 120 × (80/50) = 192 rpm |
| Overall | VR_total = 0.5 × 0.625 = 0.3125 | N₄ = 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.
Product Spotlight
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.
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.
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.
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.
| Class | Position of F, E, L | MA | Example | 中文示例 |
|---|---|---|---|---|
| First class | F between E and L | Can be >1 or <1 | Crowbar, scissors, seesaw | 撬棍、剪刀、跷跷板 |
| Second class | L between F and E | Always >1 | Wheelbarrow, nutcracker | 独轮车、胡桃夹 |
| Third class | E between F and L | Always <1 | Bicep curl, tweezers, shovel | 肱二头肌弯举、镊子、铁锹 |
Worked example (oblique force, bicep curl)
| Step | Working | Result |
|---|---|---|
| Load torque | 50 N × 0.35 m | 17.5 Nm |
| Equilibrium with oblique effort (θ = 75°) | (E × sin 75°) × 0.04 = 17.5 | E × 0.966 × 0.04 = 17.5 |
| Solve for E | E = 17.5 / (0.966 × 0.04) | E ≈ 453 N |
| Mechanical advantage | MA = 50 / 453 | MA ≈ 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 of 100 mm diameter and a driven pulley of 300 mm. The mechanical advantage and its effect are:
MA is the ratio of driven to drive diameter, 300 / 100 = 3. Torque rises by that factor and speed falls by the same one, because power cannot be created by the mechanism. An input at 300 rpm would leave the driven pulley turning at 100 rpm.
Q2 · 3.3.2 Velocity Ratios
In a crossed belt drive, the driven pulley:
Crossing the belt between the pulleys reverses the output direction without altering the diameters, so the speed and torque relationship is exactly as it would be with an open belt. An ordinary belt drive keeps both shafts turning the same way, whereas a meshing gear pair always reverses direction.
Q3 · 3.3.2 Velocity Ratios
How is the overall ratio of a compound belt drive with two stages found?
Each stage passes its output on as the input to the next, so the ratios compound. Treating the arrangement as a single pair by comparing only the first and last pulley ignores the intermediate pair entirely and is one of the most common errors in this topic. The same rule applies to compound gear trains.
Q4 · 3.3.3 Efficiency
In a belt drive, the effective force that transmits power is:
Only the net difference in tension does useful work, so power transmitted is that difference multiplied by belt speed. The slack side still carries tension, which is what keeps the belt seated on the pulley, but it contributes nothing to the power delivered.
Q5 · 3.3.3 Efficiency
A shaft transmits a torque of 1.8 N·m at 120 rpm. The power transmitted is approximately:
Convert the speed first: ω = 120 × 2π / 60 = 12.57 rad/s. Then P = τω = 1.8 × 12.57 ≈ 22.6 W. Forgetting to convert rpm into rad/s is the usual source of a wrong answer here.
Q6 · 3.3.4 Gear & Belt Systems
A compound gear train is chosen in preference to a simple gear train when:
Mounting two gears on a shared shaft lets each pair contribute its own ratio, and the ratios multiply. Achieving the same overall change with one pair would need a gear of impractical size, which is why gearboxes, drills and watches all use compound arrangements.
Q7 · 3.3.4 Gear & Belt Systems
A 20-tooth gear drives a 60-tooth gear. On the same shaft as the 60-tooth gear sits a 15-tooth gear, which drives a 45-tooth gear. With an input speed of 900 rpm, the final output speed is:
Work stage by stage. The first pair gives 900 × 20 / 60 = 300 rpm, and because the second gear shares its shaft, that speed becomes the input to the second pair: 300 × 15 / 45 = 100 rpm. Each stage reduces speed by three, so the compact two-stage train achieves an overall reduction of nine.
Q8 · 3.3.5 Cams and Followers
A cam follower stays at a constant height while the cam continues to rotate. This phase of the cam cycle is called:
Dwell occurs wherever the cam radius stays constant, so the follower is held still. Rise and fall correspond to increasing and decreasing radius. Designing the sequence of rise, dwell and fall into the profile is how a camshaft controls how far an engine valve opens, for how long, and at what moment.
Q9 · 3.3.6 Levers
A bicep curl, with the muscle attached to the forearm close to the elbow, is an example of:
The effort sits between the fulcrum and the load, so mechanical advantage is always below 1 and the muscle must exert far more force than the weight being lifted. What the body gains is range: a contraction of two or three centimetres sweeps the hand through some thirty-five, which is what makes throwing, writing and reaching possible.
Q10 · 3.3.6 Levers
A 50 N load is held 0.35 m from the elbow. The bicep attaches 0.04 m from the joint and pulls at 75° to the forearm. The effort force required is approximately:
Only the component perpendicular to the forearm creates a moment, so (E × sin 75°) × 0.04 = 50 × 0.35 = 17.5 N·m. That gives E = 17.5 / (0.966 × 0.04) ≈ 453 N. Omitting the sine term returns 437 N, which is the trap in this question.
Every Paper 2 question is attached to a product. Nothing here can be answered from memory alone: read the case study first, then answer the parts in order. The tariff tells you how many creditable points to make, and the command term tells you what kind of point counts. Numerical work appears under the command term Apply, which is how the specimen paper sets calculation: you apply a principle to the case study and state what the result means. Write your answer before you open either panel, then mark yourself against the markscheme rather than against the example. This topic is HL only.
Case study
A potter's wheel is driven by an electric motor through a flat belt. The potter needs the wheelhead to turn slowly with enough torque that pressing clay against it does not slow it down.
The motor runs at its most efficient speed and cannot usefully be run slower.
Table 1: Potter's wheel drive
| Component | Value |
|---|---|
| Motor pulley diameter | 60 mm |
| Wheelhead pulley diameter | 420 mm |
| Motor speed | 1400 rpm |
| Belt type | Flat, tensioned by a jockey pulley |
| Motor power | 250 W |
(a) State the type of motion transmitted by the belt drive, see Table 1. [1]
(b) Apply the velocity ratio to determine the wheelhead speed, see Table 1. [2]
(c) Explain why the speed reduction increases the torque available at the wheelhead, see Table 1. [3]
Example answer
(a) Rotary motion.
(b) Velocity ratio = driven diameter ÷ driver diameter = 420 ÷ 60 = 7 : 1. Wheelhead speed = 1400 ÷ 7 = 200 rpm, which is a workable throwing speed.
(c) Power is the product of torque and rotational speed, so for a given power a drive that reduces speed must increase torque in the same proportion. The belt carries the motor's 250 W to the wheelhead less whatever friction takes, and because the wheelhead turns seven times slower, the torque there is close to seven times the motor's. The physical reason is the difference in radius: the belt tension acts at 30 mm from the motor shaft and at 210 mm from the wheelhead shaft, so the same force produces seven times the moment about the larger pulley. This is what the potter needs, because throwing means pressing hard against clay off the centre of the wheel, which applies a substantial opposing torque, and a wheel that lost speed under the hands would be unusable. The motor could not deliver that torque directly at any speed, so the belt drive is converting an output the motor can produce into one the task requires.
Markscheme
(a) • Rotary ✓
Award [1] for the correct type of motion up to [1 max].
(b) Velocity ratios for gear-, pulley- and belt-driven systems can be calculated.
• VR = driven diameter ÷ driver diameter = 420 ÷ 60 = 7 ✓
• Wheelhead speed = 1400 ÷ 7 = 200 rpm ✓
Award [1] for the correct velocity ratio and [1] for the correct output speed, up to [2 max]. Award [1] only where the method is correct but the arithmetic is not.
(c) Mechanical systems are used to increase or decrease the speed, direction or power of a motion, and can provide a mechanical advantage.
• Power is the product of torque and rotational speed ✓
• For a given power, reducing speed must increase torque in proportion ✓
• The belt carries 250 W to the wheelhead, less friction losses ✓
• The wheelhead turns seven times slower, so torque is close to seven times the motor's ✓
• Belt tension acts at 30 mm from the motor shaft and 210 mm from the wheelhead shaft ✓
• The same belt force produces seven times the moment about the larger pulley ✓
• Throwing applies a substantial opposing torque off the wheel centre ✓
• A wheel that lost speed under the potter's hands would be unusable ✓
• The motor cannot deliver that torque directly at any speed ✓
• The drive converts an output the motor can produce into one the task requires ✓
Award [1] for each relevant reason / cause explaining the torque increase up to [3 max]. Award a maximum of [2] where the response does not link power, torque and speed.
Case study
A nutcracker is two hinged arms. A walnut placed near the hinge is cracked by squeezing the far ends of the arms together.
Table 2: Nutcracker dimensions and forces
| Quantity | Value |
|---|---|
| Distance from hinge to nut | 30 mm |
| Distance from hinge to hand | 150 mm |
| Force needed to crack a walnut | 240 N |
| Typical adult grip force available | 250–400 N |
| Grip force available, reduced hand strength | 60–100 N |
(a) State the class of lever formed by each arm of the nutcracker, see Table 2. [1]
(b) Apply the principle of moments to determine the hand force needed to crack the walnut, see Table 2. [2]
(c) Evaluate the nutcracker's suitability for a user with reduced hand strength, see Table 2. [3]
Example answer
(a) Second class, because the load sits between the fulcrum and the effort.
(b) Taking moments about the hinge, effort × 150 = 240 × 30, so effort = 7200 ÷ 150 = 48 N. The mechanical advantage is 150 ÷ 30 = 5.
(c) The nutcracker suits this user well. The 48 N required sits comfortably inside the 60 to 100 N a user with reduced hand strength can produce, so a task needing 240 N directly, which is far beyond them, becomes achievable. The mechanical advantage of 5 is what makes that possible, and it costs only distance: the hands travel five times as far as the nut is squeezed, which is not a limitation on a movement of a few millimetres. Two things qualify this. The margin is narrower than it appears, because a user at the bottom of the 60 N range has only 12 N of headroom, and a larger or harder nut needing more than 240 N would exceed what they can produce. The bigger problem is not the force but the grip: a user with reduced strength usually has reduced control too, and this design requires both hands to hold two arms in position while squeezing a rounded nut that can shoot out. A lever-and-cup design that holds the nut and needs only one hand would serve the same user better, so the mechanism is well suited and the way the user has to hold it is not.
Markscheme
(a) Levers comprise a beam acting on a fulcrum and are classed based on the relative position of the fulcrum to an applied load and effort.
• Second class ✓
Award [1] for the correct lever class up to [1 max].
(b) Mechanical advantage of a system can be calculated.
• Moments about the hinge: effort × 150 = 240 × 30 ✓
• Effort = 7200 ÷ 150 = 48 N ✓
• Mechanical advantage = 150 ÷ 30 = 5 ✓
Award [1] for correct application of moments and [1] for the correct hand force, up to [2 max]. Award [1] where the method is correct but the arithmetic is not.
(c) Levers reduce the effort needed to exert a force and move a load.
Strengths:
• 48 N sits comfortably inside the 60 to 100 N available to this user ✓
• A task needing 240 N directly, far beyond them, becomes achievable ✓
• A mechanical advantage of 5 is what makes it possible ✓
• The cost is distance, and the hands travelling five times as far is no burden over a few millimetres ✓
• No power, mechanism or maintenance required ✓
Limitations:
• A user at the bottom of the 60 N range has only 12 N of headroom ✓
• A larger or harder nut needing more than 240 N would exceed their capability ✓
• Reduced strength usually accompanies reduced control ✓
• The design requires both hands to hold two arms in position ✓
• A rounded nut can shoot out when it cracks ✓
• Sustained squeezing is harder than a brief peak for a user with arthritis ✓
Judgment:
• The mechanism suits the user; the way it must be held does not ✓
• A lever-and-cup design holding the nut and needing one hand would serve better ✓
Award [1] for each distinct strength / limitation, leading to an appraisal of the nutcracker's suitability, up to [3 max]. Award a maximum of [2] where only strengths or only limitations are given.
Case study · part 1
A sailing boat trims its sails with a hand-cranked winch. A rope is wrapped around a drum, and the crew turns a removable handle to pull the rope in against the load of the wind in the sail.
The winch has two gear ratios, selected by reversing the direction the handle is turned.
(a) Identify two types of motion present when the crew hauls a rope in on the winch. [2]
Case study · part 2
Table 3: Winch specification
| Quantity | Value |
|---|---|
| Handle length | 250 mm |
| Drum radius | 40 mm |
| Gear ratio, first speed | 1 : 1 |
| Gear ratio, second speed | 4.4 : 1 reduction |
| Efficiency | 90 % |
| Rope load when close-hauled | 2400 N |
(b) Apply the concept of mechanical advantage to outline why second speed is needed when close-hauled, see Table 3. [2]
Case study · part 3
The winch is 90 % efficient, so 10 % of the crew's input is lost within the mechanism.
(c) Describe where the lost 10 % goes in a winch of this kind, see Table 3. [2]
Case study · part 4
The winch contains a pawl that allows the drum to turn only in the hauling direction. An owner proposes an electric winch that removes the handle and the two speeds, driving the drum directly from a motor at a single speed.
(d) Explain the consequences of replacing the two-speed hand winch with a single-speed electric one, see Table 3. [4]
Example answer
(a) Rotary, at the handle, gears and drum; and linear, in the rope being hauled in.
(b) In first speed the mechanical advantage is just the handle-to-drum ratio, 250 ÷ 40 = 6.25, so holding 2400 N of rope load needs about 384 N at the handle, which is more than a crew member can sustain. Second speed multiplies that by the 4.4 : 1 gear reduction to give a mechanical advantage of about 27.5, bringing the handle force down to roughly 87 N before losses, which is a force a person can turn repeatedly.
(c) Most of it is friction between sliding surfaces: the gear teeth sliding against each other as they mesh, the drum and gear shafts turning in their bearings, and the pawl dragging over its ratchet on every revolution. A smaller part goes into deforming the rope as it bends around the drum and into the seals keeping salt water out, and all of it ends up as heat in the winch body.
(d) The change trades away the crew's control over force and speed in exchange for removing their effort, and whether that is worthwhile depends on what the winch is for.
The gain is straightforward. A 2400 N rope load takes real physical work to haul, and second speed only reduces the force by increasing the number of turns, so trimming a sail close-hauled is slow and tiring. A motor removes that entirely, which matters most for a short-handed or elderly crew, and it lets one person trim a sail that would otherwise need two.
The loss is the two speeds. The hand winch is not two-speed for convenience; it exists because the load varies enormously through a single operation. Most of the rope is pulled in with almost no tension, and first speed hauls it quickly, then the last part is taken up under full load in second speed at high mechanical advantage. A single-speed motor must be geared for the worst case, so the whole operation runs at the slow ratio and the fast phase disappears, which makes routine trimming slower than it was by hand.
The more serious consequence is the loss of feedback. A crew member turning a handle feels the load directly, and that is how they know the sail is sheeted correctly or that the rope has jammed. A motor feels nothing and will keep pulling, so a rope caught on a cleat, or a hand caught in a turn on the drum, is pulled with the full force of the motor rather than stopping when a person cannot turn the handle. That safety property was a by-product of the hand winch and has to be replaced deliberately with a load sensor and cut-out.
A boat also has to keep working when its electrics fail, so the electric winch needs to retain the handle socket and the pawl so it can be cranked manually. In practice the right answer is not to replace the mechanism but to add a motor to it, keeping both speeds and the manual fallback.
Markscheme
(a) There are four types of motion involved in mechanical systems.
• Rotary, at the handle, gears and drum ✓
• Linear, in the rope being hauled ✓
• Oscillating, at the pawl ✓
• Reciprocating, if the pawl is described over its full cycle ✓
Award [1] for each relevant type of motion up to [2 max].
(b) Mechanical advantage of a system can be calculated.
• First speed MA = handle length ÷ drum radius = 250 ÷ 40 = 6.25 ✓
• Handle force in first speed ≈ 2400 ÷ 6.25 = 384 N, more than a crew member can sustain ✓
• Second speed multiplies by the 4.4 : 1 gear reduction ✓
• Overall MA ≈ 6.25 × 4.4 = 27.5 ✓
• Handle force ≈ 2400 ÷ 27.5 ≈ 87 N before losses, which a person can turn repeatedly ✓
Award [1] for correctly applying mechanical advantage and [1] for relating the result to what a crew member can produce, up to [2 max]. Accept answers that account for the 90 % efficiency.
(c) Efficiency can be calculated, and real systems have efficiency less than 100 %.
• Friction between gear teeth sliding as they mesh ✓
• Friction in the bearings supporting the drum and gear shafts ✓
• The pawl dragging over its ratchet on every revolution ✓
• Deformation of the rope as it bends around the drum ✓
• Friction in the seals keeping salt water out ✓
• Viscous drag in the grease ✓
• All losses appear as heat in the winch body ✓
• Wear and corrosion increase these losses over the winch's life ✓
Award [1] for each detail, leading to an account of where the lost energy goes, up to [2 max]. Credit responses that identify heat as the final form.
(d) Mechanical systems convert an input into an output, and gear ratios are selected to match a varying load.
Gains:
• A 2400 N rope load takes real physical work, and second speed only trades force for turns ✓
• A motor removes the crew's effort entirely ✓
• Most valuable for a short-handed or elderly crew ✓
• One person can trim a sail that would otherwise need two ✓
• Consistent hauling speed regardless of crew fatigue ✓
Loss of the two speeds:
• The two speeds exist because the load varies enormously through one operation ✓
• Most of the rope is pulled in at almost no tension, then the last part under full load ✓
• A single-speed motor must be geared for the worst case ✓
• The whole operation runs at the slow ratio and the fast phase disappears ✓
• Routine trimming becomes slower than it was by hand ✓
Loss of feedback:
• A crew member turning a handle feels the load directly ✓
• That is how they know the sail is sheeted correctly or the rope has jammed ✓
• A motor feels nothing and keeps pulling ✓
• A jammed rope or a trapped hand is pulled with the motor's full force ✓
• The hand winch's safety property was a by-product and must be replaced deliberately ✓
• A load sensor and cut-out become necessary ✓
Other consequences:
• A boat must keep working when its electrics fail ✓
• The handle socket and pawl should be retained for manual operation ✓
• Electrical load on the boat's batteries increases ✓
• Motor and wiring must survive a salt water environment ✓
Judgment:
• The better solution is to add a motor to the existing mechanism rather than replace it ✓
• That keeps both ratios and the manual fallback ✓
Award [1] for each relevant detail / reason / cause relating to the consequences of the substitution up to [4 max]. Award a maximum of [3] where the response does not address the loss of the second ratio or the loss of feedback. Credit responses that reach a judgment.
Ismail al-Jazari, Wikipedia
en.wikipedia.org/wiki/Ismail_al-Jazari
The 12th century engineer whose camshaft driven automata predate European use of the camshaft by centuries. Includes the manuscript illustrations from the Book of Knowledge of Ingenious Mechanical Devices.
Giovanni Alfonso Borelli, Wikipedia
en.wikipedia.org/wiki/Giovanni_Alfonso_Borelli
The founder of biomechanics, whose 1680 De Motu Animalium showed that human joints work as third class levers that trade force for speed and range. Context for the lever calculations here.
The Engineering Mindset, YouTube channel
youtube.com/c/Theengineeringmindset
Animated coverage of belt drive calculations and gear train analysis, including velocity ratio, mechanical advantage and belt tension.
Lesics, YouTube channel
youtube.com/c/Lesics
3D animations of compound gear trains, overdrive gearboxes and camshafts. The animations make it clear why a compound train reaches a higher velocity ratio than a single pair of the same 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)
Product Spotlight: The Bicycle Derailleur
A bicycle doesn't have a gearbox in the way a car does. Instead, a rear derailleur is a small linkage that physically pushes the chain sideways off one sprocket and onto another, each one a different diameter, giving a different velocity ratio, while a spring and a parallelogram linkage keep the derailleur's jockey wheels tracking the chain at a constant distance no matter which sprocket it's sitting on.
The sprockets themselves are just the gear side of the velocity-ratio formula from this section, VR = T_driven / T_driver, applied a dozen times over on a single wheel: a modern cassette might run from a 10-tooth sprocket up to a 50-tooth one, so shifting from the smallest to the largest changes the velocity ratio by a factor of 5, without a single gear tooth ever meshing with another gear tooth. The chain is really acting as a flexible belt drive between the crankset sprocket and whichever rear sprocket the derailleur has selected, and the rider changes gear simply by choosing where on that range the chain sits.