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# Cranks and sliders

Linkages and cams · Changing motion

<!-- Speaker note: Students turn a crank and read the slider's position on a graph, find that the stroke is twice the crank radius whatever the rod's length, work out a pump's flow from stroke, piston area and turns a minute, and size a crank for a pump brief. About 40 minutes (Predict and Try 8 min · Worked example 12 min · Vary 6 min · Practise and Make 14 min). Sources: TeachEngineering: K-12 engineering design challenges (benchmark; the briefs and wording are our own), https://www.teachengineering.org/ (checked 12 October 2026). -->

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## Today

**We are learning to** a crank and slider turns rotation into back-and-forth motion: one turn is one stroke out and back, and the stroke is twice the crank's radius, whatever the rod's length. Students find the stroke by turning, work out a pump's flow, and size a crank to a brief.

- I can explain how a crank and slider turns rotation into back and forth.
- I can work out a crank's stroke from its radius, and the radius from a stroke.
- I can work out how much a crank-driven pump moves in a minute.

<!-- Speaker note: Before class: Do Mechanisms first if you can: this course assumes the idea of a brief and of a mechanism trading one thing for another. A bicycle upside down makes a good start: the pedal is a crank, and a pencil taped to the pedal traces the circle. Sewing machines, toy steam engines and old pump jacks are good pictures. Simplifications: rigid parts, no friction, and the crank turns slowly and steadily. Real pumps leak a little, so they move slightly less than the model says. Curriculum hooks: Design and Technologies, engineering principles and systems (primary); Maths, measurement, volume and rates. Teacher: Try: after a longer rod gives the same stroke, ask what the rod does change. Teacher: Worked example: keep the units in centimetres for the volume line. Teacher: Vary: the engine question flips the mechanism; ask who has seen a car engine diagram. -->

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## Why it matters

> A motor turns round and round. A pump needs a piston that goes in and out. How do you get one from the other?

<!-- Speaker note: Motors spin. Many machines need something to go **back and forth** instead: a pump's piston, a saw blade, a sewing machine's needle. A **crank** (an arm or a pin on a wheel) and a **connecting rod** turn one into the other: the **slider** at the end of the rod runs along a straight track. The **stroke** is how far the slider moves from one end of its travel to the other. -->

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## Predict

**The crank's pin is 20 mm from the centre, and the rod is 80 mm long. How far does the slider move from one end of its travel to the other?**

- 100 mm
- 40 mm
- 20 mm
- 80 mm

<!-- Speaker note: What happens: 40 mm: twice the radius. The pin goes from 20 mm on one side of the centre to 20 mm on the other, and the slider follows it. Take two or three predictions before anyone runs it. -->

<!-- Speaker note: Teacher: draw the crank at 0° and 180° on the board, and mark where the pin is. -->

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## Try it

{{embed: https://studio.learningwaypoints.com/tutorials/_playground-board/index.html?id=deck%2Fengineering%2Flinkages%2Fcrank%2Fstep-2&code=widget%3A+motion%0Amode%3A+crank%0Aradius%3A+20%0Arod%3A+80%0Achoose%3A+on%0Anumbers%3A+off%0A&fresh=1}}

[Open this step in the tutorial](https://studio.learningwaypoints.com/tutorials/engineering/linkages/crank/?full=step-2#step-2)

<!-- Speaker note:  -->

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## What happened?

* 40 mm: twice the radius. The pin goes from 20 mm on one side of the centre to 20 mm on the other, and the slider follows it.

<!-- Speaker note: Reveal after running. Ask who predicted it, and what made them think so. -->

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<span style="background:#dbeafe; color:#1a1a1a; border:1.5px solid #3b82f6; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">💬 Discussion</span>

## A 25 mm crank, then a 30 mm crank

Now a 30 mm crank and a piston 5 cm² in area, at 40 turns a minute. Start with the water per stroke.

$$\text{stroke} = 2 \times 25 = 50 \text{ mm} = 5 \text{ cm}$$

1) $\text{per stroke} = 4 \times 5 = 20 \text{ cm}^3$ — A piston 4 cm² in area moving 5 cm sweeps 20 cm³ of water.
2) $60 \text{ turns a minute} = 60 \text{ strokes a minute}$ — One turn of the crank is one stroke.
3) $20 \times 60 = 1200 \text{ cm}^3 \text{ a minute}$ — That is 1.2 litres a minute.

<!-- Speaker note: Model this live on the board before revealing anything: write each line and say why, pausing to ask the class what comes next. Then click to reveal the lines and compare. Script. Start: write stroke = 2 × 25 = 50 mm = 5 cm and say "The pin goes 25 mm one side of the centre and 25 mm the other." Line 1: write per stroke = 4 × 5 = 20 cm^3 and say "A piston 4 cm² in area moving 5 cm sweeps 20 cm³ of water." Line 2: write 60 turns a minute = 60 strokes a minute and say "One turn of the crank is one stroke." Line 3: write 20 × 60 = 1200 cm^3 a minute and say "That is 1.2 litres a minute." -->

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## A 25 mm crank, then a 30 mm crank

**A 30 mm crank drives a piston 5 cm² in area. How much water does it push each stroke, in cm³?**

* Answer: 30

<!-- Speaker note: Stroke 2 × 30 = 60 mm = 6 cm, so 5 × 6 = 30 cm³. -->

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## A minute

**At 40 turns a minute, how much water a minute, in cm³?**

* Answer: 1200

<!-- Speaker note: 30 cm³ × 40 strokes = 1200 cm³ (1.2 litres) a minute. -->

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## Find the mistake

**Noah thinks his pump gives 600 cm³ a minute. Click the line where his working first goes wrong.**

```
 1  Noah's pump: a 30 mm crank and a piston 5 cm² in area.
 2  The piston moves as far as the crank is long: 30 mm = 3 cm.
 3  Water per stroke = 5 × 3 = 15 cm³.
 4  At 40 turns a minute: 15 × 40 = 600 cm³ a minute.
```

* Answer: line 2

<!-- Speaker note: Line 2: the piston moves twice the crank's length, 6 cm, because the pin goes from one side of the centre to the other. So 30 cm³ a stroke and 1200 cm³ a minute. -->

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## Vary: size the crank



- **A 60 mm stroke:** 
- **Backwards:** 

[Open this step](https://studio.learningwaypoints.com/tutorials/engineering/linkages/crank/?full=step-4#step-4)

<!-- Speaker note: What this step is for: Reasoning from the stroke back to the radius, then a question that reverses the mechanism: in an engine the slider (piston) drives the crank. For students who need more support: Give a drawing of the crank at four positions (0°, 90°, 180°, 270°) with the slider; students measure the slider's position at each. For early finishers: A pump has two pistons on one crank, half a turn apart. How does that change the flow, and why might it be smoother? The hard part, on purpose: Students take the radius as the stroke, and mix up the rod and the crank. Keep pointing at the pin: where is it at each end of the stroke? When students get stuck: A student takes the radius as the stroke. Ask: Where is the pin at each end of the stroke? Fix: From one side of the centre to the other: twice the radius. / A student changes the rod to change the stroke. Ask: Turn it with a longer rod. Did the stroke change? Fix: Only the crank radius sets the stroke. / A student mixes millimetres and centimetres in a volume. Ask: What units is the piston area in? Fix: Convert the stroke to cm before multiplying by cm². -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Stroke

**A crank of radius 35 mm. What is the stroke, in mm?**

* Answer: 70

<!-- Speaker note: 2 × 35 = 70 mm. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Radius

**A saw blade must move 80 mm each stroke. What crank radius does it need, in mm?**

* Answer: 40

<!-- Speaker note: 80 ÷ 2 = 40 mm. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Speed

**A sewing machine's crank turns 900 times a minute. How many times does the needle go down each second?**

* Answer: 15

<!-- Speaker note: 900 strokes a minute ÷ 60 = 15 a second. -->

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## Flow

**A pump with a 4 cm stroke and a 3 cm² piston turns 50 times a minute. How much water a minute, in cm³?**

* Answer: 600

<!-- Speaker note: 3 × 4 = 12 cm³ a stroke, × 50 = 600 cm³. -->

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## Make: a pump to a brief



- **A garden pump:** 
- **Evaluate:** 

[Open this step](https://studio.learningwaypoints.com/tutorials/engineering/linkages/crank/?full=step-6#step-6)

<!-- Speaker note: What this step is for: Students work back from the flow to the stroke and the radius, then test it on the board: the same brief-first reasoning as the other Mechanisms lessons. Hints, in order: Per stroke = flow ÷ strokes a minute: 900 ÷ 30. / Stroke = water per stroke ÷ piston area. / 30 cm³ ÷ 6 cm² = 5 cm = 50 mm, so the radius is 25 mm. -->

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<span style="background:#dcfce7; color:#1a1a1a; border:1.5px solid #22c55e; padding:6px 18px; border-radius:8px; font-size:1em; font-weight:500;">✅ Check</span>

## Exit ticket

**A crank of radius 15 mm drives a piston 4 cm² in area at 50 turns a minute. What is the stroke, and how much water a minute?**

* Stroke 30 mm = 3 cm; 4 × 3 = 12 cm³ a stroke; 12 × 50 = 600 cm³ a minute.

<!-- Speaker note: Students answer on paper or a mini whiteboard before the reveal. Anyone who cannot answer needs the main idea again next lesson. Where this leads: Next: Cams and followers, where the shape of a wheel sets the path. Real examples: piston engines (backwards), pumps, sewing machines, jigsaws, steam locomotives. -->