Conversion Graphs
A conversion graph does one job extremely well: it turns a value in one unit into a value in another, in either direction, with no arithmetic at all. Most of them are direct variation, which is why a single straight line suffices, and the exceptions are informative.
Water freezes at $0$ °C and $32$ °F. It boils at $100$ °C and $212$ °F. Work out the ratio of the Fahrenheit value to the Celsius value at the boiling point. Now try the same at the freezing point. Say what goes wrong, and what that tells you about the type of relationship.
A conversion graph is a straight line drawn so that a value in one unit can be read off against the equivalent in another. Because most unit conversions are direct variation, the line passes through the origin and its gradient is the conversion factor.
$$\text{miles} \approx 0.62 \times \text{kilometres}$$
Read it in whichever direction you need: up from one axis and across to the other, or across then down. The same line serves both, which is what makes a conversion graph worth drawing once and using many times.
Know
- That a conversion graph is a straight line relating two units
- That most unit conversions are direct variation, so the line passes through the origin
- That temperature conversion between Celsius and Fahrenheit is linear but not proportional
Understand
- Why one line can be read in both directions
- Why a graph reading is approximate while the equation is exact
Can Do
- Draw a conversion graph from a known equivalence
- Read a conversion in either direction from a graph
- Decide whether a given conversion is direct variation
A conversion graph puts one unit on each axis and draws the line relating them. Once drawn, converting a value takes no arithmetic: go up from one axis to the line, then across to the other.
Most unit conversions are direct variation, and it is worth seeing why. Converting kilometres to miles multiplies by a fixed factor, so doubling the distance doubles both readings, and zero kilometres is zero miles. Both conditions from Lesson 3 hold, so the graph is a straight line through the origin.
The same is true of kilograms to pounds, litres to gallons, dollars to another currency at a fixed rate, and centimetres to inches. In each case:
$$\text{second unit} = k \times \text{first unit}$$
and the conversion factor $k$ is the gradient of the line.
For kilometres to miles the factor is about $0.62$, so $50$ km reads as about $31$ miles, which is the reading shown in the diagram. Reading it the other way, $31$ miles reads back as about $50$ km, using exactly the same line.
A conversion graph needs one known equivalence and, for a proportional conversion, the origin.
1. Choose which unit goes on each axis, and label both with their units.
2. Mark the origin, which is on the line for any proportional conversion.
3. Plot the known equivalence, choosing a large value so the point sits well away from the origin.
4. Draw the straight line through both and extend it across the useful range.
To build a kilogram-to-pound graph from the equivalence $10$ kg $\approx 22$ lb: mark $(0,0)$, mark $(10, 22)$, and join. The gradient is $2.2$ pounds per kilogram, which is the conversion factor.
Two choices worth making deliberately. Pick scales that use most of the page, since a cramped graph cannot be read accurately. And choose the known point far from the origin, for the reason given in Lesson 3: a small plotting error near the origin swings the whole line, while the same error far out barely moves it.
Converting between Celsius and Fahrenheit is not direct variation, and it is the standard example of the distinction that has run through this whole focus area.
$$F = 1.8C + 32$$
The graph is a straight line, so a conversion graph still works perfectly well. But it crosses the vertical axis at $32$ rather than at the origin, since $0$ °C is $32$ °F rather than $0$ °F.
The consequences are exactly those from Lesson 1. The ratio $\tfrac{F}{C}$ is not constant: at $100$ °C it is $\tfrac{212}{100} = 2.12$, at $10$ °C it is $\tfrac{50}{10} = 5$, and at $0$ °C it is undefined. Doubling the Celsius value does not double the Fahrenheit value: $10$ °C is $50$ °F but $20$ °C is $68$ °F, not $100$ °F.
So a Celsius-to-Fahrenheit graph needs two known points to draw, because the origin is not one of them. Any two equivalences will do, and the freezing and boiling points of water are the usual choice.
Linear is not the same as proportional, and temperature is the clearest everyday case. Currency with a fixed fee, and taxi fares with a flag fall, are the same phenomenon.
A value read off a graph is an estimate, and saying so is part of a complete answer.
The precision depends on the scale. On a graph where one centimetre represents $10$ km, a reading is good to perhaps half a kilometre, and quoting an answer to three decimal places would be dishonest. Round to a precision the graph can actually support, and use words such as "about" or "approximately".
When exactness matters, use the equation. The graph gives $50$ km as about $31$ miles; the factor $0.6214$ gives $31.07$ miles. Both are right for their purpose, and the graph is faster for repeated rough conversions while the equation is better for a single precise one.
Interpolating, reading a value from inside the drawn range, is reliable. Extrapolating, extending the line beyond its range, is less so. For a pure unit conversion extrapolation is safe, because the relationship really does continue unchanged. For a conversion fitted to data, or one with a stepped structure such as the printing charges in the previous lesson, it may not, and the honest answer is to say over what range the graph is trusted.
Watch Me Solve It · 3 examples
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1(a) Check the two conditionsZero kilometres is zero miles, so the line passes through the origin, and doubling one doubles the other since the units differ only by a fixed multiplier. Both conditions hold.
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2(b) Find the gradient$k = \frac{5}{8} = 0.625$The units are miles per kilometre. The equation is $M = 0.625K$.
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3(c) Choose the points and scalesMark the origin, then plot $(80, 50)$, which is the given equivalence scaled up by $10$. Choosing a point at the far end of the range fixes the line firmly. Scale the horizontal axis to $80$ km and the vertical to $50$ miles so the line runs corner to corner.
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4(d) Convert in both directions$M = 0.625(60) = 37.5$$20 = 0.625K \ \Rightarrow \ K = 32$So $60$ km is $37.5$ miles, and $20$ miles is $32$ km. On the graph these would be read as about $37\tfrac{1}{2}$ and about $32$, which the equation confirms.
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1(a) Test the ratio at two points$\frac{212}{100} = 2.12, \qquad \frac{32}{0} \ \text{undefined}$The ratio is not constant, and at $0$ °C the Fahrenheit value is $32$ rather than $0$, so the line does not pass through the origin. It is linear but not proportional.
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2(b) Find the gradient from the two points$m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = 1.8$Two points are needed here, because the origin is not on the line and so cannot be used as a free second point.
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3(b) Use one point to find the intercept$F = 1.8C + 32$At $C = 0$ the value is $32$, which is the vertical intercept directly.
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4(c) and (d) Substitute, then solve for equality$F = 1.8(25) + 32 = 77$$x = 1.8x + 32 \ \Rightarrow \ -0.8x = 32 \ \Rightarrow \ x = -40$So $25$ °C is $77$ °F, and $-40$ °C is exactly $-40$ °F. That is the single point where the conversion line crosses the line $F = C$.
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1(a) Write the equation and check the origin$U = 0.66A$Zero Australian dollars converts to zero US dollars, and doubling the amount doubles the converted value, so both conditions for direct variation hold and the line passes through the origin.
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2(b) Convert, then round to what a graph can support$U = 0.66(350) = 231$On a graph covering $\$500$, a reading is good to perhaps the nearest $\$5$, so a sensible answer read from the graph is about $\$230$ USD.
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3(c) Judge the over-precise readingA student quoting $\$231.42$ has the arithmetic right, but no graph at that scale distinguishes $\$231.42$ from $\$232$. Quoting a graph reading to the cent claims a precision the method does not have; if that precision is needed, use the equation and say so.
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4(d) Add a fixed fee and see what breaks$U = 0.66A - 8$With an $\$8$ fee charged in USD the line no longer passes through the origin, so the conversion stops being direct variation: the ratio $\tfrac{U}{A}$ now changes with the amount, and two points are needed to draw the graph. The effective rate is worse for small amounts and approaches $0.66$ for large ones.
Brain Trainer · 5 problems
Five items on conversion graphs. Work each one, then reveal the answer.
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1 $1$ kg is about $2.2$ lb. Convert $15$ kg to pounds.
Multiply by the conversion factor.$33$ lb -
2 $1$ kg is about $2.2$ lb. Convert $11$ lb to kilograms.
Divide by the factor, or read the graph the other way.$5$ kg -
3 Does a kilometres-to-miles conversion graph pass through the origin?
Zero of one is zero of the other.Yes -
4 Does a Celsius-to-Fahrenheit graph pass through the origin?
$0$ °C is $32$ °F.No -
5 How many known points are needed to draw a proportional conversion graph?
The origin is free.One, plus the origin
Multiple Choice · 5 questions
A conversion graph between two units that are related by a fixed multiplier is:
On a graph converting kilograms to pounds, the gradient represents:
The conversion $F = 1.8C + 32$ is:
One conversion graph can convert in both directions because:
A value read from a conversion graph should be quoted:
Short Answer · 3 questions
(a) Explain why this conversion is direct variation, checking both conditions.
(b) Write the equation converting cups to millilitres, and state the conversion factor with its units.
(c) Describe how you would draw a conversion graph covering up to $6$ cups, naming the two points you would plot.
(d) Use your equation to convert $2.5$ cups to millilitres, and $600$ mL to cups.
(a) Explain why the origin cannot be used as a free point, unlike in the previous question.
(b) Using $0$ °C $= 32$ °F and $100$ °C $= 212$ °F, find the equation of the line.
(c) Use the equation to find the Fahrenheit values at both ends of the required range, and state the two points you would plot.
(d) A student says "$30$ °C is twice as hot as $15$ °C, so it must be twice as many Fahrenheit degrees". Show that this is false and explain the underlying error.
(a) Find the equation converting litres to gallons, giving the factor to three decimal places.
(b) Use it to convert $65$ litres to gallons.
(c) State how you would quote the answer if it had been read from this graph rather than calculated, and justify the precision you chose.
(d) Explain the difference between interpolating and extrapolating on this graph, and say whether extrapolation would be safe here.
(b) A currency graph converts AUD to USD at $0.66$, and a second converts USD to euros at $0.92$. Find the single factor converting AUD directly to euros, and explain why the combined relationship is still direct variation.
(c) Fuel efficiency in Australia is measured in litres per $100$ km, and in the United States in miles per gallon. Explain why a graph converting between these two is not a straight line, and describe its shape.
One line, both ways
Up and across, or across and down
Usually proportional
Through the origin, gradient is the factor
The exception
Celsius to Fahrenheit is linear, not proportional
Readings are estimates
Quote to what the scale supports
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