Graphs, conclusions and improvements

Make the axes do useful work

Choose axes that represent the relationship you intend to test and label each with its quantity and unit. Put the deliberately varied quantity, or its chosen transform, on the horizontal axis unless the task specifies otherwise. Use simple scales that spread the data across much of the plotting area; a scale need not begin at zero if doing so would waste most of the graph, but the chosen range must be clear.

Plot points carefully and draw a line or smooth curve appropriate to the physical expectation and observed trend. A best-fit line should not connect every neighbouring point with sharp segments. To find a straight-line gradient, choose two well-separated points on the fitted line and form a large triangle. Those points need not be actual measurements; using the fit rather than one short noisy interval better represents the overall trend.

Calculate vertical change divided by horizontal change, with units following the axis quantities. A force-versus-extension gradient in N per mm must be converted if stiffness is required in N m⁻¹. An inverted axis choice gives a reciprocal physical quantity, not the same slope value with relabelled units. State the coordinates used so another reader can reproduce the calculation and check signs, scale multipliers and conversions.

Gradient = Δy/Δx; 1 N mm⁻¹ = 1000 N m⁻¹.

Symbols and units

• Δx and Δy are differences in axis coordinates, with their respective units
• gradient units are y-unit divided by x-unit.

How the equation explains the idea

1. Use two well-separated points on the best-fit line: gradient = (y₂ − y₁)/(x₂ − x₁). A large triangle reduces the effect of coordinate reading uncertainty.
2. For force against extension in mm, a slope of 0.20 N mm⁻¹ is 200 N m⁻¹ because each millimetre is 0.001 m. Axis scaling belongs in the conversion.

Worked example

A stiffness from a large triangle

1. A fitted force–extension line passes through (2.0 mm, 0.80 N) and (12.0 mm, 4.80 N).
2. Gradient = (4.80 − 0.80)/(12.0 − 2.0) = 0.400 N mm⁻¹.
3. Convert to 400 N m⁻¹ for the stiffness.

The axis scale contributes the factor of 1000.

Keep this idea. Choose readable axes and calculate a large fitted-line gradient with its actual units.

A straight line need not be proportional

A direct proportion has the form y = kx: doubling x doubles y and the ideal graph passes through the origin. A general straight line y = mx + c can have an intercept c, so y/x changes with x even though the gradient stays constant. Inspect both the shape and intercept before using the word proportional. The physical model and the uncertainty in the observations determine whether an apparent small offset is significant.

A transformed graph can test a nonlinear raw relationship. For free fall from rest, s = ½gt² means s against t² should be linear with gradient g/2. Plotting s against t instead would give a curve. An intercept can reveal a possible position offset or a limitation in the assumed release/time origin, but it does not identify a unique cause without further evidence. Do not force a zero intercept merely to make a preferred law look correct.

Extrapolation extends the model beyond measured data and can be useful, as with the zero-current e.m.f. intercept, but its reliability depends on whether the same physical relation remains valid. A straight fit over a limited range cannot prove behaviour far outside it. Examine whether residuals show a systematic curve or time-order trend, which may suggest changing conditions rather than random scatter around a valid line.

Direct proportion: y = kx. General line: y = mx + c. Free-fall transformation: s = (g/2)t².

Symbols and units

• m is gradient and c is intercept in y = mx + c
• s is fall distance in m, t is time in s and g is acceleration in m s⁻².

How the equation explains the idea

1. A straight line with nonzero c is not direct proportion: doubling x gives 2mx + c, not twice mx + c.
2. For release from rest, s = (g/2)t². Plot s against t² to obtain gradient g/2, with units m s⁻²; multiply the fitted gradient by two to estimate g.

Worked example

An offset resistance relation

1. A measured relation is R/Ω = 5.0(L/m) + 0.20.
2. At L = 0.20 m, R = 1.20 Ω; at L = 0.40 m, R = 2.20 Ω.
3. Doubling length does not double measured resistance because the 0.20 Ω offset remains.

The measured relation is linear but not directly proportional.

Keep this idea. Test the predicted shape, slope and intercept, and distinguish interpolation from assumptions beyond the data.

An improvement should solve an observed limitation

A conclusion should answer the original question with a statement supported by the data, including the tested range and relevant uncertainty. If two estimates differ, compare the size of the difference with their stated uncertainty intervals before calling the experiment a failure. Overlapping intervals can indicate consistency under the chosen estimates, but do not prove the absence of systematic bias or validate every model assumption.

Separate random scatter, systematic effects and model limitations. More repeats can help characterise random variation. A checked zero correction addresses an additive calibration offset. A deeper liquid column may allow a falling sphere to reach terminal speed, solving a model-validity problem that repeated timing of the accelerating region would not repair. The best improvement names the cause, changes the method and explains how the change improves the final quantity.

Avoid vague phrases such as human error or use a better instrument unless you identify the mechanism. Manual timing may involve start–stop reaction delay; a ruler may be read with parallax; a wire may warm during a current sweep. A more automated instrument can introduce its own calibration, threshold and sampling limitations. Report what remains uncertain after the proposed improvement and distinguish an internally plausible analysis from actual independent replication or subject review.

Compare discrepancy with stated uncertainty; target the largest justified limitation, not merely the easiest action.

Worked example

Compare two sound-speed estimates

1. One method gives 338 ± 5 m s⁻¹ and another gives 344 ± 4 m s⁻¹.
2. Their intervals are 333–343 and 340–348 m s⁻¹, which overlap.
3. They are consistent at this simple interval level, while shared temperature or calibration assumptions still need checking.

The 6 m s⁻¹ difference alone does not establish a contradiction.

Keep this idea. Connect evidence to a qualified conclusion and choose improvements that address identified causes.