Tables and Graphs

Recent additions in red

Table headings and graph axes must be in the form 'quantity divided by unit'. The unit must be the SI unit or a sub-multiple - thus x/mm and x/10^-3m would be equally acceptable. Note that, although it would be logically consistent to head the column x and insert the unit after all the numbers in the body of the table, this is not acceptable for exam purposes.

How many readings?

• At least 10 sets of readings for a graph. (That may involve care in thinking what the interval between readings is to be.) You don't necessarily have to take them in 'axis' order - though it is often inconvenient if you don't. Remember the technique of taking extra readings to fill in 'holes in the graph.
• At least 3 repeats, where appropriate, e.g. when timing something, measuring a dimension, etc.
• At least 20 oscillations (or as many as are feasible if heavily damped) when timing one (remember always to say "zero" (rather than "one"!) as the fiducial mark is passed the first time. You don't have to use the same number of swings each time - though you will need to think about your table design if you don't)

Here is a possible way of setting out a table

Note the consistent number of decimal places (NOT sig figs!) in any one column, including trailing zeros if necessary. The number of decimal places is set by the smallest scale division, unless those divisions are quite big, in which case you might go one further. It would be a good idea to comment (under the table) on the criteria that led you to adopt the choice you have.

Note the use of overaching 'banner' headings for groups of columns to do with the same variable, and that the unit is only included in the banner headings. Check that you understand all features of this table, that you agree the figures in the last three columns, and that the fractional uncertainty in the times for 20 swings are the same as the fractional uncertainties in the periods.

Drawing graphs

For examination purposes, axis scales must be chosen so that at least half of each axis is used. If it is possible, subject to this constraint, to show the origin as well, without having to resort to a 'broken' axis, then do so. Scale divisions and sub-divisions should represent 1, 2 or 5. (Under some circumstances, a scale division might represent 4 - but never 3, 7, etc).

When you have plotted the points (with their uncertainty bars), hold the grid up to your eye, and squint over the surface to see more clearly whether you have a curve or a straight line. Use a plastic ruler to draw the appropriate line (or lines if you are using the 'steepest and shallowest lines' method of determining the fractional uncertainty in the gradient).

Note that once you have drawn the line, it takes over as the authoritative version of the data for all subsequent operations (e.g. interpolations, gradient triangles, etc). Ignore the data points completely: they are only not rubbed out so as to be able to see where the now-operational data came from.

Use the mirror method to draw lines at right angles to curves. Then draw a line perpendicular to this line, which will be parallel to the tangent and, therefore suitable for finding the gradient. DO NOT attempt to do scaled-base-over-scaled-height for the mirror line, and DO NOT attempt to find the inverse of the gradient of the mirror line. These techniques (which work in mathematics) can go terribly wrong with differently-scaled axes that incorporate units in their labels! Any line used for calculating a gradient must cover at least half of one of the axes. Make sure you includ      e a calculation like

gradient = scaled height/scaled base = (8m - 2m)/(0.6s - 0.4s) = 3 m/s

DO NOT include an axis in your gradient triangle. ALWAYS draw an independent gradient triangle 'hanging' off the line. The smallest side of the gradient triangle should be at least 8 cm long, even if the means extending the straight line off which it is hanging beyond the experimental data (use a dotted line for this)

Interpreting graphs

Make sure you read How to draw Conclusions and More advanced graph-plotting

Note that examiners expect you to give the reason (i.e. what's in the bullet point) as well as the conclusion itself.

A straight line indicates linearity (y=mx+c). It's only (directly) proportional if the lines goes through the origin. Direct proportion is also indicated by a constant value of y/x for all data points

Inverse proportion is suggested by a decreasing curve touching neither axis and proved by

• a straight line through the origin for y against 1/x. (A straight line of y against 1/x not through the origin indicates y+c=m/x)
• constant values of yx for all data points
• a straight log-log plot of gradient -1 (log or ln both give this)

A decreasing (or increasing) curve that crosses the y-axis could well be exponential. The point may be tested by

• demonstrating a half-life (or doubling-life). To establish the point, the relevant period needs to persist over several halving (or doubling) cycles. A tripling life, a one third life, a 0.8 life, etc will do equally well.
• showing that ln(y) plotted against x gives a straight line. The gradient is l in y=ke^lx. (Technically, it is possible to do this by plotting log(y), but this in unacceptable for examination purposes.) The intercept on the vertical axis is ln(k), so k=e^intercept.

An increasing curve passing through the origin could well be a power law of the form y=kxⁿ. The point may be tested by

• guessing the value of n and then showing that a graph of y against xⁿ is a straight line through the origin
• showing that log(y) plotted against log(x) is a straight line. The gradient of such a graph gives the value of n, while the vertical axis intercept is log(k), so that k=10^intercept.

When you have done this basic analysis of a graph, you may have discovered a gradient, a power, an intercept, etc. There is then the question of extracting information from that. Suppose you have been connecting wires of various length up to a 2 V cell and measuring the current. You have plotted a graph of length L (vertical) against 1/i, where i is the current. The gradient has turned out to be 5. (The units will be A m.) You are asked to calculate the resistivity r, given that i = VA/rL, where A is the cross-sectional area of the wires (0.1 mm²). Here is what to do

• Re-arrange the equation in the form y=mx (no c in this case): L = VA/r ´ 1/i. 
• Extract the gradient from this equation: gradient = VA/r.
• Make the target quantity the subject of this new equation: r = VA/gradient
• Substitute the values:  2V ´ 0.1 ´ 10^-6 m² / 5 A m = 4 ´ 10^-8 W m

You might also read the opening section of MSD - getting started.