Significant Figures

The term significant figures (sf) refers to the number of digits in a measured value. This is determined by the precision of the measuring device. Measurements are recorded in such a way as to indicate the degree of precision (no more, no less), by always reading the device to the nearest tenth of the smallest scale division (also known as the least count). This last digit must be estimated. Measured quantities then consist of digits that are known exactly and one inexact digit.

I. Rules to follow

When you see data which have already been recorded, there are some simple rules to use to tell how many digits are significant.

We will assume that this is the rule to follow unless we know more about the number. 310,000 for example could really have 3 or more significant figures. Where the use of rule A will give the wrong value, we must write the number in scientific notation, and then rule B applies.

For our example from A, if we knew that we really had 3 sf in the number, we would write:
    3.10 x 10⁵
    or for 4 sf 3.100 x 10⁵ 4 sf.

Exact numbers are those not obtained by measurement but by definition or by counting small numbers of objects. They are assumed to have an unlimited number of significant digits. Examples:
6 dogs
the 2 and π in 2πr
1 km is defined to be 1000 m
1 minute is defined to be 60 sec

II. Using significant figures in calculations.

If a result is used in another calculation, one extra digit can be carried into the next calculation to avoid large round-off errors. Most of the time the result of an intermediate calculation will still be displayed on your calculator screen, and can be utilized in the next calculation without retyping. When you get the next result, round to the appropriate number of "sig figs."

For example,
Force = 11.3 lbs. width = 2.34 in. length = 4.59 in.
Pressure = Force/Area

If you are only looking for the pressure you could do this calculation all in one step:
P = F/A = 11.3 lb/(2.34 in. x 4.59 in) = 1.052 p.s.i. = 1.05 p.s.i.
(answer can only have 3 sf)

If you had to report the area separately, then
A = l x w = 2.34 in. x 4.59 in. = 10.74 in² = 10.7 in²
P = F/A = 11.3 lb./10.74 in2 = 1.052 in² = 1.05 in²
(Here I typed in just 10.74.
Usually I would just use the number that's in my calculator display, hit the 1/x key and multiply by 11.3)
note: 11.3 lb./10.7 in² = 1.056 in² = 1.06 in²

As long as your reported results have the correct number of significant figures, it will not matter how you do the calculation. You may differ from your lab partner for example in the last digit, but it should not be more than that when you each start with the same numbers.

Results can contain at most one inexact digit. The last digit retained in an answer shall correspond to the last significant figure in any of the measured quantites. (The final answer stops on the right where any of the data stop.)

ex. 43.4 + 22.35 = 65.75 = 65.8 (add or subtract the numbers and round to the appropriate decimal place.)

III. Cautions

Of course the prime caution is, to quote my former nuclear physics professor, "Don't blow up the lab! Even if you do want to see what happens with a chain reaction." Otherwise, in lab your data (measured values) govern how many significant figures you retain in subsequent calculations. If you don't read the measuring instruments to the proper number of digits (i.e. as precisely as you can) you will lose accuracy (and probably some points on your lab grade). Don't ever round off your data! Changing to a multiple or submultiple of a unit within the metric system will not change the number of significant digits in your data.

For example,
346.36 g = 0.34636 kg not 0.346 kg
32.45 cm = 0.3245 m not 0.32 m

To see how precision can affect accuracy, consider the following example. Observed in a lab as recorded data:
length: 8 cm
width: 4 cm
height: 2.4 cm

Writing the data this way causes the reader to assume that the values were not measured with a standard meter stick or metric ruler, and further that the length and width were measured with one device and the height with another. (You should recall that meter sticks are to be read to the nearest tenth of a millimeter.)

Calculating the volume then gives:
V = lwh = (8 cm) (4 cm) (2.4 cm) = 80 cm³ (can only keep 1 sf)

If the student had recorded the data properly and tried to guess between the millimeter marks as instructed, the measurements might have been 8.01 cm x 4.02 cm x 2.43 cm = 78.2 cm³ (3 sf) Which answer is more accurate?

If the edges of the block had exactly lined up with marks then the measurements should have been recorded as 8.00 cm, 4.00 cm, and 2.40 cm. The calculation would then give 76.8 cm³ Again, which is more accurate 80 cm³ or 76.8 cm³?

Useful Information.

For the Planet Psi insider with the will to succeed in College Physics....

Conversion factors
between British and metric units.

1 inch = 2.54 cm exactly

1 m/s = 3.6 km/h

1 kg weighs 2.2 lbs. on Earth

1 mi = 1.609 km

1 qt. = 946 mL

1 m = 3.28 ft.

I expect you to know how to convert within the British system and within the metric system. See the scientific notation page for metric prefixes.

1 kg of water occupies a volume of 1 L

density of water = 1 g/cm³

1 mL = 1 cm³

sine = opposite/hypotenuse

cosine = adjacent/hypotenus

tangent = opposite/adjacent

Remember SohCahToa!