Errors and Uncertainty

Errors affect the reliability of your data, and it’s important that you are aware of the sources of these errors.  Errors arise from:

• Wrong technique, for example holding a ruler in your hand to measure something that you are holding in the other hand.  The ruler will wobble.

• Positioning your eye to read the scale.  Parallax errors can occur.

• Reaction time in stopping a timer.

• Uncertainty in a reference point.  It is worth having a clear reference point when something is moving.  This is often called a fiducial mark.

The picture shows error from parallax.

Image from Physics for You

 Write down the reading shown by positions 1, 2, and 3 in the picture above.

Parallax errors can be minimised by use of a mirror behind the pointer.  When you cannot see the image of the pointer, you can be sure that you are reading the correct result.

All of these errors are random.  Random errors arise from faulty technique or faults in the equipment.

Random errors can be reduced by taking repeat readings.

 Sometimes random errors are called human error.  Do not use this term in the examination as no credit is given.  This is because the phrase human error is too vague and covers not only faulty technique, but also a whole range of other human frailties, e.g. the pilot who stalls his aeroplane on take-off, or “having one’s fingers in the till”, or “playing away”.   To err is human, to forgive divine.

Systematic errors result in readings all being shifted too high, or too low.  This can be the result of the wrong calibration of the instrument, or a change in the zero point of the instrument.  Look at the picture.

In this case, the zero has shifted.  The zero point on many instruments can be adjusted, for example on this ammeter.

 How much has it shifted by?   How would you reduce the error from this ammeter?

 A zero error does NOT mean that the instrument is super-accurate with no error whatsoever.  It means that all the results are off-set by the same small amount.  If the needle starts at 0.01 A, the readings would all be 0.01 A too high.

Quantification of errors

We have looked at how errors can arise.  Now we need to see how we can quantify the errors, i.e. turn the errors into numbers.

Suppose several students measured the diameter of the same ball bearing with a micrometer.  The results were:

1.21 mm; 1.20 mm; 1.18 mm; 1.25 mm; 1.24 mm; 1.19 mm

The average of these readings is:

1.21 + 1.20 + 1.18 + 1.25 + 1.24 + 1.19 = 1.21 mm

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The range of these readings can be worked out using:

Range = maximum value – minimum value

Range = 1.25 – 1.18 = 0.07 mm

The error (or uncertainty) is found by taking half the range:

Uncertainty = 0.5 × range = 0.5 × 0.07 = 0.035 mm

So we can write:

Diameter = 1.21 ± 0.04 mm (to 2 d.p.)

The diameter of the ball bearing can be confidently measured as:

1.17 mm to 1.25 mm

When you write down such an error remember to write in the units.

If we have several different variables, we need to combine the errors for each one.  Clearly it does not make sense to combine quantities of different units, e.g. millimetres, amps, ohms.

However we can work out the fractional error for all quantities.  These are numbers, so they can easily be combined.

We can work out the fractional error as:

Fractional error = error in the measurement ÷ average value

In this case:

Fractional error = 0.035 ÷ 1.21 = 0.029

You can convert that to a percentage.  In this case, it’s 2.9 %.

When combining errors, the following rules apply:

• Errors add up, regardless of whether the quantities are added, subtracted, multiplied, or divided;

• If a quantity is squared, the error is multiplied by 2.  If it’s cubed, the error is multiplied by 3.

• If a square root is taken, the error is halved.

Let’s look at what happens in a calculation.  Suppose the voltage has an error of 2.5 %, and the current has an error of 3.6 %.  If we calculate the resistance using R = V/I, the error will be:

Error = 2.5 + 3.6 = 6.1 %

And if we calculate the power, we use P = IV:

Error = 2.5 + 3.6 = 6.1 %

 Do not put the error into a formula in this way:  Error in the resistance = V/I = 2.5 ÷ 3.6 = 0.64 %          Error in the power = VI = 2.5 × 3.6 = 9 %