Errors in a single measurement

Errors in a large number of measurements of the same quantity

Errors in derived data

Adding or subtracting quantities

Multiplying or dividing quantities

Raising a number to a power
 
 

Here we are dealing with the quantifiable errors, either random or systematic, which occur during any experimental work. These errors are not to be confused with mistakes. For example, a badly plotted point on a graph or a transcription error when taking data from one table to another is a mistake NOT an error.

Errors in a single measurement

Imagine that some sort of super being was capable of measuring the diameter of a disc as 10.34526765867m. If you or I try to measure the diameter of the same disc, we are limited by our ability to read the instrument being used and the smallest division on that instrument.

If we record the diameter to be 10.345 m we are saying that it lies between 10.3445 m and 10.3455 m that is an error of ± 0.0005 m.

The value would then be recorded as 10.345 ± 0.0005 m.

Errors in a large number of measurements of the same quantity

When a number of repeat readings are taken then fluctuations due to the ability of the observer and the limitation of the apparatus can be shown and an estimation of the error made.

Imagine that the following measurements were made for the diameter of a wire with a standard wire gauge of 36.

0.220mm, 0.201mm. 0.190mm. 0.190mm, 0.190mm. 0.191mm. 0.190mm. 0.200mm. 0.191mm. 0.195mm.

First find the mean value, add up all the values and divide by the number of values, this gives a value of 0.1958mm.

To calculate the error we need to calculate the absolute, i.e. ignore any negative signs, difference between the mean value and each individual value. This is the deviation from the mean and the error estimate is the mean of these deviations. A spreadsheet can easily be set up to do this for you.

For our values we get a mean deviation of 0.006720. Since this value is an estimation it is a good idea to keep a careful eye on the number of significant figures. In this instance the value would be best recorded as;

0.1958 ± 0.0067 mm

Errors in derived data

Adding or subtracting quantities
Multiplying or dividing quantities
Raising a number to a power

Once readings have been recorded they are usually put into some formula or equation to generate what is called derived data. We cannot know if the error in the individual quantities are going to cancel each other or compound each other. To be sure we take the more pessimistic of the two. To calculate these errors simple formulae can be used.

 

Adding or subtracting quantities

When adding or subtracting quantities the combined error is found by adding the absolute error.

If c = a + b or c = a - b and the absolute errors in a and b are ± =Δa and ±Δb

then Δc = Δa + Δb where Δc is the absolute error in c

e.g. The four sides of a soccer pitch are measured to be 75.1m. 75.9m, 33.2m and 33.7m each with an error of ± 0.1 m. The perimeter of the pitch is then 217.9 ± 0.4m

Multiplying or dividing quantities

When multiplying or dividing quantities the combined error is found by adding the percentage or fractional error.

If c = a x b or c = a/b and the percentage errors in a and b are ± Da/a and ± Db/bthen Dc/c = Da/a + Db/b where Dc/c is the percentage error in from which Dc the absolute error in c can be calculated. e.g.

A current of 5.0 A which is read to a precision of ± 0.1A flows through a resistor of nominal value 1000. The resistor is said to be accurate to ± 10%. The potential difference across the resistor would be found by;

V = I x R

V = 5.0 x 100 volts

= 500 volts

The percentage error in the value of V would be found by

DV / V= DI /I + DR / R

DV / 500 =0.1 / 5+ 10 / 100

\ DV=60 volts

Therefore we record V = 500 + 60 V as the final answer

Raising a number to a power

If we raise a number to a power it is a special case of multiplying a number by itself and you can easily show that if the percentage error in z is ± Dz / z then the error in z² is ± 2Dz /z. In general we can say that if the percentage error in z is ± Dz then the percentage error in zⁿ is nDz / z where n is a positive number.

e.g. If the mean value for the diameter, d, of a wire is 0.194 ± 0.004mm then the cross-sectional area, A can be found from:
A = pd² / 4
A = (3.142 x O.194² ) /4 = 0.0296 mm²

The easiest way to work out the error is to work out the max and min of the value…

MAX diameter is 0.198 mm
Therefore A(max) is 0.0308 mm²

Similarly MIN diameter is 0.190 mm
Therefore A(min) is 0.0284 mm²

Error is therefore +0.0012 mm²

So the percentage error is (0.0012/0.0296) x 100 = 4.1%

Alternatively you can argue that:

The percentage error in A is equal to the percentage error in d²

Dd² /d = 2Dd /d (two lots because it is squared!) = 2 x 0.004 / 0.194 = 0.0412
= 0.0412 x 100% = 4.12%

Therefore A / A = 0.0412
Which gives A (the absolute error in A) = 0.0012 mm²

Therefore we record A = 0.0296 ± 0.0012 mm²