When considering how an uncertainty in a measurement will affect
the final result, it is important to remember that what really
matters is that the uncertainty in a given measurement should be a
small fraction of the measurement itself. 

For example, if you write, "I measured the
time to a precision of 0.01s", it
sounds good, unless you then inform your reader that the time
measured was 0.02s! 

The uncertainty is 50% of the measured time so, in reality, the
measurement is useless. 



We will define the quantity relative uncertainty as
follows: 



and to emphasize the difference, we use the term "absolute
uncertainty" where we simply said "uncertainty". 



The most common way to express a relative uncertainty is as a
percentage, in which case the fraction above is simply multiplied by
100. 

Thus, the result of a measurement of, say,
length would be stated as 

L = 15.2mm
±1% 

which would be the equivalent of saying 

L = 15.2mm
±0.152mm 

Therefore we can write 

L = 15.200mm
±0.152mm 



We will now see how to answer the question in the title of this
page. 



It is always possible, in simple situations, to find the effect
on the final result by straightforward calculations but the
following rules can help to reduce the number of calculations needed
in more complicated situations. 



Rule 1 
If a measured quantity is multiplied or divided by a
constant then the absolute uncertainty is multiplied or
divided by the same constant. (In other words the relative
uncertainty stays the same.) Example 
Rule 2 
If two measured quantities are added or subtracted then
their absolute uncertainties are added. Example 
Rule 3 
If two (or more) measured quantities are multiplied or
divided then their relative uncertainties are
added. Example 
Rule 4 
If a measured quantity is raised to a power then the
relative uncertainty is multiplied by that power. (If
you think about this rule, you will realize that it is just
a special case of rule 3.) Example 




A few simple examples might help to illustrate the use of these
rules. (Rule 2 has already been used in the section Using a
Ruler.) 



Example to Illustrate Rule 1 

Suppose that you want to find the average thickness of a page of
a book. We might find that 100 pages of the book have a total
thickness of 9mm. If this
measurement is made using an instrument having a precision of 0.1mm,
we can write 

thickness of 100 pages, T = 9.0mm
± 0.1mm 

and, the average thickness of one
page, t, is obviously given by 

t = T/100 

but the uncertainty is also divided by
100 so our measurement is t = 9/100mm
± 0.1/100mm, or 

t = 0.090±0.001mm 



Example to Illustrate Rule 2 

To find a change in temperature,
ΔT, we find an initial
temperature, T_{1}, a final temperature, T_{2} and
then use ΔT = T_{2}  T_{1} 

If T_{1} is found to be 20°C and if T_{2} is
found to be 40°C then ΔT= 20°C. 

But if the temperatures were measured to a precision of ±1°C
then we must remember that 

19°C < T_{1} <
21°C and 39°C < T_{2} <
41°C 

The smallest difference between the two temperatures is
therefore (39  21) = 18°C and the biggest difference
between them is (41  19) = 22°C 

This means that 

18°C < ΔT
< 22°C 

Therefore, ΔT
= 20°C±2°C 



Example to Illustrate Rule 3 

To measure a surface area, S, we measure two dimensions, say, x
and y, and then use S = xy 

Using a ruler marked in mm, we measure x = 50mm
± 1mm and y = 80mm
± 1mm 

This means that the area could be anywhere between 

(49 × 79)mm^{2}
and (51 × 81)mm^{2} 

that is 

3871mm^{2}
< S < 4131mm^{2} 



To state our answer we now choose the number halfway between
these two extremes and for the indeterminacy we take half of the
difference between them. 

Therefore we have 



so, S = 4000mm^{2}
±130mm^{2} 



Example to Illustrate Rule 4 

To find the volume of a sphere, we first find its radius, r,
(usually by measuring its diameter). 

We then use the formula 



Suppose that the diameter of a sphere is measured (using an
instrument having a precision of ±0.1mm)
and found to be 50mm. 

Diameter = 50.0mm
± 0.1mm 

so 

r = 25.00mm
± 0.05mm 

This means that V could be between 

(4/3)(24.95)^{3}
and (4/3)(25.05)^{3} 

As in the previous example we
now state the final result as 



which gives, V = 65451mm^{3}
± 393mm^{3} 



Check: 

Relative uncertainty in r is
0.05/25 = 0.002 

Relative uncertainty in V is 393/65451 = 0.006 so,
again the theory is verified 
