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How Does an Uncertainty in a Measurement Affect the Final Result?
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 = 0.0900.001mm  
   
Example to Illustrate Rule 2  
To find a change in temperature, ΔT, we find an initial temperature, T1, a final temperature, T2 and then use ΔT = T2 - T1  
If T1 is found to be 20C and if T2 is found to be 40C then ΔT= 20C.  
But if the temperatures were measured to a precision of 1C then we must remember that  
19C < T1 < 21C    and    39C < T2 < 41C  
The smallest difference between the two temperatures is therefore (39 - 21) = 18C and the biggest difference between them is (41 - 19) = 22C  
This means that  
18C < ΔT < 22C  
Therefore, ΔT = 20C2C  
   
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)mm2    and    (51 81)mm2  
that is  
3871mm2 < S < 4131mm2  
   
To state our answer we now choose the number half-way between these two extremes and for the indeterminacy we take half of the difference between them.  
Therefore we have  
 
so, S = 4000mm2 130mm2  
   
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 = 65451mm3 393mm3
   
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  
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