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How does an uncertainty in a measurement affect the FINAL result?

The measurements we make during an experiment are usually not the final result; they are used to calculate the final result.

When considering how an uncertainty in a measurement will affect the final result, it will be helpful to express uncertainties in a slightly different way. Remember, what really matters is that the uncertainty in a given measurement should be much smaller than the measurement itself. For example, if you write, "I measured the time to a precision of 001s", it sounds good: unless you then inform your reader that the time measured was 002s ! The uncertainty is 50% of the measured time so, in reality, the measurement is useless. We will define the quantity Relative Uncertainty as follows

(to emphasise the difference, we use the term "absolute uncertainty" where previously we simply said "uncertainty").

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

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 realise 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, in fact, already been used in the section "Using a Ruler" on page 3.)

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 01mm, we can write

thickness of 100 pages, T = 90mm 01mm

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

t = T/100

therefore our result can be stated as t = 9/100mm 01/100mm

or t = 0090mm 0001mm

Example to illustrate rule 2

To find a change in temperature, DELTA01T, we find an initial temperature, T1, a final temperature, T2, and then use DELTA01T = T2 - T1

If T1 is found to be 20C and if T2 is found to be 40C then DELTA01T= 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 < DELTA01T < 22C

In other words

DELTA01T = 20C 2C

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 and (51 81)mm

that is

3871mm < S < 4131mm

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 = 4000mm 130mm

(wellactually 4001mm but the "1" is irrelevant when the uncertainty is 130mm).

Now, lets look at the relative uncertainties.

Relative uncertainty in x is 1/50 or 002mm.

Relative uncertainty in y is 1/80 or 00125mm. So, if the theory is correct, the relative uncertainty in the final result should be (002 + 00125) = 00325.


Relative uncertainty in final result for S is 130/4000 = 00325

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: V = (4/3)PIr3

Suppose that the diameter of a sphere is measured (using an instrument having a precision of 01mm) and found to be 50mm.

Diameter = 500mm 01mm

so, ......... r = 250mm 005mm

This means that V could be between

(4/3)PI(2495)3 and (4/3)PI(2505)3

so ...... 65058mm3 < V < 65843mm3

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

V = 65451mm3 393mm3


Relative uncertainty in r is 005/25 = 0002

Relative uncertainty in V is 393/65451 = 0006 so, again the theory is verified

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