Transcript Uncertainties in calculated results

Uncertainties in
calculated results
©Gioko
®2007
Uncertainties.

Absolute uncertainty is the value of random
uncertainty reported in a measurement (the
larger of the reading error or the standard
deviation of the measurements). This value
carries the same units as the measurement.

Fractional uncertainty equals the absolute
error divided by the mean value of the
measurement (it has no units).

Percentage uncertainty is the fractional
uncertainty multiplied by 100 to make it a
percentage (it has no units).
©Gioko
®2007
Determine the uncertainties in results.
A simple approximate method rather than root mean squared
calculations is sufficient to determine maximum uncertainties

For functions such as addition and subtraction
absolute uncertainties may be added
When adding or subtracting measurements, the
uncertainty in the sum is the sum of the absolute
uncertainty in each measurement taken
An example: suppose the length of a rectangle is measured
as 26 ± 3 cm and the width is 10 ± 2 cm. The perimeter
of the rectangle is calculated as L + L + W + W = (26 +
26 + 10 + 10) ± (3 + 3 + 2 + 2) = 72 ± 10 cm
©Gioko
®2007
Determine the uncertainties in results.
Cont..
A simple approximate method rather than root mean squared calculations is
sufficient to determine maximum uncertainties
 For multiplication, division and powers
percentage uncertainties may be added
When multiplying and dividing, add the fractional or percentage uncertainties
of the measurements. The absolute uncertainty is then the fraction or
percentage of the most probable answer.
An example: find the area of the same rectangle. Area = L x W. The most probable
answer is 26 x 10 = 260 cm2.
To find the absolute uncertainty in this answer, we must work with fractional
uncertainties.
The fractional uncertainty in the length is 3/26. The fractional uncertainty in the
width is 2/10.
The fractional uncertainty in the area is the sum of these two = 3/26 + 2/10 = 41/130.
The absolute uncertainty in the area is found by multiply this fraction times the
most probable answer = (41/130) x 260 cm2 = 82 cm2. Finally, the area is 260 ±
82 cm2.
©Gioko
®2007
Determine the uncertainties in results.
Cont..
A simple approximate method rather than root mean squared calculations is
sufficient to determine maximum uncertainties
 For other functions (for example, trigonometric functions)
the mean, highest and lowest possible answers may be
calculated to obtain the uncertainty range.
 For functions of the form xn, the fractional uncertainty in the
result is equal to n x fractional uncertainty in x.
 An example: find the volume of a cube with a side of length 5
± .25 cm. Volume = L3. So, the most probable value is 53 =
125 cm3. The fractional uncertainty in the length is .25/5, so
the fractional uncertainty in the volume is 3 x .25/5 = .75/5.
Then, the absolute uncertainty in the volume is (.75/5) x 125 =
18.75 cm3. Finally, the volume is reported as 125 ± 19 cm3.
If one uncertainty is much larger than others, the approximate uncertainty in the
calculated result may be taken as due to that quantity alone.
©Gioko
®2007
Uncertainties in graphs

Based on the ideas above about drawing trend lines it is
possible to draw several different trend lines through the
data. To find the range of values for the slope & intercept
you should draw the trend line with the largest value for
the slope that still crosses all of the error bars and the
trend line with the smallest value for the slope. Then
simply measure the slope and intercept for each of these
new lines. These represent the range of values for the
slope and intercept. Take the average of these two values
as the absolute uncertainty for the value.
©Gioko
®2007
Transform equations into generic straight line
form (y = mx + b) and plot the corresponding graph


To use the slope, intercept, or area under a graph
to determine the value of physical quantities, it
is easiest to work with a straight-line graph.
Most graphs can be transformed from a curved
trend to a linear trend by calculating new
variables.
If the relationship between two variables is
known in theory, then the known relationship
can be used to transform a graph into a straightline graph.
©Gioko
®2007
An example: Suppose that the speed of a falling ball is measured at various
distances as the ball falls from a height of 10 m toward the ground on the
moon. From energy conservation, we can predict that the speed of the ball is
related to the height above the ground as v2 = 2gh. If we were to plot v vs. h we
would get a sideways facing parabola. How can we make this a linear equation?
If we calculate a new quantity (call it w) so that w = v2, then our relationship
becomes w = 2gh. If we plot w vs. h, we will get a straight line. (see the graphs
below). This straight line can be used to find g on the moon (the slope of the
graph = 2g).

speed^2 vs distance fallen
speed^2 (m^2s^2)
speed (ms^-1)
speed vs distance fallen
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5
4
3
2
1
0
40
30
20
10
0
0
2
4
6
distance fallen (m)
8
0
10
©Gioko
®2007
2
4
6
8
distance fallen (m)
10
Analyze a straight-line graph to determine the equation relating the
variables

Once the straight-line graph has been achieved, using
the intercept and slope the equation of the trend line can
be written (y = mx + b). Then substitute into the linear
equation for the quantity that was used as the y and/or x
variable to produce the straight-line graph. In our
example, the equation of the trend line is y = 3.2x. This
means that our relationship is v2 = 3.2h. And, the slope
of the trend line (3.2) should be twice the acceleration
of gravity on the moon, so 2g = 3.2, or g on the moon =
1.2 ms-2.
©Gioko
®2007
Transform equations involving power laws and
exponentials
-into the generic straight line form y = mx + b
plot the corresponding
-log-log
-semi-log graphs from the data
©Gioko
®2007
Analyze log-log and semi-log graphs to
determine the equation relating two variables

The relationship between two variables may
not be readily seen and if the relationship
between the variables is not linear, then it
might be difficult to find the correct
relationship. One method to find the
relationship is guess and check. If y vs. x is
not a straight line, then maybe y vs. x2 or y
vs. x3 might work. Luckily, mathematics
provides us a more reliable method.
©Gioko
®2007
Analyze log-log and semi-log graphs to
determine the equation relating two variables



If the relationship between two quantities, x and y,
is of the form y = kxn, where k is a constant and n
represents some unknown exponent, then we can
use logs (either base 10 logs or ln may be used) to
transform this equation into a linear relationship as
follows.
 take the log of both sides of the equation:
log y = log (kxn)
 use the properties of logs to simplify:
log y = log k + n log x

 substitute log y = Y and log x = X:
Y = log k + n X
©Gioko
®2007
Graph of Y vs. X, we would get a straight
line. AND, the slope of that line is equal to
n, the exponent in our original relationship.
The vertical intercept (b) = log k. The value
of the vertical intercept, b, can be used to
find the constant k: b = log k, so 10b =
10log k = k.

Simply substitute the values of n and k
determined from the transformed log-log graph
into the original relationship to get the equation
relating the two variables, y and x.
©Gioko
®2007
y = kenx
Again, this will produce a curved graph and we can use logs (this time it
has to be ln) to transform the graph into a straight line. Let's see what
logs will do for us this time.
 
take the ln of both sides of the equation:
ln y = ln (kenx)
 
used the properties of ln to simplify:
ln y = ln k + (nx)ln e = ln k + nx
 
substitue ln y = Y
Y = ln k +
nx.
 if we were to make a graph of Y vs x, we would get a straight line.
AND, the slope of that line is equal to n (the constant in the exponent of
our relationship). The vertical intercept (b) = ln k. The value of the
vertical intercept, b, can be used to find the constant k: b = ln k, so eb =
eln k = k.
©Gioko
®2007
This type of graph is called a semi-log graph
because we only had to use the ln function to
transform one of the variables.
Semi-log graphs will transform exponential
relationships into straight-line graphs.
Simply substitute the values of n and k determined
from the transformed semi-log graph into the
original relationship to get the equation relating
the two variables, y and x.
©Gioko
®2007
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©Gioko
®2007