Appendix C: Propagation of Errors
In many experiments, the quantities measured are not the quantities of final interest. Since all measurements have uncertainties associated with them, clearly any calculated quantity will have an uncertainty that is related to the uncertainties of the direct measurements. The procedure used to estimate the error for the calculated quantities is called the propagation of errors. Consider the general case first. Suppose that the variables A, B, C, ... represent independent measureable quantities that will be used to obtain a value for some calculated quantity U. Since U is a function of A, B, C, ..., it can be written as The measurements of the quantities A, B, C, ... yield estimates of the true values written asA, B, C, ...
and the associated uncertainties, ΔA, ΔB, ΔC, ... for each variable. To find the best estimate for the uncertainty U, each (average) measured value is substituted into the equation for U:
If the errors of A, B, C, ... are independent, random, and sufficiently small, it can be shown that the uncertainty for U is given by
where the partial derivatives are evaluated using the best estimates A, B, C, ...
as the values for the independent variables.
Note:
Don't worry if you are not familiar with the concept of partial derivative. All you will need for the lab write-ups are the results given in Equations (5)
Don't worry if you are not familiar with the concept of partial derivative. All you will need for the lab write-ups are the results given in Equations (5)
ΔU =
, (7)(ΔA)2 + (ΔB)2 + (ΔC)2 + ... |
ΔU = U
, and (9)
|
ΔU = U
n
of this Appendix.
ΔX |
X |
1. Addition and Subtraction
Suppose two quantities A and B are added. If the uncertainties associated with each variable are ΔA and ΔB, then Equation (3)ΔU =
gives
|
U = A + B | ||||
| ||||
ΔU =
| ||||
ΔU =
|
U = A − B
∂U |
∂A |
∂U |
∂B |
ΔU =
(1)2 + (ΔA)2 + (−1)2(ΔB)2 |
ΔU =
(ΔA)2 + (ΔB)2 |
2. Multiplication and Division
Suppose two quantities A and B are multiplied. Then the uncertainty for the resulting value is given byU = AB
∂U |
∂A |
∂U |
∂B |
ΔU =
(B)2(ΔA)2 + (A)2(ΔB)2 |
ΔU = AB
|
ΔU = U
.
|
ΔU =
results in an uncertainty
|
U =
A |
B |
∂U |
∂A |
1 |
B |
∂U |
∂B |
A |
B2 |
ΔU =
|
ΔU =
A |
B |
|
ΔU = U
.
|
3. Exponents and Roots
Suppose a calculation involves the use of exponents. IfU = kX2,
where k is a constant,
then the uncertainty of U is given by
| ||||
ΔU =
| ||||
ΔU = 2kXΔX | ||||
ΔU = U
|
U = KX1/2,
then
| ||||||||
ΔU =
| ||||||||
ΔU =
| ||||||||
ΔU = U
|
Note:
When uncertainties are presented in numerical form, they should be recorded to the same decimal place as the best estimate (i.e., average). The usual rules for significant figures do not apply when calculating uncertainties.
When uncertainties are presented in numerical form, they should be recorded to the same decimal place as the best estimate (i.e., average). The usual rules for significant figures do not apply when calculating uncertainties.
Examples
Example 1
The distance between two points on a straight line is found by recording the position of each point on a meterstick. Let pointX ± ΔX
be located at 23.10 ± 0.05 cm
and point Y ± ΔY
be located at 84.91 ± 0.05 cm.
Then the best estimate for the distance, D, is given by
D = X − Y.
ΔU =
is
(ΔA)2 + (ΔB)2 + (ΔC)2 + ... |
ΔD =
.
(ΔX)2 + (ΔY)2 |
D = 84.91 − 23.10 = 61.81 | ||
ΔD =
| ||
D ± ΔD = 61.81 ± 0.07 cm. |
Example 2
The dimensions of a tabletop are recorded asI = 96.12 ± 0.07 cm
and w = 25.04 ± 0.07 cm.
The surface area of the table is given by
SA = lw.
ΔU = U
is
|
ΔSA = SA
.
|
SA = (96.12)(25.04) = 2407 cm2 | ||||||||||||||
ΔSA = 2407
| ||||||||||||||
ΔSA = 7 cm2 | ||||||||||||||
SA ± ΔSA = 2407 ± 7 cm2. |
Example 3
The mass and volume of an object are measured and reported asM = 121.03 ± 0.08 grams
and V = 13.60 ± 0.09 cm3.
The density is given by
ρ =
M |
V |
ΔU = U
|
Δρ = ρ
| ||||||||||||||
ρ =
| ||||||||||||||
Δρ = 8.899
| ||||||||||||||
ρ ± Δρ = 8.899 ± 0.059
|
Example 4
The volume of a sphere with radiusr = 0.141 ± 0.002 mm
is given by the equation
V =
πr3.
4 |
3 |
ΔU = U
n
, is
ΔX |
X |
ΔV = V
3
.
Δr |
r |
V =
| ||||
ΔV = (0.0117)(3)
| ||||
ΔV = 0.0005 mm3 | ||||
V ± ΔV = 0.0177 ± 0.0005 mm3. |
Example 5
Suppose the acceleration due to gravity can be calculated from experimental data, using the equationa =
2xL |
ht2 |
x = 74.11 ± 0.05 cm |
L = 100.1 ± 0.2 cm |
h = 1.10 ± 0.1 cm |
t = 3.708 ± 0.003 sec. |
ΔU = U
and (9)
|
ΔU = U
n
is
ΔX |
X |
Δa = a
| ||||||||||||||||||||||||||
a =
| ||||||||||||||||||||||||||
Δa = 981
| ||||||||||||||||||||||||||
Δa = 9
| ||||||||||||||||||||||||||
a ± Δa = 981 + 9
|