SME
Measurements & Instrumentation
Uncertainty and error propagation
Use this for “A steel cylinder weighs 152.3 ± 0.1 g, is 25.4 ± 0.02 mm across and 40 ± 0.05 mm long. Find its density and the uncertainty.”
Chart
Ringed: Largest term, rₘₐₓ = 0.001575
Mass m = 152.3 g
Mass uncertainty uₘ = 0.1 g
Diameter D = 25.4 mm
Diameter uncertainty uD = 0.02 mm
Length L = 40 mm
Length uncertainty uL = 0.05 mm
Density ρ = 7.5142 g/cm³
Density uncertainty uρ = 0.01589 g/cm³
This is an example. Type your own values to start a new problem.
m
Mass
example
uₘ
Mass uncertainty
example
D
Diameter
example
uD
Diameter uncertainty
example
L
Length
example
uL
Length uncertainty
example
ρ
Density
calculated
uρ
Density uncertainty
calculated
rₘ
Mass term
worked out (not typed)
rD
Diameter term
worked out (not typed)
rL
Length term
worked out (not typed)
rρ
Relative uncertainty
worked out (not typed)
rₘₐₓ
Largest term
worked out (not typed)
Formulas
ρ
=
4
×
m
π ×
D
2
×
L
7.5142
=
4
×
152.3
π ×
2.54
2
× 4
rₘ
=
uₘ
m
0.0006566
=
0.1
152.3
rD
=
2
×
uD
D
0.001575
=
2
×
0.002
2.54
rL
=
uL
L
0.00125
=
0.005
4
rρ
=
√
rₘ
2
+
rD
2
+
rL
2
0.002115
=
√
0.0006566
2
+
0.001575
2
+
0.00125
2
uρ
=
ρ
×
rρ
0.01589 = 7.5142 × 0.002115
rₘₐₓ = max(rₘ, rD, rL)
0.001575 = max(0.0006566, 0.001575, 0.00125)
Formulas are worked in cm. The step-by-step shows the conversions.
Assumptions
•
The errors are random and independent of each other, so they add in quadrature.
•
Each u is one standard uncertainty of its measurement.
•
For y = x₁ᵃx₂ᵇ… each relative term is scaled by its power: D² counts uD ÷ D twice.
Step-by-step
WE KNOW
m = 152.3 g, uₘ = 0.1 g, D = 25.4 mm, uD = 0.02 mm, L = 40 mm, uL = 0.05 mm
FIND
density (ρ), mass term (rₘ), diameter term (rD), length term (rL), relative uncertainty (rρ), density uncertainty (uρ), largest term (rₘₐₓ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to cm).
D = 25.4 mm = 2.54 cm (1 cm = 10 mm)
uD = 0.02 mm = 0.002 cm (1 cm = 10 mm)
L = 40 mm = 4 cm (1 cm = 10 mm)
uL = 0.05 mm = 0.005 cm (1 cm = 10 mm)
Step 2 · Find density (ρ)
Use ρ = 4 × m ÷ (π × D² × L)
The
cylinder’s
volume
is
πD²
L
4,
and
density
is
mass
over
volume.
ρ
=
4
×
152.3
π ×
2.54
2
× 4
ρ
=
4
×
152.3
6.4516 × 4 × π
ρ
=
4
×
152.3
25.8064 × π
ρ
=
4
×
152.3
81.07
ρ
=
609.2
81.07
ρ
=
7.5142
g/cm³
Step 3 · Find mass term (rₘ)
Use rₘ = uₘ ÷ m
m
enters
ρ
to
the
first
power,
so
its
term
is
just
its
relative
uncertainty.
rₘ
=
0.1
152.3
rₘ = 0.0006566
Step 4 · Find diameter term (rD)
Use rD = 2 × uD ÷ D
D
is
squared
in
ρ
,
so
its
relative
uncertainty
counts
twice.
rD
=
2
×
0.002
2.54
rD
=
0.004
2.54
rD = 0.001575
Step 5 · Find length term (rL)
Use rL = uL ÷ L
L
enters
ρ
to
the
first
power,
so
its
term
is
its
relative
uncertainty.
rL
=
0.005
4
rL = 0.00125
Step 6 · Find relative uncertainty (rρ)
Use rρ = √(rₘ² + rD² + rL²)
Independent errors add in quadrature: square each term, add them, take the root.
rρ
=
√
0.0006566
2
+
0.001575
2
+
0.00125
2
rρ
=
√
4.311 ×
10
-7
+ 2.48 ×
10
-6
+ 1.563 ×
10
-6
rρ
=
√
4.474 ×
10
-6
rρ = 0.002115
Step 7 · Find density uncertainty (uρ)
Use uρ = ρ × rρ
The
relative
uncertainty
is
uρ
over
ρ
,
so
scale
it
by
the
density.
uρ = 7.5142 × 0.002115
uρ = 0.01589 g/cm³
Step 8 · Find largest term (rₘₐₓ)
Use rₘₐₓ = max(rₘ, rD, rL)
Squaring makes the largest term dominate the total: improve that measurement first.
rₘₐₓ = max(0.0006566, 0.001575, 0.00125)
rₘₐₓ = 0.001575
Check
Put the numbers back in. Both sides should match.
7.5142
=
4
×
152.3
π ×
2.54
2
× 4
✓
0.0006566
=
0.1
152.3
✓
0.001575
=
2
×
0.002
2.54
✓
0.00125
=
0.005
4
✓
0.002115
=
√
0.0006566
2
+
0.001575
2
+
0.00125
2
✓
0.01589 = 7.5142 × 0.002115 ✓
0.001575 = max(0.0006566, 0.001575, 0.00125) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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