SME
Climatology · Problem type
Warming trend from yearly anomalies
Use this for “Ten yearly temperature anomalies run 0.32, 0.41, 0.35, 0.47, 0.44, 0.52, 0.50, 0.58, 0.55, 0.63 °C. What is the warming trend per decade?”
Chart
Least-squares line: y = 0.0308x − 61.8
Residual = actual − predicted: 5 positive (above the line), 5 negative
Sum of squared residuals ≈ 0.01
The least-squares line makes the sum of the squared residuals as small as it can be
This is an example. Type your own values to start a new problem.
Yearly anomalies
10 numbers · commas between
Y₀
First year
example
m
Trend per year
worked out (not typed) · °C/yr
D
Trend, °C per decade
calculated
Formulas
m
=
12
×
Σ(i
−
n
+ 1
2
)
×
aᵢ
n
× (
n
2
− 1)
0.03085
=
12
×
Σ(i
−
10 + 1
2
)
×
aᵢ
10 × (
10
2
− 1)
D
=
10
×
m
0.3085 = 10 × 0.03085
Assumptions
•
One anomaly a year, in order, with no year missing: the years are evenly spaced.
•
The trend is the least-squares line’s slope: it makes the squared misses from the line as small as they can be.
•
An anomaly is the year’s mean temperature less a fixed base period’s mean, so a trend in anomalies is a trend in temperature.
Step-by-step
WE KNOW
n = 10, a₁ = 0.32 °C, a₂ = 0.41 °C, a₃ = 0.35 °C, a₄ = 0.47 °C, a₅ = 0.44 °C, a₆ = 0.52 °C, a₇ = 0.5 °C, a₈ = 0.58 °C, a₉ = 0.55 °C, a₁₀ = 0.63 °C, Y₀ = 2,015
FIND
trend per year (m), trend, °C per decade (D)
Step 1 · Find trend per year (m)
Use m = 12 × Σ(i − (n + 1) ÷ 2) × aᵢ ÷ (n × (n² − 1))
With one reading a year, the least-squares slope weights each anomaly by its year’s distance from the middle year: the late years pull up, the early years down.
m
=
12
×
((−4.5)
×
a₁
+
(−3.5)
×
a₂
+
(−2.5)
×
a₃
+
(−1.5)
×
a₄
+
(−0.5)
×
a₅
+
0.5
×
a₆
+
1.5
×
a₇
+
2.5
×
a₈
+
3.5
×
a₉
+
4.5
×
a₁₀
)
÷
(
n
×
(
n
2
−
1))
m
=
12
×
((−4.5)
×
0.32
+
(−3.5)
×
0.41
+
(−2.5)
×
0.35
+
(−1.5)
×
0.47
+
(−0.5)
×
0.44
+
0.5
×
0.52
+
1.5
×
0.5
+
2.5
×
0.58
+
3.5
×
0.55
+
4.5
×
0.63)
÷
(10
×
(
10
2
−
1))
−1.44 − 1.435 − 0.875 − 0.705 − 0.22 + 0.26 + 0.75 + 1.45 + 1.925 + 2.835 = 2.545
10
×
(
10
2
−
1)
=
990
12
×
2.545
990
=
0.03085
m
=
0.03085
°C/yr
Step 2 · Find trend, °C per decade (D)
Use D = 10 × m
A decade is ten years of the yearly trend.
D
=
10
×
0.03085
D
=
0.3085
Check
Put the numbers back in. Both sides should match.
0.03085
=
12
×
((−4.5)
×
0.32
+
(−3.5)
×
0.41
+
(−2.5)
×
0.35
+
(−1.5)
×
0.47
+
(−0.5)
×
0.44
+
0.5
×
0.52
+
1.5
×
0.5
+
2.5
×
0.58
+
3.5
×
0.55
+
4.5
×
0.63)
÷
(10
×
(
10
2
−
1))
✓
0.3085 = 10 × 0.03085 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.