SME
Climatology · Problem type
Cost of a finer model grid
Use this for “A climate model has a 1° grid and 50 levels. How many columns and boxes does it have, and how much more does a 0.5° grid cost?”
Table
Δ
(°)
K
Nc
Nb
Δ: Grid spacing
K: Cost, times a 1° grid
Nc: Columns
Nb: Grid boxes
This is an example. Type your own values to start a new problem.
Δ
Grid spacing
example · °
L
Levels
example
Nc
Columns
calculated
Nb
Grid boxes
calculated
K
Cost, times a 1° grid
calculated
Formulas
Nc
=
(
360
Δ
)
×
(
180
Δ
)
64,800
=
(
360
1
)
×
(
180
1
)
Nb
=
Nc
×
L
3,240,000 = 64,800 × 50
K
=
(
1
Δ
)
3
1
=
(
1
1
)
3
Assumptions
•
A latitude–longitude grid of cells Δ degrees on a side: 360 ÷ Δ around, 180 ÷ Δ from pole to pole.
•
Halving Δ gives 4 times the columns, and the time step must halve too, so the cost is (1 ÷ Δ)³ times a 1° grid’s.
•
Whole cells fit only when 180 ÷ Δ is whole (2°, 1°, 0.5°, 0.25°); the cost compares grids with the same levels.
Step-by-step
WE KNOW
Δ = 1°, L = 50
FIND
columns (Nc), grid boxes (Nb), cost, times a 1° grid (K)
Step 1 · Find columns (Nc)
Use Nc = (360 ÷ Δ) × (180 ÷ Δ)
Count the cells around a circle of latitude and from pole to pole, then multiply.
Nc
=
(
360
1
)
×
(
180
1
)
360
1
=
360
180
1
=
180
Nc = 64,800
Step 2 · Find grid boxes (Nb)
Use Nb = Nc × L
Each column is a stack of boxes, one for each level.
Nb = 64,800 × 50
Nb = 3,240,000
Step 3 · Find cost, times a 1° grid (K)
Use K = (1 ÷ Δ)³
Columns
grow
as
(
1
Δ
)
2
and
the
time
steps
as
1
Δ
,
so
the
cost
grows
as
their
product.
K
=
(
1
1
)
3
1
1
=
1
K
=
1
Check
Put the numbers back in. Both sides should match.
64,800
=
(
360
1
)
×
(
180
1
)
✓
3,240,000 = 64,800 × 50 ✓
1
=
(
1
1
)
3
✓
Topic check
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