SME
Climatology · Problem type
Age of ice at a depth (Nye)
Use this for “An ice sheet 3,000 m thick gets 0.25 m of ice a year. How old is the ice 1,500 m down?”
Chart
t(z) = 3,000/0.25·ln(3,000/(3,000 − z))
The vertical axis grows downward.
This is an example. Type your own values to start a new problem.
t (yr)
=
H (m)
b (m/yr)
×
ln
H (m)
H (m)
−
z (m)
λ
Yearly layer at depth
worked out (not typed) · m
Formulas
t
=
(
H
b
)
×
ln(
H
H
−
z
)
8,317.7662
=
(
3,000
0.25
)
×
ln(
3,000
3,000 − 1,500
)
λ
=
b
×
H
−
z
H
0.125
=
0.25
×
3,000 − 1,500
3,000
Assumptions
•
Steady state: snow adds b meters of ice a year at the top, and the thickness H stays the same.
•
Each year’s layer thins in step with its height above the bed, to nothing at the bed (Nye 1963).
•
Good in the upper part of the sheet; near the bed, melting and folding make the ages unreliable.
Step-by-step
WE KNOW
H = 3,000 m, b = 0.25 m/yr, z = 1,500 m
FIND
age of the ice (t), yearly layer at depth (λ)
Step 1 · Find age of the ice (t)
Use t = (H ÷ b) × ln(H ÷ (H − z))
H
b
is
how
long
the
whole
sheet
takes
to
build;
the
log
stretches
the
years
as
the
layers
thin.
t
=
(
3,000
0.25
)
×
ln(
3,000
3,000 − 1,500
)
3,000
0.25
=
12,000
3,000
3,000 − 1,500
=
2
ln(2) = 0.6931
t
=
8,317.766
yr
Step 2 · Find yearly layer at depth (λ)
Use λ = b × (H − z) ÷ H
The
layer
keeps
the
share
of
its
first
thickness
that
its
height
above
the
bed
is
of
H
.
λ
=
0.25
×
3,000 − 1,500
3,000
λ
=
0.25
×
1,500
3,000
λ
=
375
3,000
λ
=
0.125
m
Check
Put the numbers back in. Both sides should match.
8,317.766
=
(
3,000
0.25
)
×
ln(
3,000
3,000 − 1,500
)
✓
0.125
=
0.25
×
3,000 − 1,500
3,000
✓
Topic check
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