Chapter
5
Glacier ow
5.1 Glacier ow theories
Wie ist es möglich, wie kommt es zustande, dass dieser scheinbar starre
Körper, das Eis, thalabwärts iesst? (Heim, 1885, p. 290)
Which processes are responsible for glacier ow was a highly debated topic until
about 60 years ago. First written explanations stem from 1705 (by J. J. Scheuchzer).
Many contradicting theories tried to explain how a glacier moves (thorough discussion in Heim (1895), pp. 293 337). According to these theories, a glacier moves
because of ...
•
Dilatation theories
•
water freezing in crevasses and veins within the glacier
growing ice grains push glacier downward
dilatation and contraction due to temperature variations
Gravitation theories
sun light penetrates the ice and uidizes it for a short moment
plasticity, uidity without fracturing
plasticity, uidity with fracturing and regelation
sliding motion over the bedrock
It is interesting to notice that for us most of the above processes sound amusing,
even if all of them (and more) are real.
Now that we have a good theory with
predictive power (although far from perfect) we know which processes dominate,
and which ones are negligible or just boundary eects (e.g. penetration of sunlight).
To understand the foundation of modern glacier ow theories, we rst look at the
structure and deformation of individual ice crystals and small crystal assemblies.
47
Chapter 5
Glacier flow
5.2 Ice physics
Solid ice comes in twelve (!) dierent phases with fundamentally dierent crystal
structure, and two amorphous states, depending on temperature, pressure and crystallization history (Fig. 5.1). Under usual atmospheric conditions only
ice Ih (phase
I in a hexagonal lattice) is formed. (There is evidence that ice Ic with a cubic lattice
is formed in very cold, high altitude clouds.) Glacier ice is a polycrystalline material
Figure 5.1:
Phase diagram of the solid phases of ice (from Petrenko and Withworth,
1999)
that consists of many crystals of dierent sizes and orientation. A single ice crystal
mainly deforms by gliding on its basal planes (which is described by the theory of
dislocations). Typical crystal sizes in glaciers range from the sub-millimeter scale in
freshly formed or highly deformed ice to decimeters (tens of centimeters) in very old
and slow moving ice. If not aected by deviatoric stresses, the cross sectional area
D
of ice grains increases linearly with time
D − D0 = kt,
where the growth factor
k
is mainly temperature dependent. Laboratory experiments of natural and articial
◦
2 −1
ice at −2 C gave k = 85 − 260 mm a
(Azuma and Higashi, 1983).
Polycrystalline ice deforms in creeping ow because of processes within and between
the grains. The most important among them are dislocation climb (within grains),
grain boundary migration, grain rotation, and dynamic recrystallization (formation
48
Physics of Glaciers I
HS 2013
of new grains). Figure 5.2 gives an impression of the behavior of an assembly of ice
crystals under dierent applied forces.
Figure 5.2:
Model of a deforming aggregate of polycrystalline ice. The lines within
c-axis orientations. (a) axial shortening by 29 % (b) axial extension
33 % (c) pure shear by 38 % (d) simple shearing γ = 0.72 (from Zhang et al., 1994)
crystals indicate
by
If stress is applied to a block of glacier ice, the ice immediately deforms elastically.
Permanent deformation termed creep or viscous ow then begins and continues
as long as stress is applied. If the stress is removed, only the elastic deformation
can be recovered.
Viscous ow (creep) is a dissipative process that transforms
mechanical energy into heat which cannot be turned back into mechanical energy.
Depending on the initial crystal structure, measured strain rates vary in laboratory
experiments.
After
1 − 3%
of strain the crystal is completely recrystallized, and
initial anisotropies are no more discernible. Figure 5.3 shows typical creep curves
measured in laboratory experiments. Due to the long time spans and high strains,
only secondary and tertiary creep are usually considered for glacier ow.
49
Chapter 5
Figure 5.3:
Glacier flow
Ice creep curve and rheological model. (a) An idealised creep curve for
ice shows the various stages of strain e versus time t as follows: A-B: the initial
impulsive elastic strain, B-C, C-D, D-E, E-F the primary, secondary, accelerating
and tertiary stages.
(b) The corresponding log strain rate versus log time plot is
shown indicating the minimum and "steady-state" tertiary strain rates.
(c) The
corresponding log strain rate ( log ˙) versus log strain ( log ) shows the same features.
(d) A simple rheological model is indicated which results in the various stages and
components as described above and indicated by the corresponding strains in (e) and
strain rates in (f ). (From Budd and Jacka, 1989).
50
Physics of Glaciers I
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5.3 Flow relation for polycrystalline ice
The most widely used ow relation for glacier ice is (Glen, 1952; Nye, 1957)
(d)
ε̇ij = Aτ n−1 σij .
with
n ∼ 3.
The rate factor
A = A(T )
(5.1)
depends on temperature (Fig. 5.4; Table
A1) and other parameters like water content, impurity content and crystal size. The
quantity
τ
is the second invariant of the deviatoric stress tensor in Equation (4.14).
Several properties of Equation (5.1) are noteworthy:
•
Elastic eects are neglected. This is reasonable if processes on the time scale
of days and longer are considered.
•
Stress and strain rate are collinear, i.e. a shear stress leads to shearing strain
rate, a compressive stress to a compression strain rate, and so on.
•
Only deviatoric stresses lead to deformation rates, isotropic pressure alone cannot induce deformation. Ice is an incompressible material (no volume change,
except for elastic compression). This is expressed as
ε̇ii = 0
•
∂vx ∂vy ∂vz
+
+
=0
∂x
∂y
∂z
⇐⇒
A Newtonian viscous uid, like water, is characterized by the viscosity
ε̇ij =
1 (d)
σ .
2η ij
η
(5.2)
By comparison with Equation (5.1) we nd that viscosity of glacier ice is
η=
•
1
.
2Aτ n−1
Polycrystalline glacier ice is a viscous uid with a
ity
stress dependent viscos-
(or, equivalently, a strain rate dependent viscosity).
Such a material is
called a non-Newtonian uid, or more specically a power-law uid.
•
Polycrystalline glacier ice is treated as an isotropic uid. No preferred direction
(due to crystal orientation fabric) appears in the ow relation. This is a crude
approximation to reality, since glacier ice usually is anisotropic, although to
varying degrees.
Many alternative ow relations have been proposed that take into account the compressibility of rn at low density, the anisotropic nature of ice, microcracks and
damaged ice, the water content, impurities and dierent grain sizes. Glen's ow law
is still widely used because of its simplicity and ability to approximately describe
most processes relevant to glacier dynamics at large scale.
51
Chapter 5
Figure 5.4:
Glacier flow
Composite ow law for minimum strain rates from shear and compres-
sion tests. The minimum octahedral shear strain rate
octahedral shear stress
τ0
-5, -10, -20, -30, -40, -50
˙0
is shown as a function of
on log-log coordinates for the temperatures -0.05, -1, -2,
◦
C over the range of τ0 from 0.2-10 bars. The slopes of
the lines correspond to a power law index of about
1989).
52
n = 3.
(From Budd and Jacka,
Physics of Glaciers I
HS 2013
Inversion of the ow relation
The ow relation of Equation (5.1) can be inverted so that stresses are expressed in
terms of strain rates. Multiplying equation (5.1) with itself gives
(d) (d)
ε̇ij ε̇ij = A2 τ 2(n−1) σij σij
(multiply
by
1
)
2
1
1 (d) (d)
ε̇ij ε̇ij = A2 τ 2(n−1) σij σij
|2 {z }
|2 {z }
˙2
τ2
where we have used the denition for the eective strain rate
the eective shear stress
˙ = ε̇e ,
in analogy to
τ = σe
r
˙ =
1
ε̇ij ε̇ij .
2
(5.3)
This leads to a relation between tensor invariants
˙ = Aτ n .
(5.4)
Coincidentally this is also the equation to describe simple shear, the most important
part of ice deformation in glaciers
(d) n
˙xz = Aσxz
.
(5.5)
Now we can invert the ow relation Equation (5.1)
(d)
σij = A−1 τ 1−n ε̇ij
n−1
n
(d)
σij = A−1 A
(d)
σij
1
−n
=A
˙−
− n−1
n
˙
n−1
n
ε̇ij
ε̇ij .
(5.6)
The above relation allows us to calculate the stress state if the strain rates are known
(from measurements). Notice that only deviatoric stresses can be calculated. The
mean stress (pressure) cannot be determined because of the incompressibility of the
ice. Comparing Equation (5.6) with (5.2) we see that the shear viscosity is
n−1
1
1
η = A− n ˙− n .
2
(5.7)
Polycrystalline ice is a strain rate softening material:
viscosity decreases as the
strain rate increases.
Notice that the viscosity given in Equation (5.7) becomes innite at very low strain
rates, which of course is unphysical. One way to alleviate that problem is to add a
small quantity
ηo
to obtain a nite viscosity
η
−1
=
1 − 1 − n−1
A n ˙ n
2
53
−1
+ η0−1 .
(5.8)
Chapter 5
Glacier flow
5.4 Simple stress states
To see what Glen ow law of Equation (5.1) describes, we investigate some simple,
yet important stress states imposed on small samples of ice, e.g. in the laboratory.
a) Simple shear
(d) 3
3
ε̇xz = A(σxy
) = Aσxy
(5.9)
This stress regime applies near the base of a glacier.
b) Unconned uniaxial compression
along the vertical
z -axis
σxx = σyy = 0
2
(d)
σzz
= σzz ;
3
ε̇xx = ε̇yy
1
(d)
(d)
σxx
= σyy
= − σzz
3
1 3
1
= − ε̇zz = − Aσzz
2
9
2 3
ε̇zz = Aσzz
9
(5.10)
This stress system is easy to investigate in laboratory experiments, and also applies
in the near-surface layers of an ice sheet. The deformation rate is only
22 %
of the
deformation rate at a shear stress of equal magnitude (Eq. 5.9).
c) Uniaxial compression
conned in the
y -direction
σxx = 0;
ε̇yy = 0;
ε̇xx = −ε̇zz
1
1
(d)
σyy
= (2σyy − σzz ) = 0;
σyy = σzz
3
2
1
1
(d)
(d)
σxx
= −σzz
= − (σyy + σzz ) = − σzz
3
2
1 3
ε̇zz = Aσzz
8
(5.11)
This stress system applies in the near-surface layers of a valley glacier and in an ice
shelf occupying a bay.
54
Physics of Glaciers I
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d) Shear combined with unconned uniaxial compression
σxx = σyy = σxy = σyz = 0
2
(d)
(d)
(d)
σzz
= σzz = −2σxx
= −2σyy
3
1 2
2
τ 2 = σzz
+ σxz
3
2
ε̇zz = −2ε̇xx = −2ε̇yy = Aτ 2 σzz
3
ε̇xz = Aτ 2 σxz
(5.12)
This stress conguration applies at many places in ice sheets.
5.5 Field equations
To calculate velocities in a glacier we have to solve eld equations. For a mechanical
problem (e.g. glacier ow) we need the continuity of mass and the force balance
equations. The
mass continuity
equation for a compressible material of density
ρ mass continuit
is (in dierent notations)
∂ρ ∂(ρu) ∂(ρv) ∂(ρw)
+
+
+
=0
∂t
∂x
∂y
∂z
∂ρ
+ ∇ · (ρv) = 0
∂t
(5.13a)
(5.13b)
∂ρ
= 0) and constant (incompressible material ∂ρ
∂xi
∂t
0) we get, in dierent, equivalent notations
If the density is homogeneous (
tr ε̇ = ε̇ii
∇ · v = vi,i
ε̇xx + ε̇yy + ε̇zz
∂u ∂v ∂w
+
+
∂x ∂y
∂z
The
force balance
=
=0
=0
=0
(5.14b)
=0
(5.14d)
(5.14a)
(5.14c)
equation describes that all forces acting on a volume of ice,
including the body force
b = ρg
(where
g
is gravity), need to be balanced by forces
acting on the sides of the volume. In compact tensor notation they read
∇σ + b = 0 ,
(5.15a)
The same equations rewritten in index notation (summation convention)
σij,j + bi =
∂σij
+ bi = 0
∂xj
55
(5.15b)
force balance
Chapter 5
Glacier flow
and in full, unabridged notation
∂σxx ∂σxy ∂σxz
+
+
+ ρgx = 0
∂x
∂y
∂z
∂σyx ∂σyy ∂σyz
+
+
+ ρgy = 0
∂x
∂y
∂z
∂σzx ∂σzy ∂σzz
+
+
+ ρgz = 0
∂x
∂y
∂z
(5.15c)
These three equations describe how the body forces and boundary stresses are balanced by the stress gradients throughout the body.
Recipe
A recipe to calculate the ow velocities from given stresses, strain rates,
symmetry conditions and boundary conditions
1. determine all components of the stress tensor
σij
exploiting the sym-
metries, and using the ow law (5.1) or its inverse (5.6)
2. calculate the mean stress
σm
3. calculate the deviatoric stress tensor
(d)
σij
4. calculate the eective shear stress (second invariant)
5. calculate the strain rates
ε̇ij
from
τ
and
(d)
σij
6. integrate the strain rates to obtain velocities
7. insert boundary conditions
56
τ
using the ow law
Physics of Glaciers I
HS 2013
5.6 Parallel sided slab
Now we are in a position to calculate the velocity of a slab of ice resting on inclined
bedrock. We assume that the inclination angle of surface
same.
For simplicity we choose a coordinate system
x-Axis
is along the bedrock (Figure 5.5).
K
α
and bedrock
β
are the
that is inclined, i.e. the
z
α
w
H
u
ρgh sin α
0
z′
ρgh
τb
β
x′
Figure 5.5:
x
Inclined coordinate system for a parallel sided slab.
Body force
In any coordinate system the body force (gravity) is b = ρg. In the
0
0
untilted coordinate system K the body force is vertical along the ê3 direction
b = −ρg ê03 .
The rotation matrix describes the transformation from
(5.16)
K0
to
K
(Appendix C.1)


cos α 0 − sin α
1
0 
[aij ] =  0
sin α 0 cos α
and therefore
bi = aij b0j
with the components
b1 = a11 b01 + a12 b02 + a13 b03
= cos α · 0 + 0 · 0 − sin α(−g)
= ρg sin α
b2 = 0
b3 = −ρg cos α .
57
(5.17)
Chapter 5
Symmetry
Glacier flow
The problem has translational symmetry in the
x
and
y
direction. It
follows that none of the quantities can change in these directions
∂ (·)
=0
∂x
and
Furthermore no deformation takes place in the
pens in the
x, z
∂ (·)
= 0.
∂y
y
(5.18)
direction, i.e. all deformation hap-
plane. This is called plane strain and leads to the constraints
ε̇yx = ε̇yy = ε̇yz = 0.
Boundary conditions
!
⇐⇒
Σ(n̂) = 0
(5.19)
The glacier surface with the face normal
!
⇐⇒
σ n̂ = 0
 
0
!

σ 0 = 0
1
The boundary conditions at the glacier base are
Solution of the system

n̂ is traction free

σxz
σyz  =! 0 .
σzz
⇐⇒
vx = ub , vy = 0
and
(5.20)
vz = 0.
We now insert all of the above terms into the eld
equations (5.14) and (5.15).
The mass continuity equation (5.14d) together with
(5.18) leads to
0+0+
The velocity component
the base (vz
= 0)
vz
∂vz
= 0.
∂z
(5.21)
is constant. Together with the boundary condition at
we conclude that
vz = 0
everywhere.
Most terms in the momentum balance equation (5.15c) are zero, and therefore
∂σxz
= −ρg sin α ,
∂z
∂σyz
= 0,
∂z
∂σzz
= ρg cos α .
∂z
(5.22a)
(5.22b)
(5.22c)
Integration of Equation (5.22a) and (5.22c) with respect to
z
leads to
σxz = −ρgz sin α + c1 ,
σzz = ρgz cos α + c2 .
The integration constants
c1
c2 can be determined
(zs = H ) and lead to
with the traction boundary
σxz (z) = ρg(H − z) sin α,
σzz (z) = −ρg(H − z) cos α.
(5.23a)
and
conditions (5.20) at the surface
58
(5.23b)
Physics of Glaciers I
HS 2013
We see that the stresses vary linearly with depth.
To calculate the deformation rates we exploit Glen's Flow Law (5.1), and make use
of the fact that the strain rate components are directly related to the deviatoric
(d)
stress components. Obviously we need to calculate the deviatoric stress tensor σ
.
Because of the plain strain condition (Eq. 5.19) and the ow law, the deviatoric
(d)
(d)
stresses in y -direction vanish σiy = 0. By denition σyy = σyy − σm so that
σyy = σm , where use has been made of the denition of the mean stress σm :=
1
σ = 13 (σxx + σyy + σzz ).
3 ii
Using again Glen's ow law we also see that (Eq. 5.18a)
ε̇xx =
∂vx !
=0
∂x
(d)
σxx
=0
leads to
and
σxx = σm .
Therefore all diagonal components of the stress tensor are equal to the mean stress
σm = σxx = σyy = σzz = −ρg(H − z) cos α
(using Eq. 5.23b). The eective stress
τ
can now be calculated with the deviatoric stress components determined above
(d)
(d)
(d)
σxx
= σyy
= σzz
= 0,
(d)
(d)
σxy
= σyz
= 0,
(d)
σxz
= ρg(H − z) sin α,
so that
1 (d) (d) 1
(d) 2
(d) 2
2(σxz
) = (σxz
).
τ 2 = σij σij =
2
2
(5.24)
With Equation (5.23a) we obtain
(d) = ρg(H − z) sin α.
τ = σxz
(5.25)
After having determined the deviatoric stresses and the eective stress, we can
calculate the strain rates. The only non-zero term of the strain rate tensor is
(d)
ε̇xz = Aτ n−1 σxz
= A (ρg(H − z) sin α)n−1 ρg(H − z) sin α
= A (ρg(H − z) sin α)n .
59
(5.26)
Chapter 5
Glacier flow
The velocities can be calculated from the strain rates
ε̇xz
1
=
2
∂vx ∂vz
+
∂z
∂x
where the second term vanishes. Integration with respect to
Z
z
leads to
z
ε̇xz (z) dz
vx (z) = 2
0
=−
2A
(ρg sin α)n (H − z)n+1 + k
n+1
With the boundary condition at the glacier base
constant
k
as
k=
vx (0) = ub
we can determine the
2A
(ρg sin α)n H n+1 + ub
n+1
and nally arrive at the velocity distribution in a parallel sided slab
u(z) = vx (z) =
2A
(ρg sin α)n H n+1 − (H − z)n+1
|n + 1
{z
}
deformation velocity
+
ub
|{z}
(5.27)
sliding velocity
This is Equation (2.11) that we have used before. It is also known as shallow ice
equations , since it can be shown by rigorous scaling arguments that the longitudinal
∂σyi
∂σxi
and
are negligible compared to the shear stress for shallow
stress gradients
∂xi
∂xi
ice geometries such as the inland parts of ice sheets (except for the domes).
60
Physics of Glaciers I
HS 2013
5.7 Flow through a cylindrical channel
R
∆x
r
x
Figure 5.6:
ϕ
Coordinate system for a cross section through a valley glacier.
Most glaciers are not innitely wide but ow through valleys. The valley walls exert
a resistance, or drag on the glacier ow. To see how this alters the velocity eld we
consider a glacier in a semi-circular valley of radius
R
and slope
α.
The body force
from a slice has to be balanced by tractions acting on the circumference in distance
r
from the center (in a cylindrical coordinate system with coordinates
σrx πr∆x = −ρg
x, r
and
ϕ)
πr2
∆x sin α ,
2
1
σrx = − r ρg sin α .
2
Since the only non-zero velocity component is in
(5.28)
x-direction, most components of the
x-direction
strain rate tensor vanish. Under the assumption of no stress gradients in
and equal shear stress along the cylindrical perimeter this leads to
ε̇rr = ε̇xx = ε̇ϕϕ = 0
=⇒
and also
(d)
(d)
(d)
σrr
= σxx
= σϕϕ
=0
(5.29)
(d)
(d)
σrϕ
= σxϕ
=0
(5.30)
The second invariant of the deviatoric stress tensor is
τ = |σxr | = 21 rρg sin α.
To
calculate the velocity we use the ow law (5.1)
1 du
= ε̇xr = Aτ n−1 σxr = −A
2 dr
1
ρg sin α
2
n
rn
0 and R gives
n n+1
1
R
u(0) − u(R) = vx (0) − vx (R) = 2A
ρg sin α
.
2
n+1
Integration with respect to
r
between the bounds
61
(5.31)
Chapter 5
Glacier flow
The channel radius
R
is equivalent to the ice thickness
thus
udef, channel
H
in Equation (5.27) and
n
1
=
udef, slab .
2
(5.32)
The ow velocity on the center line of a cylindrical channel is
eight times slower
than in an ice sheet of the same ice thickness. For further reference we also write
down the velocity at any radius
u(r) = u(0) − 2A
1
ρg sin α
2
The ice ux through a cross section is (for
R
Z
q=
0
n
rn+1
.
n+1
(5.33)
uR = 0)
n Z R
πR2
2A
1
u(r) πr dr = u(0)
rn+2 dr
−
ρg sin α π
2
n+1 2
0
2
2
πR
2
πR
= u(0)
−
u(0)
2 n+3 2
πR2
2
πR2 n + 1
= u(0)
1−
= u(0)
2
n+3
2 n+3
The mean velocity in the cross section is dened by
ū¯ = u(0)
2
q = ū¯ πR2
(5.34)
and is
n+1
n+3
(5.35)
The mean velocity at the glacier surface is
1
ū =
R
Z
0
R
2A
u(r) dr = u(0)
n+1
1
ρg sin α
2
n
Rn+1
n+1
= u(0)
n+2
n+2
(5.36)
5.8 Flow through a parabolic or elliptic channel
Again we assume no stress gradients in
(d)
(d)
(d)
(d)
x-direction, and σxx = σyy = σzz = σyz = 0.
The equation system (5.22c) reduces to
∂σxy ∂σxz
+
= −ρg sin α
∂y
∂z
∂σzz
= −ρg cos α
∂z
From
τ =
p 2
2
σxy + σxz
and the ow law the stresses and the ow velocity
u
can be
calculated. Figures 5.7 and 5.8 show numerical results for ow velocity and stress
from Nye (1965) in dimensionless quantities.
62
Notice that Nye's
Z
is across the
Physics of Glaciers I
glacier, his
Y
is along
z
HS 2013
with opposite sign, and
H
is the maximum ice thickness or
the channel depth
z
H
H −y
Z=
H
u(y, z)
4uslab
σxz
σxz
TY =
=
σxz, slab (z = 0)
ρgH sin α
σxy
TZ =
ρgH sin α
Y =
W =
U=
half width
The shape factor
H
Sf
has been chosen so that the ice ow velocity in the channel
center at the surface is
ucenter =
2A
(Sf ρg sin α)n z n+1 + ub .
n+1
Comparison with the velocity of a circular channel (Eq. 5.33) gives
value (for an ellipse with
(5.38)
Sf = 0.5.
W = 1) and many more for dierent shapes are tabulated
Sf ρg sin α is close to σxz (y = 0, z), i.e. Sf ' Ty .
in Table 5.1. The quantity
SF
Table 5.1:
This
W
parabola
semi-ellipse
rectangle
1
0.445
0.500
0.558
2
0.646
0.709
0.789
3
0.746
0.799
0.884
4
0.806
0.849
∞
1
1
1
Shape factor for the shear stress (from Nye, 1965).
63
Chapter 5
Figure 5.7:
channel (n
Glacier flow
Distribution of velocities and stresses in an elliptical and a parabolic
= 3).
In each graph the top left quadrant is the surface velocity across
the glacier (Z -axis), the top right quadrant is the horizontal velocity with respect
to depth (Y -axis). The bottom left quadrant shows the
glacier, the bottom right quadrant the
TY
TZ
shear stress across the
shear stress with depth. From Nye (1965).
5.9 Example: Black Rapids Glacier
As an example of the real world we look at Black Rapids Glacier, Alaska (all plots
are from Amundson et al. (2006)). Measured surface velocities varied from winter to
summer with lowest velocities in winter and highest velocities in summer (Fig. 5.9,
left).
Inclinometers (tilt sensors) were deployed in ve bore holes along a transsect through
the glacier. They showed periods of unexpected, jerky movement superimposed on
the steady increase of tilt angle (similar observations have been made on Unteraargletscher, (Gudmundsson et al., 1999), Gornergletscher, and others).
Interpretation of the measurements with a numerical model of the ice ow show that
the observed tilt rates can be matched by varying the boundary condition at the
glacier base. Figure 5.9 (right) and Figure 5.10 show a stress redistribution by the
switch from winter conditions to summer conditions. Exceptional speedup events
could be matched by decoupling the glacier around the center line which leads to
highly increased basal shear stresses at sides (Fig. 5.9, right).
The ow velocities change from winter to summer conditions by a great increase
of basal motion, which is clearly visible in Figure 5.11 (left column).
The basal
shear stress is reduced in the center and is transferred to the sides, and maybe also
longitudinally.
64
Physics of Glaciers I
Figure 5.8:
HS 2013
Velocity distribution in a parabolic channel. The velocity at the base is
zero, and maximum at the channel center (left side of graphs). From Nye (1965)
65
Chapter 5
Left: Surface velocity proles on a transsect at Black Rapids Glacier
Figure 5.9:
(Alaska).
Glacier flow
Winter (triangles), spring (sqares) and summer (circles) velocities are
shown. Rigth: Model-derived basal shear stress on the same transsect. There is a
redistribution from winter (dotted) to summer (dashed). Solid lines show the transfer
of shear stress to the sides during a decoupling event. From Amundson et al. (2006)
Figure 5.10:
Stress dierence plots. The winter octahedral stress eld is subtracted
from (a) the spring octahedral stress eld and (b) the summer octahedral stress eld.
The shaded regions indicate areas where stresses were lower in spring and summer
than in winter. The values given on the contours are in units of
100 kPa;
the plots have dierent contour intervals. From Amundson et al. (2006)
66
note that
Physics of Glaciers I
HS 2013
(a-c) Mean modeled-derived winter, spring and summer velocity on a
−1
transsect (units of m a ). (d-f ) Mean winter, spring and summer octahedral stress
Figure 5.11:
elds (×100 kPa) (our
τ ).
From Amundson et al. (2006)
67