Übung zu Globale Geophysik I 01
Übung zu Globale Geophysik I: Wednesday, 16:00 – 18:00, Theresienstr. 41, Room C419
Lecturer: Heather McCreadie
Companion class to Globale Geophysik I: Monday, 12:00 – 14:00, Theresienstr. 41, Room C419
Lecturer: Rocco Malservisi
(1) from Q1: Fowler page 28 (old edition)
All plates A-D shown here move rigidly
without rotation. All ridges add at equal
rates to the plates on either side of them;
the rates given on the diagram are half
the plate separation rates. The trench
forming the boundary of plate A does
not consume A.
Use the plate velocities and
directions to determine by graphical
means or otherwise (a) the relative
motion between plates B and D and (b)
the relative motion between the triple
junction J and plate A. Where and when
will J reach the trench? Draw a sketch
of the geometry after this collision
showing the relative velocity vectors
and discuss the subsequent evolution.
Answer:
Please note: significant figures are to 1 cm or 1 degree and should only be used in the final
answer.
On the diagram below the blue arrows indicate velocity vectors. The velocity of plate D
with respect to C is labelled CVD and so forth. The magnitudes of the three vectors shown
are CVD=10 cm/yr, BVC = 6 cm/yr and BVD= unknown.
C
VD
VD
B
VC
B
25
o
Figure 1. Map of part of a flat planet with velocity vectors.
Page 1 of 11
Übung zu Globale Geophysik I 01
Answer to Q1 continued…
Figure 2. Addition of vectors
(a) BVD = BVC + CVD This is shown graphically in Figure 2.
[5/45]
Obtain BVD using the cosine rule:
2
2
2
B VD = B VC + C VD − 2 B VC CVD cos ( 25 )
= ( 6 ) + (10 ) − 2*6*10*cos ( 25 )
2
2
= 27.24
B VD = 5.22 cm / yr
Obtain the angle from the horizontal in a clockwise motion using the cosine rule:
⎛ V 2+ V 2− V 2⎞
φ = cos −1 ⎜ B D B C C D ⎟
2 B VD BVC
⎝
⎠
⎛ 27.24 + (6) 2 − (10) 2 ⎞
= cos ⎜
⎟
2*5.22*6
⎝
⎠
= 125.93 degrees
−1
(1.1)
(1.2)
Therefore BVD = 5 cm/year at an angle 126 degrees clockwise from the vector
BVC. In terms of a compass rose BVD = 5 cm/year 36 deg NE.
[10/45]
(b) The triple junction J is the point where the three plates (B, C, and D) meet. It is also the
intersection of the boundaries the three pairs BC, CD and BD. The velocity of any point
moving along one of these boundaries will lie on a line in velocity space. The velocity space
diagram for the triple junction J is given in Figure 3. This diagram describes J as the
circumcenter of the triangle BCD.
Page 2 of 11
Übung zu Globale Geophysik I 01
Figure 3. Velocity space showing triple junction position (J). [not to scale]
[10/45]
[You will be given the equations in blue if they are required in the exam.]
Definition: The circumcenter of a triangle can be found as the intersection of the
three perpendicular bisectors. (A perpendicular bisector is a line that forms a
right angle with one of the triangle's sides and intersects that side at its
midpoint.) This is because the circumcenter is equidistant from any pair of the
triangle's points, and all points on the perpendicular bisectors are equidistant
from those points of the triangle.
A circle going through all corners of the triangle with its centre the circumcenter
is called the circumcircle.
We wish to know the position and amplitude (radius) of the circumcenter with
respect to a known reference point in our velocity space.
The diameter of the circumcircle is given by the length of any side of the
triangle, divided by the sine of the opposite angle (the law of sines). The radius
is half this.
The general case for any three non-collinear points in a 2-dimensional plane P1,
P2, and P3 is
⎡ x3 ⎤
⎡ x1 ⎤
⎡ x2 ⎤
⎢
⎥
⎢
⎥
(1.3)
P1 = ⎢ y1 ⎥ , P2 = ⎢ y2 ⎥ , P3 = ⎢⎢ y3 ⎥⎥
⎢⎣ z3 ⎥⎦
⎢⎣ z1 ⎥⎦
⎢⎣ y2 ⎥⎦
The position of the center of the circle is given by
Pc = α P1 + β P2 + γ P3
where
Page 3 of 11
(1.4)
Übung zu Globale Geophysik I 01
( P1 − P2 )i( P1 − P3 )
2
2 ( P1 − P2 ) × ( P2 − P3 )
2
P1 − P3 ( P2 − P1 )i( P2 − P3 )
2
2 ( P1 − P2 ) × ( P2 − P3 )
2
P1 − P2 ( P3 − P1 )i( P3 − P2 )
2
2 ( P1 − P2 ) × ( P2 − P3 )
P2 − P3
α=
β=
γ=
2
(1.5)
If the z-component is always zero then
for α
( x2 − x3 ) + ( y2 − y3 )
( P1 − P2 )i( P1 − P3 ) = ( x1 − x2 )( x1 − x3 ) + ( y1 − y2 )( y1 − y3 )
P2 − P3 =
2
2
( P1 − P2 ) × ( P2 − P3 ) = ( ( x1 − x2 )( y2 − y3 ) − ( y1 − y2 )( x2 − x3 ) )
∴
(( x − x )
α=
2
+ ( y2 − y3 )
2
3
1
(( x − x )
γ=
1
2
3
2
2
) (( x − x )( x − x ) + ( y − y )( y − y ))
1
2
1
3
1
2
2 ( ( x1 − x2 )( y2 − y3 ) − ( y1 − y2 )( x2 − x3 ) )
Likewise
(( x − x )
β=
2
+ ( y1 − y3 )
2
2
1
3
2
(1.6)
) (( x − x )( x − x ) + ( y − y )( y − y ) )
2
1
2
3
2
1
2 ( ( x1 − x2 )( y2 − y3 ) − ( y1 − y2 )( x2 − x3 ) )
+ ( y1 − y2 )
2
2
3
2
) (( x − x )( x − x ) + ( y − y )( y − y ))
3
1
3
2
3
1
2 ( ( x1 − x2 )( y2 − y3 ) − ( y1 − y2 )( x2 − x3 ) )
3
2
2
Substituting in values for our system from Figure 3
P1 = ( x1 , y1 ) = (0.00, 0.00)
P2 = ( x2 , y2 ) = (−6.00, 0.00)
(1.7)
P3 = ( x3 , y3 ) = (3.06, 4.23)
yields
because P1 = 0 there is no need to calcuate α
(( 0 − 3.06)
β=
(( −3.06)
=
2
( ( 0 − −6 )
γ=
2
+ ( 0 − 4.23)
2
) (( −6 − 0)( −6 − 3.06) + ( 0))
2 ( ( 0 − −6 )( 0 − 4.23) − ( 0 ) )
+ ( −4.23)
2
) (( −6)( −9.06) ) = 1.15
2 ( ( 6 )( −4.23) )
2
+ ( 0)
2
2
(1.8)
2
) ((3.06 − 0)(3.06 − −6) + ( 4.23 − 0)( 4.23 − 0))
2 ( ( 0 − −6 )( 0 − 4.23) − ( 0 ) )
(( 6) ) ((3.06)(9.06) + ( 4.23)( 4.23) ) = 1.27
=
2
2 ( ( 6 )( −4.23) )
2
Page 4 of 11
2
Übung zu Globale Geophysik I 01
Inserting these values into (1.4) yields the position of the circumcenter in the velocity space:
Pc = 1.15*[−6.0, 0] + (1.27) *[3.06, 4.23]
= [−6.9, 0] + [3.89,5.37]
= [−3.01, 5.37]
(1.9)
Pc = −3.012 + 5.37 2 = 5.9
⎛ 5.37 ⎞
⎟ = 60.73deg
⎝ 3.01 ⎠
The vector describing this is the red vector shown in Figure 3. You must know how to
DRAW this vector for any system. Therefore you must know how to make Figure 3. The
length and direction of the red vector can also be found using a diagram that is to scale and
measuring the length and angle using a ruler and a protractor. This vector corresponds to
BVJ because the chosen axis intersection is on the vertex between BVC and BVD. If you
redraw your triangle from the perspective of the other plates you would find the triple
junction motion with respect to the other plates.
ϕ = tan −1 ⎜
Therefore BVJ = 6 cm/year at an angle 61 degrees clockwise from the vector
BVC or in terms of a compass rose BVJ = 6 cm/year 29 deg NW.
[5/45]
However, the plate B is subducting towards A at a rate of 8.5 cm/year at an angle of 45
degrees South East. Adding all the vectors pertaining to the motion of the triple junction J
with respect to A gives AVJ = AVB + BVJ. This is shown graphically in Figure 4.
[5/45]
Figure 4. Triple junction with respect to A.
Obtain AVJ using the cosine rule:
2
2
2
A VJ = B VJ + A VB − 2 B VJ AVB cos (α )
= ( 5.9 ) + ( 8 ) − 2*5.9*8*cos ( 60.73 − 45 )
2
2
=8
A VJ = 2.8
Obtain the angle κ using the sine rule:
Page 5 of 11
Übung zu Globale Geophysik I 01
⎛ VJ sin α ⎞
⎟
⎝ AVJ ⎠
κ = sin −1 ⎜ B
⎛ 5.9sin ( 60.73 − 45 ) ⎞
= sin −1 ⎜
⎟
2.8
⎝
⎠
= 35deg
The angle AVJ makes with the horizontal is 90-45-κ = 10 deg. J sits 2500 km due west of
plate A. The total distance J must move to reach the plate A is (2500x105 cm)/cos(10) =
2538.6x105 cm. At a rate of 2.8 cm/year. It will take
7
2538.6x105 (cm) / 2.8 (cm/year ) = 9x10 years (or 90 Million Years) for the
triple junction to reach plate A from its present position.
[5/45]
A sketch of the geometry just after the collision is shown in Figure 5. With this geometry
the ridge at plate C will continue to grow in one direction. Whilst plates B and D will
continue to be consumed until they no longer exist.
[5/45]
Figure 5. Sketch of geometry at collision of triple junction with plate
A.
Page 6 of 11
Übung zu Globale Geophysik I 01
(2) Show a great circle and the polar rotation on the plot.
Total mark 20
[10/20] for each
Answer
Page 7 of 11
Übung zu Globale Geophysik I 01
(3) The rotation vectors for the Australian NZ plate are [latitude (deg), longitude(deg),
angular velocity (deg/Myr)] : [61.40 6.16 1.080]. Below is a map of New Zealand. At
latitude -43 and longitude 170 a plate boundary exists. Determine the relative motion here.
Discuss the nature of the plate boundary.
Total mark (35)
Page 8 of 11
Übung zu Globale Geophysik I 01
Answer
Diagram describing coordinates
[5/35]
Translating the rotation vectors for the Australian NZ plates [lat=61.40 lon=6.16 ω=1.080]
into the above coordinate system gives
Ω x = 1.080*cos ( 61.40 ) *cos(6.16) = 0.5140
Ω y = 1.080*cos ( 61.40 ) *sin(6.16) = 0.0555
Ω x = 1.080*sin ( 61.40 ) = 0.9482
(1.10)
Hence
AU Ω NZ = [0.5140, 0.0555, 0.9482]
[5/35]
Translating the coordinate vectors for the point on the Earth [lat=-43, lon=170 ω=1] into the
above coordinate system gives
Px = 1*cos ( −43) *cos(170) = −0.7202
Py = 1*cos ( −43) *sin(170) = 0.1270
Px = 1*sin ( −43) = −0.682
(1.11)
Hence
P = [−0.7202, 0.1270, − 0.682]
[5/35]
The vector describing the motion at point P is given by
xˆ
yˆ
zˆ
⎛
⎞
⎜
⎟
V = AU Ω NZ × P = det ⎜ 0.5140 0.0555 0.9482 ⎟
⎜ −0.7202 0.1270 −0.6820 ⎟
⎝
⎠
= ( 0.0555* −0.6820 ) − (0.9482*0.1270) xˆ
+ ( 0.9482* −0.7202 ) − (0.5140* −0.6820) yˆ
+ ( 0.5140*0.1270 ) − (0.0555* −0.7202) zˆ
= −0.1583xˆ − 0.3323 yˆ + 0.1052zˆ
Page 9 of 11
(1.12)
Übung zu Globale Geophysik I 01
[5/35]
Now change the coordinate system from an Earth Centred one to a Cartesian one anchored
at our point on the Earth.
[5/35]
Vε 1 = (−0.1583) *cos ( −43) *cos(170)
+(−0.3323) *cos ( −43) *sin(170)
+(0.1052) *sin ( −43)
= 6.7e − 5 ≈ 0
Vε 2 = (−0.1583) *sin ( −43) *cos(170)
+(−0.3323) *sin ( −43) *sin(170)
+(0.1052) *cos ( −43)
(1.13)
= 0.01
Vε 3 = (0.1583) *sin (170 ) / cos(−43) + (−0.3323) *cos (170 ) / cos(−43)
= 0.4850
Hence
Vε = [0.0, 0.01, 0.49]
[5/35]
Azimuth from the North is given by
Page 10 of 11
Übung zu Globale Geophysik I 01
Az = 90 + tan −1 (Vε 2 / Vε 3 )
= 90 + tan −1 (0.01/ 0.49)
= 91.2 deg
The speed is given by the magnitude of the vector V (equation (1.12)).
V = (−0.1583) 2 + ( 0.3323) + ( 0.1052 )
2
(1.14)
2
(1.15)
= 0.3828
The velocity is in units of degrees/Myr. V = 0.3828*111(km / deg) = 42.5km / Myr
Therefore the Australian plate is moving towards the NZ plate with a
velocity 42.5km/Myr in the direction 91 deg E. It is a transpressive
Regime. i.e. strike slip with some compressive component.
[5/35]
There is an error in the formular Vε3
The direction of the East component ε3 is constant with latitude. It only depends on
longitude. Therefore
Vε 3 = −Vx sin(long ) + Vy cos(long )
= (0.1583) *sin (170 ) + (−0.3323) *cos (170 )
(1.16)
= −0.992
Etc…. Please see method #2 for the correct answer to this question.
No marks deducted for following the given formula.
Page 11 of 11
Alternative
method
for
the
solution
of
the
problem
Given
the
Eulerian
pole
of
rotation
of
Australia
with
respect
to
Pacific
Lat
Ω
61.4N
Lon
Ω
6.16E
and
its
magnitude
1.08
°/Myr
First
of
all
we
need
to
transform
the
magnitude
in
radiant’s:
Mag
Ω=
1.08*3.1415/180.0=
0.0188
rad/Myr
(be
sure
to
write
the
units
so
you
can
check
your
results!!!)
Note
that
if
you
do
not
use
radiant’s
and
for
R
the
radius
of
the
Earth,
you
will
have
your
final
answer
not
in
mm/yr
but
in
°/Myr,
To
get
the
answer
in
mm/yr
you
will
need
to
multiply
your
result
by
the
length
in
km
of
1
degree
of
latitude
(111
km).
The
result
will
be
at
that
point
in
km/Myr
that
is
equivalent
to
mm/yr.
Then
we
need
to
write
the
vector
on
its
component
x,y,z
so
that
PA Ω AU
Lat = 61.4N
Lon = 6.16E
Ω = 0.0188
rad
Myr
rad
Ωy = Ωcos(Lat)sin(Lon) = 0.0010
Myr
rad
Ωz = Ωsin(Lat) = 0.0165
Myr
rad
Myr
Ωx = Ωcos(Lat)cos(Lon) = 0.0090
Then
we
have
the
position
of
a
point
R
R : Lat = 43S
€
Lon = 170E
Mag = 6371km
Rx = Rcos(Lat)cos(Lon) = −4588.7km
Ry = Rcos(Lat)sin(Lon) = 809.1km
Rz = Rsin(Lat) = −4345.0km
The
velocity
vector
can
be
computed
as:
  
km
mm
V = Ω × R = (−17.5973, − 36.9613,11.7014)
or
Myr
yr
€
As
seen
in
class
we
have
that
unit
length
vectors
N,u,n,e
can
be
expressed
as
€
€
Nˆ = (0,0,1)
uˆ = (cos(lat)cos(lon), cos(lat)sin(lon), sin(lat))
Nˆ × uˆ
ˆ
e=
Nˆ × uˆ
nˆ =
uˆ × eˆ
uˆ × eˆ
lat = 43S lon = 170E
uˆ = (−0.7202, 0.1270, − 0.6820)
eˆ = (−0.1736, − 0.9848, 0.0)
nˆ = (−0.6716, 0.1184, 0.7314)
The
velocity
in
the
east
and
north
direction
can
be
derived
by:
€
 
mm
Ve = V • e = 39.45
yr
 
mm Vn = V • n = 16.00
yr
Magnitude
and
Azimuth
will
be:

mm
2
2
V = Ve + Vn = 39
yr
 Ve 
Azi = atan  = N68E  Vn 
We
can
plot
the
vector
on
the
map
(since
is
Mercator
projection
the
angle
are
correct!!!)
We
see
that
the
vector
is
mainly
parallel
to
the
plate
boundary
so
it
is
a
transform
fault.
In
reality
when
we
think
at
the
movement
of
the
Western
Province
(that
belong
to
Australian
plate)
with
respect
to
the
Eastern
Province
that
belong
to
the
Pacific
plate
we
see
that
there
is
a
bit
of
compressive
component.
This
is
one
of
the
best
known
examples
in
the
world
of
transpressive
plate
boundary.
The
Alpine
fault
(as
it
is
called
the
fault
that
marks
the
plate
boundary
in
South
Island
of
New
Zealand)
is
mainly
a
strike
slip
fault.
The
compressive
component
is
responsible
for
the
growing
of
the
Southern
Alps.
Note
that
the
direction
of
the
motion
is
not
enough
to
define
the
type
of
the
plate
boundary.
It
is
important
to
look
at
both
at
the
direction
of
motion
and
plate
boundaries.
In
Fiordland
for
example
(the
red
region
in
the
south
west
of
South
island,
the
plate
motion
direction
is
essentially
the
same
but
the
plate
boundary
is
oriented
almost
perpendicular
to
the
motion
of
the
Australian
plate.
Indeed,
this
region
is
a
compressive
region
with
the
Australian
plate
subducting
beneath
Fiordland.