Homework 4 Solution:
Age
0 Ma
15 Ma
30 Ma
40 Ma
Latitude
90
81.133
61.795
50.663
Longitude
0
102.682
121.784
143.865
First rotation matrix for 15 Ma is
$ 0.992546 0 0.121869 '
)
!
=&
(1.1)
0
1
0
&
)
&% #0.121869 0 0.9925461 )(
Now we need to keep track of logic a little bit because when B was back at this position, the pole
was at 80N 60E. When the plate moved forward, the apparent pole position moved with it, so we
need to use the inverse of our rotation matrix—we need the one that rotates B forward.
Fortunately this is just negating the angle:
$ 0.992546 0 #0.121869 '
15"0
)
! AB = &
(1.2)
0
1
0
&
)
%& 0.121869 0 0.9925461 ()
0"15
AB
We now multiply this by the vector representing the location of the APW pole on A
[0.0868240888334652
0.150383733180435 0.984807753012208] so pB = !15"0
AB p A = [0.033841 0.15038373 0.9880483] which is 81.133°N 102.682°E.
We repeat for the other two times:
# 0.98159401 -0.06371913 -0.18003629 &
!
= % 0.07850084 0.99399734 0.07620303 (
(1.3)
%
(
%$ 0.1741000 -0.08893343 0.98070384 ('
So the pole at [-0.0593911746138847
0.336824088833465 0.939692620785908] goes to
[-0.248938931494551 0.401747414934245 0.881265239856879] which is latitude 61.795°N
longitude 121.784°E.
30"0
AB
# 0.96449761 -0.13199541 -0.22873908 &
!
= % 0.16037630 0.98088333 0.11021494 (
(1.4)
%
(
%$ 0.20981848 -0.14298638 0.96722857 ('
and the pole is [-0.2716537822741840.323744370967065 0.906307787036650] going to [0.512050200578924 0.373877291834696 0.773317763107403] which is 50.663°N 143.865°E
40"0
AB
To evaluate the second problem, you need to know the directions of the magnetic field at that
point were it on plate A or B. Because we now have the VGPs in modern coordinates, we don't
have to move things back and forth. On plate A, the direction would be D=16.0, I=19.3. On
plate B, D=16.3 and I=-12.0. So the direction reported would be on plate B.