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that is, d x and g differ only by a proportionality factor. This is best expressed
in the form
dx
W x
dy
W y
dz
W z
=
=
.
(2-43)
In the coordinate system of Fig. 2.4, the curvature of the projection of
the plumb line onto the xz -plane is given by
κ 1 = d 2 x
dz 2 ;
(2-44)
this is equation (2-26) applied to the present case. Using (2-43), we have
dx
dz
= W W z
.
(2-45)
We differentiate with respect to z , considering that y =0:
d 2 x
dz 2
W z
W xz + W xx
W zz + W zx
.
1
W 2
dx
dz
dx
dz
=
W x
(2-46)
z
In our particular coordinate system, the gravity vector coincides with the
z -axis, so that its x -and y -components are zero:
W x = W y =0 .
(2-47)
Figure 2.4 shows that we also have
dx
dz
=0 .
(2-48)
Therefore,
d 2 x
dz 2
= W z W xz
W 2
= W xz
W z
= W zx
W z
.
(2-49)
z
Considering W z =
g , we finally obtain
∂g
∂x
κ 1 = 1
g
(2-50)
and, similarly,
κ 2 = 1
g
∂g
∂y .
(2-51)
These are the curvatures of the projections of the plumb line onto the xz -and
yz -plane, the z -axis being vertical, that is, coinciding with the gravity vector.
The total curvature κ of the plumb line is given, according to differential
geometry (essentially Pythagoras' theorem), by
κ = κ 1 + κ 2 = 1
g 2
x + g 2
y .
(2-52)
g
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