Geology Reference
In-Depth Information
20
30
e
2 1
0.97
(20) =
40
30
e
2 1
1.47
(40) =
60
30
e
21
1.73
(60) =
80
30
e
2 1
1.86
(80) =
100
30
e
(100) =
2 1
1.93
Note:
The exponential variogram, in comparison to spherical variogram,
rises very quickly in the beginning. On the contrary it increases very
slowly and reaches the sill value very slowly. Since it is an exponential
function, theoretically it reaches the sill value at infinity only. Therefore
a practical range is defined below.
Question 3.6
Calculate the practical range of the model, i.e. the value of
h
, for which
(
h
) attains 95% of its sill value.
Answer
h
h
a
a
.
( )
h
95%
c
0.95
c
c
1
e
0.95
1
e
h
a
e
1
0.95
0.05
. C B
Note:
One important point to be noted is that in spherical model,
a
represents the sill value at which the (
h
) reaches the maximum value.
Whereas in exponential model, theoretically, the sill value is reached
when
h
=
ha
log(0.05)
ha
3 .
. That is why we are forced to define a working sill value
of 95% value.
Question 3.7
Calculate the slope of the variogram near the origin. Show that the
tangent near the origin cuts the line
Y
=
c
at
h
=
a
.
Answe
r
We can get the slope near the origin by differentiating the Gamma
function and substituting
h
= 0 in the differential.
