Geoscience Reference
In-Depth Information
20.2.1.2 Shoreline Development Index
The shoreline development index (
D
L
) is a comparative figure relating the shoreline length to the
circumference of a circle that has the same area as the still water body. The smallest possible index
would be 1.0. For the following formula, both
L
and
A
must be in consistent units for this compari-
son—meters and square meters:
L
sA
D
L
=
(20.3)
π
where
D
L
= Shoreline development index.
L
= Length of shoreline (miles or m).
A
= Surface area of lake (acre, ft
2
, m
2
).
20.2.1.3 Mean Depth
The still water body volume divided by its surface area will yield the mean depth. Remember to
keep the units the same. If volume is in cubic meters, then area must be in square meters. The equa-
tion is as follows:
V
A
D
=
(20.4)
where
D
= Mean depth (ft, m).
V
= Volume of lake (ft
3
, acre-ft, m
3
).
A
= Surface area (ft
2
, acre, m
2
).
■
EXAMPLE 20.1
Problem:
A pond has a shoreline length of 8.6 miles. Its surface area is 510 acres. Its maximum
depth is 8.0 feet. The areas for each foot depth are 460, 420, 332, 274, 201, 140, 110, 75, 30, and 1.
Calculate the volume of the lake, shoreline development index, and mean depth of the pond.
Solution
: Compute the volume of the pond:
n
h
(
)
∑
3
V
=
AA
+ +×
AA
i
i
+
1
i
i
+
1
i
=
0
(
)
+
(
)
+
(
)
510
++ ×
460
510
46
0
460
+
420
+
460
×
420
420
+
332
+
420
×
332
(
)
+
(
)
+
(
)
+ 32
3
++ ×
274
332
274
274
+
201
+
274
×
201
201
+ 40
1
+
201
×
140
1
3
=
(
)
+
(
)
+++×
(
)
+++
140
110
140
×
110
110
+
75
+
1
10
×
75
75
30
75
30
(
)
+++×
30
1 00
=× =
13 6823
/
2274
acre-ft
Compute the shoreline development index:
2
1m
640 acres
m
2
A
=
510 acres
=
510 acres
×
=
0.7969
L
860
2314
.
860
316
.
.
D
L
=
=
=
=
272
.
2
π
A
.
×
0 7969
.
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