Equalization is a mathematical analysis of static load-sharing (also called load-distributing) 2-point anchor systems.
Consider the node, where the two anchor legs join with the main line. At this node, the sum of all forces in the x-direction must equal zero since the system is in mechanical equilibrium.
F
x
1
=
F
x
2
F
1
S
i
n
(
α
)
=
F
2
S
i
n
(
β
)
The net force in the y-direction must also sum to zero.
F
y
1
+
F
y
2
=
F
l
o
a
d
Substitute
F
1
from equation (1) into equation (2) and factor out
F
2
.
F
2
S
i
n
(
β
)
S
i
n
(
α
)
C
o
s
(
α
)
+
F
2
C
o
s
(
β
)
=
F
l
o
a
d
F
2
[
S
i
n
(
β
)
S
i
n
(
α
)
C
o
s
(
α
)
+
C
o
s
(
β
)
]
=
F
l
o
a
d
Solve for
F
2
and simplify.
F
2
=
F
l
o
a
d
[
S
i
n
(
β
)
S
i
n
(
α
)
C
o
s
(
α
)
+
C
o
s
(
β
)
]
F
2
=
F
l
o
a
d
[
S
i
n
(
β
)
C
o
s
(
α
)
S
i
n
(
α
)
+
C
o
s
(
β
)
S
i
n
(
α
)
S
i
n
(
α
)
]
F
2
=
F
l
o
a
d
[
C
o
s
(
α
)
S
i
n
(
β
)
+
S
i
n
(
α
)
C
o
s
(
β
)
S
i
n
(
α
)
]
F
2
=
F
l
o
a
d
S
i
n
(
α
)
C
o
s
(
α
)
S
i
n
(
β
)
+
S
i
n
(
α
)
C
o
s
(
β
)
Use a trigonometric identity to simplify more and arrive at our final solution for
F
2
.
Then use
F
2
from equation (3) and substitute into (1) to solve for
F
1
F
1
=
F
l
o
a
d
S
i
n
(
α
)
S
i
n
(
α
+
β
)
S
i
n
(
β
)
S
i
n
(
α
)
Let us now analyze a specific case in which the two anchors are "symmetrical" along the y-axis.
Start by noticing
α
and
β
are the same. Let us start from equation (4) and substitute
β
for
α
and simplify.
F
e
a
c
h
A
n
c
h
o
r
=
F
l
o
a
d
S
i
n
(
α
)
S
i
n
(
α
+
α
)
F
e
a
c
h
A
n
c
h
o
r
=
F
l
o
a
d
S
i
n
(
α
)
S
i
n
(
2
α
)
And using another trigonometric identity we can simplify the denominator.
F
e
a
c
h
A
n
c
h
o
r
=
F
l
o
a
d
S
i
n
(
α
)
2
S
i
n
(
α
)
C
o
s
(
α
)
F
e
a
c
h
A
n
c
h
o
r
=
F
l
o
a
d
2
C
o
s
(
α
)
Note that
α
is half of the angle
θ
between the two anchor points. To express the force at each anchor using the entire angle
θ
, we substitute
α
=
θ
/
2
.
F
e
a
c
h
A
n
c
h
o
r
=
F
l
o
a
d
2
C
o
s
(
θ
2
)
