Global Positioning System Reference
In-Depth Information
Where for
m
=
2, 3;
b
m
+
1
b
2
m
+
1
−
b
m
b
2
m
−
F
m
=
d
2
m
e
m
−
d
2
m
+
1
e
m
+
1
=
F
m
F
m
f
m
d
m
+
1
b
2
m
+
1
−
d
m
b
2
m
−
1
F
m
=
u
m
d
2
m
+
1
−
d
2
m
The unit vector
f
2
lies in the plane through satellites 1, 2 and 3. This plane is spanned by
e
2
and
f
3
is in the plane determined by satellites 1, 3 and 4.
Equation 25 determines the cosine of the two unit vectors
e
3
. Similarly
f
3
with the desired unit vector
e
1
. It will have two solutions, one above and one below the plane spanned by
f
2
and
f
3
. In case
these vectors are parallel their inner product is zero and there are infinitely many solutions
and hence the position cannot be determined.
f
2
and
The algebraic solution to equation 25 can be derived using vector triple product identity,
e
1
×
(
f
1
×
f
2
) =
f
1
(
e
1
·
f
2
)
−
f
2
(
e
1
·
f
1
)
All the terms in the right hand of the above equation is readily computed using
u
2
,
u
3
.
Substituting
h
for the right hand side and
g
for
f
1
×
f
2
,weget
e
1
×
g
=
h
(26)
Multiplying both sides of the equation by
g
and applying the vector triple product identity,
e
1
(
g
·
g
)−
g
(
g
·
e
1
) =
g
×
h
(27)
The scalar product in the second term of the left-hand side can be written in terms of the angle
θ
between unit vector
e
1
and
g
as follows
1
2
cos
= [
g
·
g
]
g
·
e
1
θ
The sine value of the angle can be found from equation 26 as follows:
1
2
1
2
1
2
sin
[
h
·
h
]
= [(
e
1
×
g
)·(
e
1
×
g
)]
= [
g
·
g
]
θ
Using the sine value in the cosine formula above, we obtain,
2
1
1
2
−
h
·
h
g
·
g
1
1
2
= ±[
g
·
g
]
= ±[
g
·
g
−
h
·
h
]
g
·
e
1
Substituting the above into equation 27, we obtain the desired solution:
2
g
×
h
±
g
g
·
g
−
h
·
h
1
e
1
=
(28)
The two values can be put in equation 24 to check the correctness of the value. The correct
parameter will result in a intersection point that lies on the earth's surface and hence must have
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