Satellite Station
Satellite Station
False, eccentric
Sometimes it is impossible to setup an
instrument exactly over the station as well
objects such as.
• Flag Poles etc, which are selected as
• Towers triangulation stations in order to
serve well shaped and well
conditioned triangles or because
of there visibility.
(Cont…)
In such a case a subsidery station is established
as near the true or principal station as possible,
the station is so established is called satellite
station.
Angles are measured at satellite station and
then reduced to what they would have been if
the true station were occupied.
Cases
Case – I
When Satellite Station is in the ∆ A B C
Measure distances CS, AS, BS
Measure Angles α, B Ŝ A, β
Now <A Ĉ B = < A Ŝ B – (α + β)
φ= θ – (α + β)
(Cont…)
Case – II
When Satellite Station is outside the ∆ A B C
Observe distances AS, CS and BS
Observe Angles A Ŝ B, α, β
Now <A Ĉ B = < A Ŝ B + (α + β)
φ= θ + (α + β)
Case – III
(Cont…)
When Satellite Station “s” is outside the ∆ A B C and
angle “β” is negative
φ
θ
φ= θ + (α - β)
Observe distances CS, AS and BS
Observe Angles A Ŝ B, α, β
Now <A Ĉ B = < A Ŝ B + (α - β)
α = Angle between Satellite Station and Flag Station. From
point “ A ”
β = Angle between Satellite Station and Flag Station. From
point “ B ”
Due requirement is to compute angle < A Ĉ B. For that purpose,
look at the ∆ A S B if any angle “ α ” or “ β ” is out of that
triangle. This angle would be negative.
So
<AĈB=<AŜB+(α+β)
If both angles are inside the ∆ A S B.
If α is outside the ∆ A S B, then.
AĈB=<AŜB+(β–α)
If β is outside the ∆ A S B, then.
AĈB=<AŜB+(α–β)
If both outside the ∆ A S B, then.
AĈB=<AŜB–(α+β)
Problem # 1
From an eccentric station S, 10.21 m from
station C, the angles measured to three
trignometrical stations A, B,C are as follows,
the station B and S being on opposite sides of
the line AC:
1. < A Ŝ B = 59° 29’ 40”
2. < C Ŝ B = 72° 23’ 20”
The lengths of AC and BC are 3530 m and 8700
m respectively, Calculate the angle A Ĉ B
Problem # 1
< A Ŝ B = 59º 29’ 40”
AC = 3530 m
BC = 8700 m
AĈB=?
(Cont…)
From Fig. 1
10.21 = 3530
Sin α Sin (131° 53’)
Sin α = 10.21 x Sin (131° 53’)
3530
= .00289 x 0.7445
Sin α = 0.00215
α = Sin-1 (0.00215)
= 0° 7’ 23.8”
(Cont…)
From Fig. 2
10.21 = 8700
Sin β Sin β 72 ° 23’ 30”
Sin β = 10.21 x Sin β 72° 23’ 20”
8700
Sin β = 0.0011185
β = Sin-1 (0.0011185)
= 0° 3’ 50.71”
Now According to Case III
<A Ĉ B = < A Ŝ B + (α - β)
= 59° 29’ 40” + [(0° 7’ 23.8”) – (0° 3’ 50.71”)]
<A Ĉ B = 59° 33’ 13.09”
Problem #2
Directions were observed from a satellite
station, 68 m from station C with following
results;
A, 0o-0’-0”
B, 71o-54’-32.25”
C, 296o-12’-00”
The approximate lengths of AC and BC are
respectively 18024 and 23761 m, compute the
angle subtended at station C.
Problem #2
A
B
α
β
23761 m
φ
C θ
68 m
S
< A Ĉ B = 71° 49’ 46.23”
Class Question
Directions were observed from a satellite
station D , 58.5 m from station B with
following results;
A, 0o-0’-0”
C, 69o-14’-24”
B, 108o-26’-49”
The approximate lengths of AB and BC are
respectively 5771.4 and 11017.8m, compute the
angle ABC. (Ans = 69o-46’-0.94”)
Class Question
(Cont…)
Problem # III
From an eccentric station E, 13.8 m from
station A, the angles measured to three
trignometrical stations A, B,C are as follows,
the station C and E being on opposite sides of
the line AB:
1. < B E C = 68° 26’ 36”
2. < C E A = 32° 45’ 48”
The lengths of AC and AB are 5588.4 m and
4371.0m respectively, Calculate the angle BAC
Problem – III
1. < B E C = 68° 26’ 36”
2. < C E A = 32° 45’ 48”
< B E A = 1 + 2 = 101° 12’ 24”
Sin α = Sin < B E A
EA AB
α = 0° 10’ 38.8”
β = 0° 4’ 35.65”
<BAC=<BEC+(α–β)
= 68° 32’ 39.15”
Assignment
Q No: 1
Directions were observed from a satellite station,
80 m from station C with following results;
A, 0o-0’-0”
B, 72o-50’-44”
C, 290o-22’-00”
The approximate lengths of AC and BC are
respectively 17 Km and 24.15 Km, compute the
angle ACB.
Assignment
Q No: 2 .
From an Satellite station S, 15 m from triangulation
station A, the angles measured to three stations A,
B,C are as follows:
the station C and E being on opposite sides of the line
AB:
1. < C SA = 35° 12’ 55”
2. < B SC = 66° 38’ 40”
The lengths of AC and AB are 5806 m and 1633 m
respectively, Calculate the angle BAC
Thanks
