1. Single Point Fixing - Resection
often interchangeably called three-point problem (special case of simple triangulation.)
locates a single point by measuring horizontal angles from it to three visible stations whose positions are known.
weaker solution than intersection

2. Single Point Fixing - Resection
extremely useful technique for quickly fixing position where it is best required for setting-out purposes.
theodolite occupies station P, and angles  and  are measured between stations A and B, and B and C.

3. Single Point Fixing - Resection (Analytical Method)
Let BAP = , then
BCP = (360° -  -  - ) - 
= S - 
 is computed from co-ordinates of A, B and C
 S is known
From PAB,
PB = BA sin  / sin  (1)
From PAB
PB = BC sin(S - ) / sin  (2)

4. Single Point Fixing - Resection (Analytical Method)
Equating (1) and (2)
sin S cot  - cos S = Q
 cot  = (Q + cos S) / sin S
knowing  and (S - ), distances and bearings AP, BP and CP are solved

5. Single Point Fixing - Resection (Analytical Method)
co-ordinates of P can be solved with the three values.
this method fails if P lies on the circumference of a circle passing through A, B, and C, and has an infinite number of positions.

6. Resection - Danger Circle
 + B + ABC (obtuse) = 180  (sum of opposite angles of cyclic quad.)
Accordingly u + v = 180
sin u = sin v, and
(sin u / sin v ) = 1; tan v = 0
at any position along the circumference,the resected station P will have the same angles  and  of the same magnitudes.

7. Resection - Danger Circle
though the computations will always give the x and y coordinates of the resected station, those co-ordinates will be suspect in all probability.
In choosing resection station, care should be exercised such that it does not lie on the ircumference of the "danger circle".

8. Ideal Selection of Existing Control Stations
The best position for station P will be
1) inside the  ABC,
2) well outside the circle which passes through A, B and C,
3) closer to the middle control station.

9. Example: Resection Refer to Figure,  = 41 20’ 35”  = 48 53’ 12”
Control points:
XA = 5,721.25, YA = 21,802.48
XB = 12,963.71, YB = 27,002.38,
XC = 20,350.09, YC = 24,861.22
Calculate the coordinates of P.

10. Example: Resection Dist. BC =7690.46004 Brg. BC = 106-09-56.8
Dist. AB =
Brg. AB =
 = (( )+( )) =
S = (360 -  -  -) =
Q = AB sin /BC sin  =

11. Example: Resection cot = (Q + cos S) / sin S  = 49 -04-15.5
BP = AB sin /sin  =
BP = BC sin (S - ) / sin 
= (checks)
 CBP = [ + (S - ) ]
= °
Brg BP = Brg. BC +  CBP =

12. Example: Resection Ep = EB + BP sin (BRG BP) = 18851.076
Np = NB + BP cos (BRG BP) =
Checks can be made by computing the coordinates of P using the length and bearing of AP and CP.

13. Intersection
used to increase or densify control stations in a particular survey project
enable high and inaccessible points to be fixed.
the newly-selected point is fixed by throwing in rays from a minimum of two existing control stations
these two (or more) rays intersect at the newly-selected point thus enabling its co-ordinates to be calculated.

14. Intersection
field work involves the setting up of the theodolite at each existing control station, back-sighting onto another existing station, normally referred to as the reference object (i.e. R.O.), and is then sighted at the point to be established.
normally a number of sets of horizontal angle measurements made with a second-order theodolite (i.e. capable of giving readings to the nearest second of arc) will be required to give a good fix.
intersection formulae for the determination of the x and y co-ordinates of the intersected point may be easily developed from first principle:

15. Intersection
Let the existing control stations be A(Xa, Ya) and B(Xb,Yb) and from which point P(X, Y) is intersected.
 = bearing of ray AP
 = bearing of ray BP.
It is assumed that P is always to the right of A and B. ( &  is from 0 to 90)

16. Intersection
Similarly

17. Intersection
Similarly

18. Intersection

19. Intersection If the observed angles into P
are used, the equation become
The above equation are also used in the direct solution
of triangulation. Inclusion of additional ray from C,
affords a check on the observation and computation.

20. Where do you want to go ? Global Positioning System
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