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Single Point Fixing - Resection

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Presentation on theme: "Single Point Fixing - Resection"— Presentation transcript:

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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