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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.
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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
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.
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)
Single Point Fixing - Resection (Analytical Method) sin S cot - cos S = Q cot = (Q + cos S) / sin S • knowing and (S - ), distances and bearings AP, BP and CP are solved Equating (1) and (2)
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.
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.
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".
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.
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.
Example: Resection Dist. BC =7690.46004 Brg. BC = 106-09-56.8 Dist. AB =8915.8391 Brg. AB = 54-19-21.5 = 180 - ((106-09-56.8)+(54-19-21.5)) = 128-09-24.6 S = (360 - - -) = 141-36-48.4 Q = AB sin /BC sin =1.322286
Example: Resection cot = (Q + cos S) / sin S = 49 -04-15.5 BP = AB sin /sin = 10197.4831 BP = BC sin (S - ) / sin = 10197.4831 (checks) CBP = 180 - [ + (S - ) ] = 38.5708769° Brg BP = Brg. BC + CBP = 144 - 44 - 12.0
Example: Resection Ep = EB + BP sin (BRG BP) = 18851.076 Np = NB + BP cos (BRG BP) = 18676.061 • Checks can be made by computing the coordinates of P using the length and bearing of AP and CP.
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.
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:
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)
Intersection Similarly
Intersection Similarly
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.
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