engineering surveying ii - dawood woo resection & intersection: 1 single point fixing -...

Engineering Surveying II - Dawood Woo Resection & Intersection: 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

A

B

C

P


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

A

B

C

P


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)

A

B

C

P


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)

sin(S - )

sin

BA sin

BC sin Q

(sin S cos - cos S sin )

sin Q


A

B

C

P


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.


A

B

C

P


Resection - Danger Circle

Accordingly u + v = 180sin 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.

+ B + ABC (obtuse) = 180 (sum of opposite angles of cyclic quad.)


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".

Resection - 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.

A

B

CP

P

CA

B


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.

A

B

C

P


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

A

B

C

P

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

A

B

C

P

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.

A

B

C

P

Example: Resection


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.


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)

Xa, Ya) B(Xb, Yb)

P(X, Y)N N

Intersection


Xa, Ya) B(Xb, Yb)

P(X, Y)N N tanX - X

Y - Y

Y - Y = X - X

X X Ytan - Y

A

A

A A

A A

tan tan

tan

Similarly

tanX - X

Y - Y

Y - Y = X - X

X X Ytan - Y

B

B

B B

B B

tan tan

tan

Intersection


Similarly

X Ytan - Y = X Ytan - Y

Y(tan - tan X X Ytan - Y

YX X Ytan - Y

(tan - tan

A A B B

B A B

B A B

tan tan

) tan

tan

)

cot = Y - Y

X - X;Y Y Xcot - X cot A

AA A

cot = Y - Y

X - X;Y Y Xcot - X cot B

BB B

Intersection


Y Xcot - X cot = Y Xcot - X cot

X(cot - cot = Y Y + X cot - X cot

X = Y Y + X cot - X cot

(cot - cot

A A B B

B A A B

B A A B

)

)

Intersection


X = Y Y + X cot + X cot

(cot + cot

Y = X X + Y cot + Y cot

(cot + cot

B A A B

A B A B

)

)

A

B

P

C

If the observed angles into Pare 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.

Intersection


Where do you want to go ?

Back to Control Survey - Main Menu

Global Positioning System

engineering surveying ii - dawood woo resection & intersection: 1 single point fixing -...

Documents