Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Kastners-Burkhardt Method - Surveying Computation | SURE 215, Study notes of Engineering

Material Type: Notes; Class: Surveying Computation; Subject: Surveying Engineering; University: Ferris State University; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/07/2009

koofers-user-9q1
koofers-user-9q1 🇺🇸

5

(1)

10 documents

1 / 27

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
THREE POINT RESECTION PROBLEM
Surveying Engineering Department
Ferris State University
INTRODUCTION
The three-point resection problem in surveying involves occupying an unknown point
and observing angles only to three known points. Today, with the advent of total
stations/EDMs, the problem is greatly simplified. If the unknown point P lies on a circle
defined by the three known control points then the solution is indeterminate or not
uniquely possible. There are, theoretically, an infinite number of solutions for the
observed angles. If the geometry is close to this, then the solution is weak. In addition,
there is no solution to this problem when all the points lie on a straight or nearly straight
line. There are a number of approaches to solving the resection problem.
KAESTNER-BURKHARDT METHOD
In the Kaestner-Burkhardt approach [Blachut et al, 1979, Faig, 1972, Kissam, 1981,
Ziemann, 1974] (also referred to as the Pothonot-Snellius method [Allan et. al., 1968])
the coordinates of points A, B, and C are known and the angles α and β measured at
point P. Inversing between the control points we can compute a, b, AzAC, and AzBC using
the following relationships:
Figure 1. Three point resection problem using the Kaestner-Burkhardt method.
8
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b

Partial preview of the text

Download Kastners-Burkhardt Method - Surveying Computation | SURE 215 and more Study notes Engineering in PDF only on Docsity!

THREE POINT RESECTION PROBLEM

Surveying Engineering Department

Ferris State University

INTRODUCTION

The three-point resection problem in surveying involves occupying an unknown point

and observing angles only to three known points. Today, with the advent of total

stations/EDMs, the problem is greatly simplified. If the unknown point P lies on a circle

defined by the three known control points then the solution is indeterminate or not

uniquely possible. There are, theoretically, an infinite number of solutions for the

observed angles. If the geometry is close to this, then the solution is weak. In addition,

there is no solution to this problem when all the points lie on a straight or nearly straight

line. There are a number of approaches to solving the resection problem.

KAESTNER-BURKHARDT METHOD

In the Kaestner-Burkhardt approach [Blachut et al, 1979, Faig, 1972, Kissam, 1981,

Ziemann, 1974] (also referred to as the Pothonot-Snellius method [Allan et. al., 1968])

the coordinates of points A, B, and C are known and the angles α and β measured at

point P. Inversing between the control points we can compute a, b, Az (^) AC , and Az (^) BC using

the following relationships:

Figure 1. Three point resection problem using the Kaestner-Burkhardt method.

( ) ( )

( ) ( )

2 C B

2 C B C B

1 C B BC

2 C A

2 C A C A

1 C A AC

b X X Y Y Y Y

X X

Az tan

a X X Y Y Y Y

X X

Az tan

Compute γ

Compute the auxiliary angles ϕ and θ. First, recognize that the sum of the interior angles

is equal to 360

o [the sum of interior angles of a polygon must equal (n – 2)

o ].

Rearrange

From the sine rule, compute the distance s

Combining these relationships yields

where λ is an auxiliary angle with an uncertainty of ± 180

o

. We then have

or

AC BC

CA CB

Az Az

Az Az

γ= −

o φ+α+β+θ+γ= 360

( ) ( ) (^1)

o

2

φ+θ = − α+β+γ =δ

β

θ

α

φ

sin

bsin and s sin

asin s

= λ β

α

θ

φ cot sin

sin

a

b

sin

sin

= λ θ

φ cot sin

sin

cot 1

cot 1

sin sin

sin sin

λ +

λ−

φ+ θ

φ− θ

Finally, compute the coordinates of point P.

An example, prepared using Mathcad is presented as follows.

Three Point Resection Problem

Kaestner-Burkhardt Method

dd ang( ) degree ←floor ang( )

mins ←(ang −degree) 100.0⋅

minutes ←floor mins( )

seconds ←(mins −minutes) 100.0⋅

degree

minutes

seconds

:= radians ang( ) d ←dd ang( )

d

π

:=

dms ang( ) degree ←floor ang( )

rem ←( ang −degree) 60⋅

mins ←floor rem( )

rem1 ←(rem −mins)

secs ←rem1 60.0⋅

degree

mins

100

secs

10000

:=

trad

π

180

:= (^) tdeg

180

π

:=

________________________________________________________________________

Given

= − θ

= +φ

BP BC

AP AC

Az Az

Az Az

P A 1 AP B 2 BP

P A AP B 2 BP

Y Y c cosAz Y c cosAz

X X csinAz X c sinAz

X (^) A := 1000.00 Y (^) A :=5300.

X (^) B := 3100.00 Y (^) B :=5000.

X (^) C := 2200.00 Y (^) C :=6300.

α := 109.3045 β := 115.

Solution - Find the coordinates of point P using the Kaestner-Burkhardt Method. Begin

by computing the azimuths and distances between the known points.

AzAC atan2 (^) (Y (^) C −YA) ,( XC −XA)    := dms (^) (Az (^) AC) ⋅( tdeg)   =50.

Az := (^)  atan2 (^) ( Y (^) C −YB) ,( XC −XB) Az =−0.

AzBC := Az +( 2 ⋅π) dms (^) ( Az (^) BC) ⋅tdeg =325.

a (^) ( XC −XA)

2 (Y (^) C −YA)

2 := + a =1562.

b ( XC −XB)

2 ( YC −YB)

2 := + b =1581.

The angle at point C is computed as are the auxiliary angles

γ := (^) ( AzAC −AzBC) ⋅ ( tdeg)+ (^360) dms ( )γ (^) =84.

δ 1 180

1

2

:= − ⋅(^ dd (α^ )^ + dd ( )β +γ) dms (δ 1 ) =25.

λ 0

b

a

sin  radians ( (α^ ))

sin  radians (^ ( )β)

:= ⋅ λ 0 =1.

λ tdeg atan

1

(^ λ 0 )

:= dms ( )λ =43.

Note that λ has an uncertainty of 180 degrees

δ 2 atan ( tan radians dms( ( (δ 1 ))))

1

tan radians dms 45(^ (^ (^ +λ)))

:= ⋅tdeg

dms (δ 2 ) =0.

φ := δ 1 +δ 2 dms ( )φ^ =25.

= ϕ ϕ

ϕ

  • = sinScot sin

sinScos K cosS

From which,

Solve for ϕ and then compute c 1 and the azimuth to determine the coordinates of point P.

Alternatively, use line-line intersection to find the coordinates of the unknown point.

Another modification of the Kaestner-Burkhardt Method is that reported by the United

States Coast and Geodetic Survey (USC&GS, now the National Geodetic Survey, NGS)

[Hodgson, 1957; Reynolds, 1934]. Figure 2 identifies three cases of the three point

resection problem. This is a modification of the USC&GS method presented in Kissam

(1981) and with a slight modification in Anderson and Mikhail (1998).

The solution can be broken down into a few steps, given here without derivation.

sinS

K cosS cot

ϕ=

P

(a)

B

C

A

a

P

(b)

b

i

j h

g

A

B

C

b

a

g h

i (^) j

P

(c)

g

i

j

A

B

C

a h b

Figure 2. Three scenarios for the three-point resection problem.

(a) Compute ( g h) 360 ( i j)

o

  • = −α+β+ + if the problem is the same as that

indicated in figure 2(a) and (b). For the configuration depicted in figure 2(c),

(g + h) =(i +j) −(α +β).

(b) Then, define,

( )

1 bsin

asin

bsin

asin

cot 45

o

β

α

β

α

+θ =

where,

α

β θ =

a sin

bsin tan

1

(c) Further,

( ) ( ) (g h) 2

g h cot 45 tan 2

tan

o − = +θ +

(d) Then,

( ) ( ) ( ) ( )

g h g h and h 2

g h g h g

(e) Finally,

i = 180 − (g +α) and j= 180 −(h +β)

o o

Now that all of the angles are known, the lengths of the different legs of the triangles can

be found using the sine law.

From the previous example, we can see that this follows the Case 2 situation shown in

figure 2. For this example we will renumber the points so that they coincide with the

figure for Case 2. Thus, from the original example, point C is now designated as point B

and the original B coordinate is now C. Therefore, the coordinates are:

XA = 1,000.00 YA = 5,300.

XB = 2,200.00 YB = 6,300.

XC = 3,100.00 YC = 5,000.

α = 109° 30' 45" β = 115° 05' 20"

It was already shown that the azimuths are

From the geometry of a circle, shown in figure 4, one can state that the angle formed at a

point on the circumference of a circle subtending a base line on the circle is the same

anywhere on the circle, provided that it is always on the same side of the base line. This

property is exploited in the Collins’ Method.

The solution involves five distinct steps:

  1. Compute the coordinate of the Collins’ Auxiliary Point, H, by intersection

from both control points A and B.

  1. Compute the azimuth Az (^) HC which will also yield the azimuth between C and

P since Az (^) HC = Az (^) CP.

  1. Compute the azimuth of the lines AP and BP
  2. The coordinates can be computed by intersection from A and C and also

from B and C.

  1. If desired, the solution can be performed using the auxiliary angles ϕ and ψ.

Then, using the sine law,

This gives

P A AP AP B BO BP

P A AP AP B BP BP

Y Y D cosAz Y D cosAz

X X D sinAz X D cosAz

Following is a MathCAD program that solves the same problem as presented earlier but

this time using the Collins method.

= + β

= −α

BP CP

AP CP

Az Az

Az Az

BC BP

AP AC

Az Az

Az Az

ψ= −

ϕ= −

( )

( )

β

β+ψ

α

α+ϕ

sin

D sin D

sin

D sin D

BC BP

AC AP

Three Point Resection Problem

Collins Method

See the same functions as defined in the Kaestner-Burkhardt MathCAD program.


Given

X (^) A := 1000.00 Y (^) A :=5300.

X (^) B := 3100.00 Y (^) B :=5000.

X (^) C := 2200.00 Y (^) C :=6300.

α := 109.3045 β := 115.

Solution - Find the coordinates of point P using the Collins Method. Begin by looking at

the triangle ABH Angles are designated by the variable "a" with subscript showing

backsight, station, and foresight lettering.

a (^) BAH := 180 −dd ( )β dms a( (^) BAH) =64.

a (^) ABH := 180 −dd (α ) dms a( (^) ABH) =70.

DAB ( XB −XA)

2 ( YB −YA)

2 := + DAB =2121.

AzAB := atan2 Y( (^) B − YA,XB −XA) dms Az( (^) AB ⋅tdeg) =98.

a (^) AHB := 180 −( 180 −dd ( )β)^ +( 180 −dd (α^ )) dms a( (^) AHB) =44.

CASSINI METHOD

The Cassini approach [Blachut et al, 1979, Faig, 1972, Klinkenberg, 1955, Ziemann,

1974] to the solution of the three-point resection problem is a geometric approach. It

breaks the problem down to an intersection of two circles where one of the intersection

points is the unknown point P while the other is one of the three control points. This is

depicted in figure 5.

The solution is shown as follows:

Compute the coordinates of the auxiliary points H 1 and H 2. First the azimuths between

A and H 1 and B and H 1 are determined.

From triangle ACH 1 , the distance from A to H 1 can be computed.

Figure 5. Three point resection problem as proposed by Cassini.

o BH BC

o AH AC

Az Az 90

Az Az 90

1

1

α

α

α

α=

cosAz tan

Y Y

sinAz tan

X X

tan

D

D

D

D

tan

AC

C A

AC

AC C A AH

AH

AC

1

1

Since the angle at A is 90

o ,

Then,

The coordinates for H 2 are computed in like fashion.

An alternative approach to coming up with the formulas for XH and YH can also be

presented. This approach breaks the solution of the Cassini Method down to 5 equations.

From the equation of the intersections of two lines, we can write:

This can also be written as

But,

Solving these last two equations can be done by subtracting the last equation from the

preceding equation resulting in

sin AzAH 1 =cosAzac ; cosAzAH 1 =−sinAzAC

( )

= + = −( − ) α

= + = + − α

Y Y D cosAz Y X X cot

X X D sinAz X Y Y cot

H A AH AH A C A

H A AH AH A C A

1 1 1

1 1 1

BH BC BH bc

bc

C B

bc

BC C B BH

sinAz cosAz ; cosAz sinAz

cosAz tan

Y Y

sinAz tan

X X

tan

D

D

2 2

2

β

β

β

( ) X (^) C −XB= YC−YBtanAzBC

XC −XB=( YC−YA) tanAzBC+(Y (^) A−YB) tanAzbc

X (^) C −XA=(Y (^) C−YA) tanAzAC

( )

= + = −( − ) β

= + = − − β

Y Y D cosAz Y X X cot

X X D sinAz X Y Y cot

H B BH BH B C B

H B BH bh B C B

2 2 2

2 2 2

The coordinates for H 2 can be developed in a similar fashion and they are given above.

Next, compute the azimuth between the two auxiliary points, H 1 and H 2.

As before, one can write the equation of intersection containing the unknown point P as:

or,

But,

Thus,

where: (^) H P 1

n =tanAz

N =n+ ( 1 /n)

The X-coordinate of the unknown point can be expressed in a similar form as:

2 1

2 1 1 2 H H

1 H H H H Y Y

X X

Az tan

( )

( )

CP H P

C CP H HP C H

CP HP

H CP H HH C CP H CP C H P

1

1 1 1

1

1 1 1 2 1 1

tanAz tanAz

Y tanAz Y tanAz X X

tanAz tanAz

Y tanAz Y tanAz Y tanAz Y tanAz X X Y

H P

CP 1

tanAz

tan Az =−

( ) ( )

CP H P

H C C H CP P H 1

1 1 1 tanAz tanAz

X X Y Y tanAz Y Y −

( )

N

Y X X

n

nY^1 Y

H 1 C C H 1 P

The same problem used in the previous methods follows showing the application of the

Cassini method to solving the resection problem.

Three Point Resection Problem

Cassini Method

See the same functions as defined in the Kaestner-Burkhardt MathCAD program.


Given

X (^) A := 1000.00 Y (^) A :=5300.

X (^) B := 3100.00 Y (^) B :=5000.

X (^) C := 2200.00 Y (^) C :=6300.

α := 109.3045 β :=115.

Solution - Find the coordinates of point P using the Cassini Method.

XH1 := XA +(Y (^) C −YA)cot dd(^ (^ α)^ ⋅trad) XH1 =645.

YH1 := YA +( XA −XC) ⋅cot dd( ( α) ⋅trad) YH1 =5725.

XH2 := XB +(Y (^) B −YC)cot dd( ( )β ⋅trad) XH2 =3708.

YH2 := YB +( XC −XB) ⋅cot dd( ( )β ⋅trad) YH2 =5421.

AzH1H2 := atan2 Y( (^) H2 − YH1,XH2 −XH1) dms Az( (^) H1H2 ⋅tdeg) =95.

n :=tan Az( (^) H1H2)

N n

1

n

:= +

YP

n Y⋅ (^) H

1

n

  • ⋅YC+ XC−XH

N

:= YP =5578.

( )

N

X Y Y

n

nX^1 X

C H 1 C H 1 P

Then,

which upon further manipulation yields

or

Since lines AF and BG are perpendicular to line CF, one can write

From these relationships, equate DAF

and equating the distance DBG

From figure 6 we can also write

( )

( ∠ − θ)

∠ − θ

D cot cot

D cot cot

D D

D D

D

D

n

m

CE B

CE A

DE EB

AE DE

DB

AD

∠ + θ= ∠ − θ

∠ + θ

∠ − θ

mcot mcot ncot n cot

cot cot

cot cot

n

m

B A

B

A

(m +n) cotθ=ncot∠A −mcot∠B

ψ

ψ= ⇒ =

θ

θ

θ= ⇒ =

χ

χ= ⇒ =

cot

D

D

D

D

cot

cot

D

D

D

D

cot

D

D

D

D

cot

cot

D

D

D

D

cot

CG BG BG

CG

GD BG BG

GD

DF AF AF

DF

CF AF AF

CF

θ

χ ⇒ = θ

χ cot

cot

D

D

cot

D

cot

D

DF

CF DF CF

ψ

θ ⇒ = ψ

θ cot

cot

D

D

cot

D

cot

D

CF

GD CG GD

Also, we have,

From above one can see that the distance from C to D can be expressed as

But from figure 6 we can write the following two relationships

Substitute these values for DDF and DDG into the relationships derived above. This is

shown as:

θ

χ− θ

θ

χ − =

θ

χ = − =

cot

cot cot

D

D D

cot

cot D D D

D

cot

D cot D D D

DF

CF DF

CF DF DF

DF

DF CD CF DF

θ

ψ+ θ

θ

ψ

θ

ψ = + =

cot

cot cot

D

D D

cot

cot D

D

cot

D cot D D D

DG

CG DG

DG

DG

DG CD CG DG

θ

χ = 1 cot

cot D (^) CD DDF

θ= ⇒ = θ

θ= = ⇒ = θ

D n cos n

D

cos

D mcos m

D

D

D

cos

DG

DG

DF

DF

AD

DF