8 three point resection problem -...

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

SURE 215 – Surveying Calculations Three Point Resection Problem Page 176

( ) ( )

( ) ( )2BC

2BC

BC

BC1BC

2AC

2AC

AC

AC1AC

YYXXbYYXX

tanAz

YYXXaYYXX

tanAz

−+−=

−−

=

−+−=

−−

=

−

−

Compute γ

Compute the auxiliary angles ϕ and θ. First, recognize that the sum of the interior angles is equal to 360o [the sum of interior angles of a polygon must equal (n – 2)180o].

Rearrange

From the sine rule, compute the distance s

Combining these relationships yields

where λ is an auxiliary angle with an uncertainty of ±180o. We then have

or

BCAC

CBCA

AzAz

AzAz

−=

−=γ

o360=γ+θ+β+α+φ

( ) ( ) 1o

21180

21 δ=γ+β+α−=θ+φ

βθ=

αφ=

sinsinbsand

sinsinas

λ=βα=

θφ cot

sinsin

ab

sinsin

λ=θφ cot

sinsin

1cot1cot

sinsinsinsin

+λ−λ=

θ+φθ−φ


Since 1

cotcot λ=λ and using trigonometric theorems, one can write

But, recognizing that cot 45o = 1 and

Therefore,

Then,

Recall that δ2 has an uncertainty of ±180o due to the uncertainty in λ. Next, using the sine rule, compute the distances c1 and c2.

If λ was picked in the right quadrant then γ2 is in the right quadrant and c1 and c2 are positive. If they turn out to be negative, δ2, φ, and θ have to be changed by 180o. As a check, recall that α + β + γ + φ + θ =360°. The next step is to compute the azimuths to point P.

( ) ( )

( ) ( )o

o

45cotcot1cot45cot

21cos

21sin2

21sin

21cos2

+λ−λ=

θ−φθ+φ

θ−φθ+φ

( ) ( ) ( ) ( )λ+δ=λ+θ−φ=θ−φ o1

o 45cottan45cot21tan

21tan

( ) ( )[ ] 2o

11 45cottantan

21 δ=λ+δ=θ−φ −

21

21

δ−δ=θ

δ+δ=φ

( )[ ] ( )

( )[ ] ( )β

θ+β=β

θ+β−=βγ=

αφ+α=

αφ+α−=

αγ=

sinsinb

sin180sinb

sinsinbc

sinsina

sin180sina

sinsinac

o2

2

o1

1


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⋅←

degreeminutes

60.0+

seconds3600.0

+

:= radians ang( ) d dd ang( )←

dπ

180.0⋅

:=

dms ang( ) degree floor ang( )←

rem ang degree−( ) 60⋅←

mins floor rem( )←

rem1 rem mins−( )←

secs rem1 60.0⋅←

degreemins100

+secs10000

+

:=

tradπ

180:=

tdeg

180π

:=

________________________________________________________________________ Given

θ−=

φ+=

BCBP

ACAP

AzAz

AzAz

BP2BAP1AP

BP2BAPAP

AzcoscYAzcoscYY

AzsincXAzsincXX

+=+=

+=+=


XA 1000.00:= YA 5300.00:= XB 3100.00:= YB 5000.00:= XC 2200.00:= YC 6300.00:= α 109.3045:= β 115.0520:= 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 YC YA−( ) XC XA−( ), := dms AzAC( ) tdeg( )⋅ 50.11399= Az atan2 YC YB−( ) XC XB−( ), := Az 0.60554−= AzBC Az 2 π⋅( )+:= dms AzBC( ) tdeg⋅ 325.18174= a XC XA−( )2 YC YA−( )2+:= a 1562.04994= b XC XB−( )2 YC YB−( )2+:= b 1581.13883= The angle at point C is computed as are the auxiliary angles γ AzAC AzBC−( ) tdeg( )⋅ 360+:= dms γ( ) 84.53225=

δ1 18012

dd α( ) dd β( )+ γ+( )⋅−:= dms δ1( ) 25.15163=

λ0ba

sin radians α( )( ) sin radians β( )( )

⋅:=

λ0 1.053482162=

λ tdeg atan1λ0( )

:= dms λ( ) 43.30291=

Note that λ has an uncertainty of 180 degrees

δ2 atan tan radians dms δ1( )( )( )( ) 1tan radians dms 45 λ+( )( )( )

⋅

tdeg⋅:=

dms δ2( ) 0.4214= φ δ1 δ2+:= dms φ( ) 25.57303=


θ δ1 δ2−:= dms θ( ) 24.33022= Compute the distances between the point P and control points A and B

c1 asin radians α( ) φ trad⋅( )+

sin radians α( )( )⋅:= c1 1162.1655=

c2 bsin radians β( ) θ trad⋅( )+

sin radians β( )( )⋅:= c2 1130.60883=

The azimuths between the control points A and B are now determined AzAP AzAC φ trad⋅+:= dms AzAP tdeg⋅( ) 76.09102= AzBP AzBC θ trad⋅−:= dms AzBP tdeg⋅( ) 300.45152= Finally, the coordinates of the unknown point are computed from both points for a check XP XA c1 sin AzAP( )⋅+:= XP 2128.390=

YP YA c1 cos AzAP( )⋅+:= YP 5578.144= Check XP XB c2 sin AzBP( )⋅+:= XP 2128.390=

YP YB c2 cos AzBP( )⋅+:= YP 5578.144= Allan et. Al. [1968] present a slightly different approach called the Pothonot-Snellius method. Recall that the distance from C to P was designated as s and was expressed

asβθ==

αφ

sinsinbs

sinsina . From this there are two methods of solving this problem. The first

method is basically that already presented above. The second method is described as follows. Write the ratio of ϕ to θ by a constant K as:

where ( )γ+β+α−= o360S . This relationship is based on the fact that the sum of the interior angles in polygon ACBPA must equal 360o. Thus, one can write from this basic relationship (refer to figure 1): ( )[ ] ϕ−=ϕ−γ+φ+α−=θ S360o . S represents the known angles. Manipulation of this last relationship yields

( ) Scossin

cosSsinsin

sinScoscosSsinsin

SsinsinsinK −

ϕϕ=

ϕϕ−ϕ=

ϕϕ−=

ϕθ=


ϕ=ϕ

ϕ=+ cotSsinsin

cosSsinScosK

From which,

Solve for ϕ and then compute c1 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.

SsinScosKcot +=ϕ

P(a)

BC

A

a

P

(b)

b

i

jh

g

A

B

C

b ag

h

i j

P

(c)

g

ij

AB

C

hab

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


(a) Compute ( ) ( )ji360hg o ++β+α−=+ if the problem is the same as that

indicated in figure 2(a) and (b). For the configuration depicted in figure 2(c), ( ) ( ) ( )β+α−+=+ jihg .

(b) Then, define,

( )1

sinbsina

1sinbsina

45cot o

+βα

−βα

=θ+

where,

αβ=θ −

sinasinbtan 1

(c) Further,

( ) ( ) ( )hg21tan45cothg

21tan o +θ+=−

(d) Then,

( ) ( ) ( ) ( )2

hghghand2

hghgg −−+=−++=

(e) Finally,

( ) ( )β+−=α+−= h180jandg180i oo

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 = 5,300.00 XB = 2,200.00 YB = 6,300.00 XC = 3,100.00 YC = 5,000.00 α = 109° 30' 45" β = 115° 05' 20" It was already shown that the azimuths are


COLLINS METHOD The Collins (or Bessel’s) method [Blachut et al, 1979, Faig, 1972, Klinkenberg, 1955, Zeimann, 1974] is different in that the problem is broken down into two intersections. A circle is drawn through two control points and the occupied point (as A, B, and P in figure 3). The line from P to C is extended until it intersects the circle at a point labeled H. This point is called the Collins’ Auxiliary Point.

Figure 3. Three point resection problem using the Collins method.

Figure 4. Geometry of circle showing that an angle on the circle subtending a base line is equal.


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.

2. Compute the azimuth AzHC which will also yield the azimuth between C and P since AzHC = AzCP.

3. Compute the azimuth of the lines AP and BP

4. The coordinates can be computed by intersection from A and C and also

from B and C. 5. If desired, the solution can be performed using the auxiliary angles ϕ and ψ.

Then, using the sine law,

This gives

BPBOBAPAPAP

BPBPBAPAPAP

AzcosDYAzcosDYY

AzcosDXAzsinDXX

+=+=

+=+=

Following is a MathCAD program that solves the same problem as presented earlier but this time using the Collins method.

β+=

α−=

CPBP

CPAP

AzAz

AzAz

BPBC

ACAP

AzAz

AzAz

−=ψ

−=ϕ

( )

( )β

ψ+β=

αϕ+α=

sinsinDD

sinsinDD

BCBP

ACAP


Three Point Resection Problem

Collins Method See the same functions as defined in the Kaestner-Burkhardt MathCAD program. ________________________________________________________________________ Given XA 1000.00:= YA 5300.00:= XB 3100.00:= YB 5000.00:= XC 2200.00:= YC 6300.00:= α 109.3045:= β 115.0520:= 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.

aBAH 180 dd β( )−:= dms aBAH( ) 64.5440= aABH 180 dd α( )−:= dms aABH( ) 70.2915= DAB XB XA−( )2 YB YA−( )2+:= DAB 2121.32034= AzAB atan2 YB YA− XB XA−,( ):= dms AzAB tdeg⋅( ) 98.07484= aAHB 180 180 dd β( )−( ) 180 dd α( )−( )+ −:= dms aAHB( ) 44.3605=


AzAH AzAB trad 180 dd β( )−( )⋅ +:= dms AzAH tdeg⋅( ) 163.02284=

DAHDAB

sin trad aAHB⋅( )

sin 180 dd α( )−( ) trad⋅ ⋅:=

DAH 2847.58555=

DBHDAB

sin aAHB trad⋅( )

sin 180 dd β( )−( ) trad⋅ ⋅:= DBH 2736.05413=

XH XA DAH sin AzAH( )⋅+:= XH 1830.59443= YH YA DAH cos AzAH( )⋅+:= YH 2576.24223= Az atan2 YH YC− XH XC−,( ):= AzCH if Az 0> Az, Az 2 π⋅+,( ):= dms AzCH tdeg⋅( ) 185.39552= Az atan2 YA YC− XA XC−,( ):= AzCA if Az 0> Az, Az 2 π⋅+,( ):= dms AzCA tdeg⋅( ) 230.11399= aACP AzCA AzCH−:= dms aACP tdeg⋅( ) 44.31447= φ 180 dd α( ) aACP tdeg⋅+( )−:= dms φ( ) 25.57303= AzAP AzCA π−( ) φ trad⋅+:= dms AzAP tdeg⋅( ) 76.09102= DAC XC XA−( )2 YC YA−( )2+:= DAC 1562.04994= From the sine law:

DAPDAC

sin dd α( ) trad⋅( )

sin aACP( )⋅:= DAP 1162.1655=

XP XA DAP sin AzAP( )⋅+:= XP 2128.390= YP YA DAP cos AzAP( )⋅+:= YP 5578.144= For a check, compute the coordinates from point B by solving for the elements in triangle BCP.


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 H1 and H2. First the azimuths between A and H1 and B and H1 are determined.

From triangle ACH1, the distance from A to H1 can be computed.

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

oBCBH

oACAH

90AzAz

90AzAz

1

1

−=

+=

α−=

α−=

α=

=α

tanAzcosYY

tanAzsinXX

tanDD

DDtan

AC

AC

AC

ACACAH

AH

AC

1

1


Since the angle at A is 90o,

Then,

The coordinates for H2 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

ACAHacAH AzsinAzcos;AzcosAzsin11

−==

( )

( ) α−−=+=

α−+=+=

cotXXYAzcosDYY

cotYYXAzsinDXX

ACAAHAHAH

ACAAHAHAH

111

111

bcBHBCBH

bc

BC

bc

BCBCBH

AzsinAzcos;AzcosAzsin

tanAzcosYY

tanAzsinXX

tanDD

22

2

=−=

β−=

β−=

β=

( ) BCBCBC AztanYYXX −=−

( ) ( ) bcBABCACBC AztanYYAztanYYXX −+−=−

( ) ACACAC AztanYYXX −=−

( )

( ) β−−=+=

β−−=+=

cotXXYAzcosDYY

cotYYXAzsinDXX

BCBBHBHBH

BCBbhBHBH

222

222


Rearranging yields

Using the form of this last equation, one can write express the Y-coordinate of the Cassini auxiliary point, H1 as

But,

and

then the Y-coordinate for H1 becomes, after multiplication by tan AzCA

The X-coordinate can also be developed in a similar fashion yielding

But ( )α−−=− o

CACH 90AzAz1

. Then,

( ) ( )( )[ ]

( )( ) ( ) BCBAACBCACBA

ACACAC

BCBABCACBC

AztanXXAztanAztanYYXX

AztanYYXXAztanYYAztanYYXX

−+−−=−

−=−−−+−=−

( ) ( )ACBC

bcBABAAC AztanAztan

AztanYYXXYY−

−+−+=

( ) ( )11

1

1AHCH

ACCHACAH AztanAztan

XXAztanYYYY

−−−−

+=

( ) ( )ACCAAC XXAztanYY −=−

1AztanAztan CAAH1−=

( ) ( )CACHACAH AzAztanXXYY11

−−+=

( ) ( )CACHACAH AzAztanYYXX11

−−−=

( )

( ) α−−=+=

α−+=+=

cotXXYAzcosDYY

cotYYXAzsinDXX

ACAAHAHAH

ACAAHAHAH

111

111


The coordinates for H2 can be developed in a similar fashion and they are given above. Next, compute the azimuth between the two auxiliary points, H1 and H2.

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

or,

But,

Thus,

where: PH1

Aztann =

( )n/1nN += The X-coordinate of the unknown point can be expressed in a similar form as:

−−

= −

12

12

21HH

HH1HH YY

XXtanAz

( )

( )PHCP

HCPHHCPC

PHCP

HCCPHCPCHHHCPHP

1

111

1

112111

AztanAztanXXAztanYAztanY

AztanAztanXXAztanYAztanYAztanYAztanY

Y

−−−−

=

−−−−+−

=

PHCP

1Aztan1Aztan −=

( ) ( )PHCP

CPHCCHHP

1

11

1 AztanAztanAztanYYXX

YY−

−+−=−

( )N

XXYn1Yn

Y 11 HCCHP

−++=


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 XA 1000.00:= YA 5300.00:= XB 3100.00:= YB 5000.00:= XC 2200.00:= YC 6300.00:= α 109.3045:= β 115.0520:= Solution - Find the coordinates of point P using the Cassini Method. XH1 XA YC YA−( )cot dd α( ) trad⋅( )+:= XH1 645.63588= YH1 YA XA XC−( ) cot dd α( ) trad⋅( )⋅+:= YH1 5725.23694= XH2 XB YB YC−( )cot dd β( ) trad⋅( )+:= XH2 3708.6571= YH2 YB XC XB−( ) cot dd β( ) trad⋅( )⋅+:= YH2 5421.378= AzH1H2 atan2 YH2 YH1− XH2 XH1−,( ):= dms AzH1H2 tdeg⋅( ) 95.39552= n tan AzH1H2( ):=

N n1n

+:=

YP

n YH1⋅1n

YC⋅+ XC+ XH1−

N:= YP 5578.14421=

( )N

YYXn1Xn

X 11 HCHCP

−++=


XP

n XC⋅1n

XH1⋅+ YC+ YH1−

N:=

XP 2128.3902=

TIENSTRA METHOD

The Tienstra method [see Bannister et al, 1984] is also referred to as the Barycentric method. An easy to understand proof is given in Allan et al [1968]. Figure 6 shows a triangle formed from the known control points. Line CD divides the angle at C into two components: χ and ψ. Line AB is also divided into two components: m and n. The angle θ is formed by the intersection of the line CD with the line AB. From figure 6 one can also see that line CE is perpendicular to line AB. Thus,

CE

DE

CE

EBB

CE

AEA

DDcot

DDcot

DDcot

=θ

=∠

=∠

Figure 6. Basic geometry outlining the principles of the Tienstra Method.

SURE 215 – Surveying Calculations Three Point Resection Problem 3

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−∠=+−==

cotcotDcotcotD

DDDD

DD

nm

BCE

ACE

EBDE

DEAE

DB

AD

θ−∠=θ+∠

θ+∠θ−∠=

cotncotncotmcotm

cotcotcotcot

nm

AB

B

A

( ) BA cotmcotncotnm ∠−∠=θ+

ψ=⇒=ψ

θ=⇒=

θ=⇒=θ

χ=⇒=χ

cotDD

DDcot

cotDD

DD

cotDD

DDcot

cotDD

DDcot

CGBG

BG

CG

GDBG

BG

GD

DFAF

AF

DF

CFAF

AF

CF

θχ=⇒

θ=

χ cotcot

DD

cotD

cotD

DF

CFDFCF

ψθ=⇒

ψ=

θ cotcot

DD

cotD

cotD

CF

GDCGGD


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:

θθ−χ=−

−

θχ=−

−θ

χ=−=

cotcotcot

DDD

1cotcotDDD

Dcot

cotDDDD

DF

DFCF

DFDFCF

DFDF

DFCFCD

θθ+ψ=+

+

θψ=

+θ

ψ=+=

cotcotcot

DDD

1cotcotD

Dcot

cotDDDD

DG

DGCG

DG

DGDG

DGCGCD

−

θχ= 1

cotcotDD DFCD

θ=⇒=θ

θ=⇒==θ

cosnDn

Dcos

cosmDm

DDDcos

DGDG

DFDF

AD

DF


Equating the two values for DCD yields

The three-point resection problem is shown in figure 7. Point P is the occupied point and points A, B, and C are the control points that are observed. The measured angles are α, β, and γ. The other angles are numbered in a clockwise manner from point A. Recall that from the intersection problem, the coordinates of a point, such as point C, can be computed as:

−

θχθ=

+

θψθ=

−

θχ=

+

θψ=

1cotcotcosm1

cotcotcosn

1cotcotDD1

cotcotDD DFCDDGDG

( ) ψ−χ=θ+

θ−ψ=θ+ψ

θθ−χθ=

θθ+ψθ

cotncotmcotnm

cotmcotmcotncotn

cotcotcotcosm

cotcotcotcosn

Figure 7. Three point resection problem using the Tienstra Method.


( )

( ) BABAC

BABAC

YYcotXcotXcotcotX

cotcotcotXcotXYYX

−+α+β=β+α

β+αα+β+−

=

where α is the angle at A and β is the angle at B. Using this basic relationship, the X-coordinate at point P can be computed as follows: Adding these three equations yields:

This is usually represented as

where: L1 = cot 3 + cot 6 L2 = cot 2 + cot 5 L3 = cot 4 + cot 1 The X-coordinate is computed as

In a similar fashion, the Y-coordinate can be written, from the intersection problem

which can be shown, after the same manipulation performed on the X-coordinate, as

( )( )( ) ACACP

CBCBP

BABAP

YY6cotX1cotX6cot1cotXYY4cotX5cotX5cot4cotXYY2cotX3cotX3cot2cotX

−++=+−++=+−++=+

( ) ( )( ) ( )1cot4cotX5cot2cotX

6cot3cotX6cot5cot4cot3cot2cot1cotX

CB

AP

+++++=+++++

( ) C3B2A1321P XLXLXLLLLX ++=++

321

C3B2A1P LLL

XLXLXLX++

++=

( )β+α

α+β+−=cotcot

cotYcotYXXY BAABC

321

C3B2A1P LLL

YLYLYLY++

++=


From figure 7, the line BP was extended until it intersected the line AC at a point labeled Q. This divides the line into two parts: m and n. Recall that the angle ∠ CPQ = 180O - α and ∠ APQ = 180o - γ. Recall that we wrote earlier: ( ) ψ−χ=θ+ cotncotmcotnm . Using the geometry from figure7, this becomes,

Recall that earlier we wrote the relationship: ( ) BA cotncotmcotnm ∠−∠=θ+ which can be written as (considering the geometry in figure 7)

Equating these last two formulas yields the following formula,

Using ( ) ψ−χ=θ+ cotncotmcotnm and ( ) BA cotncotmcotnm ∠−∠=θ+ again, write

Equating these last two equations gives

Using this formula, equate it with ( ) ( )6cot3cotn1cot4cotm +=+ giving us the next equation

or

where:

γ−∠=

β−∠=

α−∠=

cotcotK1

cotcotK1

cotcotK1

C3

B2

A1

3cotn4cotmcot)nm( −=θ+

( ) 1cotm6cotncotnm −=θ+

( ) ( )6cot3cotn1cot4cotm +=+

( )( ) AC cotmcotncotnm

cotncotmcotnm∠−∠=θ+γ+α−=θ+

( ) ( )γ−∠=α−∠ cotcotncotcotm CA

γ−∠α−∠==

++

cotcotcotcot

mn

6cot3cot4cot1cot

C

A

1

3

1

3

KK

LL =


In a similar fashion, one can easily show that

Therefore,

from which,

and

Thus,

Substituting these relationships back into the equations for Xp and YP which were expressed in terms of L1, L2, and L3 that were presented earlier yields the final form for computing the coordinates using the Tienstra method.

2

3

2

3

KK

LL =

WKL

KL

KL

3

3

2

2

1

1 ===

WKLWKLWKL

33

22

11

===

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An example using MathCAD follows:

Three Point Resection Problem Tienstra Method

See the same functions as defined in the Kaestner-Burkhardt MathCAD program _______________________________________________________________________________ This MathCAD example is the same example used in the other methods. There is a slight difference in that the triangle is lettered in a clockwise manner and α is the clockwise angle from line PB to line PC, β is the clockwise angle from line PC to line PA, and γ is the clockwise angle from line PA to line PB. See the following figure.

Given: X A 1000.00:= YA 5300.00:= α 115.0520:= X B 2200.00:= YB 6300.00:= β 135.2355:= X C 3100.00:= YC 5000.00:= γ 109.3045:=

Solution - Find the coordinates of point P using the Tienstra Method. Az AB 50.1140:= AzBA AzAB 180+:= Az CB 325.1817:= AzBC AzCB 180−:= Az AC 98.0748:= Az CA Az AC 180+:= A dd Az AC( ) dd Az AB( )−:= dms A( ) 47.5608= B dd AzBA( ) dd AzBC( )−:= dms B( ) 84.5323= C dd AzCB( ) dd AzCA( )−:= dms C( ) 47.1029= Place the angles into radians Ar A trad⋅:= Br B trad⋅:= Cr C trad⋅:= αr radians α( ):= βr radians β( ):= γr radians γ( ):=


Solve for the constants used in the Tienstra Method K1 cot Ar( ) cot αr( )−( ) 1−:= K1 0.72959=

K2 cot Br( ) cot βr( )−( ) 1−:= K2 0.90626=

K3 cot Cr( ) cot γr( )−( ) 1−:= K3 0.78052=

The solution is:

XPK1 XA⋅ K2 XB⋅+ K3 XC⋅+( )

K1 K2+ K3+:= XP 2128.391=

YPK1 YA⋅ K2 YB⋅+ K3 YC⋅+( )

K1 K2+ K3+:= YP 5578.1451=

REFERENCES Allan, A., Hollwey, J., and Maynes, J., 1968. Practical Field Surveying and Computations, American Elsevier Publishing Co., Inc., New York. Anderson, J. and Mikhail, E., 1998. Surveying: Theory and Practice, 7th edtion, WCB/McGraw-Hill, New York. Bannister, A., Raymond, S., and Baker, R., 1984. Surveying, 6th edition, Longman Scientific & Technical, Essex, England. Blachut, T., Chrzanowski, A., and Saastamoinen, J., 1979. Urban Surveying and Mapping, Springer-Vrlag, New York. Faig, W., 1972. “Advanced Surveying I (Preliminary Copy), Department of Surveying Engineering Lecture Notes No. 26, University of New Brunswick, Fredericton, N.B., Canada, 225 p. Hodgson, C., 1957. Manual of Second and Third Order Triangulation and Traverse, USC&GS Special Publication No. 145 (Reprinted, 1957), U.S. Government Printing Office, Washington, D.C. Kissam, P., 1981. Surveying for Civil Engineers, 2nd edition, McGraw-Hill, New York. Klinkenberg, H., 1955. “Coordinate Systems and the Three Point Problem”, The Canadian Surveyor, XII(8):508-518.


Reynolds, W., 1934. Manual of Triangulation Computation and Adjustment, USC&GS Special Publication No. 138 (Reprinted, 1955), U.S. Government Printing Office, Washington, D.C. Ziemann, H., 1974. “Terrestrial Surveying Methods”, Proceedings of ACSM Fall Convention, Washington, D.C., September, pp 222-233.

8 three point resection problem -...

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