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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
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:
- Compute the coordinate of the Collins’ Auxiliary Point, H, by intersection
from both control points A and B.
- Compute the azimuth Az (^) HC which will also yield the azimuth between C and
P since Az (^) HC = Az (^) CP.
- Compute the azimuth of the lines AP and BP
- The coordinates can be computed by intersection from A and C and also
from B and C.
- 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
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
