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Abstract: 8 THREE POINT RESECTION PROBLEMSurveying Engineering DepartmentFerris State UniversityINTRODUCTIONThe threepoint resection problem in surveying involves occupying an unknown pointand observing angles only to three known points. Today, with the advent of total

8 THREE POINT RESECTION PROBLEM
Surveying Engineering Department
Ferris State University
INTRODUCTION
The threepoint 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.
KAESTNERBURKHARDT METHOD
Figure 1. Three point resection problem using the KaestnerBurkhardt method.
In the KaestnerBurkhardt approach [Blachut et al, 1979, Faig, 1972, Kissam, 1981,
Ziemann, 1974] (also referred to as the PothonotSnellius 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:
SURE 215 – Surveying Calculations Three Point Resection Problem Page 176
X − XA
Y −Y
Az AC = tan −1 C a= (X C − X A )2 + (YC − YA )2
C A
X − XB
Az BC = tan −1 C
Y −Y
b= (X C − X B )2 + (YC − YB )2
C B
Compute γ
γ = Az CA − AzCB
= Az AC − Az BC
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].
φ + α + β + θ + γ = 360o
Rearrange
1
(φ + θ) = 180o − 1 (α + β + γ ) = δ1
2 2
From the sine rule, compute the distance s
a sin φ b sin θ
s= and s=
sin α sin β
Combining these relationships yields
sin φ b sin α
= = cot λ
sin θ a sin β
where λ is an auxiliary angle with an uncertainty of ±180o. We then have
sin φ
= cot λ
sin θ
or
sin φ − sin θ cot λ − 1
=
sin φ + sin θ cot λ + 1
SURE 215 – Surveying Calculations Three Point Resection Problem Page 177
cot λ
Since cot λ = and using trigonometric theorems, one can write
1
2 cos
1
(φ + θ)sin 1 (φ − θ) cot 45o cot λ − 1
2 2 =
2 sin (φ + θ)cos (φ − θ) cot λ + cot 45
o
1 1
2 2
But, recognizing that cot 45o = 1 and
tan
1
( ) (
(φ − θ) = tan 1 (φ − θ)cot 45o + λ = tan δ1 cot 45o + λ )
2 2
Therefore,
1
2
[ ( )]
(φ − θ) = tan −1 tan δ1 cot 45o + λ = δ2
Then,
φ = δ1 + δ 2
θ = δ1 − δ 2
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.
c1 = a
sin γ1
=a
[
sin 180 o − (α + φ)
=a
]
sin (α + φ)
sin α sin α sin α
c2 = b
sin γ 2
=b
[
sin 180o − (β + θ)
=b
]
sin (β + θ)
sin β sin β sin β
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.
SURE 215 – Surveying Calculations Three Point Resection Problem Page 178
Az AP = Az AC + φ
Az BP = Az BC − θ
Finally, compute the coordinates of point P.
X P = X A + c sin Az AP = X B + c 2 sin Az BP
YP = YA + c1 cos Az AP = YB + c 2 cos Az BP
An example, prepared using Mathcad is presented as follows.
Three Point Resection Problem
KaestnerBurkhardt Method
dd ( ang ) := degree ← floor( ang ) radians ( ang ) := d ← dd ( ang )
mins ← ( ang − degree ) ⋅ 100.0 π
d⋅
minutes ← floor( mins) 180.0
seconds ← ( mins − minutes ) ⋅ 100.0
minutes seconds
degree + +
60.0 3600.0
dms( ang ) := degree ← floor( ang )
rem ← ( ang − degree ) ⋅ 60
mins ← floor( rem)
rem1 ← ( rem − mins)
secs ← rem1⋅ 60.0
mins secs
degree + +
100 10000
π 180
trad := tdeg :=
180 π
________________________________________________________________________
Given
SURE 215 – Surveying Calculations Three Point Resection Problem Page 179
X A := 1000.00 YA := 5300.00
X B := 3100.00 YB := 5000.00
X C := 2200.00 YC := 6300.00
α := 109.3045 β := 115.0520
Solution  Find the coordinates of point P using the KaestnerBurkhardt 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 := 180 −
1
⋅ (dd (α ) + dd (β ) + γ ) ( )
dms δ1 = 25.15163
2
λ0 :=
sin radians ((α ))
b λ0 = 1.053482162
⋅
a sin radians ((β ))
λ := tdeg atan
dms(λ) = 43.30291
1
(λ0)
Note that λ has an uncertainty of 180 degrees
δ2 := atan tan radians dms δ1
( ( ( ( )))) ⋅ ( ⋅ tdeg
1
tan radians (dms(45 + λ)))
( )
dms δ2 = 0.4214
φ := δ1 + δ2 dms(φ) = 25.57303
SURE 215 – Surveying Calculations Three Point Resection Problem Page 180
θ := δ1 − δ2 dms(θ ) = 24.33022
Compute the distances between the point P and control points A and B
sin radians (α ) + (φ⋅ trad )
c1 := a⋅ c1 = 1162.1655
sin (radians (α ))
sin radians (β ) + (θ ⋅ trad )
c2 := b⋅ c2 = 1130.60883
sin (radians (β ))
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 PothonotSnellius
method. Recall that the distance from C to P was designated as s and was expressed
a sin φ b sin θ
as =s= . From this there are two methods of solving this problem. The first
sin α sin β
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:
sin θ sin (S − ϕ) sin S cos ϕ − cos S sin ϕ sin S cos ϕ
K= = = = − cos S
sin ϕ sin ϕ sin ϕ sin ϕ
where S = 360o − (α + β + γ ) . 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): θ = 360o − (α + φ + γ ) − ϕ = S − ϕ . S represents the
known angles. Manipulation of this last relationship yields
SURE 215 – Surveying Calculations Three Point Resection Problem Page 181
sin S cos ϕ
K + cos S = = sin S cot ϕ
sin ϕ
From which,
K + cos S
cot ϕ =
sin S
Solve for ϕ and then compute c1 and the azimuth to determine the coordinates of point P.
Alternatively, use lineline intersection to find the coordinates of the unknown point.
Another modification of the KaestnerBurkhardt 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.
B B a
j C
h
b
a
i
i j
A g P
h C b
g
P A
(a) (b)
C
A a h
b
g
B
j
i
P
(c)
Figure 2. Three scenarios for the threepoint resection problem.
SURE 215 – Surveying Calculations Three Point Resection Problem Page 182
(a) Compute (g + h ) = 360 o − (α + β + i + j) 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,
a sin α
−1
b sin β
(
cot 45 + θ =
o
)
a sin α
+1
b sin β
where,
b sin β
θ = tan −1
a sin α
(c) Further,
tan
1
2
( )
(g − h ) = cot 45 o + θ tan 1 (g + h )
2
(d) Then,
g=
(g + h ) + (g − h ) and h=
(g + h ) − (g − h )
2 2
(e) Finally,
i = 180 o − (g + α ) and j = 180 o − (h + β)
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
SURE 215 – Surveying Calculations Three Point Resection Problem Page 183
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.
SURE 215 – Surveying Calculations Three Point Resection Problem Page 184
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
Az AP = AzCP − α
Az BP = AzCP + β
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.
ϕ = Az AP − Az AC
ψ = Az BC − Az BP
Then, using the sine law,
D AC sin (α + ϕ)
D AP =
sin α
D BC sin (β + ψ )
D BP =
sin β
This gives
X P = X A + D AP sin Az AP = X B + D BP cos Az BP
YP = YA + D AP cos Az AP = YB + D BO cos Az BP
Following is a MathCAD program that solves the same problem as presented earlier but
this time using the Collins method.
SURE 215 – Surveying Calculations Three Point Resection Problem Page 185
Three Point Resection Problem
Collins Method
See the same functions as defined in the KaestnerBurkhardt MathCAD program.
________________________________________________________________________
Given
X A := 1000.00 YA := 5300.00
X B := 3100.00 YB := 5000.00
X C := 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
SURE 215 – Surveying Calculations Three Point Resection Problem Page 186
AzAH := AzAB + trad ⋅ (180 − dd (β ))
dms(AzAH⋅ tdeg ) = 163.02284
DAB
DAH := ⋅ sin (180 − dd (α ))⋅ trad
DAH = 2847.58555
sin (trad ⋅ aAHB)
DAB
DBH := ⋅ sin (180 − dd (β ))⋅ trad
DBH = 2736.05413
sin (aAHB⋅ trad )
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:
DAC
DAP := ⋅ sin (aACP ) DAP = 1162.1655
sin (dd (α )⋅ trad )
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.
SURE 215 – Surveying Calculations Three Point Resection Problem Page 187
CASSINI METHOD
The Cassini approach [Blachut et al, 1979, Faig, 1972, Klinkenberg, 1955, Ziemann,
1974] to the solution of the threepoint 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:
Figure 5. Three point resection problem as proposed by Cassini.
Compute the coordinates of the auxiliary points H1 and H2. First the azimuths between
A and H1 and B and H1 are determined.
Az AH1 = Az AC + 90o
Az BH1 = Az BC − 90o
From triangle ACH1, the distance from A to H1 can be computed.
D AC
tan α =
D AH1
D AC XC − XA YC − YA
D AH1 = = =
tan α sin Az AC tan α cos Az AC tan α
SURE 215 – Surveying Calculations Three Point Resection Problem Page 188
Since the angle at A is 90o,
sin Az AH1 = cos Az ac ; cos Az AH1 = − sin Az AC
Then,
X H1 = X A + D AH1 sin Az AH1 = X A + (YC − YA )cot α
YH1 = YA + D AH1 cos Az AH1 = YA − (X C − X A )cot α
The coordinates for H2 are computed in like fashion.
D BC XC − XB YC − YB
D BH 2 = = =
tan β sin Az bc tan β cos Az bc tan β
sin Az BH 2 = − cos Az BC ; cos Az BH 2 = sin Az bc
X H 2 = X B + D BH 2 sin Az bh 2 = X B − (YC − YB )cot β
YH 2 = YB + D BH 2 cos Az BH 2 = YB − (X C − X B )cot β
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:
X C − X B = (YC − YB )tan Az BC
This can also be written as
X C − X B = (YC − YA ) tan Az BC + (YA − YB ) tan Az bc
But,
X C − X A = (YC − YA )tan Az AC
Solving these last two equations can be done by subtracting the last equation from the
preceding equation resulting in
SURE 215 – Surveying Calculations Three Point Resection Problem Page 189
X C − X B = (YC − YA )tan Az BC + (YA − YB )tan Az BC
− [X C − X A = (YC − YA tan Az AC ) ]
X A − X B = (YC − YA )(tan Az BC − tan Az AC ) + (X A − X B )tan Az BC
Rearranging yields
YC = YA +
(X A − X B ) + (YA − YB )tan Az bc
tan Az BC − tan Az AC
Using the form of this last equation, one can write express the Ycoordinate of the
Cassini auxiliary point, H1 as
(YC − YA )tan AzCH − (X C − X A )
YH1 = YA + 1
tan AzCH1 − tan Az AH1
But,
(YC − YA )tan AzCA = (X C − X A )
and
tan Az AH1 tan AzCA = −1
then the Ycoordinate for H1 becomes, after multiplication by tan AzCA
(
YH1 = YA + (X C − X A ) tan Az CH1 − Az CA )
The Xcoordinate can also be developed in a similar fashion yielding
(
X H1 = X A − (YC − YA )tan AzCH1 − AzCA )
But Az CH1 − Az CA = −(90 o − α ) . Then,
X H1 = X A + D AH1 sin AzAH1 = X A + (YC − YA )cot α
YH1 = YA + D AH1 cos Az AH1 = YA − (X C − X A )cot α
SURE 215 – Surveying Calculations Three Point Resection Problem Page 190
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.
X H − X H1
Az H1H 2 = tan −1 2
YH 2 − YH1
As before, one can write the equation of intersection containing the unknown point P as:
YP − YH1 =
(X H1 ) ( )
− X C + YC − YH1 tan AzCP
tan Az CP − tan Az H1 P
or,
YP =
(
YH1 tan Az CP − YH1 tan Az H1H 2 + YC tan Az CP − YH1 tan Az CP − X C − X H1 )
tan Az CP − tan Az H1P
=
(
YC tan Az CP − YH1 tan Az H1P − X C − X H1 )
tan Az CP − tan Az H1P
But,
YP =
( )
n YH1 + 1 YC + X C − X H1
n
N
Thus,
1
tan Az CP = −
tan Az H1P
where: n = tan Az H1P
N = n + (1 / n )
The Xcoordinate of the unknown point can be expressed in a similar form as:
SURE 215 – Surveying Calculations Three Point Resection Problem Page 191
XP =
( )
n X C + 1 X H1 + YC − YH1
n
N
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 KaestnerBurkhardt MathCAD program.
________________________________________________________________________
Given
X A := 1000.00 YA := 5300.00
X B := 3100.00 YB := 5000.00
X C := 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)
1
N := n +
n
n ⋅ Y + 1 ⋅ Y + X − X
H1 n C C H1
YP :=
YP = 5578.14421
N
SURE 215 – Surveying Calculations Three Point Resection Problem Page 192
1 ⋅ X + Y − Y
n ⋅ XC + n H1 C H1
XP :=
XP = 2128.3902
N
TIENSTRA METHOD
Figure 6. Basic geometry outlining the principles of the 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,
D AE
cot ∠ A =
D CE
D EB
cot ∠ B =
D CE
D DE
cot θ =
D CE
SURE 215 – Surveying Calculations Three Point Resection Problem Page 193
Then,
m D AD D AE − D DE DCE (cot ∠ A − cot θ)
= = =
n D DB D DE + D EB D CE (cot ∠ B − cot θ)
which upon further manipulation yields
m cot ∠ A − cot θ
=
n cot ∠ B + cot θ
m cot ∠ B + m cot θ = n cot ∠ A − n cot θ
or
(m + n ) cot θ = n cot ∠A − m cot ∠B
Since lines AF and BG are perpendicular to line CF, one can write
D CF D CF
cot χ = ⇒ D AF =
D AF cot χ
D DF D DF
cot θ = ⇒ D AF =
D AF cot θ
D GD D GD
= ⇒ D BG =
D BG cot θ
D CG D CG
cot ψ = ⇒ D BG =
D BG cot ψ
From these relationships, equate DAF
D CF D D CF cot χ
= DF ⇒ =
cot χ cot θ D DF cot θ
and equating the distance DBG
D GD D D GD cot θ
= CG ⇒ =
cot θ cot ψ DCF cot ψ
From figure 6 we can also write
SURE 215 – Surveying Calculations Three Point Resection Problem Page 194
D DF cot χ
D CD = D CF − D DF = − D DF
cot θ
cot χ
D CF − D DF = D DF − 1
cot θ
D CF − D DF cot χ − cot θ
=
D DF cot θ
Also, we have,
D DG cot ψ
D CD = D CG + D DG = + D DG
cot θ
cot ψ
= D DG + 1
cot θ
D CG + D DG cot ψ + cot θ
=
D DG cot θ
From above one can see that the distance from C to D can be expressed as
cot χ
D CD = D DF − 1
cot θ
But from figure 6 we can write the following two relationships
D DF D DF
cos θ = = ⇒ D DF = m cos θ
D AD m
D DG
cos θ = ⇒ D DG = n cos θ
n
Substitute these values for DDF and DDG into the relationships derived above. This is
shown as:
SURE 215 – Surveying Calculations Three Point Resection Problem Page 195
cot ψ cot χ
D DG = D DG + 1 D CD = D DF − 1
cot θ cot θ
cot ψ cot χ
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