Surveying I.
Lecture 9.
Plane surveying. Fundamental tasks of
surveying. Intersections. Orientation.
The coordinate system
Northing
N
Easting
E
Northing axis is the projection of the starting meridian
of the projection system, while the Easting axis is
defined as the northing axis rotated by 90° clockwise.
The whole circle bearing
How could the direction of a target from the station be
defined?
N
||N
WC
B
AB
B
A
E
Whole circle bearing: the local north is rotated
clockwise to the direction of the target. The angle
which is swept is called the whole circle bearing.
0  WCB AB  360
Transferring Whole Circle Bearings
WCB of reverse direction:
WCB BA  WCB AB  180
Transferring WCBs: WCBAB is known,  is measured,
how much is WCBAC?
N
||N
WC
B
B
AB
A
A

WCBAC
C
E
WCB AC  WCB AB  
WCB AB
or
 WCB AC  
1st fundamental task of surveying
N
||N
N
E
WC
B
B
AB
d AB
A
E
A(EA, NA), WCBAB and dAB is known,
B(EB,NB)=?
E AB  E B  E A  d AB  sin WCB AB
N AB  N B  N A  d AB  cosWCB AB

E B  E A  d AB  sin WCB AB ,
N B  N A  d AB  cosWCB AB .
2nd fundamental task of surveying
N
||N
NB-NA
E-E
B
A
WC
B
AB
B
d AB
A
E
A(EA, NA), B(EB,NB) is known,
WCBAB=? and dAB=?
d AB 
EB  E A 2  N B  N A 2
  arctan
EB  E A
,
NB  N A
WCB AB    c
2nd fundamental task of surveying
||N
B
B
0
WCBA B
WCBA B


A
> 0
A
d AB 
EB  E A 2  N B  N A 2
  arctan
WCB AB    c
WCBAB
WCBA B
EB  E A
,
NB  N A
A

A

0
B
B
Quadrant
EB-EA
NB-NA
c
I.
+
+
0
II.
+
-
+180°
III.
-
-
+180°
IV.
-
+
+360°
Intersections
Aim: the coordinates of an unknown point should be
computed. Measurements are taken from two
different stations to the unknown point, and the so
formed triangle should be solved.
P

A

d AB
B
Foresection with inner angles
AP
d
WCBAP

A
A (EA,NA)
B (EB,NB) are known
,  are observed
=?
d BP
=?
P
WCBAB

d AB
B
WCBBA
WCBBP
1.
2.
Compute WCBAB, dAB using the 2nd fundamental task of
surveying.
Using the sine theorem compute dAP and dBP!
d AP  d AB
sin 
sin    
d BP  d AB
sin 
sin    
3. Compute WCBAP and WCBBP:
WCB AP  WCB AB  
WCB BP  WCB BA  
Foresection with inner angles
d
AP
=?
d BP
=?
P
WCBAP

WCBAB

A
d AB
4. Compute the coordinates of P:
B
WCBBA
WCBBP
From A :
EPA  E A  d AP sin WCB AP
N PA  N A  d AP cosWCB AP ,
From B :
EPB  EB  d BP sin WCB BP
N PB  N B  d BP cosWCB BP ,

EPA  EPB
EP 
2
N PA  N PB
NP 
2
Foresection with WCBs
What happens, when B is not observable from A?
P
C
d
WCB’AP

AP
=
=?
d BP
?
D
WCBAB
A
d AB
B

WCBBA
WCB’BP
A,B,C and D are known points,  and  are measured.
Foresection with WCBs
P
C
d
WCB’AP

AP
=
=?
d BP
?
D
WC BAP  WCB AC  
  WCB BD  
WC BBP
WCBAB
A
d AB
B

WCBBA
WCB’BP
The equations of the lines AP and BP:
N1  N A  E  E A  cotWCB AP
N 2  N B  E  EB  cotWCB BP
Foresection with WCBs
P
C
d
WCB’AP

AP
=
=?
d BP
?
D
WCBAB
A
d AB
B

WCBBA
WCB’BP
Let’s compute the intersection of the lines AP and BP:
N1  N 2
E cotWCB AP  cotWCB BP   N B  N A  E A cotWCB AP  EB cotWCB BP
EP 
N B  N A  E A cotWCB AP  EB cotWCB BP
cotWCB AP  cotWCB BP
N P  N A  EP  E A  cotWCB AP
Different types of intersections
How can we use intersections, when A or B is not
suitable for setting up the instrument:
P
AP
d
WCB’AP
=?
d BP
=?

WCBAB
A
d AB

B
WCBBA
WCB’BP
 can be computed by =180°--. => Foresection.
Resection
E-E
B
A
B
C
NB-NA
A
O2
O1
NA-NT1




T2
P
T1
ET1-EA
A,B,C are known control points
 and  are observed angles
Aim: compute the coordinates of P (the station)
Resection
E-E
B
A
B
C
NB-NA
A
O2
O1
NA-NT1




T2
P
T1
ET1-EA
Compute the coordinates of T1 and T2!
EB  E A N B  N A

 cot
N A  N T1 ET1  E A

E  E A  N A cot
N T1  B
,
cot
N  N A  E A cot
ET1  B
.
cot
Resection
WCBBP
B
C
A
O2
O1
WCBT1P
T2
P
T1
Since T1, P and T2 are on a straight line:
WCBT1P  WCBT1T2
WCBBP  WCBT1T2  90
Foresection with WCBs
Resection – the dangerous circle
What happens, if all the four points are on one
circumscribed circle?
B
B
A
P
C
A
 
P

C

P
Arcsection
A, B are known control
points,
DAP and DBP are measured.
P
P
DB
D
AP
Aim: compute the
coordinates of P!

A
B
d AB
Using the cosine theorem, compute the angle :
2
2
2
DBP
 DAP
 d AB
 2 DAP d AB cos

2
2
2
DAP
 d AB
 DBP
  arccos
.
2 DAP d AB
Arcsection
P
P
D
AP
DB
WCBAP
WCBAB

A
B
d AB
Compute WCBAB from the coordinates of A and B,
WCBAP = WCBAB-
1st fundamental task of surveying
Orientation
How can the WCB be determined from observations?
Recall the definition of mean direction:
All the angular observations refer
to the index of the horizontal
circle, but they should refer to the
Northing instead!
0
P
MDAP
A
ZA
0
P
Orientation
zA – orientation angle
MDAP
A
Orientation
How to find the orientation angle?
A,B are known points,
MDAP and MDAB are
observed.
0
ZA
WCB’AP
P
MDAP
Aim: Compute WCB’AP
MDAB
WCBAP
B
A
Compute the orientation angle:
zA=WCBAB-MDAB
Computing the WCB’AP:
WCBAB=zA+MDAB
Computing the mean orientation angle
In case of more orientations, as many orientation
angles can be computed as many control points are
sighted:
zAB=WCBAB-MDAB
zAC=WCBAC-MDAC
zAD=WCBAD-MDAD
zAB, zAC and zAD are usually slightly different due
observation and coordinate error.
However, the orientation angle is constant for a
station and a set of observations.
Mean orientation angle:
z AB  d AB  z CA  d AC  z AD  d AD
zA 
d AB  d AC  d AD
WCB vs provisional WCB
N
||N
WC
B
AB
B
A
E
Whole circle bearing (WCBAB): computed from
coordinates, between two points, which coordinates are
known.
Provisional whole circle bearing (WCB’AB): an
angular quantity, which is similar to the whole circle
bearing. However it is computed from observations, by
summing up the (mean) orientation angle and the
mean direction.
Thank You for Your Attention!