# Surveying for Builders

Document Sample

```					                              Maps and Plans
Background                    Basic Computations
Geometrical Concepts
Errors and Statistics
Distances               coordinates
Angles
Measurements                                          heights
Height differences

Mid Semester examination
New Technologies           GPS

Practical Surveying           Setting Out

Analysis                      Geographical Information Systems
1

Computer assisted surveying
The Department of Civil Engineering

CE-209 Fundamentals of Surveying

Lecture 2 Maps and Plans in Surveying/Basic
Computations/Geometrical Concepts

Dr.Orhan ERCAN

2
Maps and Plans in Surveying…..
Surveys are carried out to make maps and
plans.

Maps and plans are used to carry out
surveys.
Survey Types
*   Detail
*   Control
*   Setting Out
*   Heighting

4
Surveying Terminology
*   Survey area
*   Coordinates
*   Control Points
*   Datums
*   North

5
Elements of a map eg
*   North - directions
*   Grid
*   Coordinates
*   Scale - distances
*   Heights

6
What is involved in conducting a survey?
* What are the measurements made?
* What do these measurements mean?
* What further computations are
required?
* How good are our measurements?

7

* distances (m, km)

* Height differences (m)

1000m = 1km
100cm = 1m                      8
10mm = 1cm
1000mm = 1m
What do these measurements mean?
* Angles
* angles between points (eg)
* bearings

9
Basic Computation #1 Converting degrees,
minutes, seconds to decimal degrees and
Angle measured : 28o 31' 25"

1o = 60'
1' = 60"
therefore 1o = 3600"

To convert to decimal degrees = 28 + 31/60 + 25/3600
= 28.5236o

p radians = 180o where p ~ 3.1416
10
To convert decimal degrees to radians = 28.5236 x p/180
Basic Computation #2 Converting radians to
decimal degrees, and degrees, minutes and
seconds
To convert radians to decimal degrees = 0.497831 x 180/p
= 28.5236o

To convert decimal degrees to degs, mins, secs :

degs = 28
mins = (28.5236 - 28 ) x 60 = 31.416 = 31
secs = (31.416 - 31) * 60 = 24.96 = 25

28o 31' 25"
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North Directions and Whole Circle Bearings
* True, magnetic, arbitrary, grid
N
N
B    D

f
C
A
12
a
Further Computations from the
Measurements
* Compute the distance and direction
between two points given their
coordinates.

* Computing the coordinates of an unknown
point given the coordinates of a known
point and the direction and distance
between them.
13
Basic Computation #3 Computing the
distance between two points given their
coordinates - Chart 3.xls

distance    E2  N2

14
Basic Computation #4 Computing the
bearing between two points given their
coordinates - Chart 2 .xls
E
tan  
N

 E 
  tan 1     
 N 

15
Basic Computation #5 Computing the
coordinates of a point given the bearing and
distance from a known point Chart 4.xls

E  distance x sin 

N  distance x cos

E B  E A  E

N B  N A  N
16
Worked Example -
Computation of Rectangular Coordinates

The coordinates of a point A are 311.617m E, 447.245m N. Calculate
the coordinates of point B where qAB = 37o 11’ 20” and sAB = 57.916m.

EB  E A  s sin  AB                   N B  N A  s cos  AB
 311.617  57.916 sin 37 11 20
o               447.245  57.916 cos 37 o11 20
 311.617  35.007                        447.245  46.139
 346.624m                                493.384m

16
Worked Example -
Computation of bearing and distance
The coordinates of point A are 469.72m E, 338.46 N and point B are
268.14m E and 116.19mN. Compute the bearing and distance between
them.                      N               1  E AB 
 AB  tan 
       

 N AB 
A                        E  E A  
 tan 1  B           
 N B  N A 
AB       NAB                 268.14  469.72  
 tan 1                     
sAB                                         116.19  338.46  
  201.58 
 tan 1 
  222.27 

B                                        tan 1 0.906915               18
EAB             E                         +   180o
 42o1219
AB = 222o 12’ 19”
Inverse Calculations
4   E -ve                   1
E +ve
N +ve
N +ve
+360o
270o                                90o
E -ve          E +ve
N -ve          N -ve

+180o           +180o          19
3            180o            2
s AB    E 2  N 2
N                  EB  E A 2   N B  N A 2
E AB    N AB
         
A                sin  AB cos  AB
 201.58
AB       NAB       
sAB                                 sin 222 o1219
 300.06m

B         EAB             E                                   20
Maps and Plans
Background                    Basic Computations
Geometrical Concepts
Errors and Statistics
Distances               coordinates
Angles
Measurements                                          heights
Height differences

Mid Semester examination
New Technologies           GPS

Practical Surveying           Setting Out

Analysis                      Geographical Information Systems
21

Computer assisted surveying
How good are our measurements?
* precision

* accuracy

22
Precision refers to how good our observations
are with respect to each other.

Accuracy refers to how good our results are
to the true value

23
When we talk about precision and
and more specifically standard deviation.

24
Simple Statistics
30.615
Mean  x 
 x  306 .150  30.615
30.618                      n        10
30.614
30.615
 x  x 2
30.616      S tan dard Deviation                       0.002m
30.614                                          n
30.613
30.614
 x  x  True 2

 0.002m
30.616
n                                25
30.618
The difference between x  x is how
we measure how good our observations
are with respect to each other -
precision.

If we replace X with the true value we
get a measure of accuracy.

Chart1.xls
26
15                                                            15
.
. 10    .. .
.                                                                   10
. 5 . .            .                                                5
.                   .
.                                                                    . ... .. .
. . 5 . 10 15
.                                                 . .. .... ... 5
....            10 15
.       .                                                                       . .
... ....
.. ... ..
.   . .     .   .
.       .     .     .
. . .         .
. .

(a)                                                               (b)
Highly dispersed observations, therfore low precision Closely grouped observations indicating high precision

15                                                               15

10                                                               10
. ... .. .                                            .
. . .... ...
....                                                                         .
5 .... .... .
. .
.. ... .
5
.            . ... .. .
. .. .... ...
....     5           10 15
5          10 15                                           . .
... ....
.. ... ..
.
.        .
.

(c)                                                               (d)
Very precise observations, however poor          Main distribution shows high precision, with several observations
accuracy                        which are signiicantly different. These observations can be considered
outliers and subsequently rejected

27
Errors in Survey Measurements
* Gross - chart 5.xls
* Systematic - chart 6.xls
* Random

28
Errors in derived quantities
We have measured two distances d1 and d2 in a straight line. What is the total distance (D) and its
standard deviation?

d1 = 154.26m and has a SD of 0.01m, d2 = 175.34m and has a SD of 0.05m

D = d1 + d2 = 154.26 + 175.34 = 329.60m

D = (d1 + e2) + (d2 + e2) = (154.26 +.01) +(175.34 +.05 )
= 154.27 + 175.39
= 329.66

difference = 0.06m                                                                         29
The coordinates of a point A are 311.617m E, 447.245m N. Calculate
the coordinates of point B where qAB = 37o 11’ 20” and sAB = 57.916m.

What are the coordinates of B. What effect would there be of an error
in the bearing of 1o and in the distance of 0.5m.

EB  E A  s sin  AB                   N B  N A  s cos  AB
 311.617  57.916 sin 37 11 20
o               447.245  57.916 cos 37 o11 20
 311.617  35.007                        447.245  46.139
 346.624m                                493.384m

30
Maps and Plans
Background                    Basic Computations
Geometrical Concepts
Errors and Statistics
Distances               coordinates
Angles
Measurements                                          heights
Height differences

Mid Semester examination
New Technologies           GPS

Practical Surveying           Setting Out

Analysis                      Geographical Information Systems
31