Cirlce left Circle right
96º 43’ 20” 276º 43’ 00”
6.3.4 The following directions were observed from P toQ. Reduce the mean direction
Cirlce left Circle right 7º 01’ 00” 187º 00’ 40”
5.2 CO – ORDINATES (JOIN CALCULATIONS)
SA co-ordinate system –x (180º)
((θ + 90) (θ + 180)
+y (90º) – y (270º)
θ (θ + 270)
+x (0º and 360º) FORMULAE
∆yAB =yB −yA
∆xAB =xB −xA
Note: If given AB ( meaning from A to B) then its B minus A If given BA ( meaning from B to A) then its A minus B
Distance or length
AB = (∆y)2 +(∆x)2
Direction or angle
θ =
∆
− ∆ x
1 y tan NOTE:
2. To get the direction ( D) use the following
2.1 If is positive and ∆ is positive, we are in the 1 quadrant
The co-ordinates of point A and B are Y X
A – 248,17 – 58,47
B – 150,27 – 260,80
Calculate the orientated direction and distance between A and B (
Distance from A to B
2
∴
Direction = 25º 49′ 14,24″ + 270 = 295º 49′ 14,24″Question
The co-ordinates of point A and B are Y X
A – 248,17 – 58,47
B – 150,27 – 260,80
Calculate the orientated direction and distance between B and A (
Distance from A to B
2
= 115º 49′ 14,24″
EXERCISE 7
7.1 The co-ordinates of point A and B are Y X
A +310 248,17 – 1 058,47
B +309 295,17 – 1 688,04
Calculate the orientated direction and distance between A and B
7.2 The co-ordinates of point A and B are Y X
A –467,89 +120,45
B –120,45 +467,89
Calculate the orientated direction and distance between B and A
7.3 Study the information below showing point that formulate boundaries of a property. Calculate the lengths of all the side of the property
SIDES (Metres) ANGLES OF
DIRECTION
B
728
C Scale 1/1000
The figure A B C D E F represents 2100 square metres of land being
ERF PORTION OF ERF WALMER
Situated in the municipality and Administrative District of Port Elizabeth surveyed in June 1981
This diagram is annexed to No
Dated i.f.o
Registrar of Deeds
The original diagram is
No 7494/1981 annexed to Transfer/Grant
No
File No s/7902/94 S.R No
Comp. BO-8CC/x43
CHAPTER 6
SETTING OUT
6.1
How a traveller is used with profiles to control excavation and foundation levelsa) Two profiles ate place at each end of the excavation.
b) A small cross bar is fixed at each profile at a level equal to the invert level of the
excavation plus the traveller (follower)
c) The depth if the excavation is therefore controlled by dipping the traveller such
that the line of sight between the two bars of the profiles is in line with that of the
cross bar of the traveller
6.2
How to set up a rectangular site along a road and use the road as a reference on the site plana) Offsets must be taken from the road reserve to the boundary of the site that is required
b) A baseline parallel to the road drawn.
c) The corresponding chainage of the site boundary are then identified
d) the required site shape is then set out from the baseline
6.3
On site, how to set out two points A and B stationed on a third point known co-ordinatesa) Set up the instrument and calculate the direction between the point on which the instrument
is set up and point A and the distance between the two points b) Also calculate the direction between the point on which the instrument is set up and point B
and the distance between the two points
c) Orientate direction from instrument to point A and measure the distance and put a peg
d) Swing the instrument to point B using the calculated direction of point B and measure the
distance from the instrument to point B and put a peg
6.4
How to set out a rectangular proposed building using simple surveying instrument forearthworks
Equipment: 100m tape, Four steel pegs or droppers ±2m long, Lime, Fish line,
Levelling instrument. Traveller ±2m high
Measure the distance of the proposed structure from all the four corners and make it ±1m less
From each side of the building.
Put the steel pegs or droppers on these new found points and mark them for a 2m traveller
taking the 150m depth of the top soil into consideration
The area to be removed top soil is the one that is indicated by ABCD
A B
C D
6.5 CUT AND FILL
Ground level
Depth of excavation at A A
Depth of excavation at B
Invert level at A Δh
B Invert level at B
Steps to follow:
1. Calculating the Fall ( difference in height)
Δh = gradient x distance from first to the last point
2. Calculating the invert level of the first point Invert level at A = ground level – excavation
3. Calculating the invert level of the last point
Invert level at B = invert level at A – Δh
4 Calculating invert level at other points e.g. chainage is 00 20 40 55 60
invert level at 00 is the invert level of the first point invert level at 60 is the invert level of the last point Invert level at 20 = invert level at A – gradient (20) Invert level at 40 = invert level at A – gradient (40) Invert level at 55 = invert level at 40 – gradient (55) 5. How to determine Cut or Fill
Ground level of a point – invert level of that point = Cut or Fill If the answer is positive then it is a Cut
If the answer is negative then it is a Fill
EXAMPLE
Given :
Chainage Ground level A 0 50,76
20 50,00
40 45,75
45 47,28
B 60 49 20
The information above refers to a drain between point A and B. The depth of excavation at A is 1,96m
and the fall from A to B is 1:40. Calculate the Cut and Fill in metres 1. Δh = gradient x distance
Δh = 0,025 ×60 NOTE:
Δh = 1,5m
2. Invert level at A = 50,76 – 1,96
= 48,8 m (write this figure in the invert level column as shown in table 1)
3. Invert level at B = 48,8 – 1,5
= 47,3 m (write this figure in the invert level column as shown in table 1)
4. Invert level at 20 = 48,8 – 0,025(20)
= 48,3 m (write this figure in the invert level column as shown in table 1)
5. Invert level at 40 = 48,8 – 0,025(40)
= 47,8 m (write this figure in the invert level column as shown in table 1)
6. Invert level at 45 = 48,8 – 0,025(45)
= 47,675 m (write this figure in the invert level column as shown in table 1)