Introduction
Learning outcomes
Grid levels
Purpose of grid levelling
Contours and contouring from grid levels
Areas
Area of regular figures
Area of irregular shapes using graphical methods
Volumes
Volumes of regular figures
Volumes from contours
Volumes from grid levels
Introduction
Pick up a plan of a residential allotment. What do you see on the plan?
Dimensions of the lot, bearings and distances. If there are any easements
these will be shown also. Anything else? Probably not unless there are
some building improvements either proposed or in existence.
The normal house allotment plan usually does not give any indication of
the physical profile of the land. Is it a steep lot? Is it flat? How can this
type of information be shown on a plan?
In this unit you will be seeing how to illustrate on a plan the profile of the
land using contours.
We can use contours for:
• determining height of building foundations
• determining slope of the land for drainage
• identifying low points
• calculating earthwork quantities.
In this unit you will see how to use contours to determine volumes for fill
or excavation, and how to determine volumes from a grid of levels.
Learning outcomes
On completion of this unit you will be able to set out a grid for levelling.
Using the reduced levels of each grid point you will be able to:
• prepare a contour plan
• use the contours to determine areas
• use the contours to determine earthwork volumes
• determine earthwork quantities using just the grid points.
Grid levels
Purpose of grid levelling
The purpose of grid levelling is to establish a grid or pattern of levelled
points over the site. As we shall see a little later in this unit, this grid of
levels can be used for several purposes. The first step is to actually
establish the grid on which levels are to be observed.
Imagine that we have a building site for which some excavation is
required for a garage or basement for a proposed building. How much
material is to be dug out? We cannot dig the hole and count each
shovelful because we cannot as yet start digging. Before even the first
shovel is produced we must estimate the final total volume of excavation.
This means the actual volume will be not so much an estimate but a
considered ‘guesstimate’. With reasonable care in the calculations we can
be sure that our ‘estimate’ will be within about 10% of the actual value.
We obtain our estimate of the volume from plans of the site showing the
extent of the excavation and surface levels over the site. The plan will be
drawn to a scale of say 1:100 or 1:500. At a scale of 1:100 we have
immediately introduced a built in error factor of 1:100.
A pencil line of the plan 0.5 mm thick represents a line on the ground 50
mm wide. 50 mm on the ground is not much, but is the plot of the site
accurate to one millimetre? Now we are talking about a point on the
ground which may be up to 150 millimetres off its actual position. Note
that I am referring to scaled positions on the plan, not figured dimensions.
Now to the actual survey. Unless the point is tied to a physical boundary
mark by an actual dimension, the grid point to be ‘marked’ on the ground
can be up to 150 mm off true position. Will this affect the levels?
We are not marking the grid points by pegs. The staff will be placed on
the ground. Moving the staff position 150 mm or even 200 mm will not
make a significant difference to the reduced level of the point.
All this may sound a bit long-winded. The point that I am trying to make
is when setting out the grid, while every care should be taken, accuracy is
not the overriding requirement. There are times when fieldwork accuracy
is important. I have mentioned in an earlier unit how surveyors correct
measurements down to the nearest millimetre in order to accurately place
corner marks. The setting out of a grid is one occasion when accuracy is
not important. Take care, but do not waste time trying to be superaccurate.
Establishing grid levels
To establish a grid pattern over a site, the four outside limits of the grid
are marked by stakes. Let us look at an example.
Figure 1: Building site with extremities of grid
It is not necessary to mark the internal points, although if there are only
two of you and you have a ready supply of stakes, marking the internal
points does make life a lot easier for the unfortunate person who has to do
all the walking with the staff.
Step 1
At the first internal line, one person stands behind the first mark and lines
the person with the stakes through to the opposite side mark. Measure the
grid distance and place the stake. Move up to this stake and repeat the
process. This will eventually complete the marking of the first internal
line.
Step 2
Now you can dispense with the tape. For the second internal line, the
person placing the stakes has two stakes to line in. The other person lines
across to the two external marks. In this way the whole of the internal
grid points can be rapidly marked.
Figure 2: Marking internal points of grid
Remember I mentioned a little earlier that the grid points need j 91. only
to be placed to an accuracy of about 150 mm. Lining II&II through
between marks by eye is sufficiently accuracy.
Step 3
Now dust off the level and staff and read observations to each grid point.
Step 4
Next problem is identifying each grid point. Draw a sketch in your field
book showing the grid. Mark the grid lines in one direction with
alphabetic characters. Number the grid lines at right angles to the first
direction. Now you have a cross- reference for each point.
Step 5
Finally, don’t forget to complete the remarks column on the level pages
for each grid point giving the grid reference.
Figure 3: Site plan showing grid and grid reference
Contours and contouring from grid levels
One of the uses of a grid of levels is for the preparation of a contour plan.
A contour line is a line drawn on a map or plan joining all points of the
same height elevation or level. Contour lines indicate the physical shape
or form of the land.
A simple example
To see how contours are prepared from grid levels, let us take a very
simple example. Figure 4 shows a three-by-three grid with levels at each
grid point. I will use this simple grid to develop contours over the grid. At
this point, I am not concerned about the grid dimensions, only the levels
at each grid point.
Figure 4: Grid levels
For the contours over this grid, I am going to assume a contour interval of
one half metre—that is, as the lowest point on the grid has an RL of 38.7,
the first contour will be at 39 with subsequent contours at 39.5,40,40.5
and so on. Contours are always on the even metre and part metre.
Step 1
The first step in preparing a contour plan is to locate where each contour
will intersect each grid line. To illustrate this I will use the grid line from
grid point al to grid point a2.
We need to know the difference in level between each grid point. In this
example:
RL of a2 =412
RL of a1 =39.6
difference = 1.6
Step 2
Now measure the length of the grid line on the plan. Let us say that the
length of the line 10 cm. If we now divide the length of the line by the
difference in level, we have how far along the line an RL difference of
one metre will make.
10+1.6 = 6.25
The first level on the grid line is 39.6. The next contour interval above
39.6 is 40, a difference of 0.4 metres. Multiply this difference by 6.25 and
we have the distance along the grid line from the grid point to RL 40.
To RL4O 6.25x0.4=2.5
Measure 2.5 cm along the grid line and make a pencil mark on the line.
The next contour interval is at 40.5, half a metre from 40. Multiply 6.25
by 0.5 to find the next distance along the line.
to RL 40.5 6.25 x 0.5 = 3.25
Measure 3.25 cm along the grid line from the 40 mark to 40.5.
Step 3
Proceed the same way to 41. We cannot do 41.5, because this is in the
next grid. Figure 5 illustrates the line from alto a2 with the contour levels
marked.
Figure 5: Even contour intervals marked on grid line
Step 4
Proceed around each side of the grid, using the same method. Of course
the actual values will change as there will be different differences in the
levels between each grid point. Figure 6 shows the grid with the contour
levels marked on each side of the grid square.
Figure 6: Grid square with contour levels on each side
Step 5
Now join up the corresponding levels with a straight line.
Figure 7: Grid square with contour lines ruled in
Step 6
Repeat this process for each grid square (see Figure 8).
Figure 8: Four-element grid with contours ruled in
Figure 8 shows what it would look like if the contours were a series of
straight lines across the grid. As you know very well, contours are not
straight lines—they curve around hills and valleys. We need to ‘smooth’
the straight lines shown in the figure. This is done by sketching the
contour lines in freehand.
For instance, the contour line 41.5 will not come to a sharp point near
grid point b3. It will need to be flattened off. To do this means a reflex
curve on each side.
When we start to freehand the contours, we can also extend the contour
lines a little way outside the grid boundaries. This will give us a better
indication of the shape of the land.
Figure 9: Grid with smoothed contours
A final contour plan will not show the grid lines.
Figure 10: Contour plan
In Figure 10, I have shown the grid points only for your benefit. Grid
points are not normally shown on the plan. I have also numbered each
contour line. Note that contour lines are numbered facing up the hill. This
is the exception to standard drawing practice where all dimensions and
lettering should be read from the bottom of the page. Some contour
dimensions may be upside down when viewed from the bottom of the
page.
Check your progress 1.
Using the site plan shown in Figure 1 and in Figure 11, and from the table
of reduced levels, prepare a contour plan on Figure 11 with a contour
interval of 0.25 metres.
Figure 11: Building site boundary dimensions
When you have completed this activity, check your answers with those
given at the end of this unit.
Areas
Before we can attack a figure to find the area, we need a little revision of
some schoolwork. This is to find the area of regular- shaped figures for
which there are standard formulae.
Area of regular figures
Triangle
The area of any triangle is half the base times the perpendicular height.
Figure 12: Area of a triangle
In Figure 12 I have shown two triangles. Note that the formula is the
same for both even though the perpendicular height of the oblique
triangle is outside the triangle.
Rectangle
The area of a rectangle is given by the base times the width.
Figure 13: Area of a rectangle
Trapezium
The third figure we are concerned with is the trapezium.
A trapezium is a figure where two sides are parallel.
A parallelogram is a unique form of trapezium where the
opposite sides are parallel. In any trapezium, the ends do not
have to be parallel.
Figure 14: Area of a trapezium
The area of a trapezium is given by the square distance between the
parallel sides times the mean of the two parallel sides.
To summarise these formulae:
Area of triangle: A=h x b÷2
Area of rectangle: A = a x b
Area of parallelogram A = (w1 + w2) ÷2 x h
Area of irregular shapes using graphical methods
There are several methods for determining areas. Mathematical methods
and mechanical devices are outside the scope of this subject.
Unfortunately in real life, very few shapes are what we can call ‘regular’,
to which we can apply standard formula.
We can find the area of an irregular figure by applying the formula above
with a little bit of help and imagination. This method is called ‘graphical’
because we use a scale plan of the area and a ruler.
Example
Obviously determination of areas using graphical methods are not
accurate. For the purposes to which they are used, the areas are more than
adequate.
Figure 15: Irregular quadrilateral
Figure 15 illustrates an irregular-shaped quadrilateral. If we were to know
the lengths and bearings of each side, it is possible to calculate the area to
several decimal places. In this illustration we do not know the accurate
dimensions of the figure.
To find the area, simply divide the figure into triangles. I have chosen to
draw a line between A and C. On the diagram in Figure 15, I have also
shown the perpendicular heights from this dividing line to points B and
D. By measuring the lengths of the line AC and to B and D using our
scale ruler, we can calculate the area of the figure using the standard
formula for a triangle, half the base times the height.
length AC = 102 mm
on triangle ABC, height =49 mm
area =102 ÷ 2 x 49
= 2499 m2
on triangle ACD, height =47 mm
area=102÷2x47
=2397m2
Area of figure = 2499 + 2397
= 4896 mm2
This is easy. For a ‘regular’ irregular figure, simply divide the
figure into triangles.
What happens when we come to an ‘irregular’ irregular figure?
A typical example would be the surface area of a lake or of a quarry.
There are several techniques of which I will show three.
Areas by squares
Figure 16 shows an irregular shape for which the area is required.
Figure 16: An irregular-shaped figure with squares
Step 1
On the figure I have superimposed a grid of squares. You can use a
prepared sheet of tracing paper or draw the grid directly on the plan of the
site. The grid should be to an even dimension—for example, the grid
lines are 1 cm apart.
Step 2
Now count the squares but only the whole ones. In Figure 16 there are 37
whole squares within the figure. Go ahead and count. Do you agree with
me, or have you got only 35 squares?
Step 3
Now count the squares where more than half is within the figure. In
Figure 16, I make it 11 squares. If you had 35 squares for whole squares,
then you will have 13 part squares. Either way, when you add the two
together, you will come up with a total of 48 whole squares and squares
where more than half is within the figure. If you were working on a onecentimetre grid, the area of the figure is 48 square centimetres!
But what about all those squares of which less than half of the square was
in the figure? Forget them! if we take those squares where more than half
were in the figure and count them as whole squares, the difference is
cancelled out.
Areas by strips
The diagram in Figure 17 shows the same irregular shape as in Figure 16.
Figure 17: Irregular-shaped figure with strips
Across this figure I have drawn a series of parallel lines, all the same
distance apart.
Step 1
To find the area of the figure, measure the length of the dashed lines
which are along the centre of each strip. This length approximates the
mean of the lengths of each side of each strip. In other words, the length
of the dashed line is (w1 + w2) ÷2. Multiply this length by the width of the
strip and you have the area of each strip.
For example, if the length of the first dashed line is 4.2 cm and the width
of the strip is 2 cm, then the area of the strip is 8.4 square centimetres.
Step 2
Add up the areas of each strip and you have the area of the figure.
In Figure 17 you will note that the bottom of the figure only covers half a
strip. You will need to measure the distance midway between the top of
the strip and the dashed line. The dashed line forming the other side of a
strip which is half the width of the normal strip.
Areas by triangles with give-and-take lines
Figure 18: Irregular-shaped figure with triangles
Once again the same irregular shape. On this diagram I have drawn a
series of triangles.
Notice that where the side of a triangle is along the edge of the figure, I
have drawn the line partly inside the boundary and partly outside the
boundary. This approximates a balance of area, or a bit of ‘give and take’.
Step 1
Determine the area of each triangle
Step 2
Add up each area and you have the area of the figure.
There you have three different methods for the determination of areas by
graphical methods. The methods are not accurate, because they are
dependant on the accuracy of the plot of the figure and the size of the
squares, strips or triangles. If, for example, you are working with squares,
the smaller the square, the greater the accuracy.
Accuracy is also a function of scale. If the figure is a representation of the
extent of a lake surface and is drawn at a scale of 1:500, then a variation
in the ‘guesstimate’ of the mid- length of a strip of 1 millimetre
represents a variation of half a metre on the ground. As you have already
seen in the preparation of contours, a contour line is a freehand line which
represents an approximate location of the contour on the ground.
With any of these methods, the aim is to get within a 10% variation.
Check your progress 2
Using the figure shown in the Figures 16, 17 and 18, determine the area
using each of the three methods and compare your answers. Do you get
within a 10% variation?
Volumes
Volumes of regular figures
Cube
The volume of a cube is the length times breadth times height, or in the
diagram in Figure 19, a times b times h.
Figure 19: A cube
Trapezoid
Considering the volume in a slightly different way, the volume of a cube
is the area of the base times the height. If we can determine the area of
the base and we know the height, we can calculate the volume.
What happens with a trapezium? Same thing, area of the base times the
height.
Now we come to a different type of figure.
Figure 20: A trapezoid
A trapezoid is a block or figure where the planes making the top and
bottom surfaces are parallel. As you can see in Figure 20 the actual shape
of each plane is different. Let’s look at this same trapezoid in elevation.
Figure 21: Trapezoid, isometric and elevation
Looking at the elevation drawing, we can see that the figure
aa1c1c is a trapezium. It is only when we tip it over that we can see that it
is actually a trapezoid.
Back to the area of a cube. If we sum the area of the base and the area of
the top, divide by two we have the mean area. Of course this is nonsense,
one plus one is two divided by two is back to one. But if we consider this
in relation to the trapezoid?
Add the area of the base surface to the area of the top surface and divide
by two, and you have the mean area of the two surfaces. What is the area
of a trapezium’ The mean of the top and bottom lines times the height.
Using this same principle, the volume of a trapezoid is the mean of the
top and bottom surface area times the height.
We will now extend this simple rule further.
Volumes from contours
What if we had the area of two contours? Two contour lines actually form
the top and bottom surface of a trapezoid. The only difference between a
contour and the surface plane of the trapezoid in Figure 20 is that the
sides of the surface plane by contours are not regular.
Figure 22: Trapezoid with contour outlines as surfaces
Now we can combine the graphical methods for area determination with
volumetric calculations.
What is the volume of a gravel pit or quarry? Find the contour areas and
you have the volume. How much excavation? Find the contour areas and
you have the volume. Let me take you through an example.
Example
Figure 23: Volume in a quarry
Figure 23 shows a proposed quarry site on a contour plan. If the scale of
the plan is 1:500 (this may not be the exact scale due to reproduction
variations), what volume of material will be excavated?
Step 1
Let us look at the first two contour lines at the top of the excavation.
Figure 24: Excavation between contours 45 and 40
We can determine the surface areas at contours 45 and 40 using graphical
methods.
I will give you the areas at each contour line a little later. You can work
them out for yourself just to check me out!
From the plan:
• At contour 45 the area of excavation is 0.
• At contour 40 the area of excavation is 1085 mm2.
Step 2
Millimetres? Here is another problem, we need to convert the
measurements taken off the plan into ground measurements. Here the
problem is that the measurements we are dealing with are square units.
To find ground areas from plan areas, multiply the plan areas by the scale
squared
Here again we must be careful as the scale refers to metres and we have
measured in millimetres. Convert square millimetres to square metres
first.
1085 mm = 1085 ÷ 1000 ÷ 1000 (squared dimensions again!)
= 0.001085 m2
area = 0.001085 x 500 x 500
= 271.25, or 270 m2 (nearest 10 m2)
Step 3
Now we can determine the volume of excavation between contour 45 and
contour 40.
area at 45 = 0
area at 40 = 270
mean area =0+ 270÷2
= 135
volume = 135 x 5
= 675 m3
Step 4
Figure 25 illustrates the next calculation between contours 40 and 35.
Each contour interval must be taken individually.
Figure 25: Excavation of quarry
The calculation of volumes from contours is set out more conveniently in
a table.
Calculation check
By scale the dimensions of the floor at contour 25 are 37 metres by 37
metres. Note these are rough measurements—the aim is to check the
calculation above. This gives an approximate floor area of 1370 square
metres. The height is 20 metres. Using the triangle formula, the volume is
very approximately 1370 x 20+2= 13700 metres cubed. This gives an
indication that our calculations in the table are correct.
Check your progress 3
Use the areas shown in the millimetre column of the table above.
Determine the total volume of excavation if the plan scale was 1:1250
and not 1:500 as used in the example.
Volumes from grid levels
This is a specific application where the depth of excavation is known at
each grid point. A typical example is the excavation for an underground
car park at a commercial development site. Figure 26 shows a grid with
reduced levels on natural surface at each grid point.
Figure 26: Grid of levels
Method
If the floor level of the excavation is to be at RL 6, what wifi be the
amount of excavation? The method is to treat each grid square as a cube.
Figure 27: Grid cube
Step 1
To find the volume of the cube, average the four mean depths to find the
mean depth of the cube. In Figure 26:
mean depth=(a+b+c+d)÷4
The volume is given by the mean depth times the area of the base.
Step 2
Let me take the top left hand grid square as an example. Assume that the
grid is marked out at 2.5 metre intervals. The depths of excavation at each
corner of the grid are:
a (al) = 3.14
b (a2)= 3.25
c (b2)= 3.42
d (b1)= 3.38
total = 13.19
average = 13.19 ÷ 4
= 3.2975 or 3.3 to one decimal place
volume = 3.3 x 2.5 x 2.5 (area of square = 2.5 x 2.5)
= 2.06 m3
We could proceed over the whole grid by this method which would be a
rather tedious process. Lets shorten it up a bit.
Step 3
We must treat each grid square individually. We cannot just average all
the depths as this will give an incorrect answer. If we move on to the
second grid square, a2, a3, b3, b2:
a2=3.25
a3=3.31
b3=3.60
b2=3.42
We have already used a2 and b2. Why not combine them? Now we have
the makings of a formula:
Volume = (a1+2xa2+a3+b3+2xb2+b1)÷4xarea of individual square.
Does it work? Let’s try it and see but first the volume of the second cube.
Volume of second cube = (3.25 + 3.3 + 3.60 + 3.42) ÷ 4 x 2.5 x 2.5
=21.22m3
Total Volume for the two cubes = 41.82 m3
A simplified method
Now for the simplified method.
3.14 + 2 x 3.25 + 3.31 + 3.6 + 2 x 3.42 + 3.38 = 26.77
Divided by 4 and then multiplied by 2.5 = 41.42m3
We can even simplify it further. When working out the whole volume, all
the internal grid points will be used four times. In our example b2 and b3
are common to four grid squares. Multiply all the internal height
differences by four. In other words, multiply all height differences by the
number of time they are used. The four corners are used once.
Finally, do not try to show volumes to a decimal of a cubic metre (unless
you are working with a shovel and wheelbarrow!). In excavations
involving earthmoving equipment, something a little larger than a
barrow—you cannot guarantee volumes of excavation to even the nearest
truckload. Unless it is a very small amount, round off volumes in your
calculations to the nearest 5 or 10 cubic metres.
Check your progress 4
Complete the calculations for volume from the grid levels in Figure 26.
Check your answers with those at the end of this unit when you have
finished.
Answers to check your progress
Check your progress 1
Figure 28: Cøntour plan
Check your progress 2
Your answers may vary slightly from those given. This could be due to
reproduction variations of the diagrams and also minor variations in
scaling the measurements.
Areas by squares
As I noted in the text of this lesson there are 48 squares. The grid is 1.5
centimetres square so the area of the figure is:
48 x 1.5 x 1.5 = 108 cm2
Area by strips
Note the strips are 2 centimetres wide. My measurements for each strip
are:
4.4 + 8.5 + 12.7 + 13.6+12.2 = 51.4
area = 51.4 + 9.4
= 112 cm2
This agrees very well with the previous area.
Area by triangles
My scaled triangle dimensions, half base times height are:
2.9 x 3.0 ÷ 4.0 x 2.2 ÷ 7.0 x 5.8 ÷ 7.0 x 3.3 + 6.4 x 4.1 = total area 96 m2
Perhaps my ‘give and take’ is more take than give, although a 10%
variation from 108 (area by squares) gives 97 cm2, which is very close to
96. As I mentioned earlier, scaled areas such as these cannot be
guaranteed to be within 10%. Hence it is advisable to determine the area
of irregular figures by two different methods for comparison.
Check your progress 3
Contour Area in mm Area in metres Mean area Volume
45
0
0
40
1.085
1.695 (1 700)
850
4.250
35
2.290
3.578(1580)
2.460
1.2300
30
3.375
5.273(5270)
4.425
2.2125
25
4.380
6.844(6840)
6.055
3.0275
Total
Volume of excavation = 69 000 m3
Check 84x84÷2x20=7O500 (OK)
68.950
Check your progress 4
Levels used once:
3.14 + 3.7 + 4.45 + 4.08 = 15.37
Levels used twice
3.25 + 3.31 + 3.39 + 3.45 + 4.01 ± 4.23 + 4.26 + 4.30 + 4.38 + 4.32 +
4.21 + 4.17 ÷ 3.87 + 3.69 + 3.51 + 3.38 = 58.35
Levels used four times:
3.42 + 3.60 + 3.75 + 3.88 + 3.55 + 3.68 + 3.83 + 4.12 + 3.77 + 3.79 +
3.87 + 4.00 + 3.95 + 4.03 ± 3.15 + 4.21 = 60.6
Volume: