Introduction Learning Outcomes Grid Levels Purpose of Grid Levelling Contours and Contouring

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 super- accurate.

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.

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 one- centimetre 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.

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.

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

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.

Contou Area in Area in Mean Volume r mm metres area

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 68.950

Volume of excavation = 69 000 m3 Check 84x84÷2x20=7O500 (OK) 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: