Horticulture Maths Resource
Mathematical resource for Horticulture
Kevin Albert
Land Built Environment & Sustainability
kalbert@wodongatafe.edu.au
(02) 6055 6758
© Wodonga TAFE 2012
Horticulture Maths Resource
Table of Contents
2
Plan interpretation and Setting out
Calculation basics
Landscaping
Turf and Parks & Gardens
Water Harvesting and Irrigation
Business
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Plan interpretation & Setting out
Index
3
Select from following links:
• Setting out 900 angles using 3:4:5 triangles
• Scale
• Using Scale Rulers
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Setting out a 900 angle using “3:4:5” triangle
Task - We are going to set out an area for landscaping where we need a square corner.
On small jobs we might be able to use a builders square to start us off but there are
many situations where the old ‘3: 4: 5 triangle’ comes in handy
4
Considerations
•
So what is a 3:4:5 triangle? You may have heard of “Pythagoras theory” where the square root
of the diagonal of a triangle equals the sum of the squares of the other two sides
Eg
(a)= 3
(c) = 5
a2 + b2 = c2
32 + 42 = 52
or 9 + 16 =25
(b) = 4
•
The 3: 4: 5 is a ratio between the sides of the triangle It can be multiplied out by 2, 10 or 100
or any number so long as all sides are multiplied the same i.e. 300mm: 400mm: 500mm. So if
you refer to 300mm as 3 units in this case or in other situations you may use 3m: 4m : 5m
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Setting out a 900 angle using “3:4:5” triangle
Guidelines
5
1. Determine your starting point
2. From the starting point peg out a line in the most critical direction. This will
be your base line and shouldn’t change (this may be parallel to a building, path
or fence line)
3. Measure 3 units along this line (a unit can be 300mm or 3m depending on
the length required)and mark each end i.e. with pegs
4. Set out an approximate right angle to the base line (a right angle is 90
degrees from the base line)
5. Measure along the 2nd line a distance of 4 units from the zero end of your
first line and scribe a small arc at the required distance
6. Now measure 5m across the diagonal from the 3m end and scribe another
arc. Where the arcs cross should be the final corner and you can set out the
right angle
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Setting out a 900 angle using “3:4:5” triangle
Sample
6
• Step 1. Determine your starting point , i.e. the back corner of the house.
Place a peg (note: if using timber pegs you will need to place a nail in the
centre of the top once hammered in)
house
Start point
• Step 2. From the starting point peg out a line in the most critical direction
this will be your base line and shouldn’t change
In this case use the back of the house as the base line
• Step3. Measure 3 units along this line in this case use 3 metres and install a
peg with the nail at exactly 3m from the first peg
(a)= 3 units
3
0
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Setting out a 900 angle using “3:4:5” triangle
Sample
7
• Step 4. Set out an approximate right angle to the base line (a right angle is
90 degrees from the base line)
• Step 5. Measure along the 2nd line 4 units
In this case use 4 metres and install a peg
with the nail at exactly 4m from the first peg
house
0
(a)= 3 units
3
(b) = 4 units
4
© Wodonga TAFE 2012
Setting out a 900 angle using “3:4:5” triangle
Sample
• Step 6. Now check across the diagonal
from the peg at the end of the 3m line to
the peg at the end of the 4m line. This
should measure 5 units – if not, adjust
the 2nd line until the 4 units out is 5
units exactly on the diagonal
8
house
(a)= 3 units
(c) = 5 units
(b) = 4 units
• In this case the measurement should be
exactly 5 metres from the nail in each
peg
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Scale
Task - You have been asked to set out some landscape works and
need to use a scaled plan
9
Considerations
Scale means to reduce or increase something in its size in equal proportions.
eg 1 a matchbox cars are often models of real cars but 33 x smaller
i.e. made to a scale of 1 : 33
eg 2 House plans are usually drawn 100 times smaller than the real size
i.e the plan is drawn at 1:100 scales
The bigger the number, the smaller the scale and the more surface area of the ground can be shown, i.e.
•
A large scale covers a small area or close up item i.e.
•
A detail drawing of a planter box may have a scale of 1:10
•
A plan of an oval may be at 1:500 or 1:1000 scale
•
A small scale covers a large area of land i.e
•
The topographic map of a district may have a scale of 1:55 000
•
A map of a town may be a scale of 1: 10 000
•
A scale of 1:1 means that the item is in real size
To read scale you can use a scale ruler. To draw to scale you can also use graph paper where each square
represents a given measurement i.e. 1 m
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Scale
Sample
10
A small scale covers a large area of land i.e 1:55000
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Scale
Sample
11
Source: http://media.photobucket.com/image/recent/warren1605/My%2520Cars/P1010002.jpg
A large scale covers a small area or close up item i.e. 1: 10
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Using a Scale ruler
Task - You have been asked to set out an area of a landscape
plan and need to use a scale rule
12
Considerations
• To read scale you can use a scale rule. A scale rule has a range of different
scales that you can select from to suit the plan you are working with.
• i.e. If you were trying to find the length of say a wall of a house,
• select the appropriate scale (look in the title block of the plan or under the
particular drawing)
– 1:100 or 1:200 are common
– 1:100 means 1 cm on the plan is 100cm or 1 m on the ground.
• place the 0 end of the scale on one end of the wall on the plan and then
• read off the number that is lining up with the other end of the wall.
• You should notice that the numbers will be in metres not millimeters
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Using a Scale ruler
Sample
13
Here is an example using a house plan
The bottom scale
Is 1:100, numbers read
as 1m, 2m etc
Length of the wall is 5.2m
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Using a Scale ruler
Sample
14
• 1. Select the appropriate scale
•
2. Place the 0 end of the scale on one end
• 3. Read off the number that is lining up with the other end
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Calculation Basics
Index
15
Select from following links:
•
•
•
•
•
•
•
•
•
Surface area
Surface area – squares & rectangles
Surface area – triangles
Surface area – circles
Circumference of a circle
Volume of materials
Linear Measurements
Gradients
Terminology and Conversions
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15
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Surface area
Task - You are to work out the amount of materials such as pavers asphalt
16
or turf for the area you are working in which needs to be covered.
Considerations
•
•
•
•
There are four things which you will need to find out
The length of the area to be covered (which is written as L).
The width of the area to be covered (which is written as W)
You will need to multiply the length x the width to get the total area to be
covered
• i.e. by calculating the length (L) x width (W) of a square or rectangle or for
more involved shapes, refer to the following formulas
• The total area to be covered is written as square metres or (m2) this tells
you that it is has 2 dimensions.
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Surface area
Formulas
17
Finding the area
• Area of a rectangle = length  width
• Area of a triangle = ½ base  perpendicular height
• Area of a trapezium = ½ (a + b)  perpendicular height
• Area of a circle = 3.14  r2 (or   r2)
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Surface area – square / rectangle
18
Sample
How do you work out the area to be covered in your work area?
– You need to carry out a number of steps
• Find out the width of your area by measuring from A- B
–
This will give you 1 metre
• Find out the length by measuring from B-C
–
This will give you 1 metre
• You then use the formula of Length Lx Width W = Area m2
1.000
– Which will be 1mx1m= 1 m2
1.000
• This means the material to be ordered must cover 1 m2
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Surface area – square / rectangle
Task - You are about work out the amount of materials such as pavers asphalt or turf
19
for a triangular area you are working in which needs to be covered.
•
•
•
•
•
•
•
•
•
Finding the area
Area of a triangle
= ½ base  perpendicular height
H (the height is the
distance from tip to
base )
B(base)
Area triangle = B X H ÷ 2
There are three (3) things which you will need to find out
The length of the base of the triangular area to be covered (which is written as B).
The perpendicular height of the triangular area to be covered (which is written as
H)
You will need to (multiply the base x the height) and then divide the answer by 2
to get the total area to be covered
Note: The total area to be covered is written as square metres or (m2) this tells
you that it has 2 dimensions to the calculation.
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Surface area – Circle
Task - You are about work out the amount of materials such as pavers asphalt or
20
turf for a circular area you are working in which needs to be covered.
• Finding the area
• Formulae
• Area of a circle =3.14  r2 (or   r2)
r = centre to
outer edge
• There are three (3) things which you will need to find out
1. The radius of the circular area to be covered (which is written as r).
•
The radius is the measurement from the center of the circle to the
outside edge
2. The “square” of the radius. (to find the square you multiply the radius by
itself : r x r = r2)
•
i.e. if the radius was 2, multiply 2 x 2 = 4
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Guidelines
Surface area – Circle
21
3. You will need to multiply the answer to r2 x “Pi” () to get the total area to
be covered
• (Pi or  is a give figure which is a relationship between the circumference
of a circle and its diameter – it is expressed as 22 divided by 7 or in
decimal as 3.14)
• Area of a circle = 3.14 x r2 = m2
• using above example Area = 3.14 x 22 = m2
= 3.14 x 2 x 2 = 12.56m2
• Note: The total area to be covered is written as square metres or (m2) this
tells you that it has 2 dimensions.
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Surface area – Circle
Sample
•
•
22
How do you work out the area to be covered in your work area?
You are going to pave a circular area with a diameter of 8 metres. How many
pavers would you need to order
8m
Step 1
• Radius is radius is the length from the center out.
• Diameter is the total length from one side of the circle to the other.
• Radius is half the diameter
• Half of 8 = 4m
Step 2 The square of the radius is the radius multiplied by itself i.e. 42 = 16
• or 4m x 4m = 16m2
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Sample
23
Surface area – Circle
Step 3 Using r2 ,
where  = 3.14,
• therefore and r2 is 4 x 4 = 16
–
r=4
3.14 x 16
= 50.24 m2
•
•
•
•
•
•
Note:
Normally we would allow 5 to 10 percent for wastage and broken paversWe will allow 10 % for easier calculation in this exercise
10% of 50.24 = 5 .024m2
50.24 + 5.024 = 55.264
We will round this off and order 55m2 of pavers
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Circumference of a circle
Task - You need to find out how many pavers will be needed to
go as a border around a circular paving area
24
Consideration
• The circumference is the distance around the outside of a circle
• The formula for the circumference is d x π where:
π = 22 divided by 7 (or in decimal, 3.14)
d = the diameter of the circle
• Once you have the circumference of the circle, divide it by the width of the
pavers you will be using
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Circumference of a circle
Sample
•
25
If the circle of paving was to be 4m diameter
Using the formulae dπ (diameter x pi ) [or you may also have learnt it as 2 x pi x radius (2πr]
= 4m x 3.14
= 12.56m
In other words, it is 12.56metres around the outside of the circle
•
How many pavers are required?
The paver is 110mm or 0.11m and the gap is 5mm or 0.05
Circumference ÷ width of the paver + gap
= 12.56 ÷ (0.11 +0.05)
=12.56 ÷ 0.115
=109.2 pavers
Round up to the number of whole pavers = 110 pavers
•
Add 10% for wastage
= 110 + 10 %
= 110 + 11
=121 pavers
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Volume of material
Task - The quantities of many landscape materials such as soil, roadbase, sand, concrete
26
and mulch are worked out by calculating the volume of space that the material takes up.
Considerations
• The quantities are worked out by
• calculating the surface area of where the material is to cover (m2)
• Then multiplying by the required depth (D) or thickness
• Some of the simple shapes that we normally have to deal with in landscape
construction are
Area
Area
Area
Depth
•
Depth
The answer will be expressed in m3. The 3 represents the 3 dimensions that have
been used in the calculation
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Volume of material – rectangle / square
Sample
27
Example 1
• Problem:
• To find the volume of concrete in the simple rectangular driveway 100 mm
thick.
• Figure 1: Volume of a slab
• Volume = length x width x thickness
• = 7.850m x 2.500m x 0.100m
• Convert all dimensions to metres.
– 7850 mm length = 7.85 m
– 2500 mm width = 2.5
– 100 mm depth = 0.100 m
•
= 1.96 m3
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Volume of material
Sample
28
Example 1 continued
• Next we should add an allowance of say 10% to make sure that we have
enough concrete for the job.
• 10% of 1.96 (0.10  1.96 m3) = 0.196 m3
• Round to 0.2m3
• Now add
• 1.96 m3 + 1.96 m3 = 2.16 m3
• You could round this off to 2.2 m3.
• If you are ordering ready-mix concrete you will need to round it off to 0.2
of a cubic metre as ready-mix concrete suppliers will not accept orders in
smaller fractions of a cubic metre.
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Volume of material – trapezium
29
Sample
Example 2
• Problem: To calculate the volume of concrete required for the pedestrian pavement
75 mm thick shown below.
6750
3340
75
•
•
•
•
•
•
•
There are two different ways you could do this one.
(a) Treat it as a trapezium:
Volume = Area x thickness
= ½ (a + b) x perpendicular height x thickness
= ½ (10.09 + 3.34) x 2.1 x 0.075
= ½ x 13.43 x 2.1x 0.075
=1.06 m3
3340
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Volume of material – triangle / square
30
Sample
Example 2 cont….
• (b) Alternatively you could break the area up into a triangle and a rectangle as
shaded:
6750
3340
75
3340
Volume = ( Area of triangle + area of rectangle ) x thickness
= [ (½ base x height) + (length x breadth) ] x thickness
= [ (½ x 6.750 x 2.100) + (3.340 x 2.100) ] x 0.075
= [ 7.088 + 7.014 ] x 0.075
= 1.06 m3
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Volume of material – triangle / square
Sample
31
Example 2 continued
• With both methods of calculation it is important to remember to complete
calculations inside the brackets first, starting with the inner brackets.
• Again add 10%,
• 1.06 x 10% = 0.16
• 1.06 + 0.16 = 1.22 m3
• So order 1.2 or 1.4 m3
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Volume of material – circle
32
Sample
Example 3
• Problem: To calculate the volume of concrete required for a circular pavement slab
shown below
2700 diameter
125
•
•
•
•
•
Volume = Area x thickness
= ( 3.142 x r2 ) x thickness ( where  is roughly 3.14 and the radius is half the
diameter)
= (3.142 x 1.350 x 1.350 ) x 0.125
= 0.72 m3
Add 10% for error (which = 0.10 m3)
•
•
Answer = 0.80m3
Order at least 0.8m3 if not 1 m3.
© Wodonga TAFE 2012
Lineal Measurements
Task - You need to order materials for a project that has single line to
33
measure, i.e. border of pavers , irrigation or drainage pipes, fencing etc
Considerations
• The linear measurement is simply a measurement along a line, so weather
on the site, or scaling off a plan, measure the distance along the line where
the material is to go, i.e. each edge of the area to be paved or each line of
irrigation ( you may mark out with paint on the ground first so you know
where to measure
• Lineal measurements don’t need other dimensions of width or depth
• total up all the measurements
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Lineal Measurements
Sample
34
Example 1 - Calculating for garden borders border
• Set out and mark where gardens are to go using either or a string line for
straight edges or a garden hose for curved edges and a can of marking
paint
• Measure along each edge of each garden and record them in some
systematic way.
• Total up all measurements
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Lineal Measurements
Sample
35
Example 2 - How much drainage line for a job
• Set out and mark where drainage or lines are to be installed.
• Measure along each line with a tape measure or measuring wheel and
write the measurement for each line down on a piece of paper or in a
notebook (if you only need a rough idea, you could even step it out. to do
this more accurately you can measure out the distance of your each step in
your stride.)
• Total up all measurements making sure you have a measurement for all
lines
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Lineal Measurements
Sample
36
Example 3 - How much pipe for an irrigation job
• as above, set out and mark where irrigation lines will be installed ( you may
use different colours for if you are to use different sized pipe i.e. mains and
laterals
• Measure up each line and record the different colours separately
• Total up all measurements for each coloured line so you geta total length of
pipe required for both mains and laterals
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Lineal Measurements
Sample
37
Example 4 how much wire and how many posts for a fence
• Set out and mark where fence lines are to run
• Measure the line or each line if more than one
• Total up all measurements
• This is where it gets a little different depending on what
• type of fence you are installing
• i.e. for a rural post and wire fence, for the wire multiply the total length by
the number of strands required of each type of wire eg. if 100lm of fence x
4 strands of plain wire would = 400lm of plain wire
• or 100lm of fence with 2 strands of plain and 3 of barbed wire would be
200lm of plain wire and 300lm of barbed wire
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Lineal Measurements
Sample
38
Example 4 continued
• (allow extra for slack before straining and to wrap around stays)
• For posts divide the total length of fence by the required distance between
each post to find the number of posts
• Eg for our 100lm of fence if using star steel posts at 4lm apart, 100 / 4 = 25
posts plus a strainer and stay at each end
• For a colorbond fence which come in a panels 2.36lm long (each panel has
2 end posts+ top and bottom rail + 3 sheets of tin), divide the total length
by the length of a standard panel eg 30lm of fence / 2.36m per panel =
12.71 panels. (so you would order 13 panels and cut the last one to fit)
© Wodonga TAFE 2012
Gradients
Task – you need to run a storm water pipe drain so that water runs in the
39
right direction to the outlet or you need to construct a disabled ramp
Considerations
• Gradient is a ratio between the vertical height and the horizontal distance
• Determine the required gradient for the task
• Determine the horizontal distance to be travelled i.e. The length of a ramp or
pipe line
• Determine the vertical height available or required i.e. The difference between
the top of a deck and the surrounding ground or the depth of an available pipe
inlet to connect into and the minimum depth you can start your drain pipe
• You will need the formula F = L x G
G
• F = fall in metres
F
• L = length
• G = gradient
L
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Gradients
Sample - known length and gradient
40
Eg 1 – you have a pipe to lay at 1 : 100 gradient over 12m distance
• Apply the formula
• H=LxG
• H = 12.0 x 1/100
G = 1: 100
•
= 12/100
•
= 0.120m or 120mm
L =12.0
H=?
H = 120mm
© Wodonga TAFE 2012
Gradients
41
Sample - known height and gradient
Eg 1 – you have a ramp to install for disabled access which needs to conform
to Australian standards of a minimum 1:14 with a difference in levels of
600mm (or 0.6m)
• L=F/G
G = 1: 14
• L = 0.6 / (1/14)
H = 0.6
• L = 0.6 / 0.07
L =?
• L = 8.57m
• Therefore the ramp would need to be at least 8.6 m long to have the
required minimum gradient
L =8.6m
© Wodonga TAFE 2012
Terminology and conversions
Task - You are working out the hydraulics for an irrigation system or water feature
42
and need to know some of the important terminology as different catalogues talk
in different terms
Some Typical Terminology and abbreviations
•
psi = pounds per square inch
•
Gpm = gallons per minute
•
Head = downward pressure of gravity on water
•
Gph = gallons per hour
•
ft = foot
•
Ltr
= litre
•
m = metre
•
L/s
= litres per second
•
kPa = kilopascals
•
L/min = litres per minute
•
Hp = horse power
•
L/hr = litres per hour
© Wodonga TAFE 2012
Terminology and conversions
43
Useful Conversions - Fluids
• 2.3 ft head = 1 psi
• 10 m head = 14.5 psi = 1 bar = 100kPa
• 1 m head = 1.45 psi = 0.1 bar = 10 kPa
• 1 imperial Gpm = 4.546 L/m
• 1 USA Gpm = 3.787 L/m
(1 L/m = 0.219 Gpm)
(1 L/m = 0.264 Gpm)
Useful Facts
• 1 litre = 100mm3 (100mm x 100mm x 100mm) = 1kg
• 1000 litres = 1m3 (1m x 1m x 1m) = 1000kg
= 1 tonne
• Some pump sizes are rated by the Gph at 1 foot of height (L/hr @ 1m)
© Wodonga TAFE 2012
Landscaping
Index
44
Select from following links:
• Calculating steps Sizes
• Excavating for paving
• Calculating pavers for a border / coping
•
•
•
•
How many bricks for a wall
Quantities of ingredients for mortar
How to calculate a timber order
Selecting a suitable pump for a water feature
© Wodonga TAFE 2012
Step sizes
Task - You have been asked to construct a set of steps as part of landscaping a back yard. The steps
45
need to be a safe and comfortable size for the client, as well as their family and friends
Considerations
• The building codes and Australian standards
set out standard dimensions that must be
used for step construction so as to ensure
safety of users
• The formula used is 2r + t = between 590
and 700. In other words the height of 2
individual risers (r) plus the distance of one
tread (t) added together should give a total
dimension (=) between the accepted range
of 590 to 700 mm.
• The ideal total dimension is 650mm
© Wodonga TAFE 2012
Step sizes
Guidelines
46
• The accepted range of individual risers is between 110 and 190mm,
with 150 being optimum
• The accepted range of individual treads is between 250 and 450,
with 350 being optimum
• As you can see there is room for flexible design.
There is also a relationship to notice:
• On gentle slopes, riser heights will be smaller (<150) and tread dimension
larger (>350)
• On steeper slopes, riser heights will be
larger (>150) and tread dimension smaller (<350)
© Wodonga TAFE 2012
Step sizes
The process is as follows:
47
• 1. Find dimensions of flight
(For simple staircase without landing/s)
• 2. Find how many steps fit the flight
• 3. Calculate sizes of individual risers and treads
• 4. Check that the sizes fit into the allowed range by using the basic formula
2r + t =590 to 700
© Wodonga TAFE 2012
Step sizes
48
Sample
1. Find dimensions of flight (For simple staircase without landing/s)
• Take levels to work out the height of the rise of flight ( R)
• Measure the horizontal distance of the embankment, or desired route to
find the going of the flight (T)
• Example
R= 1200
T = 2500
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Step sizes
Sample
49
2. Find how many steps fit the flight
• To find the number of steps, use the formula and divide by the ideal total
dimensions by 650
•
i.e. 2R+T ÷ 650
• this will give us an answer of approximately how many steps will fit our
flight
•
eg 2 x 1200 + 2500 ÷ 650
•
= 2400 + 2500 ÷ 650
•
= 4900÷ 650
•
= 7.53
• The answer will rarely have a complete number, so remembering that
there is always one more riser than treads, so there would be 7 treads and
8 risers in the above flight
© Wodonga TAFE 2012
Step sizes
Sample
50
3. Calculate sizes of individual risers and treads
• 1st Divide the rise of the flight (R) by the number of risers
= the size of each riser
eg 1200 (R) / 8 = 158mm (r)
• 2nd Divide the going of the flight (T) by the number of treads = the size of
each tread
•
eg 2500 (T) / 7 = 357.14mm (t)
© Wodonga TAFE 2012
Step sizes
Sample
51
• 4. Check that the sizes fit into the allowed range by using the basic
formula 2r + t =590 to 700
• - if yes then you can proceed to construction phase,
- if no, then you would need to select the next number of steps , for both
treads and risers in the flight either up or down until dimensions do fit the
formula
•
•
eg 2 x 158 + 357
or 316 + 357 = 673 (this fits within the range)
© Wodonga TAFE 2012
Excavation for Paving - calculation
Task - Before excavating an area to be paved, we need to get an
idea of how much material will be dug out.
52
Considerations
• How much area is to be excavated? (Area m2)
• How deep will you need to excavate to.(Volume m3)
• How much soil will you need to remove from the work site? (Bulking
Factor)
• How much will the soil weigh? Convert volume (m3) to mass I.e. tonnes (t)
This is important so we can decide whether to use a machine or dig by hand
and also so we know if we need to organise a truck - if so how big a truck
© Wodonga TAFE 2012
Excavation for Paving - calculation
53
Guidelines
1.
•
How much surface area is to be excavated? (Area)
You will need to measure the area you want to excavate. You will need to know the
length (L) and the width(W) of the area
L
•
•
•
•
W
You then workout the Total Area to be excavated by multiplying the length (L) by
the width (W)
Length (l) multiplied by the width (w) = Total Area m2
l x w = A (m2)
This is referred to in square metres and is written as m2
© Wodonga TAFE 2012
Excavation for Paving - calculation
Guidelines
2.
54
How deep will you need to excavate to and what will be the Volume
•
•
•
•
•
To work this out you will need to know
How thick your paver is
How thick you want to have the bedding sand
How thick your base material needs to be
This will be the depth of the excavation
•
Depth of the excavation = paver thickness + Bedding Sand thickness + Base Material
thickness
Paver
Bedding Sand
Base Material
Natural Ground
© Wodonga TAFE 2012
Excavation for Paving - calculation
55
Guidelines
3. How much soil will you need to remove from the work site?
•
•
•
So what is a bulking factor?
- Different types of soil become larger in size once they are dug up, as they get loosened up,
they generally fluff up or become aerated. If you then try to put them back there is usually a
mound of ‘excess’
To work out how much soil you will need to remove from the site, you will need to know what
the soil type is. You will then need to use the following bulking factor chart to help you work
out the total amount of soil to be removed.
Material
Clay
Clay + gravel
Clay loam
Loam
Sandy loam
Sand
Gravel (6 – 50mm)
•
(including Bulking Factor)
Bulking factor
1.4
1.18
1.3
1.25
1.2
1.12
1.12
i.e. Total Volume of soil = Total area x Total Depth x Bulking Factor.
© Wodonga TAFE 2012
Excavation for Paving - calculation
Guidelines
56
4. How much will the soil weigh?
To find out we need to Convert volume (m3) to mass I.e. tonnes (t)
• Once we know how many cubic metres of material we need, it is sometimes useful to have an
estimate of the weight of the material so that you know how how many trucks will be needed
and how big, and so that they are not overloaded.
• If this is the case, we need to do another calculation to convert our volume to a mass.
• This calculation is based on some given facts
We are told that:
• 1m3 = 1.4 Tonne
• or in reverse;
1 tonne = 0.73m3
•
i.e. multiply the quantity of m3 to be excavated by 1.4 to find the weight in tonnes
© Wodonga TAFE 2012
Excavation for Paving - calculation
Sample
57
Step 1 - How much area is to be excavated? (Area M2)
• You are to excavate a driveway 6 metres x 4 metres
6m x 4m = 24m2
Step 2 - How deep will you need to excavate to.(Volume m3)
• The drive is to be excavated to 200mm (0.2 of a metre)
• So multiply the area that you calculated by 0.2
24m2 x 0.2 = 4.8m3
© Wodonga TAFE 2012
Excavation for Paving - calculation
Sample
58
Step 3 - How much soil will you need to remove from the work site?
(Bulking Factor)
Example
• If you worked out that a drive to be excavated had a volume of 24m3, and
the soil type was clay, then you would multiply the volume by the bulking
factor of the chart (1.4)
• i.e.
4.8 m3 x 1.4
=6.72 m3
• Therefore you would need to be prepared to remove 6.8m3 of material
© Wodonga TAFE 2012
Sample
Excavation for Paving - calculation
59
Step 4 - How much will the soil weigh?
•
Convert volume (m3) to mass I.e. tonnes (t)
• Simply multiply the bulked volume 6.8m3 by the given conversion rate of
1.4t / m3
• 6.8 x 1.4 = 9.52 tonne which is equivalent to approximately
– 1.5 loads in a 8 tonne single axle truck loads
– 1 loads in a 12 tonne dual axle truck loads
– 5 loads in a mini landscapers truck loads (2 tonne each)
© Wodonga TAFE 2012
Calculating pavers required for a border / coping
Task - You need to find out how many pavers to order for the border of a paving project
60
with contrasting colour, coping pavers for around a pool or to cap a set of steps
Considerations
•
The linear measurement is simply a measurement along a line, so weather on the
site, or scaling off a plan, measure the distance along each edge of the area to be
paved
•
total up all the measurements
•
divide the total length by the width of the pavers that you are using (remembering
to convert all measurements to either mm or metres to make them work out
properly)
•
Note We use the symbol lm to represent lineal metres
© Wodonga TAFE 2012
Calculating pavers required for a border / coping
Sample
61
Example 1
1. If the area you were to pave was 6m x 5m then you would measure each side 6 + 5 + 6 + 5
2. the total of all the measurements =22 lineal metres (22lm)
3. 22lm divided by the width of each paver (i.e. a brick paver as a header would be approximately
110mm or 0.110m
•
therefore
•
=22lm / 0.110
•
=200 pavers or
• Allowing that there would be approximately 38 pavers per m2
•
200/ 38 = 5.263m2
• Note these would be ordered with the contrasting coloured paver.
4. Normally for the main colour, work out the area for the whole section of pavers then subtract
the borders quantity off the quantity of main pavers required
© Wodonga TAFE 2012
Calculating pavers required for a border / coping
Sample
62
Example 2
1. Measure the perimeter of the pool
i.e. measure around the edge of the pool
2. Total up all measurements
3. Divide the total length by the width of each paver i.e. if using 400 x 400 pavers would be 0.4m
•
Eg if the pool was 28lm around the edge divided by 0.4
•
28lm / 0.4 = 70 coping pavers.
4. You usually allow10% extra for these as to get around the corners, there is significant wastage
due to only having one rounded edge
•
So we would need to order 70 + 10% = 77 coping pavers
© Wodonga TAFE 2012
Calculating footings
Task - You are going to build a small brick wall and need to put a concrete footing to
63
build it on so you will need to work out how big a footing will be required to be
strong enough
Consideration – Situation 1
•
For brick walls up to 5 courses high, use this formulae
1. Determine the total length of the footing - the length of the footing should be 0.25 x the
width of the wall, longer than the wall on either end - i.e. for a single skin wall = 110 so it should
be 27.5mm longer at either end or for a double skin wall which is 230mm wide, it should be
57.5mm longer either end
2. Determine the total width of the footing - The width of the footing should be 1.5 x the width
of the wall i.e. 165mm for a single skin or 345mm for a double brick wall
3. Determine the total Depth of the footing - The depth of the footing should be the same as the
width of the wall i.e. 110mm for a single skin wall or 230mm for a double brick wall
4. multiply L x W x D
© Wodonga TAFE 2012
Calculating footings
64
Sample – 1 to 5 course
You are going to build a single skin wall 10m long 4 courses high
1. Determine the total length of the footing (L)
•
10m + [ (0.25 x the width) x2]
•
=10m +[ (0.25 x 110mm) x2]
•
=10m + [27.2mm x 2]
•
=10m + 54.4mm
•
convert so both are the same unit i.e. metres
•
10.0m + 0.054m
•
=10.054m
2. Determine the total width of the footing (W)
•
1.5 x 110mm
•
=165mm
•
converted to metres (x 1000)
•
= 0.165m
Footing extends past the wall at each end
for a length of half the wall width
Width of footing 1.5 x the width of the wall
© Wodonga TAFE 2012
Calculating footings
65
Sample – 1 to 5 course
3. Determine the total Depth of the footing (D)
•
110mm for single skin
The depth of the footing is same as the width of the wall
•
converted to metres (x 1000)
•
= 0.110m
4. L x W x D
• 10.054 x 0.165 x 0.110
• = 1.659m2 x 0.110
• = 0.183m3
•
•
L
W
D
allow 10% extra
Order or mix up 0.2m3 of concrete
© Wodonga TAFE 2012
Calculating footings
Task - You are going to build a small brick wall and need to put a concrete footing to build it on
66
so you will need to work out how big a footing will be required to be strong enough
Consideration - Situation2
• For brick walls 6 – 10 courses high use this formulae
• Insert a diagram showing a small wall, footing & formulae
1. Determine the total length of the footing •
the length of the footing should be 0.5 x the width of the wall longer than the wall
on either end i.e. for a single skin wall = 110 so it should be 55mm longer at either end
or for a double skin wall which is 230mm wide, it should be 115mm longer either end
2. Determine the total width of the footing •
The width of the footing should be 2 x the width of the wall i.e. 220mm for a single
skin or 460mm for a double brick wall
3. Determine the total Depth of the footing –
•
The depth of the footing should be 1.33 x the width of the wall i.e. 147mm for a
single skin wall or 306mm for a double brick wall
4. multiply L x W x D
© Wodonga TAFE 2012
Calculating footings
Sample – 6 to 10 course
67
Footing extends past the wall at each end
For a double skinned wall 12m long:
for a length of half the wall width
1.Determine the total length of the footing
•
12m + [ (0.5 x the width) x2]
•
=12m +[ (0.5 x 230mm) x2]
•
=12m + [110mm x 2]
•
=12m + 230mm
•
convert so both are the same unit i.e. metres
•
12.0m + 0.230m
•
=12.23m
Width of footing twice the width of the wall
•
2.Determine the total width of the footing
•
2 x 230mm
•
=460mm
• converted to metres (x 1000)
•
=0.460m
© Wodonga TAFE 2012
Sample – 6 to 10 course
68
Calculating footings
3. Determine the total Depth of the footing
•
1.33 x 230mm
•
=306mm
• converted to metres (x 1000)
•
=0.306m
4. Calculate volume L x W x D
•
12.23 x 0.460 x 0.306
•
=5.62m2 x 0.306
•
=1.719m3
•
•
•
The depth of the footing is 1.33 x the width of the wall
L
W
D
add 10% extra (0.17)
= 1.892m3
you need to order to next 0.2 of a metre3 so order 2m3 of concrete
© Wodonga TAFE 2012
Calculating stepped footings
Task - You are going to build a small brick wall on sloping ground. You need to put
69
a concrete footing to build it on this time, the footing will need to have steps in it
Considerations
1. use a levelling instrument to help Calculate the difference in height of the
ground from one end to the other.
2. Divide the difference by the height of a course of bricks (86mm). This will tell
you how many steps in the footing
3. calculate the brick footing as if no steps
4. Add to the total amount of concrete required, add two brick lengths x the width
of the footing x the height of a brick - for each step required i.e. 460 x W x 86
2 brick length overlap
© Wodonga TAFE 2012
Sample
Calculating stepped footings
1.
2.
•
•
•
Measure up and find levels
If the difference in height was 222mm
222 / 86 = 2.58 steps
therefore round up to 3 steps
1.83m3
3.
•
•
•
•
Using example 1- where:
L = 10.054m
W = 0.165m
D = 0.110m
Total concrete = 1.83m3
70
4.
1.83m3 + 3 steps of overlap = 3 x [0.460 x
0.165 x 0.0 86]
= 1.83 + 3 x [0.0759 x 0.086]
= 1.83 + 3 x [0.00653]
= 1.83 + 0.199
= 1.849m3
Which in this case does not probably need to change
concrete order
Now try using example 2 in brick footings for practice
© Wodonga TAFE 2012
How many bricks for a wall
Task - To estimate the quantity of bricks required to construct a brick
71
wall from a plan so that we can order enough bricks to do the job
Considerations
•
There are two alternative methods of working this out.
•
Before we start, assume that the wall is to be constructed in stretcher bond.
There will be no difference in the quantity of bricks required if another bond is
used and this method is easier if we think of stretcher bond.
© Wodonga TAFE 2012
How many bricks for a wall
Method 1
72
Brick quantity = Number of bricks long  number of bricks high  number of
bricks thick.
• To find the number of bricks long the wall is (how many bricks end to end
along the wall), take the required length of the wall and divide it by the
length of a brick plus its mortar joint (230 + 10 = 240 mm)
• To find the number of bricks high the wall is (how many courses of bricks
stacked on top of each other) take the required height of the wall and
divide it by the thickness of a brick plus a mortar joint (76 + 10 = 86)
• To find how many bricks thick the wall is, remember that a single (1) brick
wall is 110 mm thick; a double (2) brick wall is 230 mm thick (110 + 10 +
110 = 230 mm); and a wall three bricks thick measures 350 mm (110 + 10 +
110 + 10 + 110 = 350mm).
© Wodonga TAFE 2012
How many bricks for a wall
Method 2
73
This method also works whatever bond is to be used.
• Allowing that there are approximately 50 bricks per m2 of each skin of a
wall.
• 1. Find the area (m2) of the wall off the plan by multiplying the height
(H) by the length (L).
• 2. Multiply the area by 50 bricks / m2 to get the number of bricks per
skin required.
• 3. Multiply by the number of skins to the wall.
• 4. Add 10% for wastage etc
© Wodonga TAFE 2012
How many bricks for a wall
74
Sample
3590
Method 1
We will look at a wall that is to be 3590 long,
850 high and double skinned.
•
•
•
•
•
•
850
1. To find the number of bricks long the wall is (how many bricks end to end along the
wall), take the required length of the wall and divide it by the length of a brick plus its mortar
joint (230 + 10 = 240 mm)
i.e. Number of bricks long = 3590 / 240
= 14.96 bricks
2. Round this up i.e. to 15 bricks
3. To find the number of bricks high the wall is (how many courses of bricks stacked on top
of each other) take the required height of the wall and divide it by the thickness of a brick plus
a mortar joint (76 + 10 = 86)
i.e. Number of bricks high = 850 / 86 = 9.88 bricks
4. Round this up i.e. to 10 bricks
© Wodonga TAFE 2012
How many bricks for a wall
Sample
5.
•
6.
7.
8.
•
75
To find how many bricks thick the wall is,
230mm = 2 bricks
remember that a single (1) brick wall is 110 mm thick; a double (2) brick
wall is 230 mm thick (110 + 10 + 110 = 230 mm); and a wall three bricks
thick measures 350 mm (110 + 10 + 110 + 10 + 110 = 350mm).
To find how many bricks are required for the example:
Brick quantity = 15  10  2
= 300 bricks
Add an allowance of 10% for wastage
10% of 300
= 30 bricks
Total quantity of bricks required:
= 300 + 30
= 330 bricks
© Wodonga TAFE 2012
How many bricks for a wall
76
Sample
3590
• Method 2
850
•
•
•
•
•
•
Eg. Using the same wall as in the example for method 1
1. 0.850(H) x 3590 (L) = 3.05m2 of wall face
2. 3.05m2 x 50 =152 bricks per skin
3. 152 x 2 skins (for a double brick wall) = 304 bricks
4. 304 + 10% = 304 + 30
=334
Therefore the total number of bricks required is 334 which is only 4 bricks
different to our first example, so use which ever method suites you best
© Wodonga TAFE 2012
Quantities of ingredients for Mortar
Task - You have a small wall to construct and need to work
out the how much sand, lime and cement you will need
77
Considerations
• You will need to know :
• how many bricks are in the wall (click to link to separate slide)
• The class of mortar / for ratio of sand : lime : cement
• That 1 cubic metre of sand has a mass of 1.36 tonnes
• 1m3 of cement = 60 x 20kg bags
lime
• 1m3 lime = 32 x 20kg bags
• You will also need to use 2 tables
•
•
cement
Sand
1 - The mortar required per 1000 bricks / per m2 table
2 – mix proportions by volume (from AS 3700)table
© Wodonga TAFE 2012
Quantities of ingredients for Mortar
78
Sample
Step 1 – Determine the number and type of bricks needed for the wall
Eg you are told the wall has 2500 standard metric bricks
Step 2 find the amount of mortar – using the mortar required table
Mortar required per 1000 masonry units and per m2
•
•
•
Masonry units
Mortar per
100 units
M2 of wall per
1000 units
Mortar per m2
of wall
Standard metric brick laid on edge
0.4m3
28.57m2
0.014m3
Standard metric bricks
0.7m3
20m2
0.035m3
Blocks
1. 25m3
80m2
0.015m3
The chart indicates you will need 0.7 m3 of mortar for every 1000 bricks
The wall will have 2500 bricks, so multiply above volume by 2.5
2.5 x 0.7 m3 = 1.75 m3
© Wodonga TAFE 2012
Sample
79
Quantities of ingredients for Mortar
Step 3 - determine the mortar mix and proportions
• Refer to the following table for ingredient ratio for the mix to suit your project
• Eg, we have been told we will be using M3 class mortar as it is a low garden wall
•
•
•
•
Required mix
6 part sand
1 part lime
1 part cement
Class
Mix proportions by volume
Sand
Lime
Cement
Water thickener
practice
3
1
0
optional
M2
9
2
1
No
M3
6
5
1
0
1
1
optional
yes
M4
9
1
2
optional
© Wodonga TAFE 2012
Sample
Quantities of ingredients for Mortar
80
Step 4 – Find the volume of sand you’ll need
• As sand is coarse and cement and lime fine, the smaller particles fit in the poor spaces
between sand grains so the quantity of mortar is the same as the volume of sand used
• eg we need 1.75m3 of mortar so will need 1.75m3
Step 5 – Find the volume of cement required
• The mix requires 6 : 1 : 1
• Volume of Cement = 1 part cement
1.35m3 of sand
6 parts sand
• (Make volume of cement the subject)
• Volume of cement = 1.75m3 of sand x 1 part cement
6 parts of sand
= 0.291m3
Step 6 – Find the volume of Lime required
•
(in this case the same as cement = 0.291m3
•
other wise substitute the mix part for lime into the equation as above)
© Wodonga TAFE 2012
Quantities of ingredients for Mortar
81
Sample
Step 7 – Convert the volumes to units to order
•
For the sand, it will depend on your supplier, with some 1.75m3 will suffice
•
(Or others will sell by the tonne so you may need to convert
•
as stated - 1 cubic metre of sand has a mass of 1.36 tonnes
•
so 1.75m3 x 1.36t/ m3 = 2.39tonne)
•
•
•
•
For cement Convert m3 to 20kg bags
There is one cubic metre of cement in 60 bags so multiply volume by 60
20kg bags of cement = 0.291 x 60
= 17.46 bags (round up to 18bags)
•
•
•
•
For Lime Convert m3 to 20kg bags
There is one cubic metre of lime in 32 bags so multiply by 32
20kg bags of lime = 0.291 x 32
= 9.31 bags (round up to 10 bags)
© Wodonga TAFE 2012
How to calculate a Timber order
Task - You need to find out how much timber to order to construct a project. You
have a plan to work off. You need to calculate how many pieces, what sizes and then
compile an order in a manner that the timber yard or hardware will understand
82
Considerations
1. Study the plan and count all the pieces of each type of component i.e. for a deck or
pergola, count how many posts, then count how many bearers / beams; joists / rafters;
batons; fascias and any other trimmings.
2. List them in groups according to size i.e. all the 100x 100mm pieces and all the 150 x
50mm pieces together
3. Look at what lengths are required for the different components as they are specified
and organize into groups as well, i.e. list the size and how many pieces at each length
4. Make note of what type and grade of timber is required
5. Then compile your order
For decking or timber retaining walls, you may need to calculate by working out the
area of coverage.
© Wodonga TAFE 2012
How to calculate a Timber order
83
Plan
This is a plan of a small deck that we will use for the example
© Wodonga TAFE 2012
How to calculate a Timber order
Sample
84
1. Try and work out how many posts, bearers and joists needed for the deck plan on the previous
slide.
• The plan indicates that:
• the posts are 100 x 100 x 1500 long,
• the bearers are 150 x 75 x 4000 long,
• the joists are 150 x 50 x 3000 long,
• and the fascia are 150 x 25
2. As you may have worked out by now,
• There are 12 “posts”. These are the little squares on the inside of the frame
• There are 3 ”bearers” These are the 3 pieces running across the screen
• There are 9 “joists”. These are the pieces running up and down the page at approximately 450
spacings
• There is fascia boards all the way around the deck so there would be 2 short lengths for each
end and 2 longer lengths for the sides
© Wodonga TAFE 2012
How to calculate a Timber order
Sample
85
3a. The sizes then grouped by size and length
• 100 x 100. 6 / 3.0 lengths (i.e. 12 / 1500 with 2/ 1500 pieces out of each 3000
length)
• 150 x 75. 3 / 4.2 lengths (timber is only available in increments of 300mm or the
old imperial “foot”)
• 150 x 50 9 / 3.0 lengths
• 150 x 25. 2 / 4.0 lengths
• 3b. For a deck 4000 x 3000, you would determine which direction the decking was
to run, then calculate how many pieces of the decking you are to use will fit across
the width.
• So if the deck were to run the 4200 using 90 x 22 treated pine decking boards, then
you would divide 3000 by 90 which will give you an answer of 33.3 boards
• Even though in reality you will install the decking with a small gap between them, I
would order:
Kiln dried CCA treated pine decking
90 x 22 - 34 / 4.2m
© Wodonga TAFE 2012
How to calculate a Timber order
Sample
86
4. The timber specified is F7 grade copper chromium arsenate Treated timber
(CCA)
5. You would order
• F7 CCA timber
100 x 100. 6 / 3.0
150 x 75. 3 / 4.2
150 x 50 9 / 3.0
150 x 25. 2 / 4.0
The process is similar for other timber projects it just takes a bit of careful practice.
Short cuts may come with experience, but it is best to know the correct way first.
© Wodonga TAFE 2012
How to calculate a Timber order
Sample - fixings
87
To calculate fixings, as with the timber, count up the attachment points for the
joining of each lot of components
Bolts to join Bearers to posts @ 2 per post
• 12 posts = 24 / 125mm M10gal coach bolts + nut and washer for each
Nails to join joists to bearers @ 2 per contact point and scew nailing technique
•
•
•
9 joists x 3 bearers = 54 contact points
54contacts x 2 nails per contact = 108 /100mm gal nails
Determine best way to buy i.e. buy the qty or by the weight (see below for example)
© Wodonga TAFE 2012
How to calculate a Timber order
Sample - fixings
•
•
•
•
•
•
•
•
•
•
•
•
88
Decking nails (or screws if specified)
@500 centres, with 2 nails in each joist, on each of the 34 boards, you would need (9 x 2) x 33
nails
= 612 / 50mm galvanized deck head, twist nails
Weigh out say 100 grams and count how many nails (i.e. 40.)
Then multiply by 10 to get how many nails per kilo –( in this case 400)
So you would need over 1.5 k nails
It would be best to buy a 2 k box. Bulk nails work out cheaper, but you also get them in handy
boxes.
Bugle headed Screws for facia@ 2 on each end and 2 @ approximately 800 to 900 centres using 100mm screws
3750 /900 = 4.1 + add start end = 5 contact points
and 2500 / 800 = 3.1 + add start end = 4 contact points
Add all sides = 2 x 5 + 2 x 4 = 18 contact points
Multiply total contact points by 2 screws per contact = 36 screws
© Wodonga TAFE 2012
Selecting a Water Feature Pump
Task - You have been asked to select an appropriate
pump for a waterfall feature in a new landscape.
89
Considerations
•
•
•
•
Work out your required output at spillway height
Measure the length of pipe required to reach from the pump to the outlet
Estimate the vertical height or required head of lift
Work out the required head of pressure that the pump should provide to
lift the required height
• Add 1 and 4 together
1 is how much water pressure you need at the top
4 is how much water pressure you need to get up to the top
• Select a suitable pump from catalogue charts
© Wodonga TAFE 2012
Process
Selecting a Water Feature Pump
90
1. Work out your required output at spillway height
•
Allow 568L /hr for every 25 mm of spillway width on the feature (150 Gph per inch)
2. Measure the length of pipe required to reach from the pump to the outlet
•
Allow extra 3m of head / for every 30m of pipe or 300mmH / 3m pipe
3. Estimate the vertical height or required head of lift
•
•
For a submersible pump situation – measure from the pump outlet to the top of
the spillway (include the depth in the water)
For a suction inlet set up – measure from the base of the pond to the top of the
spillway
© Wodonga TAFE 2012
Process
Selecting a Water Feature Pump
91
4. Work out the required head of pressure that the pump should provide to
lift the required height
• Add the extra head from step 2 to the height of step 3
• Using the given constant of 15L /min - per metre of head you need to lift
• multiply by 60 to convert to L/hr output required
5. Add 1 and 4 together as 1 is how much water pressure you need at the top
and 4 is how much water pressure you need to get up to the top
6. Select a suitable pump from catalogue charts
© Wodonga TAFE 2012
Sample
Selecting a Water Feature Pump
92
Example 1
•
•
•
•
you are selecting a pump for a water feature with
a 200mm wide weir,
a 5m lift
15 m pipe
1. weir width is 200mm, therefore if 568l/hr per 25mm
• 200 / 25 = 10
• 10 x 568L/hr = 5680L/hr
2. if 30m of pipe needs extra 3m head
• 15m of pipe needs extra 1.5m head
3. 5m lift + 1.5m extra head = 6.5m head required
© Wodonga TAFE 2012
Sample
Selecting a Water Feature Pump
93
Example 1 continued
4.
•
•
•
6.5 m of lift @ 15L/min per metre of head
6.5 x 15 =97.5L/min
multiply by 60 mins to get L/hr
97.5 x 60 = 5850L/hr required head of lift
5. Add step 1 to step 4
• 5680 + 5850 = 11,530 L/hr required output of pump minimum
•
Select suitable pump that will suit the situation and deliver more than the
minimum requirement. Remember you can install flow control valves to reduce
flow, but you can’t get more once its at maximum
© Wodonga TAFE 2012
Index
Turf / Parks & Gardens
94
Select from following links:
Calculating Fertiliser rates
Calculating chemical rates
Calculating seed rates
Calculating perched water table
Excavation calculations
© Wodonga TAFE 2012
Calculating fertilizers rates
Task - You are required to fertilise a garden or lawn area and are required to calculate
how much fertiliser and the best product to use for both nutrients required and cost
efficiency
95
Considerations
1. look up the percentage of the required nutrient in a couple of
recommended fertilisers
2. divide 100 by the percentage of nutrient in the fertiliser
3. Then Multiply by the recommended application rate (g / m2 ) for the time
of year
4. Multiply this figure by the area to be fertilised (m2)
5. Divide the cost of a bag of selected fertiliser by the weight of the bag and
then multiply by the desired application rate kg / 100 m2 x $ cost of a the
application for this product
6. Repeat steps 2,3 & 4 with another product
7. Compare cost of each product
© Wodonga TAFE 2012
Calculating fertilizers rates
Sample
96
Example 1 – comparison of best value / most efficient product
1. If you were to use nitrogen at a rate of 8 gram per m2 and you were covering 100m2. To
decide between sulphate of ammonium (21% nitrogen) or ammonium nitrate (34% nitrogen)
• The nitrogen rates (N) are in units of ‘actual N per 100 square metres’.
• The actual nitrogen rates are scheduled per month
• The % of each fertilizer is displayed on the fertilizing product.
E.g., Actual N rate x 100 ÷ N% = Fertilizer product Rate required
2a. First we will look at sulphate of ammonium (21%)
100 / 21 % = 4.76
3a. 4.76 x 8 g / m2 =38.08 g / m2
4a. 100 m2 x 38g = 3800g or 3.8 kg / 100 m2
5a. Check the current cost of Sulphate of ammonium per kg
Divide the cost of a bag of Sulphate of Amonium by the weight of the bag to get a $ per kg
and then multiply by the required 3.8kg = $ cost of a the application
© Wodonga TAFE 2012
Calculating fertilizers rates
Sample
97
Example 1 Continues
6. Now to compare ammonium nitrate (34%)
Repeat steps 2,3 & 4 and 5 with a different fertiliser
2b. 100 / 34 = 2.9
3b. 2.9 x 8 g / m2 = 23.2g / m2
4b. 100 m2 x 23.2g = 2300 or 2.3kg
5b check the current cost of ammonium nitrate per kg
Divide the cost of a bag of Ammonium Nitrate by the weight of the bag to get a $
per kg and then multiply by the required 2.3kg = $ cost of a the application
•
Now you could compare cost of each more accurately as you no how much of each
you would need to use
© Wodonga TAFE 2012
Calculating fertilizers rates
98
Sample
Example 2
• Fertilizer Rate kg per 100m2 x N% ÷ 100 = Actual N rate
• 6 kg fertilizer per 1000 square metres is therefore equivalent to 0.6 kg
fertilizer per 100 square metres (m2).
• always work in units of per 100 square metres (m2)
• 0.6kg urea x 46%
=
0.276 kg Actual N per 100 m2
© Wodonga TAFE 2012
Chemical application Rates
Task - You have been asked to spray a small area, and need to work out
99
how much chemical to mix into the sprayer
Considerations - liquid
• 1litre = 1000ml
• 1litre weighs 1kg
• So a 15litre knapsack will weigh 15kg plus the weight of the actual sprayer
when full
• 1000kg = 1 tonne
• So a 1000L tank weighs 1 tonne plus the weight of the tank when full
• Keep in mind a part full tank allows the water to move transferring the
weight from side to side at each bump and can be dangerous to transport
due to risk of rolling the vehicle and spilling the chemical mix
© Wodonga TAFE 2012
Chemical application Rates
Considerations continued
100
Surface area / space to cover
• 1 Hectare = 10 000 m2 or 100m x 100m
• 1000m2 = 100m x 10m or 50m x 20m etc
• 100m2 = 100m x 1m or 10m x 10m etc
• 10m2 = 10m x 1m or 5m x 2m etc
• An average front yard would be approximately 15m wide by 6m which
would = 90 m2
• An average house block would be somewhere around 800 to 100 m2
• 1 Hectare = 2.47acres
© Wodonga TAFE 2012
Chemical application Rates
Process
101
Step 1- Calculate the application rate per litre. This is how much chemical to
be mixed with how much water, (the rate is not always given as ml of
chemical per L of water).
• The rate is sometimes given as a range. The smaller quantity in the range is
for seedling weeds up to 15cm high the higher qty is for larger more
established weeds
• Rate off Label
• divide rate by number of litres given to get a per Litre rate
• (a) =
© Wodonga TAFE 2012
Chemical application Rates
Process
102
Step 2 - Now you will also need to calculate the quantity of chemical needed for size of the
job at hand.
• You will need to measure the surface area of the area to be sprayed (length by width of
which could be averaged if irregular shaped)
• Check to see if there is a recommended coverage given for hand spraying otherwise you
will need to scale down the hectare rate as follows
• (i)Litres per hectare (10 000 m2) =
•
(ii)Divide by 100 to get a rate in litres per 100 m2
•
(iii)Convert to millilitres (i.e. multiply by 1000)
•
•
(iv)Multiply by how many m2 to be covered to give total qty of chemical in ml to use
(b) =
© Wodonga TAFE 2012
Chemical application Rates
Process
103
Step 3 - Now you just need to put the two figures together to calculate how
much mix to make up for the job
• Divide the total qty of chemical required (b) by the per litre rate (a)
• (b) / (a)
• =total quantity of mix required
© Wodonga TAFE 2012
Chemical application Rates
Sample
104
• Step 1-
© Wodonga TAFE 2012
Chemical application Rates
Sample
105
© Wodonga TAFE 2012
Calculating for a Perched Water Table
Task - To calculate the materials required for a perched water table.
106
These include soil for the root zone, and pea gravel to allow drainage.
There is sometimes an intermediate layer as well.
Considerations
• The volume of material required is calculated for each section
• Steps to calculate a perched water table
• Calculate the area of the turf required
– Length x width
• Calculate the volume of each layer
– Area x depth
– Compaction must be included
• For the pea gravel section, there is a 1% fall.
– Calculate the volume as a triangle
– Height x area ÷2
© Wodonga TAFE 2012
Calculating for a Perched Water Table
107
Sample - Golf
•
•
•
•
•
•
To calculate a perched water table
Use the following example to understand the steps.
Area = length by width. 20m X 10m =200m2
300mm root zone
50mm intermediate layer
Pea gravel – 50mm top layer and 1% fall
300mm root zone Brimmin Sand
50mm intermediate layer
Pea gravel – top layer and 1% fall
Change all measurements to metres
• i.e. 300mm = 0.3m; 50mm = 0.05m
© Wodonga TAFE 2012
Calculating for a Perched Water Table
108
Sample - Golf
Root Zone
• Multiply the depth by area
– 0.300m X 200m2=60m3
• Add the compaction factor
– 40% compaction
– 60m3+ 40% = 84 m3
Intermediate Layer
• Multiply the depth by area
– 0.05 X 200m2 =10 m3
• Add the compaction factor
– 20% compaction
– 10m3+20% = 12 m3
Pea Gravel Layer
• Top section
– Multiply the depth by area
• 0.05 X 200m2 =10 m3
• Triangle – 1% fall
– 1% of the length
• 20m X 1 % = 0.2m
– Height X area ÷2
• 0.2m X 200m2 ÷2 = 20 m3
• Add top and triangle sections
together
• 10 m3+ 20 m3 = 30m3
• Add the compaction factor
• 20% compaction
• 30m3+20% = 36 m3
Calculating for a Perched Water Table
109
Sample – Bowling Green
• To calculate a perched water table for a bowling green with a ‘roof top’
design in the water table.
• Use the following example to understand the steps.
• Area = length by width. 40m X 40m =1600m2
300mm root zone – Brimmin Sand
Pea gravel – top layer and 1% fall
• Change all measurements to metres
• i.e. 300mm = 0.3m; 50mm = 0.05m
© Wodonga TAFE 2012
Calculating for a Perched Water Table
Sample – Bowling Green
Root Zone – Brimmin Sand
• Multiply the depth by area
– 0.300m X 1600m2=480m3
• Add the compaction factor
– 20% compaction
– 480m3+ 20% = 576 m3
110
Pea Gravel Layer
• Top section
– Multiply the depth by area
• 0.05 X 1600m2 =80 m3
• Triangle – 1% fall
– Bowling green is divided into 4 sections.
• 40m ÷ 4 = 10m
– 1% of the length
• 10m X 1 % = 0.1m
– Height X area ÷2
• 0.1m X 1600m2 ÷2 = 80 m3
• Add top and triangle sections together
• 80 m3+ 80 m3 = 160m3
• Add the compaction factor
• 5% compaction
• 160m3+5% = 168 m3
Excavation calculation
Task - Before excavating an area for a project, we need to get an idea of
how much material will be dug out so we can work out where to put it
and what trucks and machinery may be needed.
111
Considerations
• How much area is to be excavated? (Area m2)
• How deep will you need to excavate to.(Volume m3)
• How much soil will you need to remove from the work site? (Bulking
Factor)
• How much will the soil weigh? Convert volume (m3) to mass I.e. tonnes (t)
• This is important so we can decide whether to use a machine or dig by
hand and also so we know if we need to organise a truck - if so how big a
truck
• This calculation could be for a a garden bed, a new green, a drain, a
concrete slab or footings or any number of other projects
© Wodonga TAFE 2012
Guidelines
112
Volume calculation - Excavation
1.
How much surface area is to be excavated? (Area)
•
•
•
•
•
•
You will need to measure the area you want
W
to excavate. You will need to know the length (L)
and the width(W) of the area
You then workout the Total Area to be excavated by multiplying the length (L) by the width
(W)
Length (l) multiplied by the width (w) = Total Area m2
l x w = A (m2)
•
This is referred to in square metres and is written as m2
L
L
2. How deep (D) do you need to excavate
• You will need the Volume formula V = L x W x D
W
D
© Wodonga TAFE 2012
Volume calculation - Excavation
113
Guidelines
3. How much soil will you need to remove from the work site?
(including Bulking Factor)
• So what is a bulking factor?
- Different types of soil become larger in size once they are dug up, as they get loosened up,
they generally fluff up or become aerated. If you then try to put them back there is usually a
mound of ‘excess’.
•
To work out how much soil you will need to remove from the site, you will need to know what
the soil type is. You will then need to use the following bulking factor chart to help you work
out the total amount of soil to be removed
Material
Clay
Clay + gravel
Clay loam
Loam
Sandy loam
Sand
Gravel (6 – 50mm)
•
Bulking factor
1.4
1.18
1.3
1.25
1.2
1.12
1.12
i.e. Total Volume of soil = Total area x Total Depth x Bulking Factor.
© Wodonga TAFE 2012
Volume calculation - Excavation
Guidelines
114
4. How much will the soil weigh?
To find out we need to Convert volume (m3) to mass I.e. tonnes (t)
• Once we know how many cubic metres of material we need, it is sometimes useful
to have an estimate of the weight of the material so that you know how how many
trucks will be needed and how big, and so that they are not overloaded.
• If this is the case, we need to do another calculation to convert our volume to a
mass.
• This calculation is based on some given facts
We are told that:
• 1m3 = 1.4 Tonne
• or in reverse;
1 tonne = 0.73m3
•
i.e. multiply the quantity of m3 to be excavated by 1.4 to find the weight in tonnes
© Wodonga TAFE 2012
Volume calculation - Excavation
Sample
115
Step 1 - How much area is to be excavated? (Area M2)
• You are to excavate a garden 12 metres x 2 metres
•
12m x 2m = 24m2
Step 2 - How deep will you need to excavate to.(Volume m3)
• The drive is to be excavated to 200mm (0.2 of a metre)
• So multiply the area that you calculated by 0.2
•
24m2 x 0.2 = 4.8m3
© Wodonga TAFE 2012
Volume calculation - Excavation
Sample
116
Step 3 - How much soil will you need to remove from the work site?
(Bulking Factor)
• Example
• If you worked out that a garden to be excavated had a volume of 24m3,
and the soil type was clay, then you would multiply the volume by the
bulking factor of the chart (1.4)
• i.e.
4.8 m3 x 1.4
•
=6.72 m3
• Therefore you would need to be prepared to remove 6.8m3 of material
© Wodonga TAFE 2012
Volume calculation - Excavation
Sample
117
Step 4 - How much will the soil weigh?
•
Convert volume (m3) to mass I.e. tonnes (t)
• Simply multiply the bulked volume 6.8m3 by the given conversion rate of
1.4t / m3
•
•
•
•
•
6.8 x 1.4 = 9.52 tonne
which is equivalent to approximately
1.5 loads in a 8 tonne single axle truck loads
1 loads in a 12 tonne dual axle truck loads
5 loads in a mini landscapers truck loads (2 tonne each)
© Wodonga TAFE 2012
Water Harvesting and irrigation
Index
118
Select from the following links
• Roof Catchment
• Tank Storage
• Water pressure town / pump
• Water pressure gravity feed
© Wodonga TAFE 2012
Roof Catchment
Task - You’ve been asked to work out the roof catchment of a
building to see how much water can be collected in a tank
119
Considerations
• To work out the catchment area you will need to know
• The length of the roof area that will drain into the tank (in this case 10m)
• The width of the roof area that will drain into the tank (in this case 5m)
• You will also need the area formula L x W
10m
5m
Shed
Tank
© Wodonga TAFE 2012
Roof Catchment
120
Sample
•
•
•
•
Step 1 – find the area of the roof catchment
Apply the formulae of the rectangle – L x W
10m x 5m
10m
= 50m2
5m
Shed
Tank
• An area of 50m2 will be draining into the tank
• Each 1m2 of roof will catch 1Ltr of water per mm of rainfall
© Wodonga TAFE 2012
Tank Storage
Task - You have been asked to work out the size of a tank
to see how much water it will store
121
Considerations
• To find the volume of a round tank, you will need the formulae for the
volume of a cylinder :
π r2 x h
r = 1.8m
Where
• π = 3.143
• r = the radius of the tank (half the diameter)
• h = the height of the tank
h = 2.5m
• Note 1 m3 of space in the tank holds 1000Ltr of water
© Wodonga TAFE 2012
Tank Storage
Sample
Step 1- find the volume of the tank
Apply the formula :
π r2 x h
• If the diameter = 3.6, then the radius = half or 3.6 /2 = 1.8
• 3.143 x 1.82 x 2.5
• 3.143 x 1.8 x1.8 x 2.5
• 25.45m3
• Each m3 = 1000Ltr of water
• so 25.45 m3 x 1000
• Therefore the tank holds 25,450Ltr
122
3.6m
2.5m
© Wodonga TAFE 2012
Water pressure – town / pump
Task - You have been asked to install a small irrigation system so you will need to
123
measure the water pressure and volume available from the supply
Considerations
To work out the if the water pressure available for an irrigation system is
adequate, you will need to know:
• The water pressure and volume available at the water source (tap / valve)
you can use one of 2 methods:
•
a. Using a bucket – time the time it takes to fill a 9 ltr bucket and then x
Add calc / chart
•
b. Using a pressure gauge – attach to tap and turn tap on to get a static
reading then turn gauge valve on to get a flow reading
• The minimum recommended pressure for the system emitters
• Divide available pressure by emitter output to calculate quantity of
emitters that can be used at one time
© Wodonga TAFE 2012
Water pressure – town / pump
Sample – Bucket test
124
• At the source of the irrigation system take off i.e. Household tap close to
the water meter,
• Turn the tap on full
• Move a 9ltr bucket under the tap, holding securely under the tap and using
your watch, time the time it takes to fill the bucket
• i.e. A typical house pressure may take between 9 and 15 seconds
• Divide 9 ltr by the fill time and then multiply by 60 will give ltr/min
• Eg 9 litres divided by 12 sec = 0.75 ltr/sec
• 0.75ltr/sec x 60sec = 45 ltr/min
• If needing to convert to ltr/hr, multiply by 60min
• Eg 45ltr/min x 60 min = 2700 ltr/hr
© Wodonga TAFE 2012
Water pressure – Gravity feed
Task - You have been asked to install a small drip irrigation system so you will need
125
to measure the water pressure and volume available from the supply
Considerations
• To work out if you have sufficient pressure available and or how many emitters you
can put on each line you will need to know:
• The height of the header tank above the emitters (which can be measured on site
or off a plan)
• The height of the water in the header tank
• The minimum recommended pressure for the emitters (Which can be
obtained from your irrigation supplier)
• That gravity applies a constant force of 9.8m/sec2
Water level ?
tank
• You will need the formula
• Pressure (kPa) = height x gravity
Number of emitters and output
?
© Wodonga TAFE 2012
Water pressure – Gravity feed
Sample
126
Step 1 - find the head of water
• Find on the plan or measure on site :
• a. The height of the tank outlet above the irrigation system i.e. 1m
• b. The height of the water level at say half full i.e. 1.2m
• head of water = a + b
• eg
= 1m + 1.2m
•
= 2.2m
tank
Step 2 – Apply the formula (substitute values in)
Water
level 1.2m
• Pressure = head x gravity
• Eg
= 2.2m x 9.8m/sec2
•
= 21.56kpa
Number of emitters and output
Tank height
above system
1.0m
© Wodonga TAFE 2012
Water pressure – Gravity feed
Sample
127
Step 3 – find the output of the emitters
• eg. Typical drippers range from 2 to 8ltr / hr
• We will use 4ltr/hr
Step 4 – calculate how many emitters can be used off the system at one time
• Eg we have 21.56kpa of pressure
• If given 15ltr/min per metre of head (10kpa /m head)
• 21.56 divided by 10kpa = 2.156
• 2.156 x 15ltr/min = 32.34ltr / min
• multiply by 60min = 1940ltr/hr
• 1940 ltr/ hr divided by 4ltr/hr = 485 emitters
• To allow for friction loss in pipes, subtract say 10%
• 485 – 10%
• 485 – 48.5 = 436.5
• Therefore in theory up to 436 emitters could be used at one time while the tank is half full
• Note - pipe diametre and length of each line also needs to be considered
© Wodonga TAFE 2012
Business Administration
Index
128
Select from following links:
• Quoting
• Invoicing
© Wodonga TAFE 2012
Quote
Task - You have been asked to quote on a small landscape
project
129
Considerations
• Calculate your costs using prices that represent the cost of materials plus the cost of labour plus a
profit margin plus GST. You will need to measure up the site accurately beforehand
•
•
•
•
•
•
•
•
•
•
Measure up the site and or measure off a plan
Get up to date prices on materials for each component of the job
Estimate how many hours for how many workers (man hours) for each component of the job
Multiply man hours by your hourly rate ( it is suggested to use an average hourly rate worked out off
all workers in your team i.e. an apprentice and a tradesperson. (don’t forget to allow for on costs)
Add ‘mark up’ to materials price (this is your cut for organising and purchasing goods)
Add profit margin
Add GST
Calculate total cost
Divide the total quantity of the component by the total cost to come up with a unit rate
Prepare the quote
© Wodonga TAFE 2012
Quote
130
Sample
An example of how to set your workings out can be seen below
• Item number Part of job
measurement x cost per unit
• 2.1
Rear boundary 30.5 lineal metres (lm) of fence with each metre
costing $55.00 (@$55.00) can be shown as
30.5 lm @ $55.00 / lm = $1603.00
• 2.2
1 x single gate kit
= $235.00
• 2.3
Western boundary
25 lm @ $55.00 / lm = $1376.00
• 2.4
Eastern boundary
32.5 lm @ $55.00 / lm = $1739.00
© Wodonga TAFE 2012
Quote
131
Sample quote presentation
Business letterhead
Client details:
Date:
RE:
Colorbond Fence and Concreting
I am pleased to forward this quote for works as requested
Being for:
•
To remove old fences
$380.00
•
To supply and install new 1500 high colorbond fences
•
•
•
•
•
2.1
2.2
2.3
2.4
Rear boundary
30.5 lm
1 x single gate kit
Western boundary
25 lm
Eastern boundary 32.5 lm
Fence total
$1603.00
$ 235.00
$1376.00
$1739.00
$4953.00
© Wodonga TAFE 2012
Quote
132
Sample quote presentation continued
3.
Concrete slab to rear of building As per specification but reduced in size as discussed
•
•
•
•
•
•
Price includes:
to supply and install 1.8 m3 of concrete as required including
-excavation, set levels, fill where required,
-form up, supply and place steel, chairs and jointex,
-pour screed and finish concrete,
-supply and install top soil and turf along edge
•
•
Thank you for the opportunity to quote. I look forward to your reply.
Yours faithfully
•
•
This price may vary if we encounter rock or tree roots during excavation
This Quote is valid for 60 days
$1130.00
© Wodonga TAFE 2012
Invoicing
Task - You have just finished a small landscape job.
133
You need to give your client an invoice
Considerations
1. List items as per quote that have been completed
2. Add any extra items that were not quoted (variations to the contract)
3. Total up costs
4. Add GST
– 10%
– Show how much if already included (divided your total by 11)
5. Show total amount due
© Wodonga TAFE 2012
Invoice
134
Sample
Letter head / logo with company name and details
ABN Number
Tax Invoice
Client details
Date
Item
Description
Qty
Unit
Cost
1
Supply and install Colorbond fence plus 1 gate
88
lm
$4953.00
2
Supply & install Concrete slab
18
M2
$1130.00
variations
SubTotal
$6083.00
(Incl GST)
$ 553.00
Total Amount Due
$6083.00
Due date
and payment details
© Wodonga TAFE 2012
Well Funded Project
List names and roles
Kev, Jeni H, Phone, Email
Kevin Albert
Land Built Environment & Sustainability
kalbert@wodongatafe.edu.au
(02) 6055 6758