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Horticulture Maths Resource Mathematical resource for Horticulture Kevin Albert Land Built Environment & Sustainability [email protected] (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 © Wodonga TAFE 2012 Plan interpretation & Setting out Index 3 Select from following links: • Setting out 900 angles using 3:4:5 triangles • Scale • Using Scale Rulers © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 Scale Sample 10 A small scale covers a large area of land i.e 1:55000 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 15 © Wodonga TAFE 2012 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. © Wodonga TAFE 2012 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) © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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. © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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. © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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. © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 30 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 © Wodonga TAFE 2012 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 [email protected] (02) 6055 6758