TURFGRASS PROFESSIONAL'S GUIDE TO CALIBRATION

TURFGRASS PROFESSIONAL'S GUIDE TO CALIBRATION by John Wildmon

Table of Contents Module 1: Fractions Unit 1-1 Adding and Subtracting Fractions Unit 1-2 Adding Mixed Numbers Unit 1-3 Subtracting Mixed Numbers Unit 1-4 Multiplying with Fractions Unit 1-5 Dividing with Fractions Unit 1-6 Converting Fractions to Decimals Module 2: Decimals and Percentages Unit 2-1 Understanding Decimals Unit 2-2 Adding and Subtracting Decimals Unit 2-3 Multiplying Decimals Unit 2-4 Dividing Decimals Unit 2-5 Percentages Unit 2-6 Percentage Calculations Module 3: Simple Equations, Ratio and Proportion Unit 3-1 Solving Simple Equations Unit 3-2 Ratio, Proportion, Mean, Mode, and Median Module 4: Parts Per Million and Dilutions Unit 4-1 Parts Per Million Unit 4-2 Determining PPM When Known Quantities are Mixed Unit 4-3 Determining Solute Quantities fof Desired Concentration Unit 4-4 Mixing PPM Fertilizer Solutions Unit 4-5 Determining PPM From Known Quantities of Fertilizer Unit 4-6 Dilutions Unit 4-7 Calibrating Hose-on applicators Module 5: Area Unit 5-1 Area of Circles and Part-Circles Unit 5-2 Area of Squares, Rectangles, Parallelograms & Triangle Unit 5-3 Area by the Offset Method Unit 5-4 Area by Indirect Offsets Unit 5-5 Area by Average Radius Unit 5-6 Finding the Area of Combinations of Shapes Unit 5-7 Determining Plant Numbers by Area and Spacing Module 6: Volume and Units Unit 6-1 Determining Volumes of Cubes and Cylinders Unit 6-2 Volume of Tapering Pots Unit 6-3 Converting Units and Conversion Factors Unit 6-4 Fill Dirt and Topsoil Volumes Unit 6-5 Celsius and Fahrenheit Temperature Conversions

Module 7: Fertilizers Unit 7-1 Fertilizer Analysis, Grade, Ratio, and Comparative Cost Unit 7-2 Fertilizer Application Rates Unit 7-3 Determining Rate From Applied Fertilizer Unit 7-4 Calculating Fertilizer Cost Unit 7-5 Fertilizer From Effluent Irrigation and Fertigation Module 8: Spreader Calibration Unit 8-1 Calibrating Drop Spreaders for Seed and Pesticide Unit 8-2 Calibrating Drop Spreaders for Fertilizer Unit 8-3 Calibrating Centrifugal Spreaders for Seed and Pesticide Unit 8-4 Calibrating Centrifugal Spreaders for Fertilizer Unit 8-5 Calibrating Lg. Centrifugal Spreaders Seed and Pesticide Unit 8-6 Calibrating Large Centrifugal Spreaders for Fertilizer Unit 8-7 Pure Live Seed Calculations Module 9: Sprayer Calibration and Tank Mixing Unit 9-1 Sprayer Calibration Unit 9-2 Sprayer Calibration by the 1/128 Acre Method Unit 9-3 Tank Mixing Chemicals Unit 9-4 Selecting Nozzle Volume Module 10:Cost Equipment Unit 10-1 Cost of Ownership Unit 10-2 Cost of Operation Unit 10-3 Hourly Cost Appendix A-1 Practice Problems for Modules 1, 2, 3, & 4 A-2 Answers for A-1 A-3 Practice Problems for Module 4 A-4 Practice Problems for Module 5 A-5 Practice Problems for Module 6 A-6 Practice Problems for Module 7 A-7 Practice Problems for Module 8 A-8 Practice Problems for Module 8 & 9 A-9 Practice Problems for Module 9 A-10 Practice Problems for Module 10 A-11 Answer for A-10 A-12 Boom Sprayer Diagram A-13 Number of Seeds per Pound for Common Turfgrasses A-14 Conversion Equivalents

Unit 1-1 Adding and Subtracting Fractions ***************************************************************** Procedure: 1. Find a common denominator, preferably the least common denominator. 2. Change all the fractions into equivalent fractions that have the same denominator. 3. Add or subtract the numerators as the sign indicates and write this number over the common denominator. 4. Simplify the fraction in your answer. ***************************************************************** Example 1 Add: 3/4 + 1/6 Find a common denominator. 12 Change the fractions to equivalent 1/6 x 2/2 = 2/12 fractions with 12 as the denominator. 3/4 x 3/3 = 9/12 Add the fractions. 2/12 + 9/12 = 11/12 Example 2 Subtract: 9/16-3/8 Find a common denominator. 16 Change the fractions to equivalent fractions with 16 as the denominator. 3/8 x 2/2 = 6/16 Subtract the fractions. 9/16-6/16 = 3/16 Example 3 3/4 + 5/6-1/3 =? Find a common denominator. 12 Change the fractions to equivalent 3/4 x 3/3 = 9/12 fractions with 12 as the denominator 5/6 x 2/2 = 10/12 1/3 x 4/4 = 4/12 Perform the mathematical operations 9/12 + 10/12-4/12 = 19/12-4/12 = 15/12 Change 15/12 to a mixed number 15/12 = 1 3/12 and simplify 1 3/12 = 1 1/4

Practice Exercise 1-1 1. 1/4 + 3/8 2. 3/6 + 7/8 3. 1/2 + 1/2 4. 3/4 + 5/12 + 1/6 5. 5/8 + 3/10 + 3/4 6. 3/16 + 11/32 + 3/8 7. 11/16-1/4 8. 5/6-1/3 9. 5/8-7/16 10. 1/2-5/32-1/8 11. 15/16-1/4-3/8 12. 53/64-5/16-3/8 13. 3/4-5/8 + 5/16 14. 5/6 + 3/18-1/2 15. 3/8 + 3/16-9/16 Answers 1. 5/8 2. 1 3/8 3. 1 4. 1 1/3 5. 1 27/40 6. 29/32 7. 7/16 8. 1/2 9. 3/16 10. 7/32 11. 5/16 12. 9/64 13. 7/16 14. 1/2 15. 0

Unit 1-2 Adding Mixed Numbers 1. Add the fractions first and simplify. 2. Add the whole numbers next. 3. Combine the two. Example 1 Add: 5 1/4 + 2 3/8 Add the fractions first 1/4 + 3/8 = 5/8 Add the whole numbers 5 + 2 = 7 Combine the two 7 + 5/8 = 7 5/8 Example 2 Add: 6 9/16 + 7 5/6 Add the fractions first and simplify 9/16 + 5/6 = 67/48 = 1 19/48 Add the whole numbers 6 + 7 = 13 Combine the two 13 + 1 19/48 = 14 19/48 Practice Exercise 1-2 1. 3 3/8 + 5 1/16 2. 3 5/9 + 8 1/4 3. 23 7/8 + 7 5/16 4. 14 7/12 + 5 11/16 5. 6 1/3 + 4 1/4 6. 3 1/8 + 5 3/4 7. 1 3/4 + 6 5/8 + 5/6 8. 22 6/9 + 5 3/4 9. 3 1/3 + 5 3/8 + 1/2 Answers 1. 8 7/16 2. 11 29/36 3. 31 3/16 4. 20 13/48 5. 10 7/12 6. 8 7/8 7. 9 5/24 8. 28 5/12 9. 9 5/24

Unit 1-3 Subtracting Mixed Numbers Procedure: 1. Subtract the fractions first, borrowing 1 from the whole number if necessary. 2. Subtract the whole numbers next. 3. Combine the two. Example 1 Subtract: 4 5/8-2 3/16 Subtract the fractions first borrowing from the whole number if necessary 5/8-3/16 = 7/16 Subtract the whole numbers next 4-2 = 2 Combine the two 2 + 7/16 = 2 7/16 Example 2 Subtract: 8 1/4-3 7/8 Subtract the fractions first borrowing 8 1/4 = 7 5/4 from the whole number if necessary 5/4-7/8 = 3/8 Subtract the whole numbers next 7-3 = 4 Combine the two 4 + 3/8 = 4 3/8 Practice Exercise 1. 6 3/4-2 1/2 2. 14 9/16-2 3/8 3. 6 1/4-5 11/16 4. 22 1/3-13 7/8 5. 9 1/2-8 1/2 6. 14 11/32-5/16 7. 8 3/64-4 5/8 8. 2 9/16-1 1/8 9. 102 1/3-64 7/16 Answers 1. 4 1/4 2. 12 3/16 3. 9/16 4. 8 11/24 5. 1 6. 14 1/32 7. 3 27/64 8. 1 7/16 9. 37 43/48

Unit 1-4 Multiplying With Fractions Procedure: 1. If whole numbers are involved change them to fractions with a denominator of 1. 2. If mixed numbers are involved change them to improper fractions. 3. Try to simplify the fractions. When possible divide a numerator and a denominator by a common factor. 4. Multiply the numerators. 5. Multiply the denominators. 6. Simplify the answer if necessary. Example 1 Multiply: 3/4 x 5/8 Multiply numerator and denominators 3 x 5 = 15 4 x 8 = 32 Example 2 Multiply: 5/6 x 3/10 1 Simplify 5 x 3 6 x 10 2 1 1 x 3 6 x 2 2 Multiply numerators and denominators 1 x 1 = 1 2 x 2 = 4 Example 3 Multiply: 14 x 2/3 Change the whole number to a fraction with a denominator of 1. 14/1 Multiply numerators and denominators 14 x 2 = 28 1 x 3 = 3 Simplify 28/3 = 9 1/3

Example 4 Multiply: 4 3/8 x 1/2 Change mixed numbers to an improper fraction 4 = 32/8 32/8 + 3/8 = 35/8 Multiply numerators and denominators 35/8 x 1/2 = 35/16 Simplify 35/16 = 2 3/16 Example 5 Multiply: 5 x 3 3/5 x 1/4 Change whole numbers to fractions with denominators of 1. 5 = 5/1 Change mixed numbers to improper fractions. 3 3/5 = 18/5 Simplify the fractions if possible 5/1 x 18/5 x 1/4 1/1 x 18/1 x 1/4 Multiply numerators and denominators 1/1 x 18/1 x 1/4 =18/4 Simplify 18/4 = 9/2 = 4 1/2 Practice Exercise 1-4 1. 1/2 x 4/9 2. 20 x 3/5 3. 4 11/16 x 12 4/9 4. 16 3/4 x 5/8 5. 2.3 x 5/9 6. 13/32 x 8/9 7. 32 3/8 x 4 5/16 8. 5/6 x 1/4 x 8/32 9. 1 3/8 x 24/30 x 2/3 10. 6 x 3/36 x 5/8 11. 13 x 6 x 11/16 12. 5 2/5 x 5/8 x 4/9 Answers 1. 2/9 2. 12 3. 58 1/3 4. 10 15/32 5. 10/27 6. 13/36 7. 139 79/128 8. 5/96 9. 11/15 10. 11. 5/16 53 5/8 12. 1 1/2

Unit 1-5 Dividing With Fractions Procedure: 1. If whole numbers are involved change them to fractions with a denominator of 1. 2. If mixed numbers are involved change them to improper fractions. 3. Invert the divisor (the number that is being divided by.) 4. Multiply the inverted divisor by the dividend (the number that is being divided into.) Example 1 Divide: 3/8 by 1/4 Invert the divisor 1/4 becomes 4/1 Multiply the inverted divisor by the dividend. 3/8 x 4/1 = 12/8 = 1 1/2 Example 2 Divide: 5/16 by 5 Change whole numbers to fractions with a denominator of 1. 5 = 5/1 Invert the divisor 5/1 become 1/5 Multiply the inverted divisor by the dividend. 5/16 x 1/5 = 5/80 = 1/16 Example 3 Divide: 6 3/4 by 9/16 Change mixed numbers to improper fractions. 6 3/4 = 27/4 Invert the divisor. 9/16 becomes 16/9 Multiply the inverted divisor by the dividend. 27/4 x 16/9 = 3/1 x 4/1 = 12 Practice Exercise 1-5 Divide the following: 1. 7/16 by 1/4 4. 6 by 3/8 2. 5/6 by 3/8 5. 4 3/8 by 3/16 3. 3/5 by 2 6. 3 5/16 by 3 5/8 7. 14 by 2 5/6 8. 4 1/4 by 11/8 9. 14 5/32 by 6 1/6

Answers 1. 1 3/4 2. 2 2/9 3. 3/10 4. 16 5. 23 1/3 6. 53/58 7. 4 16/17 8. 3 1/11 9. 2 175/592

Unit 1-6 Converting Fractions to Decimals Procedure: 1. Separate the fraction from the whole numbers. 2. Divide the numerator by the denominator. 3. Put the resulting decimal to the right of the whole number. Example 1 Convert 3/4 to a decimal. Divide 3 by 4 = 0.75 Example 2 Convert 5 7/8 to a decimal. Separate the fraction from the whole number. 5 7/8 Divide the numerator by the denominator. = 0.875 Put the resulting decimal to the right of the whole number. 5.875 Practice Exercise 1-6 Convert the following fractions to decimals: 1. 1/2 2. 1/4 3. 1/8 4. 1/16 5. 1/3 6. 1/6 7. 1/9 8. 1/12 9. 1/32 10. 3/4 11. 5/8 12. 2/3 13. 7/11 14. 4/19 15. 4/7 16. 6 3/8 17. 25 5/6 18. 9 2/9 Answers 1. 0.5 2. 0.25 3. 0.125 4. 0.0625 5. 0.33 6. 0.166 7. 0.11 8. 0.833 9. 0.03125 10. 0.75 11. 0.625 12. 0.66 13. 0.63 14. 0.21 15. 0.57 16. 6.375 17. 25.833 18. 9.22

Unit 2-1 Understanding Decimals Concept: Decimals are nothing more than fractions with denominators that are in multiples of 10. As you move to the right of the decimal point each place is a fraction that is a multiple of 10 smaller than the place before it. The illustration below should help you understand the concept. Decimal Expressed as a fraction Place value in words.1 1/10 tenths.01 1/100 hundredths.001 1/1,000 thousandths.0001 1/10,000 ten-thousandths.00001 1/100,000 hundred-thousandths.000001 1/1,000,000 millionths Converting decimals to fractions is easily accomplished by writing the decimal as if it were a whole number and putting it in the place of the one in the fractions expressed above. The procedure is illustrated below. 0.7 = 7/10 0.83 = 83/100 0.649 = 649/1000 0.034 = 34/1000 0.6539 = 6539/10,000 When reading or writing decimals in words you should do the following. Read the whole number part, if any, first. Then ignoring any leading zeros you should read the digits as if they were whole numbers and say the name of the place value of the last digit. The following examples illustrate the proper procedure. 0.4 is written or spoken four tenths 0.453 is four hundred and fifty three thousandths 4.67 is four and 67 hundredths 5.0612 is five and six hundred and twelve ten-thousandths

Unit 2-2 Adding and Subtracting Decimals Procedure: 1. Line up the decimal points. 2. Fill in any blanks with zeros. 3. Perform the operation. Example 1 Add: 3.75 + 5.3 Line up the decimal point. 3.75 +5.3 Fill in any blanks with zeros. 3.75 +5.30 Perform the operation 9.05 Example 2 Subtract: 8.46-3.251 Line up the decimal point 8.46-3.251 Fill in any blanks with zeros 8.460-3.251 Perform the operation 5.209 Practice Exercise 2-2 1. 3.5 + 5.67 2. 3.456 + 23.1 3. 21.35 + 45.867 4. 4.649 + 6.251 5. 67.55 + 3.001 6. 35.02 + 30.009 7. 5.6-2.36 8. 87.739-20.6 9. 65.21-38.746 10. 8.345-7.849 11. 23.46-7.005 12. 21.06-13.004 Answers 1. 9.17 2. 26.556 3. 67.217 4. 10.9 5. 70.551 6. 65.029 7. 3.24 8. 67.139 9. 26.464 10. 0.496 11. 16.455 12. 8.056

Unit 2-3 Multiplying Decimals Procedure: 1. Multiply the numbers as if they were whole numbers. 2. Count the number of places to the right of the decimal in each number. 3. Add the number of places together. 4. Count the total number of places from the right of the product of the multiplication and put in a decimal point. Example 1 Multiply: 2.3 X 4.51 Multiply the numbers as if they were whole numbers. 23 X 451 = 10373 Count the number of places to the right 2.3 = 1 of the decimal in each number. 4.51 = 2 Add the total number of places together 1 + 2 = 3 Count the total number of places from the right of the product 10.373 Example 2 Multiply: 6.045 x 4.29 Multiply the numbers as if they were whole numbers. 6045 x 429 = 2593305 Count the number of places to the right 6.045 = 3 of the decimal in each number 4.29 = 2 Add the total number of places together 3 + 2 = 5 Count the total number of places from the right of the product 25.93305 Practice Exercise 2-3 1. 3.6 x 5.4 2. 1.9 x 0.45 3. 23.4 x 3.056 4. 0.36 x 0.17 5. 46.489 x 3.43 6. 4.69 x 5.003 7. 342.1 x 0.005 8. 76.9 x 6.5872 9. 3.578 x 97.0038 Answers 1. 19.44 2. 0.855 3. 71.5104 4. 7. 0.0612 1.7105 5. 8. 159.45727 506.55568 6. 9. 23.46407 347.07959

Unit 2-4 Dividing Decimals Procedure: 1. Move the decimal in the divisor to the right until it is a whole number and count the number of places the decimal point moved. 2. Move the decimal in the dividend the same number of places it was moved in the divisor adding zeros if needed for place holders. 3. Perform the operation. Example 1 Divide 4.46166 by 1.26 Move the decimal in the divisor to the 1.26 becomes 126 right to make a whole number and count the number of places moved. places moved = 2 Move the decimal in the dividend the same number of places 4.46166 becomes 446.166 3.541 Perform the operation 126 446.166 Example 2 Divide 3.55 by 0.005 Move the decimal in the divisor to the 0.005 become 5 right to make a whole number and count the number of places moved. places moved = 3 Move the decimal in the dividend the same number of places adding zeros for place holders if needed. 3.55 becomes 3550 710 Perform the operation 5 3550 Practice Exercise 2-4 Divide the following: 1. 4.5 by 0.01 2. 36.897 by 14.7 3. 4.8 by 3.75 4. 38.3838 by 11.1 5. 17.7741 by 26.1 6. 34.41 by 5.55 7. 8.784 by 2.4 8. 0.37758 by 0.58 9. 0.0742 by 2.8 Answers 1. 450 2. 2.51 3. 1.28 4. 7. 3.458 3.66 5. 8. 0.681 0.651 6. 9. 6.2 0.0265

Unit 2-5 Percentages Concept: Percent means per 100 and is used to represent what portion of whole quantity is involved. In other words percents less than 100 mean less than the whole quantity is involved. 100% means the whole quantity is involved and is equivalent to the fraction 100/100 or the number 1. Percents greater than 100 imply that greater than the whole quantity is involved. Percents can be used in numerous calculations if they are converted to either decimals or fractions. Procedure: 1. To convert a decimal to percent multiply by 100 and write a percent sign after the product. 2. To convert a percent to a decimal divide by 100 and remove the percent sign. Converting Fraction to Percent and Percent to Fraction 1. To convert fractions to percent multiply by 100, simplify the fraction, and write a percent sign after the product. 2. To convert percent to a fraction write the percent as a fraction with a denominator of 100, remove the percent sign, and simplify the fraction. Example 1 Convert 0.657 to a percent Multiply the decimal by 100 and write a percent sign after the product 0.657 x 100 = 65.7% Example 2 Convert 43.61% to a decimal Divide by 100 and remove the percent sign 43.61% / 100 = 0.4361 Example 3 Convert 5/8 to a percent Multiply by 100 5/8 x 100 = 500/8 Simplify and add percent sign 500/8 = 62.5%

Example 4 Convert 24% to a fraction Write the percent as a fraction with a denominator of 100 and remove the % sign 24% becomes 24/100 Simplify the fraction 24/100 = 6/25 Practice Exercise 2-5 Convert the following decimals to percentages 1. 0.57 2. 0.983 3. 0.068 4. 4.32 5. 67.739 6. 9.0024 Convert the following percentages to decimals 7. 43% 8. 28.3% 9. 22.06% 10. 207.64% 11. 1000% 12. 100.07% Convert the following fractions to percentages 13. 3/4 14. 3/8 15. 1/2 16. 9/4 17. 2 5/8 18. 14 11/16 Convert the following percentages to fractions 19. 71% 20. 101% 21. 352% 22. 45.2% 23. 307.5% 24. 176.03% Answers 1. 57% 2. 98.3% 3. 6.8% 4. 432% 5. 6773.9% 6. 900.24% 7. 0.43 8. 0.283 9. 0.2206 10. 2.0764 11. 10 12. 1.0007 13. 75% 14. 37.5% 15. 50% 16. 225% 17. 262.5% 18. 1468.75% 19. 71/100 20. 1 1/100 21. 3 13/25 22. 113/250 23. 3 3/40 24. 1 7603/10,000

Unit 2-6 Percentage Calculations Procedures: Calculating a percent of a number 1. Convert the percent to a decimal. 2. Multiply the number by the decimal. Calculating what percent one number is of another number 1. Write a fraction with the number representing 100% as the denominator and the "%" number as the numerator. 2. Simplify the fraction if possible and convert it to a percent. Calculating percent increase or decrease. 1. Find the amount of increase or decrease by subtracting the smaller number from the larger number. 2. Write a fraction in which the numerator is the amount of increase or decrease and the denominator is the original amount. 3. Simplify the fraction if possible and convert it to a percent. Example 1 Find 18% of 356 Convert the percent to a decimal 18/100 = 0.18 Multiply the number by the decimal 356 x 0.18 = 64.08 Example 2 Find what percent 42 is of 75 Write a fraction with the number representing 100% as the denominator and the % number as the numerator 42/75 Convert to a percent 42/75 x 100 = 56% Example 3 Find the percent increase from 54 to 81 Find the amount of increase 81-54 = 27 Write a fraction in which the numerator is the amount of increase and the denominator is the original amount 27/54 Simplify if possible and convert to 27/54 = 1/2 a percent 1/2 = 0.5 x 100 = 50%

Practice Exercise 2-6 Calculate the following: 1. 47% of 100 2. 56% of 2 3. 6% of $1.42 4. 201% of 450 5. 112% of 100 6. 9% of 15 What percent is 7. 34 of 100 8. 34 of 200 9. 15 of 440 10. 45 of 20 11. 29 of 29 12. 465 of 1000 Find the percent increase or decrease from 13. 25 to 25 14. 100 to 125 15. 46 to 91 16. 32 to 96 17. 50 to 15 18. 341 to 297 19. I have a class of 24 students if 3 people fail the first quiz what percent of the class passed the first quiz? 20. If I have 10 apples in my refrigerator and 3 have worms in them, what percent of the apples have worms? 21. If gas prices go from $1.07 per gallon to $1.21 per gallon what is the percent increase? 22. If enrollment in Materials Calculation goes from 80 students to 60 students what is the percent decrease? 23. If my rent is $140.00 per month and it increases by 9% what will be the new amount I will have to pay? Answers 1. 47 2. 1.12 3. $0.09 4. 904.5 5. 112 6. 1.35 7. 34% 8. 17% 9. 3.4% 10. 225% 11. 100% 12. 46.5% 13. 0% 14. 25% increase 15. 97.8% increase 16. 200% increase 17. 70% decrease 18. 13% decrease 19. 87.5% 20. 30% 21. 13% 22. 25% 23. $152.60

Unit 3-1 Solving Simple Equations Concept: Equations consist of two mathematical expressions on opposite sides of an equal sign. A very simple form of an equation would be something like 2 x 4 = 8. The expressions on each side of the equal side are equivalent, this an equation. In mathematics, as it applies to horticulture, it is very useful to be able to solve simple equations with one missing variable. An example of this type of equation is 2 x y = 8 where Y represents some number. The value of Y in this example must be 4 in order for the equality to be preserved. The idea behind solving equations is to get the missing variable in a multiple of one on one side of the equal sign and the numbers on the other. This is accomplished by manipulating the numbers on each side of the equal sign. You can do any legitimate mathematical operation you want to one side of the equation provided you do the exact same operation on the other side of the equation. Examples of the types of equations you will need to be able to solve for this course are given below. (Note: 2 x Y and 2Y mean the same thing, any number written adjacent to a letter variable means multiply.) Example 1 Solve: 2Y = 16 Divide both sides of the equation by 2 2Y/2 = 16/2 y = 8 Example 2 Solve: 3Y/14 = 12/7 Multiply both sides by 14 14 x 3Y/14 = 12/7 x 14 3Y = 24 Divide both sides by 3 3Y/3 = 24/3 Y = 8 Example 3 Solve: 21/9 = 42/5Y Cross multiply (i.e. Multiply both sides 5Y x 21/9 = 5Y x 42/5Y by the denominators, 5Y and 3) 105Y/9 = 42 9 x 105Y/9 = 42 x 9 105Y = 378 Divide both sides by 105 105Y/105 = 378/105 Y = 3.6

Example 4 Solve: 2Y + 0.25Y = 6 + 7.5 Do the addition first 2Y + 0.25Y = 2.25Y 6 + 7.5 = 13.5 2.25Y = 13.5 Divide both sides by 2.25 2.25Y/2.25 = 13.5/2.25 Y = 6 Practice Exercise 3-1 Solve the following equations: 1. 5y = 10 2. 7X = 24.5 3. 2.6Y = 6.1 + 1.7 4. Y/2 = 12 5. 5X/7 = 34/3 6. 26/4 = X/8 7. 6/5X = 12/23 8. 70/494 = 62/12Y 9. 24/3 = 15/12y Brain Teasers 10. If there are 6 women in a group of people and they comprise 20% how many people are in the group? 11. If perennial ryegrass seed cost $1.26 and this is a 5% increase over last years cost, what did it cost last year? 12. I lost 1/8 of my money at the dog track, my friend Mike lost 1/2 of his money. He started out with $350 and we now have the same amount how much money did I start with? Answers 1. Y = 2 2. X = 3.5 3. Y = 3 4. Y = 24 5. X = 15.866 6. X = 52 7. X = 2.3 8. Y = 35.286 9. Y = 0.15625 10. 30 11. $1.20 12. $200

Unit 3-2 Ratio, Proportion Mean, Mode, and Median Concept: Ratio, proportion, mean, mode, and median all tell us something about how numbers relate to one another. A ratio is a comparison between two numbers. Ratios are used to describe relationships between a myriad of things from gears to racial balance in schools. They are written several different ways. For example if I had 3 cats and 2 dogs, the cat to dog ratio could be written 3 to 2, or 3:2, or like a fraction 3/2. Ratios can also be used to express rates where the top and bottom numbers have different units. Examples of this are miles/hour (miles per hour) and pounds/acre (pounds per acre). A proportion is a statement that two ratios are equal. For example 2/1 = 4/2, that is to say that 2 is to 1 as 4 is to 2. Although the absolute quantities are different the ratio is the same. Proportions are very useful for calculations and will come up again in this book. If 3 quantities are known in a proportion and the fourth is unknown you have a simple equation. The fourth quantity can be solved for to make the proportion true. It is also possible to verify that a proportion is true by cross multiplying to see if the products are equal. The example below illustrates solving for a true proportion and checking the solution. 3/2 = X/4 Solving for X reveals that X = 6 To check if the proportion is true: 2 x 6 = 12 and 3 x 4 = 12 Mean, mode, and median are statistical terms that tell us something about a group of numbers. Mean is the arithmetic average and is calculated by adding the numbers together and dividing the total by the number of observations. Mode is simply the number that occurs most often. Median is the number which half of the observations are lower and half of the observations are higher. The three together give us a better picture of a group of numbers than any one can individually. Consider the example below. I weighed a group of nine people and obtained the following weights. 120, 165, 101, 187, 315, 141, 295, 155, 141 - To obtain the mean, I add them all up (sum = 1620) and divide by the number of observations (# of observations = 9). Mean = 180. - To obtain the median, I list the number in ascending order and look for the number in the middle. Median = 155. 101, 120, 141, 141, 155, 165, 187, 295, 315 - To obtain the mode, I find the number occurring most often. Mode= 141. These statistical terms are normally used to describe larger groups of numbers that are not as easily digested as the example given above. Notice how a few high numbers can inflate the mean and give a distorted view of the entire population. This is why median and mode are also useful in obtaining an accurate picture of a group of numbers. As the old saying goes, if your feet are in the fireplace and your head is in the freezer, on the average you are comfortable.

Example 1 The grading system in my Underwater Basket Weaving class is 90% = A, 80% = B, 7-% = C, 60% = D, and less then 60% = F. Here are the final averages: 96, 87, 43, 12, 79, 87, 82, 85, 95. What is the Mode? Mode = 87% What is the Mean? Mean = 666/9 = 74% What is the Median? Median = 85% What is the Pass/Fail ration Pass/Fail ratio = 7/2 If the pass fail ratio stays the same for next years class and 6 people fail how many people will pass? Set up a proportion 7/2 = X/6 Solve for X X = 21 people will pass Practice Exercise 3-2 Seven golf courses were surveyed to determine their capital expenditure on equipment each year. The results are listed below. $15,000, $4,500, $48,000, $11,200, $9,300, $17,800, $12,000 1. What is the median capital expenditure? 2. What is the mean capital expenditure? 3. What is the ratio of courses spending more than $10,000 to courses spending less than $10,000? I rated my golf greens for color on a scale of 1 to 10. (1 = yellow, 10 = deep green) The results are listed below. 8, 7, 8, 9, 7, 10, 8, 8, 5, 7, 8, 9, 9, 7, 8, 6, 8, 3 4. What is the mean color score? 5. What is the mode? 6. If a score of 6 or less is considered unacceptable what is the ratio of acceptable greens to unacceptable greens? Answer 1. $12,000 2. $16,828.57 3. 5/2 4. 7.5 5. 8 6. 5/1

Unit 4-1 Parts Per Million Concept: Parts per million (ppm) is a measure of concentration. In solutions it is a weight relationship. In other words 1,000,000 pounds of water that contains 1 pound of salt would have a concentration of 1 ppm salt and can be written as the ratio of salt to water - 1/1,000,000. By using the ratio of the weight of salts added to the weight of water used in a given solution and setting it equal to the ratio for ppm it is possible to make various calculations for determining and mixing ppm solutions. It is very important to remember that the weight of water (8.34 pounds per gallon or 133.44 weighed ounces per gallon) not the volume must be used in ppm calculations. The salt in the solution is referred to as the solute and the water is referred to as the solvent. The general form of the proportion used in ppm calculations is below X/y = Z/1,000,000 Where: X = the weight of the solute Y = the weight of the solvent Z = the desired or unknown concentration in ppm If any two of the variables are known the third can be calculated. (Note: Calculation with the formula above assumes the weight of the salt in the final solution to be negligible. If a very precise concentration is required the weight of the solute and solvent should be added together and plugged in as the Y variable.

Unit 4-2 Determining PPM When Known Quantities are Mixed P rocedure: 1. Convert the volume of solvent to a weight with the same units as the added solute. 2. Plug in the weight of solute as the X variable. 3. Plug in the weight of the solvent as the Y variable. 4. Solve for concentration Z. Example 1 If I add 2 pounds of NaC1 to 50 gallons of water, how many ppm NaC1 will the resulting solution be? Convert the volume of solvent to a weight with the same units as the added solute Plug in the weight of the solute added as the X variable Plug in the weight of the solvent as the Y variable Solve for concentration Z 50 x 8.34 lb/gal = 417 lb 2/Y = Z/1,000,000 2/417 = Z/1,000,000 Z = 4796 ppm Example 2 If 3 ounces of salt are added to 2 gallons of water, what will be the salt concentration in ppm of the resulting solution? Convert the volume of solvent to a weight with the same units as the added solute Plug in the weight of the solute added as the X variable Plug in the weight of the solvent as the Y variable 2 x 133.44 = 266.88 oz 3/Y = Z/1,000,000 3/266.88 = Z/1,000,000 Solve for concentration Z Z = 11,241

Practice Exercise 4-1 1. If 6 pounds of salt are mixed with 100 gallons of water how many ppm salt will the resulting solution be? 2. If 3.4 pounds of KCI are added to 500 gallons of water how many ppm KCI will the resulting solution be? 3. If 20 ounces of NaC1 are added to 15 gallons of water, what will be the concentration NaC1 in ppm of the resulting solution. 4. What will the concentration of salt in ppm be if 45 ounces of salt are added to 40 gallons of water? 5. If 5 pounds 6 ounces of salt are added to 200 gallons of water how many ppm salt will the resulting solution be? 6. If 2 ounces of NaC1 are added to 10 gallons of water, how many ppm NaC1 will the resulting solution be? Answer 1. 7194 ppm 2. 815 ppm 3. 9992 ppm 4. 8430 ppm 5. 3222 ppm 6. 1499 ppm

Unit 4-3 Determining Solute quantities for Desired Concentrations P rocedure: 1. Convert the volume of solvent to a weight with the desired units. 2. Plug in the desired concentration in ppm as the Z variable. 3. Plug in the weight of the solvent as the Y variable. 4. Solve for X. Example 1 How many pounds of NaC1 should be added to 80 gallons of water to obtain a 300 ppm NaC1 solution? Convert the volume of solvent to a weight with the desired units 80 x 8.34 = 667.2 lb Plug in the desired concentration in ppm as the Z variable X/y = 300/1,000,000 Plug in the weight of solvent as the Y variable X/667.2 = 300/1,000,000 Solve for X X = 0.2 lb Example 2 How many ounces of salt should be added to 8 gallons of water to obtain a 2,500 ppm salt solution? Convert the volume of solvent to a weight with the desired units 8 x 133.44 = 1067.52 oz Plug in the desired concentration in ppm as the Z variable X/Y = 2,500/1,000,000 Plug in the weight of solvent as the Y variable X/1067.52 = 2500/1,000,000 Solve for X X = 2.6688 oz

Unit 4-4 Mixing PPM Fertilizer Solutions Concept: Fertilizer recommendations are sometimes given in parts per million of a particular element such as nitrogen. It is possible to calculate how much of a given fertilizer should be mixed with a specific quantity of water in order to obtain the desired concentration of an element. this is done by: 1) calculating the weight of the element needed in solution to give the desired concentration, and 2) calculating the weight of fertilizer to give the desired weight of the element (no fertilizer will be 100% of an element). Step 1 is done by the same procedure used in Unit 4-3, in other words solving for the weight of the solute needed. Step 2 is done using the formula below. Where: BC = X B = the percent composition of the fertilizer of the element needed expressed as a decimal. C = the weight of the fertilizer to be added X = the weight of the element needed Procedure: 1. Solve for the weight of the solute. -Convert the volume of the solvent to a weight with the desired units. -Plug in the weight of the solvent as the Y variable. -Plug in the desired concentration of the element being mixed as the Z variable. -Solve for the X variable. (This yields the quantity of element needed) 2. Plug in % composition of the fertilizer of the element needed expressed as a decimal, as B. 3. Plug in the weight of the element (solute) needed as X. 4. Solve for C, the weight of the fertilizer to be added. Example 1 How many ounces of 20-20-20 should be added to 5 gallons of water to obtain a 350 ppm nitrogen solution? Solve for the weight of the solute X/667.2 = 350/1,000,000 X = 0.23352 oz nitrogen Plug in % composition of the fertilizer of the element as a decimal, as B Plug in the weight of the solute as X 0.20C = X 0.20C = 0.23352 oz N Solve for C C = 1.1676 oz of 20-20-20

Example 2 How many pounds of 33-0-0 should be added to 200 gallons of water to obtain a 500 ppm nitrogen solution? Solve for the weight of solute X/1668 = 500/1,000,000 X = 0.834 lbs nitrogen Plug in % composition of the fertilizer of the element as decimal, as B 0.33C = X Plug in the weight of the solute as X 0.33C = 0.834 lbs nitrogen Solve for C C = 2.527 lbs of 33-0-0 Practice Exercise 4-3 & 4-4 1. How many ounces of salt should be added to 25 gallons of water to obtain a 1500 ppm salt solution? 2. How many pounds of salt should be added to 500 gallons of water to obtain a 12,000 ppm salt solution? 3. How many ounces of urea (45-0-0) should be added to 10 gallons of water to mix a 250 ppm nitrogen solution? 4. How many ounces of muriate of potash (0-0-60) should be added to 100 gallons of water to obtain a 1000 ppm potassium solution? 5. How much 12-15-20 should be added to 5 gallons of water for the resulting solution to be 100 ppm nitrogen? 6. How many pounds of 20-20-20 should be added to 300 gallons of water to obtain a 500 ppm nitrogen solution? 7. How much ammonium nitrate (33-0-0) should be added to 200 gallons of water to obtain a 600 ppm nitrogen solution? Answers 1. 5 oz salt 2. 50 lbs salt 3. 0.74 oz urea 4. 22.24 oz muriate of potash 5. 0.556 oz 12-15-20 6. 6.255 lb 20-20-20 7. 3 lbs ammonium nitrate

Unit 4-5 Determining Elemental PPM From Known Quantities of Fertilizer Procedure: 1. Plug in the weight of the fertilizer added as C. 2. Plug in the % composition of the fertilizer of the element as a decimal, as B 3. Solve for X, the weight of the element added. 4. Solve for the concentration as in Unit 4-2. Example 1 If 15 ounces of urea are added to 20 gallons of water, how many ppm nitrogen will the resulting solution be? Plug in the weight of fertilizer as C Plug in % composition as a decimal, as B Solve for X Solve for the concentration B x 15 oz = X 0.45 x 15 = X X = 6.75 ox of nitrogen 6.75/2668.8 = Z/1,000,000 Z = 2529 ppm nitrogen Example 2 If 4 pounds of ammonium nitrate are added to 100 gallons of water, how many ppm nitrogen will the resulting solution be? Plug in the weight of fertilizer as C Plug in % composition as a decimal, as B Solve for X Solve for the concentration B x 4 lb + X 0.33 x 4 lb = X X = 1.32 lbs nitrogen 1.32/834 = Z/1,000,000 Z = 1583 ppm nitrogen Practice Exercise 4-5 1. If 35 ounces of urea are added to 50 gallons of water, how many ppm nitrogen will the resulting solution be? 2. If 2 ounces of 20-20-20 are added to 1 gallon of water, how many ppm nitrogen will the resulting solution be? 3. If 100 pounds of ammonium nitrate are added to 500 gallons of water, how many ppm nitrogen will the resulting solution be? 4. If 2 pounds of muriate of potash are added to 200 gallons of water, how many ppm potassium will the resulting solution be? Answers: 1. 2360 ppm N 2. 2998 ppm N 3. 7914 ppm N 4. 719 ppm K

Unit 4-6 Dilutions Concept: Many pesticides and other products are packaged as concentrated liquids for ease of handling. It is often desirable to mix more dilute solutions from these concentrates. When making dilutions there is an inverse relationship between the volume of the resulting solution and it's concentration. In other words as more and more diluent is added to a concentrate the volume increases but the concentration decreases. It is possible to take advantage of this relationship and calculate the proper quantities of concentrate and diluent to combine to achieve a desired concentration by using the following formula: Where: C 1 V 1 = C 2 V 2 C 1 V 1 C 2 V 2 = concentration of the concentrate = volume of the concentrate = concentration of the final solution = the final solution volume This equation is also useful in calibrating hose-on type applicators that dilute by siphoning concentrates in a ratio to the volume of diluent that passes through the hose. Procedure: 1. Plug in the concentration of the concentrate as C 1. 2. Plug in the desired concentration of the final solution as C 2. 3. Plug in the desired volume of final solution as V 2. 4. Solve for V 1. 5. Put V 1, the volume of concentrate, in a container and bring the total volume in the container to V 2, the final solution volume. Example 1 I want to mix 2 gallons of 15% clorox solution to sterilize tools, how much clorox and how much water do I mix together? Plug in the concentration of the concentrate as C 1 (Note: everything in 100 X V 1 = C 2 V 2 the bottle is clorox so the conci. = 100%) Plug in the desired concentration of the final solution 100 x V 1 = 15 x V 2 Plug in the desired volume of final solution 100 x V 1 = 15 x 256 oz (Note: 2 gallons = 256 oz) Solve for V 1 V 1 = 38.4 oz Put 38.4 oz of clorox in a container and bring the final solution volume to 2 gallons.

Unit 4-7 Calibrating Hose-on Applicators Concept: Hose-on applicators dilute and apply pesticides and fertilizers by siphoning a concentrate from a container and mixing it with water flowing through a hose. The container may be a small cup attached at the applicator nozzle or a larger bucket. In the latter situation the siphon device is usually located at the hose bib. Hose-on type applicators mix the water and concentrate in a ratio, usually 11/1. However this ratio can vary from one device to another and will vary depending on water pressure, and the viscosity and specific gravity of the concentrate being applied. The ratio of the mixture in a given situation can be determined by catching a volume of the diluted mixture and comparing it to how much concentrate was siphoned. For example if I catch 10 quarts of mixture while 1 quart of concentrate is siphoned, I have a 10/1 ratio for that situation. Ten quarts of mixture for every 1 quart was concentrate siphoned. One this ratio has been established the hose-on applicator can be calibrated using the dilution equation shown below. C 1 V 1 = C 2 V 2 Where: C 1 = The concentration of the concentrated solution in the cup. V 1 = The volume of the concentrate mixed C 2 = The final solution concentration. V 2 = The final solution volume. Calibration implies deciding on the concentration of the solution applied (i.e.c 2 ) and adjusting the device to achieve that concentration. Since the ratio at which the concentrate and water are combined is fixed, the only thing that can be changed to achieve the desired calibration is the concentration of the concentrate (i.e. C 1 ). In other words the concentrate in the cup or bucket is diluted to give the desired final solution concentration. The dilution equation above can also be used to make other dilutions that do not involve a hose-on applicator. Procedure: 1. Determine the ratio at which the hose-on mixes water and concentrate. 2. Plug in the smaller number (amount of concentrate mixed) as V 1. 3. Plug in the larger number (amount of water mixed) as V 2. 4. Plug in the desired final solution concentration as C 2. 5. Solve for C 1. (This is the concentration needed in the cup or "stock solution"). 6. Mix the proper concentration of stock solution, or dilute a concentrate to give the proper concentration of stock solution. Example 1 I have a hose-on applicator which applied 5.5 gallons of solution while siphoning 0.5 gallon of stock solution. How many ppm nitrogen should the stock solution be in order to obtain 300 ppm nitrogen final solution concentration? Example 1 cont.

Determine the ration 5.5/0.5 = 11/1 Plug in the smaller number as V 1 C 1 x 1 = C 2 V 2 Plug in the larger number as V 2 C 1 x 1 = C 2 x 11 Plug in the desired final solution conc. C 1 x 1 = 300 x 11 Solve for C 1 (the stock solution conc.) C 1 = 3300 ppm nitrogen Mix a 3300 ppm nitrogen solution for stock solution. Example 2 How would I mix 3 gallons of the stock solution in Example 1 using urea (45-0-0)? Solve for the weight of the solute X/400.32 = 3300/1,000,000 X = 1.32 oz nitrogen Solve for the weight of the urea 0.45 C = 1.32 C = 2.933 oz urea in 3 gal. Example 3 I have a hose-on sprayer which applies 13 quarts of spray while siphoning 2 quarts out of the cup. I want to apply an insecticide at a rate of 2 ounces per gallon of water. How should I mix the stock solution? Determine the ratio 13/2 Plug in the smaller number as V 1 C 1 x 2 = C 2 x V 2 Plug in the larger number as V 2 C 1 x 2 = C 2 x 13 Plug in the desired final solution conc. C 1 x 2 = 2 oz/gal x 13 Solve for C 1 (the stock solution conc.) C 1 = 13 oz/gal Put 13 ounces of insecticide in a container and add 115 oz of water to bring the final solution volume to 1 gallon. Practice Exercise 4-6 & 4-7 1. How would you mix 4 gallons of 35% Round-up solution? 2. How would you mix 6 quarts of 10% clorox solution? 3. Give a hose-on applicator that applies 6.5 gallons of solution while siphoning 0.5 gallons of stock solution, how many ppm nitrogen should the stock solution be to apply 200 ppm N in the final solution?

Practice Exercise 4-6 & 4-7 cont. 4. A hose-on sprayer applies 160 oz of spray while siphoning 16 oz from the cup. How should the stock solution be mixed to obtain 3 oz/gal herbicide in the spray. 5. Given an 11/1 hose-on sprayer, how would you mix 5 gallons of stock solution to obtain 1 oz/gallon of insecticide in the spray? 6. Given a 9/1 hose-on sprayer, how would you mix 2 quarts of stock solution to obtain 5 oz/gal of herbicide in the spray? Answers 1. 1.4 gal Round-up & 2.6 gal water 2. 0.6 qt clorox & 5.4 qt. water 3. 2600 ppm nitrogen 4. conc of herbicide = 30 oz/gal 5. 55 oz insecticide & 585 oz of water 6. 22.5 oz herbicide & 41.5 oz of water

Unit 5-1 Area of Circles and Part-Circles Concept: The area of a complete circle can be determined by using the formula A = 3.14 X r 2 Where: A = the area of the circle r = the radius of the circle With part circles the area can be determined by multiplying an expression which describes the portion of the circle present by formula above. For example the area of a half circle would be described by A = 1/2 since 180 of the entire 360 are present. (180/360 = 1/2) A more general form of this formula is shown below: A = 3.14 X r 2 X /360 Where: = the interior angle of the portion of the circle present Example 1 Find the area of the circle pictured r = 7 feet = 3.14 A = 3.14 x 7 = 153.86 square feet r = 7' Example 2 Find the area of the part circle pictured r = 4 feet = 3.14 = 265 degrees A = 3.14 x 4 2 x 265/360 = 36.98 square feet Practice Exercise 5-1 Find the area of the circles and part circles below: r = 15' r = 25' r = 4' = 195 0 = 105 0 Answers: 1) 706.5 ft 2) 1063 ft 3) 1465 ft

Unit 5-2 Area of Squares, Rectangles, Parallelograms and Triangles Concept: The area of squares, rectangles, and parallelograms can be determined by using the following formula: A = L x W Where: A = the area of the square, rectangle, or parallelogram L = the length W = the width (L and W will always be adjacent sides as in the diagrams below) W The area of a triangle can be determined by the following formula: A = 1/2 B x H Where: A= the area of the triangle B= the length of the base of the triangle H= the height of the triangle (B and H are illustrated in the diagrams below) L H B Example 1 Determine the area of the square. L = 2 feet W = 2 feet A = 2 x 2 = 4 square feet Example 2 Determine the area of the rectangle. L = 5 yards W = 2 yards A = 5 x 2 = 10 square yards

Example 3 Determine the area of the parallelogram L = 16 inches W = 4 inches A = 16 x 4 = 64 square inches Determine the area of the triangles B = 6 feet H = 3 feet A = 1/2 x 6 x 3 = 9 square feet B = 20 inches H = 5 inches A = 1/2 x 20 x 5 = 50 square inches Practice Exercise 5-2 Determine the area of the shapes below. (1) (2) 4 4" 4.5 4" 20 (3) (4) 3 6 10 9 5" 4 (5) (6) 6" 15 Answers 1. 16 square inches 2. 90 square feet 3. 30 square feet 4. 27 square feet 5. 15 square inches 6. 30 square feet

Unit 5-3 Area by the Offset Method Concept: It is possible to estimate the area of long irregular shapes such as a golf fairway by dividing the area into smaller "boxes." This is accomplished by measuring the width of the shape at regular intervals (i.e. the offsets). An estimate of the area is derived by multiplying the total of the offsets by the interval length. In effect you are multiplying the average width times the length. The closer the intervals are together the more accurate the estimate will be. The more irregular the shape is the closer the intervals need to be to obtain a reliable estimate. Procedure: 1. Select an interval which will divide evenly into the entire length of the shape. 2. Measure the offsets (i.e. the width of the shape at each interval.) 3. Add up the offsets. 4. Multiply the sum of the offsets by the interval. Example 1 Estimate the area of the golf fairway below. Interval = 50 feet Offset lengths in feet: A = 55 B = 58 C = 55 D = 43 E = 38 F = 37 G = 49 H = 61 I = 64 J = 67 K = 61 Add up the offsets Multiply the total of the offsets by the interval length ft. Sum = 588 ft. 588' x 50' = 29,400 sq.

Unit 5-4 Area by Indirect Offsets C oncept: Occasionally it is necessary to measure long irregular shapes such as lakes and ponds which do not lend themselves to easy direct measurement of the offset lengths. In these cases the length of the offsets can be obtained indirectly. Once the offset lengths are known the area can be estimated by the offset method. P rocedure: 1. Lay out a rectangle of known length and width enclosing the shape of interest. 2. Measure regular intervals along the length of each side of the rectangle. 3. Measure the portion of the offset remaining between the shape and the edge of the rectangle on each side, at each interval, and add the two together. 4. Subtract total of each set of offsets from the width of the rectangle this yields the actual offset length. 5. Calculate the area by the offset method. Example 1 Interval = 50 feet W = 100 feet (i.e. the width of the rectangle) Indirect offset lengths in feet A1 = 32 A2 = 29 B1 = 24 B2 = 28 C1 = 19 C2 = 20 D1 = 15 D2 = 33 E1 = 14 E2 = 37 F1 = 18 F2 = 37 G1 = 25 G2 = 10 H1 = 30 H2 = 10 I1 = 24 I2 = 20 The length of offset A = W - (A1 + A2) = 100' - (32' + 29') = 39 feet The length of offset B = W - (B1 + B2) = 100' - (24' + 28') = 48 feet

Example 1 cont. C = 100' - (19' + 20') = 61 feet D = 100' - (15' + 33') = 52 feet E = 100' - (14' + 37') = 49 feet F = 100' - (18' + 37') = 45 feet G = 100' - (25' + 10') = 65 feet H = 100' - (30' + 10') = 60 feet I = 100' - (24' + 20') = 56 feet Sum of the offsets = 388 feet Area of the lake = 388' x 50' = 19,400 square feet Practice Exercise 5-3 & 5-4 1. Estimate the area of the golf fairway below. Interval = 75' A = 67' B = 71' C = 65' D = 54' E = 51' F = 59' G = 60' H = 70' I = 77' J = 69' K = 66' 2. Estimate the area of the lake below. A1 = 45' A2 = 33' B1 = 35' B2 = 30' C1 = 25' C2 = 29' D1 = 22' D2 = 43' E1 = 21' E2 = 49' F1 = 24' F2 = 45' G1 = 39' G2 = 21' H1 = 25' H2 = 18' I1 = 21' I2 = 20' Answers 1) 53,175 ft 2 2) 64,400 ft 2

Unit 5-5 Area by Average Radius Concept: the area of shapes that are similar to a circle but not perfectly round can be estimated by taking several measurements from the approximate center of the shape, computing the mean of the measurements, and using that mean as the radius. As with the offset method the more measurements that are taken the more accurate the estimate is. Measurements every 10 degrees around the center will yield 36 observations and a very reliable estimate. This can be accomplished by using a steel tape and a 3' x 3' board with straight lines radiating from the center every 10 degrees. Example 1 (Note: For purposes of illustration only a few measurements are used in this example. This would not be enough measurements to give a very reliable estimate) Estimate the area of the golf green below. A = 45' B = 43' C = 39' D = 44' E = 49' F = 52' G = 47' H = 41' Sum of the radii = 360 Mean radius = 45' Area = 3.14 x 45 2 = 6358.5 square feet Practice Exercise 5-5 Estimate the area of the shape below. A = 55' B = 57' C = 60' D = 58' E = 66' F = 56' G = 69' H = 67' Answer: area = 11,684 square feet

Unit 5-6 Finding the area of Combinations of Shapes Concept: Very often areas are a combination of two or more recognizable shapes. When this situation arises a total area can be derived by breaking up the larger area into small shapes which lend themselves to easy calculation and then totaling the area of the smaller shapes. Some examples of how larger areas can be broken down into recognizable shapes are illustrated below with dotted lines.

Unit 5-7 Determining Plant Numbers by Area and Spacing Concept: It is often useful to estimate the amount of plants needed or present based on the area of the planting bed and the centers on which the plants are spaced. This can be accomplished by assuming each plant occupies an area equivalent to the centers squared. For example if the plants are on 1' centers each plant occupies 1 square foot as illustrated by the diagram below. In this situation there is one plant per square foot. With centers other than 1', it is necessary to calculate the number of plants per square foot. The number of plants per square foot can be calculated using the proportion below: 1/C 2 = x/144 Where: 1 = one plant in the area C 2 C = the centers or spacing of the plants in inches (C 2 = the area occupied by one plant) x = the number of plants per square foot 144 = the number of square inches per square foot Once the number of plants per square foot is known it can be multiplied by the total area of bed space to determine the number of plants needed or present. Procedure: 1. Plug in the plant centers in inches as C. 2. Solve for the number of plants per square foot X. 3. Multiply the number of plants per square foot times the area.

Example 1 How many junipers planted on 10" centers would be required to plant a bed with an area of 800 square feet? Plug in plant centers in inches as C 1/10 2 = X/144 Solve for X ft. X = 1.44 plants per sq. Multiply the number of plants per square foot by the area 1.44 x 800 = 1152 junipers Example 2 How many 3 gallon pots placed on 14 inch centers would be needed to put three, 20' X 100' beds into production? Plug in plant centers in inches as C 1/14 2 = X/144 Solve for X Area involved ft Multiply the number of pots per square foot by the area X = 0.735 pots per sq ft 3 x 20' x 100' = 6000 sq 0.735 x 6000 = 4410 pots Practice Exercise 5-7 1. How many crotons on 16" centers would be required to plant a bed with an area of 1700 square feet? 2. How much border grass on 1 foot centers would be required to plant a bed with an area of 600 square feet? 3. How many 30 gallon containers on 3 foot centers would be needed to fill an area of 20,000 square feet? 4. How many 1 gallon containers placed on 10" centers would be needed to put two, 40' x 200' beds into production? 5. How many St. Augustine grass plugs on 18" centers would be needed to plant an area of 36,000 square feet? 6. How many St. Augustine grass plugs on 14" would be needed for the area in Question #5? Answers 1. 956 2. 600 plants 3. 2222 4. 23,040 5. 16,000 6. 26,450