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TURFGRASS PROFESSIONAL'S GUIDE TO CALIBRATION by John Wildmon Table of Contents Module 1: Fractions Unit Unit Unit Unit Unit Unit 1-1 1-2 1-3 1-4 1-5 1-6 Adding and Subtracting Fractions Adding Mixed Numbers Subtracting Mixed Numbers Multiplying with Fractions Dividing with Fractions Converting Fractions to Decimals Module 2: Decimals and Percentages Unit Unit Unit Unit Unit Unit 2-1 2-2 2-3 2-4 2-5 2-6 Understanding Decimals Adding and Subtracting Decimals Multiplying Decimals Dividing Decimals Percentages Percentage Calculations Module 3: Simple Equations, Ratio and Proportion Unit 3-1 Unit 3-2 Solving Simple Equations Ratio, Proportion, Mean, Mode, and Median Module 4: Parts Per Million and Dilutions Unit Unit Unit Unit Unit Unit Unit 4-1 4-2 4-3 4-4 4-5 4-6 4-7 Parts Per Million Determining PPM When Known Quantities are Mixed Determining Solute Quantities fof Desired Concentration Mixing PPM Fertilizer Solutions Determining PPM From Known Quantities of Fertilizer Dilutions Calibrating Hose-on applicators Module 5: Area Unit Unit Unit Unit Unit Unit Unit 5-1 5-2 5-3 5-4 5-5 5-6 5-7 Area of Circles and Part-Circles Area of Squares, Rectangles, Parallelograms & Triangle Area by the Offset Method Area by Indirect Offsets Area by Average Radius Finding the Area of Combinations of Shapes Determining Plant Numbers by Area and Spacing Module 6: Volume and Units Unit Unit Unit Unit Unit 6-1 6-2 6-3 6-4 6-5 Determining Volumes of Cubes and Cylinders Volume of Tapering Pots Converting Units and Conversion Factors Fill Dirt and Topsoil Volumes Celsius and Fahrenheit Temperature Conversions Module 7: Fertilizers Unit Unit Unit Unit Unit 7-1 7-2 7-3 7-4 7-5 Fertilizer Analysis, Grade, Ratio, and Comparative Cost Fertilizer Application Rates Determining Rate From Applied Fertilizer Calculating Fertilizer Cost Fertilizer From Effluent Irrigation and Fertigation Module 8: Spreader Calibration Unit 8-1 Unit 8-2 Unit 8-3 Pesticide Unit 8-4 Unit 8-5 Pesticide Unit 8-6 Unit 8-7 Calibrating Drop Spreaders for Seed and Pesticide Calibrating Drop Spreaders for Fertilizer Calibrating Centrifugal Spreaders for Seed and Calibrating Centrifugal Spreaders for Fertilizer Calibrating Lg. Centrifugal Spreaders Seed and Calibrating Large Centrifugal Spreaders for Fertilizer Pure Live Seed Calculations Module 9: Sprayer Calibration and Tank Mixing Unit Unit Unit Unit 9-1 9-2 9-3 9-4 Sprayer Calibration Sprayer Calibration by the 1/128 Acre Method Tank Mixing Chemicals 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 A-2 A-3 A-4 A-5 A-6 A-7 A-8 A-9 A-10 A-11 A-12 A-13 A-14 Practice Problems for Modules 1, 2, 3, & 4 Answers for A-1 Practice Problems for Module 4 Practice Problems for Module 5 Practice Problems for Module 6 Practice Problems for Module 7 Practice Problems for Module 8 Practice Problems for Module 8 & 9 Practice Problems for Module 9 Practice Problems for Module 10 Answer for A-10 Boom Sprayer Diagram Number of Seeds per Pound for Common Turfgrasses 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 fractions with 12 as the denominator. 1/6 x 2/2 = 2/12 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 fractions with 12 as the denominator 3/4 x 3/3 = 9/12 5/6 x 2/2 = 10/12 1/3 x 4/4 = 4/12 Perform the mathematical operations = 9/12 + 10/12 - 4/12 Change 15/12 to a mixed number and simplify 19/12 - 4/12 = 15/12 15/12 = 1 3/12 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. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 5/8 1 3/8 1 1 1/3 1 27/40 29/32 7/16 1/2 3/16 7/32 5/16 9/64 7/16 1/2 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. 2. 3. 4. 5. 6. 7. 8. 9. 8 7/16 11 29/36 31 3/16 20 13/48 10 7/12 8 7/8 9 5/24 28 5/12 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 from the whole number if necessary 8 1/4 = 7 5/4 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 4. 22 1/3 - 13 7/8 7. 8 3/64 - 4 5/8 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 4 1/4 12 3/16 9/16 8 11/24 1 14 1/32 3 27/64 1 7/16 37 43/48 2. 14 9/16 - 2 3/8 5. 9 1/2 - 8 1/2 8. 2 9/16 - 1 1/8 3. 6 1/4 - 5 11/16 6. 14 11/32 - 5/16 9. 102 1/3 - 64 7/16 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 5 x 3 6 x 10 2 Simplify 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 Simplify 18/4 = 9/2 = 4 1/2 x 18/1 x 1/4 =18/4 Practice Exercise 1-4 1. 4. 7. 10. 1/2 x 4/9 16 3/4 x 5/8 32 3/8 x 4 5/16 6 x 3/36 x 5/8 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 2/9 12 58 1/3 10 15/32 10/27 13/36 139 79/128 5/96 11/15 5/16 53 5/8 1 1/2 2. 20 x 3/5 3. 5. 2.3 x 5/9 6. 8. 5/6 x 1/4 x 8/32 9. 11. 13 x 6 x 11/16 12. 4 11/16 13/32 x 1 3/8 x 5 2/5 x x 12 4/9 8/9 24/30 x 2/3 5/8 x 4/9 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 7. 14 by 2 5/6 2. 5/6 by 3/8 5. 4 3/8 by 3/16 8. 4 1/4 by 11/8 3. 3/5 by 2 6. 3 5/16 by 3 5/8 9. 14 5/32 by 6 1/6 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 1 3/4 2 2/9 3/10 16 23 1/3 53/58 4 16/17 3 1/11 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 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. 4. 7. 10. 13. 16. 1/2 1/16 1/9 3/4 7/11 6 3/8 2. 5. 8. 11. 14. 17. 1/4 1/3 1/12 5/8 4/19 25 5/6 Answers 1. 3. 5. 7. 9. 11. 13. 15. 17. 0.5 0.125 0.33 0.11 0.03125 0.625 0.63 0.57 25.833 2. 4. 6. 8. 10. 12. 14. 16. 18. 0.25 0.0625 0.166 0.833 0.75 0.66 0.21 6.375 9.22 7/8 3. 6. 9. 12. 15. 18. 1/8 1/6 1/32 2/3 4/7 9 2/9 Unit 2-1 Understanding Decimals ********************************************************************** Concept: Decimals are nothing more than multiples of 10. As you move place is a fraction that is a before it. The illustration concept. Decimal fractions with denominators that are in to the right of the decimal point each multiple of 10 smaller than the place below should help you understand the 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 9.05 Perform the operation 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 5.209 Perform the operation Practice Exercise 2-2 1. 4. 7. 10. 3.5 + 4.649 5.6 8.345 5.67 + 6.251 2.36 - 7.849 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 9.17 26.556 67.217 10.9 70.551 65.029 3.24 67.139 26.464 0.496 16.455 8.056 2. 5. 8. 11. 3.456 + 23.1 67.55 + 3.001 87.739 - 20.6 23.46 - 7.005 3. 6. 9. 12. 21.35 35.02 65.21 21.06 + + - 45.867 30.009 38.746 13.004 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 of the decimal in each number. 2.3 4.51 = 1 = 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 of the decimal in each number 6.045 4.29 = 3 = 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. 4. 7. 3.6 x 5.4 0.36 x 0.17 342.1 x 0.005 2. 5. 8. 1.9 x 0.45 46.489 x 3.43 76.9 x 6.5872 3. 6. 9. 23.4 x 3.056 4.69 x 5.003 3.578 x 97.0038 2. 5. 8. 0.855 159.45727 506.55568 3. 6. 9. 71.5104 23.46407 347.07959 Answers 1. 4. 7. 19.44 0.0612 1.7105 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 right to make a whole number and count the number of places moved. Move the decimal in the dividend the same number of places Perform the operation 1.26 becomes 126 places moved = 2 4.46166 becomes 446.166 3.541 126 446.166 Example 2 Divide 3.55 by 0.005 Move the decimal in the divisor to the right to make a whole number and count the number of places moved. Move the decimal in the dividend the same number of places adding zeros for place holders if needed. Perform the operation 0.005 become 5 places moved = 3 3.55 becomes 3550 710 5 3550 Practice Exercise 2-4 Divide the following: 1. 4. 7. 4.5 by 0.01 38.3838 by 11.1 8.784 by 2.4 2. 5. 8. 36.897 by 14.7 17.7741 by 26.1 0.37758 by 0.58 3. 6. 9. 4.8 by 3.75 34.41 by 5.55 0.0742 by 2.8 2. 5. 8. 2.51 0.681 0.651 3. 6. 9. 1.28 6.2 0.0265 Answers 1. 4. 7. 450 3.458 3.66 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. 2. To convert a decimal to percent multiply by 100 and write a percent sign after the product. 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. 4. 0.57 4.32 2. 5. 0.983 67.739 3. 6. 0.068 9.0024 Convert the following percentages to decimals 7. 10. 43% 207.64% 8. 11. 28.3% 1000% 9. 12. 22.06% 100.07% Convert the following fractions to percentages 13. 16. 3/4 9/4 14. 17. 3/8 2 5/8 15. 18. 1/2 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. 4. 7. 10. 13. 16. 19. 22. 57% 432% 0.43 2.0764 75% 225% 71/100 113/250 2. 5. 8. 11. 14. 17. 20. 23. 98.3% 6773.9% 0.283 10 37.5% 262.5% 1 1/100 3 3/40 3. 6. 9. 12. 15. 18. 21. 24. 6.8% 900.24% 0.2206 1.0007 50% 1468.75% 3 13/25 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 a percent 27/54 = 1/2 1/2 = 0.5 x 100 = 50% Practice Exercise 2-6 Calculate the following: 1. 47% of 100 2. 4. 201% of 450 5. 56% of 2 112% of 100 3. 6. 6% of $1.42 9% of 15 What percent is 7. 34 of 100 10. 45 of 20 34 of 200 29 of 29 9. 12. 15 of 440 465 of 1000 15. 18. 46 to 91 341 to 297 8. 11. Find the percent increase or decrease from 13. 25 to 25 14. 100 to 125 16. 32 to 96 17. 50 to 15 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. 4. 7. 10. 13. 16. 19. 22. 47 904.5 34% 225% 0% 200% increase 87.5% 25% 2. 5. 8. 11. 14. 17. 20. 23. 1.12 112 17% 100% 25% increase 70% decrease 30% $152.60 3. 6. 9. 12. 15. 18. 21. $0.09 1.35 3.4% 46.5% 97.8% increase 13% decrease 13% 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 Divide both sides by 3 14 x 3Y/14 = 12/7 x 14 3Y = 24 3Y/3 = 24/3 Y = 8 Example 3 Solve: 21/9 = 42/5Y Cross multiply (i.e. Multiply both sides by the denominators, 5Y and 3) 5Y x 21/9 = 5Y x 42/5Y 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 1. 5y = 10 4. Y/2 = 12 7. 6/5X = 12/23 equations: 2. 7X = 24.5 5. 5X/7 = 34/3 8. 70/494 = 62/12Y 3. 6. 9. 2.6Y = 6.1 + 1.7 26/4 = X/8 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. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Y = 2 X = 3.5 Y = 3 Y = 24 X = 15.866 X = 52 X = 2.3 Y = 35.286 Y = 0.15625 30 $1.20 $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. 2. 3. What is the median capital expenditure? What is the mean capital expenditure? 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. 5. 6. What is the mean color score? What is the mode? If a score of 6 or less is considered unacceptable what is the ratio of acceptable greens to unacceptable greens? Answer 1. 2. 3. 4. 5. 6. $12,000 $16,828.57 5/2 7.5 8 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 50 x 8.34 lb/gal = 417 lb Plug in the weight of the solute added as the X variable 2/Y = Z/1,000,000 Plug in the weight of the solvent as the Y variable 2/417 = Z/1,000,000 Solve for concentration Z 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 2 x 133.44 = 266.88 oz Plug in the weight of the solute added as the X variable 3/Y = Z/1,000,000 Plug in the weight of the solvent as the Y variable 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. 2. 3. 4. 5. 6. 7194 815 9992 8430 3222 1499 ppm ppm ppm ppm ppm 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 Solve for X X/1067.52 = 2500/1,000,000 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. BC = X Where: 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 0.20C = X Plug in the weight of the solute as 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 nitrogen 0.33C Solve for C C = 2.527 lbs of 33-0-0 = 0.834 lbs 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. 2. 3. 4. 5. 6. 7. 5 oz salt 50 lbs salt 0.74 oz urea 22.24 oz muriate of potash 0.556 oz 12-15-20 6.255 lb 20-20-20 3 lbs ammonium nitrate Unit 4-5 Determining Elemental PPM From Known Quantities of Fertilizer ********************************************************************** Procedure: 1. 2. Plug in the weight of the fertilizer added as C. 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 B x 15 oz = X Plug in % composition as a decimal, as B 0.45 x 15 = X Solve for X X = 6.75 ox of nitrogen Solve for the concentration 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 B x 4 lb + X Plug in % composition as a decimal, as B 0.33 x 4 lb = X Solve for X X = 1.32 lbs nitrogen Solve for the concentration 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. 2. 3. 4. 2360 ppm N 2998 ppm N 7914 ppm N 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: C1V1 = C2V2 Where: C1 V1 C2 V2 = = = = 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. 2. 3. 4. 5. Plug in the concentration of the concentrate as C1 . Plug in the desired concentration of the final solution as C2 . Plug in the desired volume of final solution as V2 . Solve for V1 . Put V1, the volume of concentrate, in a container and bring the total volume in the container to V2 , 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 C1 (Note: everything in the bottle is clorox so the conci. = 100%) 100 X V1 = C2V2 100 x V1 = 15 x V2 Plug in the desired volume of final solution oz (Note: 2 gallons = 256 oz) 100 x V1 = 15 x 256 Solve for V1 V1 Plug in the desired concentration of the final solution = 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. C1V1 = C2V2 Where: C1 = The concentration of the concentrated solution in the cup. V1 = The volume of the concentrate mixed C2 = The final solution concentration. V2 = The final solution volume. Calibration implies deciding on the concentration of the solution applied (i.e.C2) 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. C1). 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 V1. 3. Plug in the larger number (amount of water mixed) as V2. 4. Plug in the desired final solution concentration as C2. 5. Solve for C1. (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 V1 C1 x 1 = C2 V2 Plug in the larger number as V2 C1 x 1 = C2 Plug in the desired final solution conc. C1 x 1 = 300 x 11 Solve for C1 (the stock solution conc.) C1 = 3300 ppm nitrogen x 11 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 V1 C1 x 2 = C2 x V2 Plug in the larger number as V2 C1 x 2 = C2 x 13 Plug in the desired final solution conc. C1 x 2 = 2 oz/gal x 13 Solve for C1 (the stock solution conc.) C1 = 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. 2. 3. 4. 5. 6. 1.4 gal Round-up & 2.6 gal water 0.6 qt clorox & 5.4 qt. water 2600 ppm nitrogen conc of herbicide = 30 oz/gal 55 oz insecticide & 585 oz of water 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 r2 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 r2 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 42 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' = 1950 Answers: 1) 706.5 ft 2) 1063 ft 3) 1465 ft r = 4' = 1050 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 L The area of a triangle can be determined by the following formula: A = 1/2 B x H Where: A= the B= the H= the (B and area of the triangle length of the base of the triangle height of the triangle H are illustrated in the diagrams below) 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" (5) 6" Answers 1. 16 square inches 4. 27 square feet (6) 4’ 15’ 2. 90 square feet 5. 15 square inches 3. 30 square feet 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 Sum = 588 ft. Multiply the total of the offsets by the interval length 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 = The length of offset B = W = - (A1 + A2) 100' - (32' + 29') = 39 feet - (B1 + B2) 100' - (24' + 28') = 48 feet Example 1 cont. C D E F G H I = = = = = = = 100' 100' 100' 100' 100' 100' 100' - (19' (15' (14' (18' (25' (30' (24' + + + + + + + 20') 33') 37') 37') 10') 10') 20') = = = = = = = 61 52 49 45 65 60 56 feet feet feet feet feet feet 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 B C D E F G H I J K = = = = = = = = = = = 67' 71' 65' 54' 51' 59' 60' 70' 77' 69' 66' 2. A1 A2 B1 B2 C1 C2 D1 D2 E1 E2 F1 F2 G1 G2 H1 H2 I1 I2 Estimate the area of the lake below. = = = = = = = = = = = = = = = = = = 45' 33' 35' 30' 25' 29' 22' 43' 21' 49' 24' 45' 39' 21' 25' 18' 21' 20' Answers 1) 53,175 ft2 2) 64,400 ft2 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 B C D E F G H = = = = = = = = 45' 43' 39' 44' 49' 52' 47' 41' Sum of the radii = 360 Mean radius = 45' Area = 3.14 x 452 = 6358.5 square feet Practice Exercise 5-5 Estimate the area of the shape below. A B C D E F G H = = = = = = = = 55' 57' 60' 58' 66' 56' 69' 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/C2 = x/144 Where: 1 = one plant in the area C2 C = the centers or spacing of the plants in inches (C2 = 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 1/102 = X/144 Solve for X ft. X = 1.44 plants per sq. Multiply the number of plants per square foot by the area junipers 1.44 x 800 = 1152 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 1/142 = X/144 Solve for X X = 0.735 pots per sq ft Area involved ft 3 x 20' x 100' = 6000 sq Multiply the number of pots per square foot by the area 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. 2. 3. 4. 5. 6. 956 600 plants 2222 23,040 16,000 26,450 Unit 6-1 Determining Volumes of Cubes and Cylinders ********************************************************************** Concept: Units of volume are always cubed because they are 3 dimensional. The volume of cubes and similar shapes can be determined using the formula below. V = L x W X H Where: V = the L = the W = the H = the volume of the cube length width height The volume of a cylinder can be determined using the formula below. V =r2H Where: V = the volume of the cylinder = 3.14 r = the radius of the cylinder H = the height of the cylinder Example 1 Determine the volume of the cube. L W H V = = = = 4' 4' 4' 4' x 4' x 4' = 64 cubic feet Example 2 Determine the volume of the shape. L = 10' W = 20' H = 10' V = 10' x 20' x 10' = 2000 cubic feet Example 3 Determine the volume of the shape. L W H V = = = = 4" 6" 3" 4" x 6" x 3" = 72 cubic inches Example 4 Determine the volume of the cylinder. r H v = = = = 5' 10' 3.14 r2H = 3.14 x (5')2 x 10' = 785 cubic feet Practice Exercise 6-1 Determine the volume of the shapes below. Answer: 1) 3) 48 cubic inches 84 cubic feet 2) 4) 30 cubic yards 1017.36 cubic feet Unit 6-2 Volume of Tapering Pots ********************************************************************** Concept: It is useful to be able to calculate the volume of different pots for the purpose of determining potting media cost and estimating how much media to order. Pots usually taper from the top to bottom so they are not cones. Estimating the volume of a tapering posts can be accomplished by imagining there are two cylinders, one larger cylinder with the radius of the top of the pot, and one smaller cylinder with the radius of the bottom of the pot, as in the diagram below. The volume of the pot is the mean of the two cylinders. ********************************************************************** Procedure: 1. Compute the volume of the larger cylinder. 2. Compute the volume of the smaller cylinder. 3. Find the mean of the two cylinders. This is the pot volume. ********************************************************************** Example 1 Compute the volume of the pot. Compute volume of the larger cylinder V = r2H V = 3.14 x 4"2 x 7" V = 351.68 cubic inches Compute volume of the smaller cylinder V = 3.14 x 3"2 x 7" V = 197.82 cubic inches Determine the mean of the two cylinders Sum = 351.68 + 197.82 = 549.2 cubic inches Mean = 549.5/2 = 274.75 cubic in. Unit 6-3 Converting Units & Conversion Factors ********************************************************************** Concept: Numbers without units, as a practical matter, are meaningless. Therefore, it is essential to know the proper units on a number and be able to convert units when necessary. There are numerous situations in which units need to be changed. For example: Knowing how many cubic inches of potting medial you need is of very little use since potting media is bought in cubic yards. Computing a volume when the surface area is in square feet and the height is in inches yields meaningless units of square feet inches not cubic feet. Units have all the same properties as numbers and can be treated like numbers in an equation. Conversion from one unit to another is easily accomplished by multiplication using conversion factors which equal 1. The number 1 can be written as a fraction in a variety of ways. Any fraction in which the quantity in the numerator and denominator are equal is equal to one. For example 4/4, 12/12, Y/Y all equal 1. So it is also true that if you take equal quantities with different units and write them as a fraction, that fraction will equal 1. For example 4 quarts - 1 gallon so it follows that the fraction 4 qt/1 gal = 1. It is also true that multiplying a number by 1 does not change it's value. So you can multiply a number with units by a conversion factor which equals 1 and it will not change the value of the expression, only the units. the "trick" is to pick a conversion factor and arrange it so the old units cancel out leaving the desired units. If the factor is written upside down the existing units will be squared rather than canceling. ********************************************************************** Example 1 Change 2.5 gallons to quarts Conversion factor 4 qt = 1 gal Multiply by the factor as a fraction 2.5 gal x 4 qt/1 gal = 10 qt Example 2 How many ounces are in 5 pounds? 1 lb = 16 oz 5 lb x 16 oz/1 lb = 80 oz Example 3 Convert 9,880 square feet to acres 1 A = 43,560 sq ft 9,880 sq ft x 1 A/43,560 sq ft = 0.2268 A Example 4 Convert 6 miles per hour to feet per minute 1 mi = 5,280 ft 1 hr = 60 min 6 mi/hr x 5,280 ft/1 mi x 1 hr/60 min = 528 ft/min Example 5 How much does 4 ounces per 1000 sq ft equal in pounds per acre? 16 oz = 1 lb 1 A = 43560 sq ft 4 oz/1000 sq ft x 1 lb/16 oz x 43.56 (1000 ft2)/1 A = 10.89 lb/A Example 6 Convert 350,000 cubic inches to cubic yards 1 cubic yard = 27 cubic feet 1 cubic ft = 1728 cubic inches 350,000 cu in x 1 cu ft/1728 cu in x 1 cu yd/27 cu ft = 7.5 cu yd Practice Exercise 6-2 & 6-3 1. Compute the volume of the pot pictured. 2. Compute the volume of the pot pictured. 3. 4. 5. 6. 7. 8. Convert Convert Convert Convert Convert Convert Answers: 10 square feet to square inches. (144 sq in = 1 sq ft) 55 cubic feet to cubic yards. 4,560,000 cubic inches to cubic yards. 2 pounds per 1000 sq ft to pounds per acre. 5 ounces (fluid) per second to gallons per minutes. 880 ft per second to miles per hour. 1) 2027.7 cu in 2) 883.1 cu in 4) 2.04 cu yd 5) 97.7 cu yd 7) 2.34 gal/min 8) 600 mi/hr Practice Exercise 6-4 3) 1440 sq in 6) 87.12 lb/A Unit 6-4: Topsoil, fill, & media Volumes ********************************************************************** Concept: Volumes in general can be calculated as a surface area multiplied by a depth, or as a cross-sectional area multiplied by a length. These types of calculations are useful for a number of different practical applications involving sand, soil, and others types of growing media. The volume of fill or topsoil needed for a particular job can be estimated using the following formula: V = A x T Where: V = volume of topsoil or fill required A = surface or cross-sectional area T = thickness or length This formula will work for any situation where the thickness of the layer or the cross sectional area is reasonably uniform. Examples of this would be an application of topdressing or a section of drainage ditch. Non-uniformity can usually be overcome by using average depth or cross-sectional area. Remember when using the volume formula all measurements must be converted to the same units at some point in the calculation. For example if A is expressed in feet then T must also be expressed in feet. ********************************************************************** Procedure: 1. Convert all measurements to the same units. (Select common units that will give numbers of reasonable magnitude to work with.) 2. Compute the volume. 3. Change the answer to the desired units. ********************************************************************** Example 1 How many cubic yards of topsoil would be required to spread a layer 1" thick over an area of 9,500 square feet? Convert to common units 1" x 1 ft/12 in = 0.08333 ft Compute the volume 0.08333' x 9,500 ft2 = 791.6 Change answer to desired units 791.6 ft3 x ft3 1 yd3/27 ft3 = 29.3 yd3 Example 2 How many cubic yards of soil would be required to fill a ditch which is 50' long and has a cross-sectional area of 9 ft2 ? Convert to common units units are already the same Compute the volume 50 ft Change answer to desired units 450 ft3 x x 9 ft2 = 450 ft3 1 yd3/27 ft3 = 16.7 yd3 Practice Exercise 6-4 1. Approximately how many cubic yards of soil would be required to build a golf tee 2 feet high with a surface area of 4,600 square feet? 2. How much topsoil would be needed to top-dress 18 golf greens (total surface area 2.2 acres) with a layer 3/8" thick? 3. How much media would be needed to fill a planter box that measures 2' wide, 9' long, and 3' deep? 4. How many cubic yards of potting media would be needed to fill 5000 pots (volume of each pot = 650 cubic inches)? 5. How many cubic yards of media would be in each pot in Question #4 6. If the media in Question #4 cost $21.00 per cubic yard what would be the total cost of the media used? 7. What would be the cost of the media per pot? 8. I intend to start production on 3 new nursery beds that measure 20' x 100'. I will be using the pot below on 15" centers. The media I am using cost $18.00 per cubic yard. What will be the total cost of the potting media? Answers 1. 2. 3. 4. 5. 6. 7. 8. 341 cubic yards 111 cubic yards 2 cubic yards 70 cubic yards 0.014 cubic yards $1,470.00 $0.29 Pots per square foot = 0.64 Total area in production = 6,000 square feet Total number of pots = 3840 Volume of each pot = 514.96 cubic inches Total volume of media required = 42.4 cubic yards Total cost of the media = $763.20 Unit 6-5 Celsius and Fahrenheit Temperature Conversions ********************************************************************** Concept: Unlike most units of measure conversion of temperature measurements must be made with specific formulas. This is due to the nature of the two different scales. With most measurements zero on one scale is equal to zero on the other. For example 0 feet equals 0 meters. However, 0 degrees celsius is equal to 32 degrees fahrenheit thus a formula must be used that adjust the beginning point of the calculations. The formulas used for these conversions are shown below. To convert celsius to fahrenheit use: F = 9/5C + 32 To convert fahrenheit to celsius use: C = 5/9 (F-32) Where: F = degrees fahrenheit C = degrees celsius (i.e. degrees centigrade) Care must be taken when using these formulas to insure the mathematical operations are done in the right order. When converting celsius to fahrenheit always multiply C by 9/5 first then add 32. When converting fahrenheit to celsius always subtract 32 from F first then multiply by 5/9. ********************************************************************** Example 1 Convert 24 degrees celsius to fahrenheit. F F F F = = = = 9/5C (9/5 43.2 75.2 + 32 X 24) + 32 + 32 degrees fahrenheit Example 2 Convert 4.1 degrees fahrenheit to celsius. C C C C = = = = 5/9 (F-32) 5/9 (4.1 -32) 5/9 X 9 5 degrees celsius Practice Exercise 6-5 1. Convert the following fahrenheit temperatures to celsius: a) 105 b) 22 c) 83 2. Convert the following celsius temperatures to fahrenheit: a) 36 b) 9 c) 27 Answers: 1. a) 40.5 b) -5.5 2. a) 96.8 b) 48.2 c) 28.3 c) 80.6 Unit 7-1 Fertilizer Analysis, Grade, Ratio and Comparative Cost ********************************************************************** Concept: Fertilizers analysis refers to the percentage of N, K2O, and P2O5 in the fertilizer and appears on the fertilizer label. Grade is the total of the three numbers. ratio refer to the ratio of N to P2O5 to K2O in the fertilizer and is determined by dividing all three numbers by the lowest number. for example a fertilizer labeled 16-4-8 on the tag would have an analysis of 16-4-8, a grade of 28, and a ratio of 4-1-2. Cost comparisons based on grade (i.e. cost per unit of plant food) can be made between fertilizer materials provided they have relatively similar ratios and are composed of comparable materials. For example it might be fair to compare an 8-1-3 and a 16-4-8 based on grade if they were both composed of cheap water soluble materials. It would not be fair to compare these two fertilizers if one were composed of expensive slow release materials and the other composed of water soluble materials. Comparisons can also be based on cost per pound of nitrogen if the materials are similar. For example it would be fair to compare cost per pound of N between ammonium nitrate and urea. It would probable be fair to compare cost between IBDU and UF since these materials behave similarly. It would not e fair to compare cost per pound of N between IBDU and ammonium nitrate. ********************************************************************** Procedure: Comparing cost based on grade. 1. Determine the grade. (i.e. add up the three numbers in the analysis) 2. Convert the grade to a decimal. (i.e. divide by 100) 3. Divide the cost of the unit of fertilizer purchased by the weight of the unit purchased. this yields the cost per pound of fertilizer. 4. Divide the cost per pound of fertilizer by the grade in decimal form. This yields the cost per unit of plant food. Comparing cost per pound of N 1. Convert the % N to a decimal. 2. Divide the cost of the unit of fertilizer purchased by the weight of the unit. This yields the cost per pound of fertilizer. 3. Divide the cost per pound of fertilizer by the % N in decimal form. This yields the cost per pound of nitrogen. ********************************************************************** Example 1 Compare the cost of the two materials based on grade. Brand X is a 6-6-6 in 50 pound bags for $6.95 Brand Y is a 10-10-10 in 80 pound bags for $14.80 Determine the grade 6 + 6 + 6 = 18 10 + 10 + 10 = 30 Example 1 cont. Convert the grade to a decimal 18/100 = 0.18 30/100 = 0.30 Divide the cost of the unit by the weight $6.95/50 lb = $0.139/lb $14.80/80 lb = $0.185/lb Divide the cost per pound of fertilizer $0.139/0.18 = $0.77/lb $0.148/0.30 = $0.62/lb Cost per pound of plant food for Brand X = $0.77 Cost per pound of plant food for Brand Y = $0.62 Brand Y is a much better buy if they are comparable products. Example 2 Compare the cost per pound of N for the two materials. Urea (45-0-0) for $215.00 per ton. Ammonium nitrate (33-0-0) in 100 pound bags for $9.50 Convert % N to a decimal 45/100 = 0.45 33/100 = 0.33 Divide the cost per unit by the weight $215/2000 lb = $0.1075/lb $9.50/100 lb = $0.095/lb Divide the cost per pound of fertilizer by the % N in a decimal form $0.1075/0.45 = $0.24/lb N $0.095/0.33 = $0.29/lb N Cost per pound of N from the urea = $0.24 Cost per pound of N from the ammonium nitrate = $0.29 The urea is a better buy. Practice Exercise 7-1 1. Brand Brand Brand Brand 2. Compare the cost based on grade for the following materials: W X Y Z is is is is a 6-6-6 at $185.00 per ton an 8-8-8 at $7.95 per 50 lb bag a 10-10-10 at $17.50 per 100 lb bag a 12-8-12 at $8.99 per 40 lb bag Compare the cost per pound of nitrogen for the following: urea at $225 per ton ammonium sulfate (21-0-0) at $180 per ton ammonium nitrate at $190 per ton Answers 1. W = $0.51/lb, X = $0.66/lb, Y = $0.58/lb, Z = $0.70/lb 2. urea = $0.25/lb N ammonium sulfate = $0.43/lb N ammonium nitrate = $0.29/lb N Unit 7-2 Fertilizer Application Rates ********************************************************************** Concept: Fertilizer application rates are usually based on pounds of nitrogen per 1000 square feet or acre. However, the same calculations can also be used to determine the amount of fertilizer needed to achieve desired rates of P, K, and any other element. It is only necessary to know the percent composition of the fertilizer with regard to the element of interest and the area the fertilizer is to be applied to. The amount of an element needed to achieve a desired rate on an area can be calculated by the formula below: R = E/T Where: R = the rate of the element as a ration (weight of element/area) Examples: lb/A, lb/1000 sq ft, Kg/ha. E = the total quantity of the element needed for the desired rate. T = the total area the element is to be applied to. (Note: The units for weight and area must be the same on both sides of the ratio.) Once the total amount of the element necessary is calculated it can be determined how much a particular fertilizer is needed to supply that quantity. This is accomplished using the formula below: P x F = E Where: P = the percent composition of the fertilizer of the element of interest expressed as a decimal. F = the total quantity of fertilizer needed for the desired rate. E = the total quantity of the element needed for the desired rate. ********************************************************************** Procedure: 1. 2. 3. 4. 5. 6. 7. Change the areas to the same units if necessary. Plug in the ratio for the desired rate as R. Plug in the area the fertilizer is to be applied to as T. Solve for E the total quantity of the element needed. Plug E into the equation P x F = E Plug in the % composition of the fertilizer of the element. Solve for F. This is the total quantity of fertilizer to be distributed uniformly over the area to be fertilized. ********************************************************************** Example 1 How much 6-6-6 would be required to fertilizer 35,000 square feet at a rate of 40 pounds of nitrogen per acre? Change the area to the same units 35,000 = 0.8 A Plug in the rate as R 40 lb/1 A = E/T Plug in the total area as T 40 lb/1 A = E/0.8 A Example 1 cont. Solve for E E = 32 lb N Plug E into equation P x E P x F = 32 lb N Plug in % composition as P 0.06 x F = 32 lb N Solve for F F = 533 lb of 6-6-6 Example 2 How much 16-4-8 would be required to fertilize an area of 9,750 square feet at a rate of 1 pound of nitrogen per 1000 square feet? Convert units already the same Plug in the rate as R 1 lb N/1000 sq ft = E/T Plug in total area as T 1 lb N/1000 sq ft = E/9,750 sq ft Solve for E E = 9.75 lb N Plug E into P x F = E P x F = 9.75 lb N Plug in % composition N 0.16 x F = 9.75 lb N Solve for F F = 60.9 lb 16-4-8 Practice Exercise 7-2 1. How much 10-10-10 would be required to fertilize 15,500 square feet at a rate of 0.5 pounds of nitrogen per 1000 sq feet? 2. How much 8-1-3 would be needed to fertilize an area of 2.5 acres at a rate of 1 pound of nitrogen per 1000 square feet? 3. How much 45-0-0 would be required to fertilize 65,000 square feet at a rate of 60 pounds of nitrogen per acre? 4. how much 21-0-0 would be required to fertilize an area of 3 acres at a rate of 30 pounds of nitrogen per acre? 5. how many tons of 33-0-0 would be required to fertilize an area of 40 acres at a rate of 1 pound of nitrogen per 1000 square feet? 6. How many 40 pound bags of milorganite (6-3-2) would be required to fertilize 12,000 square feet at a rate of 0.75 pounds of nitrogen per 1000 square feet? Answers 1) 4) 77.5 lb 429 lb 2) 5) 1361 lb 5280 lb = 2.64 tons 3) 6) 200 lb 150 lb = 4 bags Unit 7-3 Determining Rate From Applied Fertilizer ********************************************************************** Procedure: Using the formulas in Unit 7-2: 1. Plug in the total weight of fertilizer applied as F. 2. Plug in % composition in decimal for as P. 3. Solve for E. this is the total weight of the element applied. 4. Plug E into the formula R = E/T 5. Plug in t the total area the fertilizer was applied to. 6. Plug in desired units as R and solve for pounds. ********************************************************************** Example 1 If 500 pounds of UF (38-0-0) is applied to an area of 95,000 what is the rate of nitrogen per 1000 square feet? Per acre? Plug in total weight of fertilizers as F P x 500 = E Plug in % composition as a decimal for P 0.38 x 500 = E Solve for E, the total weight of element E = 190 lb N Plug E into R = E/T R = 190 lb N/T Plug in T, the total area Convert to pounds per 1000 square feet R = 190 lb N/95,000 sq ft ? lb/1000 sq ft = 190 lb N/95,000 sq ft lb N = 2 R = 2 lb N/1000 sq ft Convert to pounds per acre 1 acre ? lb/43,560 sq ft = 190 lb lb N = R = 87 = 43,560 sq ft N/95,000 sq ft 87.12 lb N/A Practice Exercise 7-3 1. If 300 pounds of 16-4-8 are applied to 48,000 square feet what will be the rate of nitrogen in pounds per 1000 square feet? 2. I applied 150 pounds of 8-8-8 to my yard which measures 15,000 square feet. How many pounds of nitrogen per 1000 square feet did I apply? 3. My golf course has 45 acres of fairways. I just applied 1 ton of urea to the fairways. How many pounds per acre of nitrogen did I apply? 4. My foreman just told me he applied 150 pounds of ammonium sulfate to #8 green which has an area of 6,800 square feet. What is the rate of nitrogen in pounds per 1000 square feet? Should I pack my bags? Answers 1) 1 lb N/1000 sq ft 2) 0.8 lb N/1000 sq ft 3) 20 lb N/A 4) 4.6 lb N/1000 sq ft Probably Unit 7-4 Calculating Fertilizer Cost ********************************************************************** Concept: Cost for fertilizer and other chemicals can be looked at in a number of different ways. Cost per acre, 1000 square feet, application, and year all can be useful ways of expressing cost. It can also be useful to determine cost for specific areas such as greens or fairways. All of these are calculated in a similar manner. Multiplying the amount of fertilizer used in an area or time period by the cost of the material yields the cost. Consider the examples below. ********************************************************************** Example 1 How much would it cost per application to apply a rate of 1 lb N/1000 square feet to an area of 1.5 acres using 15-15-15 that cost $265.00 per ton. Amount of 15-15-15 required Cost 435.6 pounds $265/2000 lb x 435.6 lb = $57.72 What is the cost per acre $57.72/1.5 A = X/1 A X = $38.48 per acre Example 2 How much would it cost to fertilize a lawn with an area of 16,400 square feet at a rate of 3/4 of a pound of nitrogen per 1000 square feet using 8-1-3 that cost $7.60 for a 50 pound bag? Amount of 8-1-3 required Cost 153.75 lb $7.60/50 lb x 153.75 lb = $23.37 What would be the cost per 1000 sq ft? $23.37/16,400 = X/1000 X = $1.42 per 1000 sq ft Example 3 If it cost $31.50 to fertilize my lawn and I do it 6 times per year what will be my annual cost? Annual cost 6 app. x $31.50/app = $189.00/year Unit 7-5 Fertilizer From Effluent Irrigation and Fertigation ********************************************************************** Concept: Since ppm is a weight relationship it is possible to determine the weight of an element applied in solution from effluent irrigation or fertigation it the volume of the water applied, and the concentration of the element in ppm are known. Irrigation volumes are usually in inches. Using the volume of an acre inch of water (27,154 gal) and the concentration of the water in ppm, pounds per acre of the element can be derived. The formula below is a quick way to determine pounds per acre of an element from irrigation water. pounds per acre = 0.226464 x C x I Where: C = the concentration of the element in ppm I = the number of acre inches applied ********************************************************************** Example 1 If I irrigate with 6 acre inches of water this month and the irrigation water contains 150 ppm nitrogen, how many pounds per acre of nitrogen was applied in the water? pounds per acre = 0.226464 x 150 x 6 = 204 pounds of nitrogen Per A Practice Exercise 1. I used 2 inches of water last week from an irrigation source containing 110 ppm nitrogen. How many pounds per acre nitrogen did I apply? 2. My irrigation water contains 20 ppm phosphorus. If I irrigate with 60 inches per year, how many pounds per acre of phosphorus will I apply? 3. How many pounds per acre of nitrogen would be applied per month if the irrigation water contains 250 ppm nitrogen and 10 inches per month is used? 4. What is the answer In Question #3 equal to in pounds per 1000 square feet? Brain Teaser: Show how the number 0.226464 was derived. answers 1. 2. 3. 49.8 lb N/A 272 lb P/A 566 lb N/A Brain Teaser 1 A in x 27,154 gal water/A in x 1/1,000,000 x 8.34 lb/gal water Unit 8-1 Calibrating Drop Spreader for Seed and Pesticides ****************************************************************** Procedure: 1. Divide the desired rate by 2. this will allow the actual application to be made in two different directions giving more uniform coverage and eliminating missed spots. 2. Mark off a test run length. 3. Multiply the width of the band applied by the spreader times the length of the test run. This is the test area. 4. Set up the proportion below: (30' to 50' is enough) E/test area = Y/1000 sq. ft Where: E = the amount of the material to be applied to the test area Y = the rate in pounds of the material desired per 1000 square feet 5. Fill the spreader hopper, select a narrow gate setting, walk or drive the test run length at spreading speed with the gate open, and catch and weigh the material that passes through the gate. 6. Repeat step #5 using progressively wider gate settings until the weight of material caught is similar to the amount calculated in step #4 for two consecutive runs. In other words find a gate opening by trail and error that consistently gives approximately the amount of material needed in the test area. 7. Write down the gate setting, fertilizer used, and spreader used, and file it for future reference. ****************************************************************** Example 1 Given a 3 foot wide drop spreader and a 50 foot long test run, how much granular insecticide should pass through the spreader over the length of the test run to supply a rate of 4 pounds per 1000 square feet? Divide the rate by 2 4/2 = 2 lb/1000 sq.ft. Multiply the width of spread by the length of the test run 3' x 50' = 150 sq.ft. Set up the proportion E/150 = 2/1000 sq.ft. E = 0.3 lb.insecticide Determine the proper gate opening to give the test area. Don't forget to go two actual application. Also don't forget to it. This will save time the next time you 0.3 pounds of insecticide in directions when making the write down the data and file apply this same material. Unit 8-2 Calibrating Drop Spreaders for Fertilizer ***************************************************************** Procedure: 1. Divide the desired rate by 2. This will allow the actual application to be made in two different directions giving more uniform coverage and eliminating missed spots. 2. Mark off a test run length. 3. Multiply the width of the band applied by the spreader time the length of the test run. This is the test area. (30' to 50' is enough). 4. Set up the proportion below: E/test area = Y/1000 sq. ft. Where: E = the amount of the element to be applied to the test area Y = the rate in pounds of the element desired per 1000 square feet 5. Calculate the amount of fertilizer needed in the test area to supply the amount of element needed. Use the formula below: P x F = E Where: P = the % composition of the fertilizer of the element desired expressed as a decimal F = the weight of fertilizer needed in the test area E = the weight of the element needed in the test area 6. Fill the spreader hopper, select a narrow gate setting, walk or drive the test run length at spreading speed with the gate open, and catch and weigh the fertilizer that passes through the gate. 7. Repeat step #6 using progressively wider gate settings until the weight of fertilizer caught is similar to the amount calculated in step #5 for two consecutive runs. In other words find a gate opening by trail and error that consistently gives approximately the amount of fertilizer needed in the test area. 8. Write down the gate setting, fertilizer used, and spreader used, and file it for future reference. Example 1 Given a 3 foot wide drop spreader and a test run 40 feet long, how much 12-0-12 should pass through the spreader over the length of the test run for the spreader to be properly calibrated to apply 1.5 pounds of nitrogen per 1000 square feet. Divide the rate by 2 1.5/2 = 0.75 lb.N/1000 sq.ft. Multiply the width of spread by the test run length. 3' x 40' = 120 sq.ft Set up proportion: E/120 sq.ft. = 0.75 lb./1000 sq.ft. E = 0.09 lb. N needed in test area Calculate amount of fertilizer needed in test area: 0.12 X F = 0.09 lb.N F = 0.75 lb.12-0-12 needed Determine the proper gate opening to give 0.75 lb.12-0-12 in the test area. Don't forget to go two different directions when making the actual application. Also don't forget to write down the data and file it. this will save time the next time you fertilize with the same material. Practice Exercise 8-2 1. I have a drop spreader that is 3' 50'. How much 21-0-0 should pass run length for the spreader to be pound of nitrogen per 1000 square wide and a test run length of through the gate over the test properly calibrated to apply 1 feet? 2. Given a 5' wide drop spreader and a test run 40' long, how much 38-0-0 should pass through the gate over the test run length for the spreader to be properly calibrated to apply 2 pounds of nitrogen per 1000 square feet? 3. I have a How much test run nitrogen 4. Using the spreader and test run length in Question #3 how much 1010-10 should pass through the gate over the test run length for the spreader to be calibrated to apply 60 lbs. of nitrogen per acre? 4' wide drop spreader and a test run that measures 40'. 8-1-3 should pass through the gate over the length of the for the spreader to be properly calibrated for 1 pound of per 1000 square feet? Answers 1. 0.36 lbs.21-0-0 2. 0.53 lbs.38-0-0 3. 1 lb. 8-1-3 4. 1.1 lb Unit 8-3 Calibrating Centrifugal Spreaders for Seed and Pesticides ********************************************************************** Procedure: 1. Measure the total width of throw of the spreader with the material to be used. 2. Multiply the width of throw by 2/3, this gives the effective width. (Centrifugal spreaders apply more material in the center of the swath than they do on the outside. When doing the actual application it is necessary to overlap each side to get even distribution of the material. Passes should be spaced the effective width apart, not the total width.) 3. Mark off a test run length. 4. Multiply the effective width times the length of the test run. This is the test area. 5. Set up the proportion below: (30' to 50' is enough) E/test area = Y/1000 sq.ft. Where: E = the amount of the material to be applied to the test area Y = the rate in pounds of the material desired per 1000 square feet 6. Put a known weight of material in the hopper, select a narrow gate opening, walk or drive the test run length at spreading speed with the gate open, and weigh the remaining material in the hopper. The difference between the initial weight in the hopper and the weight of what remains in the hopper is the weight applied. 7. Repeat step #6 using progressively wider gate settings until the weight of material caught is similar to the amount calculated in step #5 for two consecutive runs. In other words find a gate opening by trial and error that consistently gives approximately the amount of material needed in the test area. 8. Write down the gate setting, material used, and spreader used, and file it for future reference. Example 1 Given a centrifugal spreader which throws 9 feet using ryegrass seed and a test run 35 feet long, how much seed should pass through the spreader over the test run length for the spreader to be properly calibrated to apply 15 pounds of seed per 1000 square feet? Find the effective width. 9' x 2/3 = 6 feet Find the test area. 6' x 35' = 210 square feet Set up the proportion: E/210 = 15/1000 E = 3.15 lb.seed Determine the proper gate opening to give 3.15 pounds of ryegrass seed in the test area. Don't forget to overlap when doing the actual application. This can be accomplished by spacing your passed the effective width apart. Also don't forget to write down the data and file it. This will save time the next time you use this material. Practice Exercise 8-3 1. Given a drop spreader which spreads a 5 foot swath and a 60 foot test run length how much Nemacur 10G should pass through the spreader over the length of the test run for the spreader to be properly calibrated to apply 4 pounds per 1000 square feet? 2. Using a drop spreader which spreads a total width of 3 feet and a test run 40 feet long how much bentgrass seed should pass through the spreader over the length of the test run for the spreader to be properly calibrated to apply 2 pounds per 1000 square feet? 3. Given a centrifugal spreader which throws a total width of 15 feet using Dursban granules, how much of this material should pass through the spreader over a 30 foot test run to give a rate of 5 pounds per 1000 square feet? 4. Using a centrifugal spreader which throws ryegrass seed a total width of 14 feet, how much seed should pass through the spreader over a 45 foot test run length to give a rate of 30 pounds per 1000 square feet? 5. Using the same spreader, seed, and test run in Question #4 how much seed should pass through the spreader to give a rate of 300 pounds per acre? Answers 1. 2. 3. 4. 5. 0.6 lb.Nemacur 10G 0.12 lb.bentgrass seed 1.5 lb.Dursban granules 12.6 lb.ryegrass seed 2.9 lb.ryegrass seed Unit 8-4 Calibrating Centrifugal Spreaders for Fertilizer ********************************************************************** Procedure: 1. Measure the total width of throw of the spreader with the material to be used. 2. Multiply the width of throw by 2/3, this gives the effective width. (Centrifugal spreaders apply more material in the center of the swath than they do on the outside. When doing the actual application it is necessary to overlap each side to get even distribution of the material. Passes should be spaced the effective width apart, not the total width.) 3. Mark off a test run length. 4. Multiply the effective width times the length of the test run. This is the test area. 5. Set up the proportion below: (30' to 50' is enough) E/test area = Y/1000 sq.ft. Where: E = the amount of the element to be applied to the test area Y = the rate in pounds of the element desired per 1000 square feet 6. Calculate the amount of fertilizer needed in the test area to supply the amount of element needed. Use the formula below: P X F = E Where: P = the % composition of the fertilizer of the element desired expressed as a decimal F = the weight of fertilizer needed in the test area E = the weight of the element needed in the test area 7. Put a known weight of fertilizer in the hopper, select a narrow gate opening, walk or drive the test run length at spreading speed with the gate open, and weight the remaining fertilizer in the hopper. The difference between the initial weight in the hopper and the weight of what remains in the hopper is the weight applied. 8. Repeat step #7 using progressively wider gate settings until the weight of fertilizer caught is similar to the amount calculated in step #6 for two consecutive runs. In other words find a gate opening by trial and error that consistently gives approximately the amount of fertilizer needed in the test area. 9. Write down the gate setting, fertilizer used, and spreader used, and file it for future reference. ********************************************************************** Example 1: Calculate the effective width Multiply the test run length by the effective width Set up proportion Calculate the amount of fertilizer 2/3 X 12’ = 8’ 8’ X 50’ = 400 sq. ft. test area E/400 sq. ft. = 1 lb.N/1000 sq. ft. E = 0.4 lb. N needed in test area 0.31 X F = 0.4 lb. N F = 1.3 lb. IBDU needed Determine the proper gate opening to give 0.75 lb.12-0-12 in the test area. Don't forget to go two different directions when making the actual application. Also don't forget to write down the data and file it. This will save time the next time you fertilize with the same material. Practice Exercise 8-4 1. Given a centrifugal spreader which throws a total width of 9 feet using 6-6-6 fertilizer and a test run 40 feet in length, how much of this fertilizer should pass through the spreader over the length of the test run to be calibrated to apply 0.5 lbs. of nitrogen per 1000 sq. ft. ? 2. Given a centrifugal spreader which throws a total length of 15 feet using 18-46-0 fertilizer and a test run of 30 feet, how much of this fertilizer should pass through the spreader over the length of the test run to be calibrated to apply 1.0 lbs. of nitrogen per 1000 sq. ft.? 3. Given a centrifugal spreader which throws a total width of 13 feet using 8-1-3 fertilizer and a test run of 40 feet, how much of this fertilizer should pass through the spreader over the length of the test run to be calibrated to apply 1.5 lbs. of nitrogen per 1000 sq. ft.? 4. Given a centrifugal spreader which throws a total width of 12 feet using 38-0-0 fertilizer and a test run of 40 feet, how much of this fertilizer should pass through the spreader over the length of the test run to be calibrated to apply 80 lbs. of nitrogen per acre? Answers: 1. 2. 3. 4. 2 lbs. 6-6-6 1.67 lbs. 18-46-0 6.5 lbs. 8-1-3 1.7 lbs. 38-0-0 Unit 8-5 Calibrating Large Centrifugal Spreaders for Seed and Pesticides ********************************************************************** Procedure: 1. Calculate the test area using a convenient predetermined weight of material using the proportion below. W/T = WM/1000 sq.ft. Where: W = the predetermined weight of material T = the test area WM = the weight of material needed per 1000 square feet 3. Calculate the effective width of the spreader. 4. Divide the calculated test area by the effective width of the spreader, this is the test run length. 5. By trial and error find a gate opening that supplies the predetermined weight of material over the calculated test run length at the selected spreading speed. 6. Write down the gate setting, tractor speed, spreader used, and fertilizer used and file it for future reference. ********************************************************************** Example 1 Given a large centrifugal spreader which throws mole cricket bait 33 feet, how long should the test run be if the bait is in 50 pound bags and the rate is 3 pounds of bait per 1000 square feet? Calculate the test area 50/T = 3/1000 sq ft T = 16,667 sq ft Calculate effective width of the spreader 2/3 x 33 = 22 feet Calculate the test run length 16,667/22 = 757 feet By trial and error find a gate opening which allows one 50 pound bag of bait to pass through the spreader over a distance of 757 feet at spreading speed. Don't forget to overlap when making the actual application. Passes should be spaced the effective width of the application. Passes should be spaced the effective width of the spreader. Also don't forget to record the data for future reference. Practice Exercise 8-5 Select a test run length for a large centrifugal spreader which throws pelletized dolomite a total width of 45 feet. The material is in 100 pound bags and the rate is 2000 pounds per acre. Answer: 72.6 feet Unit 8-6: Calibrating Large Centrifugal Spreaders for Fertilizer ********************************************************************** Concept: With large centrifugal spreaders that do not feed the material to the fan with a conveyor it is usually not possible to catch the material being applied for purposes of calibration. It is also not practical to empty the spreader contents in order to determine the amount of material spread by difference as with smaller centrifugal spreaders. Thus the need for another method of calibration. Calibrating larger centrifugal spreaders is accomplished using the same proportion described in Unit 8-2 and 8-4 but solving for a different variable. With large centrifugal spreaders rather than selecting a test run length and calculating how much material should be applied to the test area you select a convenient weight such as a 50 pound bag and calculate for a test run length. ********************************************************************** Procedure: 1. Calculate the weight of fertilizer needed to give the desired rate per 1000 square feet using the formula below. P x WF = WE Where: P = the % composition of the fertilizer of the element desired expressed as a decimal WF = the weight of the fertilizer needed per 1000 square feet WE = the weight of the element needed per 1000 square feet 2. Calculate the test area using a convenient predetermined weight of fertilizer using the proportion below. W/T = WF/1000 sq.ft. Where: W = the predetermined weight of fertilizer T = the test area WF = the weight of fertilizer needed per 1000 square feet 3. Calculate the effective width of the spreader. 4. Divide the calculated test area by the effective width of the spreader, this is the test run length. 5. By trial and error find a gate opening that supplies the predetermined weight of fertilizer over the calculated test run length at the selected spreading speed. 6. Write down the gate setting, tractor speed, spreader used, and fertilizer used and file it for future reference. Don't forget to overlap when doing the actual application. should be spaced the effective width of the spreader. Passes Example 1 I have a 3 point hitch mount centrifugal spreader which throws 10-10-10 a total width of 30 feet. The 10-10-10 is packaged in 80 pound bags. How far forward should the tractor travel while 1 bag of fertilizer passes through the spreader to achieve a rate of 1.5 pounds of nitrogen per 1000 square feet? Calculate weight of fertilizer needed for desired rate. .10 x WF = 1.5 WF = 15 lb 10-10-10 Calculate test area 80/T = 15/1000 sq ft T = 5333.3 sq ft Calculate effective width of the spreader 2/3 x 30' = 20 feet Divide the test area by the effective width of spreader 5333.3/20 = 266.6 feet (test run) By trial and error find a gate opening which allows one 80 pound bag of 10-10-10 to pass through the spreader over a distance of 266.6 feet at spreading speed. Don't forget to overlap when making the actual application. Also don't forget to record all the data for future reference. Practice Exercise 8-6 Given the following data for large centrifugal spreaders select a test run length for each situation below: 1. -spreader throws 16-4-8 a total width of 24 feet. -16-4-8 packaged in 50 pound bags -desired application rate 0.5 pounds of nitrogen per 1000 sq ft 2. -spreader throws 38-0-0 a total width of 27 feet -38-0-0 packaged in 40 pound bags -desired rate 2 pounds of nitrogen per 1000 sq ft 3. -spreader throws 6-6-6 a total width of 19 feet -6-6-6 packaged in 100 pound bags -desired rate 1 pound of nitrogen per 1000 sq ft Answers 1. 2. 3. 1000 ft 422.5 ft 473 ft Unit 8-7 Pure Live Seed Calculations ********************************************************************** Concept: When seed is purchased it will not be 100% pure but rather will contain some inert matter, weed seed, and crop seed. Additionally not all of the seed of the named variety in the bag will germinate. So one pound of any seed will not furnish one pound of pure seed that is viable. Thus the need to adjust seeding rates based on the information given on the seed tag. The two numbers on the label used to calculate the amount of pure live seed (PLS) per pound are % germination and % purity. Percent purity refers to the percentage of seed in the bag that is actually the named variety on the label. Percent germination refers to the percentage of the named variety seed which germinates normally under controlled, ideal conditions. When the two are multiplied together in decimal form it yields the fraction of the bag that is the named variety and has the potential to germinate and grow normally. This number can then be used to calculate the weight of seed that is needed to establish a desired quantity of seedlings in a given area. The exact number of seedlings per unit area varies with the species and the intent of the seeding. For purposes of initial establishment of a turf the ideal number of seedlings varies from about 8 to 18 per square inch with 10 per square inch being a good median number. For other purposes such as overseeding the numbers can vary drastically depending on the use of the area being overseeded and the species of grass being used. Pure live seed calculations are typically not done for overseeding but can be very helpful for comparative purposes and to evaluate recommended overseeding rates. ********************************************************************** Procedure: 1. 2. 3. Determine the number of seedlings per square inch and per sq. ft. desired. Determine the number of total seedlings needed based on the area to be seeded and seeding rate. (# per sq. ft.) Determine the number of pure live seed per pound using the formula below G x P x N = PLS Where: G = the percent germination expressed as a decimal P = the percent purity expressed as a decimal N = the number of seed per pound for the variety being used PLS = the number of pure live seed per pound 4. Divide the total number of seedlings needed by the number of pure live seed per pound, this yields the pounds of seed needed. Procedure for determining seedlings (PLS) per square inch when pounds of seed used and area are known: 1. 2 Determine PLS per pound. Multiply the number of PLS per pound by the total number of pounds used, this yields the total number of pure live seed used. 3 Divide the total number of pure live seed used by the total area seeded. 4 Convert the area units to square inches. ********************************************************************** Example 1 How many pounds of common bermudagrass (1,750,000 seeds per pound) 90% purity and 85% germination would be needed to plant 10 pure live seed per square inch on an area of 9,300 square feet? Determine total number of seedlings needed 10 plants/sq in x 144 sq in/sq ft = 1440 seedlings/sq ft 1440 x 9,300 = 13,392,000 seedling Determine PLS per pound 0.9 x 0.85 x 1,750,000 = 1,338,750 Divide total seedlings by PLS per pound 13,392,000/1,338,750 = 10 pounds Example 2 If perennial ryegrass (275,000 seed per pound) with 95% germination and 98% purity is planted at a rate of 30 pounds of seed per 1000 square feet, how many seedlings (PLS) will this yield per square inch? Determine total PLS used 0.95 x 0.98 x 275,000 x 30 = 7,680,750 PLS total Divide total PLS by area 7,680,750 PLS/1000 sq ft = 7,681 PLS/sq ft Convert area to square inches 7,681 PLS/sq ft x sq ft/144 sq in = 53 seedlings/square inch Practice Exercise 8-7 1. How many pounds of bahiagrass (160,000 seeds per pound) with 92% germination and 88% purity would be required to plant 25,000 square feet at a rate of 10 pure live seed per square inch? 2. How many pounds of centipedegrass (400,000 seeds per pounds) with 80% germination and 90% purity would be required to plant an area of 7,600 square feet with 10 pure live seed per square inch? 3. How many PLS per square inch would be furnished by creeping bentgrass (7,000,000 seeds per pound) 98% purity and 97% germination planted at a rate of 1 pound per 1000 square feet? Answers: 1) 278 lb 2) 38 lb 3) 46 PLS/sq in Unit 9-1 Sprayer Calibration ********************************************************************** Concept: The volume of liquid (carrier) applied in a given area by a sprayer is governed by four parameters: -the speed of the sprayer -the pressure in the boom -the size of the nozzle orifice -the viscosity of the liquid being sprayed If these four are held constant it is possible to calculate how much volume of carrier is applied per unit area. Once the relationship between volume applied and area covered is established the sprayer is calibrated. Sprayers are usually calibrated in units of gallons per acre. ********************************************************************** Procedure: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Measure the nozzle spacing and count the number of nozzles Multiply the nozzle spacing by the number of nozzles and convert to feet. This is the boom width. Measure a test run length. Drive the test run length at spraying speed and time it. Multiply the test run length over the time by the boom width. This yields an area over a time. (Example: sq ft/sec) Convert the units to acres per minute. Catch each nozzle for a period of time (15 to 30 seconds is usually sufficient) and add up the volumes. This yields a volume over a time. (Example: oz/sec) Check the nozzles for evenness. Individual nozzles should not vary over 10% above or below the mean. Convert the units on the total from step #7 to gallons per minute. Plug in the gallons per minute from step #9 and the acres per minute from step #6 into the calibration equation below. This yields gallons per acre. gal min Gal/A = A min 11. Check the nozzle alignment. The nozzle fan should be aligned slightly off-center of the boom to prevent the patterns colliding and disrupting each other. Note: These calculations are for broadcast applications. To calibrate for ban applications substitute spray band width of the nozzles for nozzle spacing.) ********************************************************************** Example 1 Given the data below calculate the gallons per acre this sprayer is applying. -9 nozzles spaced at 20", the sprayer traveled 100' in 17 seconds -the catch from each nozzle in ounces for a 20 second period follows Example 1 cont. Nozzle # #1 #2 23 24 #3 #4 #5 #6 #7 -----------ounces in 20 seconds ----------22 20 22 21 21 #8 #9 22 23 Calculate boom width and convert units to feet. 9 x 20" = 180" boom width 180 in x ft/12 in = 15 feet Multiply test run length over time by boom width. 100 ft/17 sec x 15 ft = 1500 sq ft/17 seconds Convert to acres per min 1500 sq ft/17 sec x A/43,560 sq ft = 0.0344 A/17 sec 0.0344 A/17 sec x 60 sec/min = 0.1214 A/min Compute total volume sum = 198 oz/20 sec Convert to gallons per minute 198 oz/20 sec x gal/128 oz = 1.547 gal/20 sec 1.547 gal/20 sec x 60 sec/min 4.641 gal/min Calculate gallons per acre gal/A = 4.641/0.1214 Gallons per acre = 38.2 Check nozzles for evenness Mean = 198 oz/9 = 22 oz 22 oz x 0.10 = 2.2 oz Acceptable variation = 22 oz + 2.2 oz All of the nozzles are within the acceptable range. Practice Exercise 9-1 1. Given the data below calculate the gallons per acre this sprayer is applying. Is the variation between nozzles in an acceptable range? -9 nozzles spaced at 18" -sprayer traveled 60' in 15 seconds -the catch from each nozzle in ounces for a 25 second period is as follows: Nozzle # #1 #2 #3 #4 #5 #6 #7 ----------- ounces in 25 seconds ----------18 18 19 18 17 19 16 2. #8 #9 16 18 I have a sprayer with 3 nozzles spaced 22 inches apart. The sprayer travels 75 feet in 19 seconds at spraying speed. I caught the following volumes from the nozzles in 15 seconds. #1 = 28 oz., #2 = 26 oz., #3 = 26 oz. How many gallons per acre is this sprayer applying? Answers: 1) 40 gal/A Yes 2) 83.6 gal/A Unit 9-2 Sprayer Calibration by the 1/128 Acre Method ********************************************************************** Procedure: 1. Check the nozzles for evenness. 2. Compute 1/128 of an acre. number is a constant) 3. Divide 340 square feet by the nozzle spacing in feet. 4. Drive the distance in feet from step #3 at spraying speed and time it. 5. Catch one nozzle for the length of time driven in step #4. The ounces caught equals the gallons per acre the sprayer is applying. (Note: The mean of all the nozzles will give a more accurate number than the catch from one nozzle.) (1/128 acre = 340 square feet. This The above method is for broadcast spraying but can be used for band applications by using the width of the band sprayed by a nozzle in place of the nozzle spacing. ********************************************************************** Example 1 Given a sprayer with a nozzle spacings of 18" what distance should the sprayer be driven for purposes of calibration with the 1/128 acre method? 18" = 1.5 feet 340/1.5 = 226.7 feet Drive forward 226.7 feet at spraying speed and time it. Catch the volume from a nozzle for that length of time. The ounces caught equals gallons per acre applied. Unite 9-3 Tank Mixing Chemicals ********************************************************************** Concept: Once a sprayer is calibrated it is possible to determine how much of a particular chemical should be put in the sprayer tank based on the volume of the tank and the label rates of the chemical. The most commonly used chemicals are formulated in a variety of ways and label recommendations may either be based on active ingredient (ai) in the material or on the complete product (cp). Formulations that commonly give recommendations based on active ingredient are emulsifiable concentrates (abbreviated on the label as EC or E), wettable powders (WP or W), and soluble powders (SP or S). The amount of active ingredient in these formulations can be readily determined by the number ahead of the letter on the label. For EC's the active ingredient is in pounds per gallon. For example a 2EC has 2 pounds per gallon active ingredient, a 6EC has 6 pounds ai per gallon, and so on. For WP's and SP's the active ingredient is in percent by weight. A 50WP is 50% active ingredient, a 75SP is 75% active ingredient. Other formulations such as flowables (F), dispensable granules (DG), and many fungicides give label rates based on complete product. In these instances the tank mix does not have to account for the amount of active ingredient in the product. ********************************************************************** Procedure: 1. Calculate the acres per tank by inverting the sprayer calibration (i.e. gal/A inverts to A/gal) and multiplying it times the gallons per tank (gal/tank). 2. Convert rates given in ounces per 1000 square feet to pounds or gallons per acre. 3. Convert rates given in active ingredient to pounds or gallons of complete product per acre. 4. Calculate the tank mix. Multiply pounds of complete product per acre times acres per tank. This yields the pounds of complete product per tank. ********************************************************************** Example 1 Given a sprayer that is applying 40 gallons per acre with a 100 gallon tank: Calculate acres per tank A/40 gal x 100 gal/tank = 2.5 A/tank How many gallons of wetting agent should be mixed in a full tank if the rate is 0.5 gallons per acre? Calculate the tank mix 0.5 gal cp/A x 2.5 A/tank = 1.25 gal cp/tank Example 1 cont. How many gallons of 6 EC should be put in a full tank if the label rate is 2 pounds of active ingredient per acre? Convert to gals of cp per acre gal cp/6 lb ai x 2 lb/A = 0.33 gal cp/A Calculate the tank mix 0.33 gal cp/A x 2.5 A/tank = 0.825 gal cp/tank How many pounds of 50WP would be required in a full tank to supply 4 pounds of active ingredient per acre? Convert to pounds per acre of cp lb cp/0.5 lb ai x 4 lb ai/A = 8 lb cp/A Calculate the tank mix 8 lb cp/A x 2.5 A/tank = 20 lb cp/tank How many pounds of Daconil would be required in a full tank to supply 4 ounces per 1000 square feet? Convert to pounds per acre ft/A 4 oz/1000 sq ft x 43,560 sq Calculate the tank mix 10.89 lb/A x 2.5 A/tank 27.225 lb Daconil/tank = 174.24 oz/A 174.24 oz/A x lb/16 oz = 10.89 lb/A Practice Exercise 9-3 Given a sprayer that is calibrated to apply 35 gallons per acre and has a 200 gallon tank answer the following questions: 1. How many gallons of 2EC should be put in a full tank to supply 1.5 pounds of active ingredient per acre? 2. How many pounds of 40WP should be put in a full tank to supply 3 pounds of active ingredient per acre? 3. how many pounds of fungicide should be put in a full tank to supply 6 ounces (weighed) of complete product per 1000 square feet? Given a sprayer that is calibrated to apply 28 gallons per acre and has a 100 gallon tank answer the following questions: 4. How many gallons of 4E should be put in a full tank to supply 1 pound of active ingredient per acre? 5. How many pounds of 50W should be added to a full tank to give 0.5 pounds of active ingredient per acre? Answers 1) 4) 4.275 gal EC 0.9 gal EC 5) 2) 42.82 lb cp 3.57 lb WP 3) 93.3 lb cp Unit 9-4 Selecting Proper Nozzle Volumes ********************************************************************** Procedure: 1. Determine gallons per acre of carrier (typically water) to be applied with the chemical by referring to the label recommendations. (Convert units to gallons per acre if necessary.) 2. Multiply the nozzle spacing by the number of nozzles and convert to feet. This is the boom width. 3. Select a spraying speed suited to the field conditions. test run length at spraying speed and time it. 4. Multiply the test run length over the time by the boom width. This yields an area over a time. 5. Convert the units to acres per minute. 6. Plug in acres per minute calibration equation below. and gallons per minute Drive a into the gal min Gal/A = A min 7. Solve for gallons per minute. 8. Divide gallons per minute by the total number of nozzles on the boom. this yields gallons per minute per nozzle. 9. Select a nozzle that will give the proper volume at the pressure you intend to use. (Note: If calculated nozzle volumes become excessive the volume needed can be reduced by reducing the sprayer speed. It is also possible to select a nozzle volume and calculate necessary sprayer speed using the same equation.) ********************************************************************** Example 1 Select the proper nozzle volume to apply a fungicide which calls for 10 gallons of water per 1000 square feet using the sprayer described below. -7 nozzles spaced at 24 inches -sprayer travels forwards 80 feet in 15 seconds Determine gallons per acre 10 gal/1000 sq ft x 43,560 sq ft/A = 435.6 gal/A Calculate boom width and convert the units to feet 24" x 7 = 168" 168" x ft/12" = 14 feet Example 1 cont. Multiply the test run length over time by the boom width 80 ft/15 sec x 14 ft = 1120 sq ft/15 sec Convert to acres per minute 1120 sq ft/15 sec x A/43,560 sq ft = 0.0257 A/15 sec 0.0257 A/15 sec x 60 sec/min = 0.1028 A/min Plug in gallons per acre and acres per minute into the calibration equation 435.6 + gal min 0.1028 A min Calculate gallons per minute gal/min = 435.6 x 0.1028 = 44.78 gal/min Divide gallons per minute by the number of nozzles 44.78 gal per min/7 nozzles = 6.4 gals per minute per nozzle Practice Exercise 9-4 1. 2. given the information below calculate gallons per minute necessary from each nozzle. -fungicide calls for 15 gallons of water per 1000 square feet -sprayer has 5 nozzles spaced at 15 inches -sprayer travels 50 feet in 14 seconds I have a pesticide which requires 20 gallons per 1000 square feet be applied. If my sprayer has 14 nozzles spaced at 18 inches and travels forward 100 feet in 17 seconds how many gallons per minute should the nozzles I select furnish? 3. I want to apply 400 gallons per acre of water with the pesticide I am using. Given the sprayer data below select the proper nozzle volume. -9 nozzles spaced at 16 inches -sprayer travels forward 150 feet in 20 seconds 4. If my sprayer is traveling 4 miles per hour, how many feet per minute is this? Brain Teaser Given the data below how many feet per minute will the sprayer need to travel to achieve the desired results? -6 nozzles spaced at 14 inches -pesticide needs 25 gallons of water per 1000 square feet -nozzles supply 4 gallons per minute each Answers 1. 3.9 gallons per minute per nozzle 2. 10.6 gallons per minute per nozzle 3. 5.5 gallons per minute per nozzle 4. 352 feet per minute Brain teaser 114.5 feet per minute Unit 10-1 Annual Cost of Ownership for Large Equipment ********************************************************************** Concept: Cost of ownership refers to cost which are on going regardless of whether the equipment is being utilized. Although the life of a particular piece of equipment is affected by how much use it receives and consequently can influence annual cost of ownership, it is still a fixed cost. Cost of ownership is influenced by three factors, depreciation, opportunity cost, and taxes and insurance. Depreciation refers to the decrease in the value of a piece of equipment due to wear and age. It can be calculated a number of different ways. The simplest way to calculate depreciation is call straight line depreciation and is the method demonstrated by the example in this section. Opportunity cost refers to the cost of equipment rather than something else, an interest bearing savings account for example. Taxes and insurance can vary considerably but taxes and insurance can be estimated as 4% to 6% of the purchase price. ********************************************************************** Procedure: 1. Calculate the annual depreciation cost using the formula below. annual depreciation = 2. number of years the equipment will last Calculate opportunity cost. opportunity cost = 3. purchase price - estimated salvage value purchase price x current interest rate 2 Determine cost of taxes and insurance. taxes and insurance = actual figures or 4% to 6% of purchase price 4. Add annual depreciation, opportunity cost, and taxes and insurance. This is the annual cost of ownership. ********************************************************************** Example 1 Calculate the annual cost of ownership for a $12,000.00 greens mower which will last 5 years and have a salvage value of $1,000.00. Current interest rate is 6.5%. Taxes and insurance will cost $700.00 per year. Calculate annual depreciation $12,000 - $1,000/5 = $2,200 Calculate opportunity cost $12,000 x 0.065/2 = $390 Determine taxes and insurance = $700 per year Add up all cost $2,200 + $390 + $700 = $3,290 Practice Exercise 10-1 1. Calculate the annual cost of ownership for a sod cutter given the information below. -Purchase price = $5,500 -Salvage value = $500 -Estimated life of the sod cutter = 10 years -Current interest rate = 5% -Taxes and insurance = $100 per year 2. Determine the annual cost of ownership for a tractor which cost $15,000, will last 8 years, and have a salvage value of $2,000. The current interest rate is 7%. Taxes and insurance will cost $800 per year. 3. What will be the annual cost of ownership of fairway mower which cost $50,000, will last 5 year, and will have a salvage value of $4,000. Current interest rate is 6.75%. Taxes and insurance will cost $1,600 per year. Answer 1. 2. 3. $737.50 $2950.00 $12,487.50 Unit 10-2 Annual Cost to Operate Large Equipment ********************************************************************** Concept: Costs of operation are only incurred when the equipment is utilized. Costs of operation fall into 3 categories listed below. 1) Fuel, fluids, filters, belts, tune ups, and other minor repairs. 2) Labor to run the equipment. 3) Major repair. The cost of category #1 can be estimated by multiplying the cost per gallon of fuel by the gallons of fuel used per year by a factor of 1.1 to account for fluids, filters, and etc. The number of gallons of fuel used per year can be estimated by multiplying the hours per year the equipment runs by the gallons of fuel per hour it consumes. the cost per year for category #2 can be estimated by multiplying the cost per hour for labor (including cost of FICA, fringe benefits, and etc.) times the total hours of operation per year. The cost for category #3 can be estimated either from existing records for similar pieces of equipment or by taking a percentage of the purchase price. The percentage used will be lower for equipment which has few moving parts and is typically low maintenance such as a utility vehicle. For low maintenance equipment 2% to 4% of the purchase price per year should give a good estimate. For higher maintenance equipment with many moving parts such as an aerifier or rotary mower higher percentages 5% to 7% of the purchase price should be used. ********************************************************************** Procedure: 1. Calculate annual cost of fuel, fluids, service, and minor repair using the formula below. Fuel cost = hours/year x gallons/hours x price per gallon x 1.1 2. Calculate annual labor cost by multiplying hourly cost for labor by total hours per year the equipment operated. 3. Estimate annual repair cost from existing records multiplying the purchase price by a percentage factor. or by 4. Add fuel cost, labor cost, and repair cost. This is the total annual cost of operation. ********************************************************************** Example 1 Given the data below what would be the annual cost to operate mower which burns 1.4 gallons of fuel per hour and is utilized 1500 hours per year? -Purchase price of mower = $7,000 -Price per gallon of fuel $1.02 -Labor cost = $6.70 per hour -Repair cost = 5% of purchase price Calculate annual cost of fuel, 1500 x 1.4 x $1.02 x 1.1 fluids, and service = $2,356.20 Example 1 cont. Calculate annual labor cost 1500 x $6.70 = $10,050 Estimate repair cost $7,000 X 0.05 = $350 Add fuel cost, labor cost, and repair cost $2,356.20 + $10,050 + $350 = $12,756.20 annual operating cost Practice Exercise 10-2 1. Calculate the cost to operate a tractor given the data below. -$16,000 initial purchase price -fuel consumption = 1.9 gallons per hour -fuel cost = $0.98 per gallon -labor cost = $8.50 per hour -repair rate = 3% of initial purchase price -hours per year operated = 1000 2. What will be the annual cost to operate a verticutter given the data below? -fuel consumption = 0.8 gallons per hour -labor cost = $4.75 per hour -repair cost = $150.00 per year -hours operated per year = 100 -fuel cost = $0.93 3. Calculate the cost to operate a reel mower which cost $12,000, is operated 1200 hours per year, and consumes 1.5 gallons per hour. Labor cost $6.80 per hour. Repair rate is 6% of the original purchase price. Fuel cost $1.15 per gallon. Answers 1. 2. 3. $11,028.20 $706.84 $11,157 Unit 10-3 Cost Per hour to Own and Operate Large Equipment ********************************************************************** Procedure: 1. Calculate the annual cost to own the piece of equipment. 2. Calculate the annual cost to operate the piece of equipment. 3. Add the cost to own and the cost to operate. 4. Divide the sum by the total number of hours operated per year. ********************************************************************** Example 1 Calculate the hourly cost to own and operate a front end loader given the information below. -Initial purchase price = $19,000 -Salvage value = $2,500 -Number of years the loader will last = 10 years -Current interest rate = 5.5% -Taxes and insurance = $700 per year -Hours operated per year = 450 -Fuel cost = $0.94 per gallon -Fuel Consumption = 2.1 gallons per hour -Labor cost = $9.85 per hour -Annual repair cost = 3% of purchase price Calculate depreciation cost $19,000 - $2,500/10 = $1,650 per year Calculate opportunity cost $19,000 x 0.055/2 = $522.50 Taxes and Insurance = $700 Calculate cost of ownership $1,650 + $522.50 + 700 = $2,872.50 Calculate fuel and service cost 450 x 2.1 x $0.94 x 1.1 = $977.13 Calculate labor cost $9.85 x 450 = $4,432.50 Calculate repair cost $19,000 x 0.03 = $570 Calculate cost to operate $977.13 + $4,432.50 + $570 = $5,979.63 Add cost to own and operate $2,872.50 + $5,979.63 = $8,852.13 Divide total cost to own and operate by hours of operation $8,852.13/450 $19.67 per hour to own and operate Practice Exercise 10-3 1. Calculate the hourly cost to own and operate the triplex mower below: -Purchase price $10,500 -Salvage value $1500 -Life expectancy 4 years -Insurance = $350 -Hours operated per year =1000 -Labor cost $7.20 per hour -Current interest rate = 6.75% -Fuel consumption = 1.6 gallons per hour -Repair rate = 3% 2. Calculate the hourly cost to won and operate the riding rotary mower described below: -Purchase price $6,800 -Salvage value $500 -Life expectancy = 3 years -Insurance cost $200 -Hours operated per year = 1600 -Labor cost $5.50 per hour -Current interest rate = 5 1/2% -Fuel cost $0.98 per gallon -Fuel consumption = 2.1 gallons per hour -Repair rate = 5% 3. Compute the hourly cost to won and operate the delivery pick-up truck described below: -Purchase price $8,350 -Salvage value $600 -Life expectancy 5 years -License and insurance = $800 -Hours operated per year = 500 -Labor = $6.50 per hour -Current interest rate = 6.5% -Fuel cost $1.02 per gallon -Fuel consumption = 2.5 gallons per hour -Repair rate = 6% Answers 1. Depreciation = $2,250 Opportunity Cost = $354.38 Insurance = $350 Annual cost to own = $2954.38 Fuel and service = $1848 Labor = $7200 Repair = $315 Annual cost to operate = $9363 Hourly cost = $12.32 2. Depreciation = $2100 Opportunity cost = $187 Insurance = $187 Annual cost to own = $2487 Fuel & service = $3622.08 Labor = $8800 Repair = $340 Annual cost to operate = $12,762.08 Hourly cost = $9.53 3. Depreciation = $1550 Opportunity cost = $271.38 License & Insurance = $800 Annual cost to own = $2621.38 Fuel & service = $1402.50 Labor = $3250 Repairs = $501 Annual cost to operate = $5153.50 Hourly cost = $15.55 A-1 Practice Problems for Modules 1, 2, 3, & 4 1. Addition 3/8 + 11/16 = 10/32 + 1/2 = 2 3/8 + 0.76 = 3/8 + 1.05 = 2. Subtraction 32 1/3 - 4 3/16 = 11/36 - 3/32 = 2 2/3 - 1.35 = 5 1/4 - 2.25 = 3. Multiplication 3/8 x 9/16 = 6 7/8 x 3 3/4 = 3.56 x 3 1/8 = 36.75 x 0.0039 = 4. Division .035/3.15 = 11/32 divided by 32/11 = 5. If I receive a 4% discount for paying cash and the total purchase without the discount is $146.58 how much will I owe? 6. My mechanic made $8.56 per hour last year, now he makes $8.82. What percent increase did he receive last year? 7. Bahia seed cost me $1.16 per pound this year which is a 14% increase from last year. What did it cost last year? 8. If I add 3 weighed oz of table salt to 2 gallons of water how many parts per million salt is the resulting solution? (1 gallon of water weights 8.34 lbs.) 9. I need 1 quart of 5% bleach solution. How many ounces of bleach and how many ounces of water do I combine? 10. If I put a 60% solution in the bottle of an 11/1 hose-on sprayer what per cent solution will I be spraying? How can I obtain a 3% solution from the same sprayer and stock solution? 25.1/0.1 = 3 5/8 divided by 0.125 = A-2 Answers for A-1 1. 1 1/16 13/16 3.135 1.425 2. 28 7/48 61/288 1.32 3 3. 27/128 25 25/32 11.125 0.1433 4. 0.01 121/1024 251 29 5. 140.72 6. 3% 7. $1.02 8. 11.241 9. 1.6 oz. of bleach with 30.4 oz. bleach 10. 5% Dilute the stock solution to 33% A-3 Practice Problems for Module 4 1. If I add 4 ounces of table salt to 3 gallons of water how many PPM salt will the resulting solution be? 2. How many ounces of salt would need to be mixed in 15 gallons of water to obtain a 7,500 PPM salt solution? 3. If I add 4 pounds of urea 45-0-0 to 100 gallons of water how many PPM nitrogen will the resulting solution be? 4. How many pounds of ammonium nitrate 33-0-0 would be required to mix 300 gallons of 500 PPM nitrogen solution? 5. Using an 11:1 hose on proportioner what would the concentration of the stock solution need to be in order to spray 150 PPM nitrogen? 6. How would you mix 5 gallons of 35% Round-up solution for a wick applicator? 7. How many ounces of Peters 20-20-20 would be required to mix 300 gallons of 150 PPM nitrogen solution? 8. Given a 13:2 hose-on sprayer how would you mix the stock solution to obtain a 1 ounce per gallon Diazinon spray? 9. If I apply 1.5 inches of water per week from an irrigation source containing 45 PPM nitrogen, how many pounds per acre nitrogen will this supply per week? 10. If I apply 1.5 inches of water per week from an irrigation source containing 45 PPM nitrogen, how many pounds per acre nitrogen will this supply per week? Answers (1) 9992 PPM, (2) 15 oz., (3) 2158 PPM N, (4) 3.8 lb., (5) 1650 PPM, (6) 1.75 gal Round-up 3.25 gal. water, (7) 30 oz., (8) 6.5 oz. Diazinon per gallon, (9) 15.3 lb/A N A-4 Practice Problems for Module 5 1. Compute the area for the following golf hole. Fairway Interval = 20 yds AB = 160 yds C = 30 yds D = 35 yds E = 38 yds F = 42 yds G = 43 yds H = 46 yds I = 47 yds J = 50 yds Green A = 38 B = 40 C = 51 D = 60 E = 60 F = 49 G = 47 H = 53 I = 62 J = 60 K = 56 L = 55 feet feet feet feet feet feet feet feet feet feet feet feet 2. Express the answer in Question #1 in acres. 3. If I top-dress the green in Question #1 with 3/8" of sand, how many cubic yards of sand will I need? 4. Find the area of the pond below. AB = 200', AC = 100', E1 = 25', E2 = 25', F1 = 20', F2 = 30', G1 = 10', G2 = 41', H1 = 12', H2 = 60', I1 = 20', I2 = 37', J1 = 35', J2 = 18', K1 = 56', K2 = 20', Interval = 25'. 5. If the lake above tapers uniformly from the 4 feet, with a 1/1 slope, how many gallon lake? How much pure copper sulfate would obtain 3 ppm copper? (1 cubic foot of water surface to a depth of of water are in this have to be added to = 7.5 gallons) Answers (1) Tee 1800 sq. ft., Fairway 59580 sq. ft., green 8688 sq. ft. (2) .04, 1.2, 0.2, (3) 10 cu. yds., (4) 7275 sq. ft., (5) 196,725 gal, 1.5 lbs. A-5 Practice Problems for Module 5 & 6 1. If I want to plant an area of 3,500 square feet with St. Augustine grass plugs on 10 inch centers how many plugs will be required? 2. How many containers would be required to put an area into production that measured 200 feet by 100 feet if the containers were placed on 16 inch centers? 3. What is the volume of the container below? 18’’ 17’’ 14’’ 4. How many cubic yards of potting media would be required to fill 5000 of the pots in question #3? 5. If I have 5 nursery beds measuring 20 feet by 100 feet and I want to put the containers pictured below on 14 inch centers, approximately how much potting media will be required to put the entire area into production? 9’’ 10’’ 8’’ 6. How much top-dressing material would be required to top-dress a 9,100 square foot lawn with a layer 1/2 inch thick? 7. How much top mix would be required per 1000 square feet to put the 12" layer on a USGA green? Answers: (1) 5040 plugs, (2) 11,250 pots, (3) 3469.7 in3, (4) 372 yd3, (5) 90 yd3, (6) 14 yd3, (7) 37 yd3 A-6 Practice Problems for Module 7 During the next year, I want to fertilize the fairway below 5 times with 16-4-8 at 1 lb.N per 1000 sq. ft., the tee every 4 weeks and the green every 2 weeks. On the tee and green, I will use IBDU (31-0-0) at a rate of 0.5 lb. of nitrogen in the winter (16 weeks) and 16-4-8 at a rate of 1 lb. of nitrogen for the remainder of the year (36 weeks). Cost per ton of the material is as follows: IEDU = $700 PER TON, 16-48 = $276 per ton. Fairway Interval = 20 yds. A = 22 yds. B = 24 yds. C = 26 yds. D = 29 yds. E = 32 yds. F = 29 yds. G = 24 yds. H = 25 yds. I = 27 yds. Green A = 17 B = 31 C = 25 D = 28 E = 28 F = 27 G = 24 H = 22 Fairway? yds. yds. yds. yds. yds. yds. yds. yds. 1. What is the area (in sq. ft.) of the tee? 2. How much 16-4-8 will be needed per application for the tee? Fairway? Green? 3. How much IBDU will be needed per application on the tee? 4. What will be the cost per application of IBDU on the tee? 5. What will be the total annual cost for fertilizer on this hole? 6. How much of each material will I need for this hole for the year? Answers: (1) 20,925 sq. ft., 42,840 sq. ft., 18,018 sq. ft. (2) 131 lb., 268 lb., 113 lb. (3) 34 lb., 29 lb., (4) $11.81, $10.17 (5) $756.91, (6) 16-4-8 = 4553 lb., IBDU = 368 lb. Green? Green? Green? A-7 Practice Problems for Module 8 1. A centrifugal spreader which throws a total width of 15 ft. spreads 6 lbs. of complete product after traveling a distance of 80 ft. How many pounds of complete product is this per 1000 sq. ft.? 2. If the product in Question #1 is 15-0-15 how many pounds of nitrogen per 1000 sq. ft. is being applied? 3. Given the area below and a drop spreader you are told to apply nitrogen at a rate of 1.5 lbs. per 1000 sq. ft. The drop spreader measures 3 ft. across and you measure a test area 33.3 ft. long. The fertilizer analysis is 32-0-0. How many pounds of product should you apply to the area shown below? A B C D E F G H = = = = = = = = 42 47 52 45 43 44 44 46 ft. ft. ft. ft. ft. ft. ft. ft. 4. A tractor mounted centrifugal spreader throws a total width of 60 ft. If I drive the tractor 4 mph how much material should I use per minute to obtain a 2 lb. per 1000 sq. ft. rate of nitrogen from a 16-4-8 fertilizer? 5. How much complete product from Question #4 would be required to fertilize the area below? Answers: (1) 7.5 lbs., (2) 1.125 lb. N, (3) 30.4 lbs., (4) 176 lb., (5) 1273 lbs. A-8 Practice Problems for Module 8 & 9 1. I want to apply 10-10-10 at a rate of 1 lb. of nitrogen per 1000 square feet using a drop spreader. The spreader is 3 feet wide and I have marked off a test run 33.3 feet long. How much complete product should I catch in this distance for the spreader to be properly calibrated? 2. I have a centrifugal spreader which throws a total width of 30 feet. If I want to apply Mocap at a rate of 4 pounds per 1000 square feet how much material should I calibrate for? How much material should be distributed over the length of an 80 foot test run? 3. Compute the gallons per acre for the sprayer information below. 8 nozzles spaced at 18 inches sprayer traveled 75 feet in 18 seconds volumes caught in 18 seconds as follows: 1 14oz 2 16oz 3 15oz Nozzle # 4 5 14oz 17oz 6 16oz 7 14oz 8 14oz 4. What is the speed of the sprayer in Question #3 in miles per hour? 5. You have a sprayer with 6 nozzles which you wish to calibrate. The nozzle spacing is 18 inches. The sprayer speed is 88 feet in 20 seconds. You catch the flow from each nozzle 3 times for 15 seconds and obtain the following: 1st Run 2nd Run 3rd Run Flow From Each Nozzle (oz/15 sec. Nozzle No. 1 2 3 4 5 7.2 6.8 6.7 6.8 7.0 7.2 6.9 6.9 7.0 7.1 7.0 6.8 6.8 7.2 7.0 6 6.8 6.8 6.9 a) How many gallons per acre are you applying? b) How many quarts of Lasso 4EC do you need to add to the tank to apply 2 lbs. of active ingredient per acre? Tank size = 120 gal. 6. How many pounds of AAtrex 80W ar required for 4 pounds of active ingredient? 7. How many pints of Suntan 5.7EC are required for 4 pounds of active ingredient? 8. I want to apply Daconil 2787 (a wettable powder fungicide) at a rate of 6 oz. per 1000 square feet. How many pounds per acre does this equal? Answers: (1) 0.5 lb., (2) 2 lb. per 1000 square feet, 3.2 lb., (3) 45.3 gal/A (4) 2.8 mi/hr (5a) 23.85 gal/A, (5b) 10 qts, (6) 5 lbs, (7) 5.6 pints, (8) 16.335 lbs. A-9 Practice Problems for Module 9 1. Given the data below, how many gallons per acre would this sprayer be applying? Nozzle spacing 20 inches spraying speed 4 mph catch in 17 seconds: nozzle #1 = 22 oz. nozzle #2 = 25 oz. nozzle #3 = 23 oz. nozzle #4 = 22 oz. nozzle #5 = 21 oz. nozzle #6 = 24 oz. 2. Compute gallons per acre for the sprayer data below: 11 nozzles spaced at 18 inches forward speed 50 feet in 12 seconds average catch per nozzle 14 oz. in 20 seconds 3. How many acres will a sprayer cover if it is applying 43 gallons per acre and is equipped with a 100 gallon tank? How many gallons of a 2EC should be added to this tank to obtain 1/2 pound ai per acre rate? How many pounds of a 50WP should be added to obtain 3 pounds ai per acre? How much of a WP fungicide should be added to obtain a 4 oz. per 1000 square feet rate? 4. Given a sprayer that is applying 29 gallons per acre with a 300 gallon tank, how many gallons of MSM should be added to the tank to obtain a rate of 1/2 gallon per acre MSMA? How much 2-4D should be added to obtain a rate of 2 pints per acre? Answers: (1) 46.6 gal/A, (2) 38 gal/A, (3) 2.3 A, 0.575 gal EC, 13.8 lbs. WP, 25 lbs. fungicide, (4) 5.15 gal. MSMA, 2.6 gal. 2-4D A-10 1. Practice Problems for Module 10 WHAT WILL BE THE COST TO OWN AND OPERATE THIS EQUIPMENT PER YEAR AND PER HOUR GIVEN THE FOLLOWING DATE: STUMP REMOVER New Price: Salvage: Est. Lifespan: License & Insurance: Operated 90 hours/yr. Labor (Total/Hr.) Savings Interest: Fuel: Consumption: Repair Rate: 1. $9,500.00 500.00 11 years $75.00/yr. $6.10/hr. 6% $1.20/Gal 3 gals./hr. 4% (a) Ownership: 9500-500/11 = 9500/2 x .06 = $75.00 = $818.18 Depreciation $285.00 Opportunity Cost $75 License & Ins/1178.18 per year (b) Operating: 90 x 3 x 1.20 x 1.1 = 6.10 x 90 = 356.40 Fuel, Lube 549.00 Repairs: 9500 x .04 = 380.00/1285.40 per year Own & Operate Total Cost Per Yr. 2. $2463.58/90 own & operate =$27.37 total cost per hr. PLS = .8 X .9 = .72 10 PLS/sq. in. x 144 sq. in./sq. ft. x 1000 sq. ft. = 1,440,000 1,440,000/.72 x 2,250,000 = 0.89 lbs. A-12 Boom Sprayer Diagram A-13 Number of Seeds per Pound for Common Turfgrasses Scientific Name Seeds Per Pounds Common Name Agropyron 324,000 crested wheatgrass Agrostia alba 4,990,000 redtop Agrostis canina Agrostis palustris Agrostis tenuis 11,800,000 7,890.000 8,723,000 velvet bentgrass creeping bentgrass colonial bentgrass Axonopus affinis 1,123,000 carpetgrass Bromus inermis 136,000 smooth bromegrass Buchloe dactyloides 50,000 buffalograss Cynodon dactylon (hulled) 1,787,000 common bermudagrass Dactylis glomerata 750,000 orchardgrass Eragrostis curvula 1,500,000 weeping lovegrass Eremochloa ophiuroides 400,000 centipede Festuca arundinacea 227,000 common ryegrass Festuca rubra 546,000 creeping Lolium multiflorum 227,000 common ryegrass Paspalum notatum 166,000 bahiagrass Phleum pratense 1,134,000 timothy Poa annua 2,250,000 annual bluegrass Poa pratensis 2,177,000 Kentucky bluegrass Poa trivialis 2,540,000 rough bluegrass Zoysia japonica 1,369,000 Japanese lawngrass or red fescue A-14 Conversion Equivalents Conversion of U.S. Measures 1 mile = 1760 yards = 5280 feet 1 yard = 3 feet = 36 inches 1 foot = 12 inches 1 gallon = 4 quarts = 8 pints = 128 fluid ounces 1 gallon of water = 8.34 pounds = 133.44 weighed ounces 1 acre = 43,560 square feet 1 square yard = 9 square feet = 1296 square inches 1 square foot = 144 square inches 1 1 1 1 cubic yard = 27 cubic feet = 46,656 cubic inches cubic foot = 1728 cubic inches cubic foot of water = 62.34 pounds = 7.47 gallons acre inch of water = 27, 154 gallons 1 ton = 2000 pounds 1 pound = 16 ounces av. (i.e. weighed ounces) Metric Conversions 1 kilometer = 1093.6 yards = 3280.8 feet = 1000 meters 1 meter = 1.0936 yards = 3.28 feet = 39.37 inches = 1000 millimeters 1 centimeter = 10 millimeters = 0.01 meters = 0.3937 inches 1 liter = 1.0567 quarts = 33.815 fluid ounces = 1000 milliliters 1 milliliter = 1 cubic centimeter = 1 gram of water 1 liter = 1000 cubic centimeters = 1 kilogram of water 1 hectare = 2.471 acres = 10,000 square meters 1 square kilometer = 100 hectares = 1,000,000 square meters 1 metric ton = 2,204.6 pounds = 1000 kilograms 1 kilogram = 2.2 pounds = 35.27 weighed ounces = 1,000 grams Miscellaneous 1 1 1 1 1 1 part per million = 1 milligram per kilogram pound per acre = approximately 1 kilogram per hectare acre furrow slice = approximately 2,000,000 pounds part per million (soil test value) = approx. 2 pounds per acre per cent = 10,000 parts per million pound per 1000 square feet = 43.56 pounds per acre Bibliography Anderson, Wood P., Weed Science Principles. York. West Publishing Co. New Beard, James B., Turfgrass: Science and Culture. Englewood Cliffs, N.J. 1973. Bohmont, Bert L., Teh New Pesticide Users Guide. Prentice-Hall Inc., Reston, 1983. Donahue, Roy L., Raymond W. Miller, John C. Shickluna, Soils an Introduction to Soils and Plant Growth. Prentice-Hall Inc. Englewood Cliffs, N.J. 1983. Florida Pest Control Association/IFAS. Florida Pest Control Industry Education Seminar, Volume X: Turfgrass Pest Control. Florida Pest Control Association. 1980. Golf Course Superintendents Association of America. The Mathematics of Turfgrass Maintenance. Golf Course Superintendents Association of America. Lawrence, KS. Higgs, R., Agricultural Mathematics: Problems in Production, Management, Marketing, Mechanization, and Environmental Quality. The Interstate Printers & Publishers. Danville, IL. 1973. McGee, R.V., Mathematics in Agriculture. Cliffs, N.J. 1954. Prentice-Hall Inc. Englewood Ross-Payne & Associates, Financial Management & Costing Systems. Payne & Associates. Arlington Heights, IL 1978. Ross-Payne & Associates, Equipment Costing. Arlington Heights, IL 1978. Ross- Ross-Payne & Associates. Spraying Systems Co., TeeJet Agricultural Spray Products (Catalog 39). Spraying Systems Co. Wheaton, IL 1987. University of California, Division of Agricultural Sciences. How much chemical do I put in the tank? (Leaflet 2718). USDA Cooperative Extension. Berkley, CA 1976