Download GCO 2601 Guide To Calibration

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.
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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
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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
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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
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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