G2044
Calibration of Sprayers (Also Seeders)
Methods and calculations for calibrating sprayers and seeders for accurate infield application plus a section on converting weights and measures.
Robert N. Klein, Western Nebraska Crops Specialist
Applying the correct rate of a product is an important part of obtaining good results with both seeders and pesti cide sprayers. With seeders too little seed reduces crop yields and increases weeds while too much seed increases costs and may reduce yields. With a pesticide application, too little product can mean poor control, while too much can mean crop injury, extra costs, and possible residue on the crop and/or carryover.
Many methods can be used to calibrate sprayers, including the ounce calibration and formulabased methods. With the ounce calibration method, 1/128 of an acre is sprayed and the spray is collected. When measured in ounces, the amount collected would be equal to the number of gallons applied per acre since there are 128 ounces in a gallon. (For further information on this method, see UNL Extension NebGuide FineTuning a Sprayer with the “Ounce” Calibration Method, G1756.) Other methods involve using formulas which need to be remembered or recorded for easy use. These methods also may require converting some of the information you have.
The methods discussed in this NebGuide are simple relationships and do not require remembering formulas. However, you do need a general understanding of cross multiplication. The important thing is to be consistent: If you put an item on top of an equation on one side, the same item also goes on the top on the other side.
Factors Affecting Rates
Three factors determine sprayer application rate:
1. Speed
2. Nozzle spacing
3. Nozzle output (determined by orifice size, pressure, and density of spray solution)
Where:
Speed = Length or distance covered divided by time
Nozzle spacing = Width
Nozzle output = The quantity applied/unit time
The following diagram shows how these three factors are related:
Speed = length or distance covered divided by time  
Nozzle spacing (width) 

For example, to determine speed:
1 mile per hour (mph) is:
1 mile (5,280 ft) in 1 hour (60 minutes)
Or 1 mph =  5,280 ft/hour 
= 88 ft/min 
60 min/hour 
Problem 1. Determine speed in mph.
If we travel 440 feet (ft) in 30 seconds (sec), what is our speed in mph?
The objective is to determine the distance traveled in 60 seconds (1 minute) and divide by 88 (88 feet/minute is equal to 1 mph).
for  30 sec 
60 sec 
(D is the distance we are solving 

= 

440 ft 
D 
We cross multiply to find the value of D
30 D = 60 x 440
30 D = 26,400
D 
= 
26,400 
30 
D = 880 ft/60 sec
Since every 88 ft traveled/60 sec (1 min) is equal to 1 mph, we divide 880 by 88 to get 10 mph.
Problem 2. Determine speed in mph.
If we travel 297 feet in 27 seconds, what is our speed?
27 sec 
60 sec 

= 

297 ft 
D 
27 D = 60 x 297
27 D = 17,820
D 
= 
17,820 
27 
D = 660 ft/60 sec
Divide by 88 since 1 mph = 88 ft/60 sec (1 min)
660 
= 
7.5 mph 
88 
Problem 3. Determine speed in mph.
If we travel 660 feet in 1 minute and 15 seconds, what is our speed?
First, convert 1 minute and 15 seconds to seconds: 60 + 15 = 75 seconds
75 (sec) 
= 
60 (sec) 
660 (ft) 
D 
75 D = D = 
39,600 528 ft/60 sec 
Divide by 88 since 1 mph = 88 ft/60 sec (1 min)
528 
= 
6 mph 
88 
Problem 4. Determine rate/acre.
If the sprayer is moving at 6 mph, the distance covered in one minute is 528 feet (6 mph x 88 ft/min = 528 feet).
To determine the area you cover with one nozzle in one minute if your sprayer has a 30inch nozzle spacing:
Distance traveled 6 x 88 = 528 ft/min 
30 in (2.5 ft) 
Area sprayed = 1,320 sq ft (2.5 ft x 528 ft/min) 
Collect the output of several nozzles and determine the average output per nozzle. All nozzles should be within 10 percent of the manufacturer’s rating for that nozzle. For example an XR11003 delivers 0.3 gpm at 40 psi. If it delivers more than 0.33 gpm or 42.24 (128 x .33) ounces/min at 40 psi, the nozzle should be replaced. Any nozzle delivering 5 percent above or below the average delivery rate for all the nozzles should be replaced.
For this example, the average nozzle output is 32 oz per minute
or 32 (oz/min) ÷ 128 (oz/gallon) = 0.25 gpm
What is the rate per acre? One way to calculate application rate without remembering a formula is to use a relationship: The amount applied and the area sprayed per minute are the same as the amount applied and the area sprayed per acre. R = gal/acre
Minute Box 
Acre Box 

Distance 6 x 88 = 528 ft 

Nozzle Spacing 30 in ÷ 12 = 2.5 ft 

= 


528 x 2.5 = 1320 sq ft 
43,560 sq ft 
From minute box  0.25 
= 
R 
From acre box 
1320 
43,560 
1320R = 10,890 (0.25 x 43,560)
R = 8.25 gal/acre
Problem 5. Determine the acres sprayed per minute.
Travel distance in one minute = 616 ft
Nozzle spacing = 30 in (20 nozzles on sprayer)
Nozzle output = 64 oz/minute
What is travel speed?  616 ÷ 88 = 7 mph (Remember 88 ft/min = 1 mph) 
What is sprayer width?  20 nozzles x 2.5 ft (30inch spacing) per nozzle = 50 ft 
What is application rate?  64 oz/min 
= 
0.5 gpm 
128 oz/gal 
Minute Box 
Acre Box 

Distance 616 ft 

30inch nozzle spacing (2.5 ft) 

= 


1,540 sq ft 
43,560 sq ft 
0.5 
= 
R 
1,540 
43,560 

1540 R  = 
21,780 
R  = 
14.14 gal/acre 
To determine the area covered by the sprayer in one minute:
1,540 sq ft/nozzle/minute
20 nozzles
1,540 x 20 ÷ 43,560 sq ft/acre = 0.71 acre/minute
Problem 6. Determine nozzle size needed to achieve the operational goal.
Sprayer speed = 7 mph
Nozzle spacing = 20 inches
Application rate desired = 17 gpa
Nozzle flow rate = F
Minute Box 
Acre Box 

7 x 88 = 616 ft 

Nozzle Spacing


= 


1,029 sq ft 
43,560 sq ft 
F 
=

17
43,560

1,029 
43,560 F  =  17,493 
F  =  0.40 gpm or XR8004* at 40 psi 
If we need 0.40 gpm, by design an XR8005* will give 0.5 gpm at 40 psi. Output varies by the square root of the pressure.
To solve for “P” take the result multiplied by itself.
5.056 x 5.056 = 25.6 psi
an XR8005 at 25.6 psi will give you 0.40 gpm
*Selected from TeeJet Nozzle Booklet by Spraying Systems.
Problem 7. Calibrate a hand sprayer.
First fill sprayer with water to a known level, a mark you can later refill to accurately. (Tip: It’s best to spray a test area over concrete so you can see the evenness of application.)
Spray test area 
100 sq ft = 10 ft x 10 ft  
or 
250 sq ft = 10 ft x 25 ft  
or 
500 sq ft = 10 ft x 50 ft or 20 ft x 25 ft 
Refill sprayer to same level as before, measuring amount of water it takes to refill sprayer.
If the pesticide recommendation is for 2 liquid ounces of product per 1,000 sq ft, the amount to include per 1,000 sq ft would be 1/4 cup or 4 tablespoons or 12 teaspoons. (See Weights and Measures Conversions on page 4.)
If during the test, 28 oz of water were applied over 250 sq ft, how much water and pesticide should be added to a 3gallon sprayer?
The amount of water you applied in  28 oz 
V for volume 
How much water you will apply per test area 1,000 sq ft  
= 

250 sq ft 
1,000 sq ft 
250 V 
= 
28,000 
V 
= 
112 ounces or ÷ 32 (oz/qt) = 
3.5 qt of water per 1,000 sq ft 
This indicates that 2 oz of pesticide should be added for every 3.5 qt of sprayer capacity.
With a 3gallon sprayer, 12 qt (3 x 4 qt/gal) of water should be added to the sprayer tank.
2 oz 
P for Pesticide 

= 

3.5 qt 
12 qt 
3.5 P 
= 
24 
P 
= 
6.86 oz or 0.86 cup (8 oz/cup) 
6.86/8 = 0.86 cup  This is the amount of pesticide to add to a 3gallon sprayer 
Problem 8. Determine the density of spray solution.
The rate at which a fluid flows through a spray orifice varies with its density. Since all the tabulations are based on spraying water, which weighs 8.34 lb per U.S. gallon, conversion factors must be used when spraying solutions which are heavier or lighter than water. To determine the proper size nozzle for the solution to be sprayed, first multiply the desired GPM or GPA of solution by the water rate conversion factor. The conversion factors are the square root of specific gravity. (See Weights and Measures Conversion chart on page 4 for some common fertilizers.)
For example, the specific gravity of 28% nitrogen, which weighs 10.65 lb/gal, is:
10.65 (Wt of 2800/gal) 
= 
1.28 specific gravity 
8.34 (Wt of water/gal) 
Conversion factor for 2800 fertilizer or 28% nitrogen is
Weight of Solution 
Specific Gravity 
Conversion Factors 
7.0 lb per gallon  0.84 
0.92 
8.0 lb per gallon  0.96 
0.98 
8.34 lb per gallon  water  1.00 
1.00 
9.0 lb per gallon  1.08 
1.04 
10.0 lb per gallon  1.20 
1.10 
10.65 lb per gallon  28% nitrogen  1.28 
1.13 
11.0 lb per gallon  1.32 
1.15 
11.06 lb per gallon  32% nitrogen  1.33 
1.15 
12.0 lb per gallon  1.44 
1.20 
14.0 lb per gallon  1.68 
1.30 
Example of using the conversion factor:
Desired application rate is 20 GPA of 28% N.
GPA (solution) x Conversion factor = GPA (water)
20 GPA (28%) x 1.13 = 22.6 GPA (water)A nozzle size should be selected to supply 22.6 GPA of water at the desired pressure, speed, and nozzle spacing.
Problem 9. Determine the density of a spray solution.
In this example, the following has been recommended for an ecofallow corn field:
75 lb of nitrogen from 28% UAN
Density of 28% N = 10.65 lb/gal
10.65 x .28 = 2.982 lb N/gal
75 lb N 
= 
25.15 gal of 28% solution 
2.982 lb N/gal 
Ingredient  Amount  Gallons 
28% Nitrogen  75 lb N  25.151 
Balance Pro  2.0 oz  0.016 
Fultime  2.25 qt  0.563 
Gramoxone Extra  2 pt  0.250 
Crop Oil Concentrate  1 qt  0.250 
2,4D 6 LVE  1/2 pt  0.063 
26.293 or 26.3 gal/acre 
To determine how this will spray out and what gallonage of water is needed to get 26.3 gal/acre of this spray solution, three steps are required:
1. To determine specific gravity weigh an equal amount of the spray solution and an equal amount of water.
S.S. 
Water 

13.08 lb 
10.3 lb 
To determine specific gravity weight of spray solution:
13.08 lb (wt of spray solution) 
= 
1.27 specific gravity 
10.3 (wt of water) 
2. Determine conversion factor
3. Determine the quantity of water to calibrate sprayer:
Spray Rate x Conversion Factor = Water Amount Equivalent
26.3 gal/acre x 1.13 = 29.6 gal/acre
Now you need to calibrate the equipment to apply 29.6 gal/acre of water.
Problem 10. Calibrate a seeder.
How may pounds of seed are needed to plant 18 seeds/ft in a row with 10inch spacing. Seed size is 15,000 seeds/lb and seed is collected for 500 ft.
To determine pounds of seed needed per acre:
12 in/ft 
= 
1.2 1.2 x 43,560 sq ft/acre = 52,272 ft of row/acre 
10 in/row 
52,272 x 18 seeds/ft row = 940,896 seeds/acre ÷ 15,000 seeds/lb = 62.7 lb/acre
Determine area seeded with one opener on one acre:
Test Box 
Acre Box 

10 inches per row or  
500 ft long 



= 


415 sq ft (500 x .83) 
43,560 sq ft 
Then cross multiply:
Wt 
62.7 

= 

415 
43,560 
43,560 Wt 
= 
26,020.5 (62.7 x 415) 
Wt 
= 
0.6 lb/opener or 9.6 oz/opener 
Weights and Measures Conversion
Weight
16 ounces = 1 lb = 453.6 grams
1 gallon water = 8.34 lb = 3.78 liters
1 short ton = 2,000 lb
1 long ton = 2,240 lb
1 cubic foot water = 62.4 lb
Liquid Measure
1 fluid ounce = 2 tablespoons = 29.57 milliliters
1 tablespoon = 3 teaspoons = 14.79 milliliters
1 cup = 16 T = 8 oz = 236.583 milliliters
16 fluid ounces = 1 pint = 2 cups
8 pints = 4 quarts = 1 gallon
Dry Measure
1 ounce = 28.3495 grams
Length
1 inch = 2.54 centimeters
3 feet = 1 yard = 91.44 centimeters
16.5 feet = 1 rod
5,280 feet = 1 mile = 1.61 kilometers
320 rods = 1 mile
Area
9 square feet = 1 square yard
43,560 square feet = 1 acre = 160 square rods
1 acre = 0.405 hectare
640 acres = 1 square mile
1 hectare = 2.47 acres
Speed
88 feet per minute = 1 mph
1 mph = 1.61 km/hour
1 mph = 0.477 meter/second
Volume
27 cubic feet = 1 cubic yard
1 cubic foot = 1,728 cubic inches = 7.48 gallons
1 gallon = 231 cubic inches
1 cubic foot = 0.028 cubic meters
Volume of sphere = D^{3} x 0.5236
Common Abbreviations and Terms Used
GPM = gallons per minute
GPA = gallons per acre
psi = pounds per square inch
mph = miles per hour
RPM = revolutions per minute
GPH = gallons per hour
FPM = feet per minute
T = tablespoon
t = teaspoon
Circles
Diameter x 3.1416 = circumference
Radius^{2} x 3.1416 = area
Spraying Systems Droplet Size in Microns
Very Fine = 153 and less
Fine = 154  241
Medium = 242  358
Coarse = 359  451
Very coarse = 452  740
Extensively coarse = 741 +
Fertilizer Facts
Pounds per gallon of liquid fertilizer at 60°F
10340  11.40 
11370  11.60 
7217  11.00 
2800  10.65 
2800  10.65 
3200  11.06 
8200  5.15 
120026  11.50 
Comparable Concentrations
1 ppm = 1 second in 12 days or 0.013 ounces in 100 gallons or about 8/10 of 1 teaspoon in 1,000 gallons
1 ppb = 1 second in 32 years or 0.013 ounces in 100,000 gallons or about 8/10 of 1 teaspoon in 1,000,000 gal
1 ppt = 1 second in 320 centuries
1 pint of water in ocean = 5,000 molecules in pint of water
1 psi = 2.31 feet
1 foot of lift of water = 0.433 psi
452 gpm = 1inch/1 acre/1 hour
Grain Information
Lb/bu 
Moisture % 

Corn  56 
15.5 
Soybeans  60 
13.0 
Grain sorghum  56 
14.0 
Wheat  60 
13.5 
Sunflower  25 
10.0 
Cu ft x 0.8 = bushel of grain
Cu ft x 0.4 = bushel of ear corn
1 horsepower  = 550 ft lb/second 
= 33,000 ft lb/minute  
= 746 watts 
This publication has been peer reviewed.
Disclaimer Reference to commercial products or trade names is made with the understanding that no discrimination is intended of those not mentioned and no endorsement by University of Nebraska–Lincoln Extension is implied for those mentioned. 
Visit the University of Nebraska–Lincoln Extension Publications Web site for more publications.
Index: Farm Power & Machinery
Machinery
Issued February 2011