5.4 Equipment Calibration & Mathematical Calculations

Principles of Equipment Calibration

Calibration is the process of measuring and adjusting the liquid output of pesticide application equipment to ensure that the pesticide is applied at the exact rate specified on the product label. Under Washington State Pesticide Application Act (RCW 17.21), applying pesticides in excess of maximum label rates is illegal, causes crop injury (phytotoxicity), increases environmental residue risks, and wastes money. Conversely, under-application fails to control target pests and accelerates pesticide resistance development.

The Three Key Calibration Variables

Application rate in Gallons Per Acre (GPA) is governed by three primary variables:

1. Travel Speed (MPH): Travel speed has an inverse relationship to application volume. Doubling travel speed reduces application volume (GPA) by half. Halving travel speed doubles GPA.
2. Nozzle Flow Rate (GPM): Output per nozzle has a direct relationship to GPA. Doubling GPM doubles GPA. Note that nozzle flow rate increases with pressure, but quadrupling system pressure is required to double nozzle flow rate ($GPM_2 = GPM_1 \sqrt{P_2 / P_1}$).
3. Sprayed Width per Nozzle (W): Nozzle spacing (or band width) in inches has an inverse relationship to GPA. Wider nozzle spacing dilutes output per unit area.

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The Standard Calibration Formula

The fundamental equation governing liquid broadcast boom spraying is:

$\text{GPA} = \frac{\text{GPM} \times 5,940}{\text{MPH} \times W}$

Where:

• $\text{GPA}$ = Application rate in Gallons Per Acre
• $\text{GPM}$ = Liquid flow rate per nozzle in Gallons Per Minute
• $\text{MPH}$ = Ground speed of equipment in Miles Per Hour
• $W$ = Nozzle spacing on boom in inches (or sprayed band width in inches)
• $5,940$ = Mathematical conversion constant ($60 \text{ min/hr} \times 43,560 \text{ sq ft/acre} / 5,280 \text{ ft/mi} \times 12 \text{ in/ft}$)

Rearranging the formula allows solving for required nozzle flow rate ($\text{GPM}$):

$\text{GPM} = \frac{\text{GPA} \times \text{MPH} \times W}{5,940}$

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Step-by-Step Boom Sprayer Calibration Procedure

1. Inspect & Clean Equipment: Fill spray tank half-full with clean water. Operates pump at target application pressure. Check all nozzles for uniform pattern and measure individual output into a catch container for 1 minute. Discard and replace any nozzle that varies by more than 5% from the average output of all nozzles.
2. Measure Nozzle Spacing ($W$): Measure distance between nozzle centers on the boom in inches (e.g., $W = 20 \text{ inches}$).
3. Determine True Ground Speed ($\text{MPH}$):
   • Mark a test course of known distance (e.g., $200 \text{ feet}$) in the field site.
   • Drive the sprayer across the course at normal operating gear and engine RPM. Record time in seconds.
   • Calculate ground speed using the formula: $\text{MPH} = \frac{\text{Distance (ft)} \times 60}{\text{Time (sec)} \times 88}$
4. Measure Output & Calculate GPA: Collect water from each nozzle for 1 minute, convert average fluid ounces to GPM ($1 \text{ GPM} = 128 \text{ fl oz/min}$), and calculate GPA using the calibration equation.

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Step-by-Step Worked Mathematical Calculations

+---------------------------------------------------------------------------------------+
|                         WORKED CALCULATION SUMMARY TABLE                              |
|                                                                                       |
|  Calculation Type       Given Parameters                  Formula & Result            |
|  ----------------       ----------------                  ----------------            |
|  1. GPA Output          GPM=0.34375, MPH=5.0, W=20 in     GPA = (0.34375x5940)/(5x20) |
|                                                               = 20.42 GPA             |
|  2. Required GPM        GPA=15, MPH=6.0, W=20 in          GPM = (15x6.0x20)/5940      |
|                                                               = 0.303 GPM (38.8 oz/min)|
|  3. Tank Mixing         Tank=300 gal, GPA=20, 2.5 pt/ac   Acres = 300/20 = 15 acres   |
|                                                               Product = 15x2.5 = 37.5 pt  |
|  4. Active Ingredient   Target=2.0 lb a.i., Formulation   Product = 2.0 / 0.75        |
|     (75% WDG Dry)       75% WDG                           = 2.67 lbs WDG / acre       |
+---------------------------------------------------------------------------------------+

Calculation 1: Calculating Application Rate (GPA) from Field Discharge

• Scenario: A boom sprayer has nozzles spaced $W = 20 \text{ inches}$ apart. Operating speed is measured at $\text{MPH} = 5.0 \text{ mph}$. Average nozzle output measured into a graduated cylinder is $44 \text{ fluid ounces}$ per minute.
• Step A: Convert fluid ounces per minute to GPM: $\text{GPM} = \frac{44 \text{ fl oz/min}}{128 \text{ fl oz/gal}} = 0.34375 \text{ GPM}$
• Step B: Apply calibration formula to solve for GPA: $\text{GPA} = \frac{0.34375 \times 5,940}{5.0 \times 20} = \frac{2,041.875}{100} = 20.42 \text{ GPA}$

Calculation 2: Determining Required GPM per Nozzle for Catalog Selection

• Scenario: An applicator wants to apply a herbicide at a target volume of $\text{GPA} = 15 \text{ GPA}$. Tractor speed is set at $\text{MPH} = 6.0 \text{ mph}$, and nozzle spacing is $W = 20 \text{ inches}$. What GPM nozzle orifice size should be selected?
• Step A: Solve for GPM: $\text{GPM} = \frac{\text{GPA} \times \text{MPH} \times W}{5,940} = \frac{15 \times 6.0 \times 20}{5,940} = \frac{1,800}{5,940} = 0.303 \text{ GPM}$
• Step B: Convert GPM to fluid ounces per minute for bucket verification: $\text{fl oz/min} = 0.303 \text{ GPM} \times 128 \text{ fl oz/gal} = 38.8 \text{ fl oz/min}$

Calculation 3: Tank Mixing Calculations

• Scenario: A commercial applicator has a spray rig with a 300-gallon spray tank. Calibrated application volume is $20 \text{ GPA}$. The pesticide label recommends an application rate of $2.5 \text{ pints}$ of formulated liquid product per acre. How many gallons of product must be added to a full spray tank?
• Step A: Calculate acres treated per full spray tank: $\text{Acres per Tank} = \frac{\text{Tank Capacity (gal)}}{\text{GPA}} = \frac{300 \text{ gallons}}{20 \text{ GPA}} = 15 \text{ acres}$
• Step B: Calculate total pints of product required per tank: $\text{Total Product (pints)} = 15 \text{ acres} \times 2.5 \text{ pints/acre} = 37.5 \text{ pints}$
• Step C: Convert pints to gallons ($8 \text{ pints} = 1 \text{ gallon}$): $\text{Gallons Product} = \frac{37.5 \text{ pints}}{8 \text{ pints/gal}} = 4.6875 \text{ gallons} \text{ (4 gal + 5.5 pints)}$

Calculation 4: Active Ingredient (a.i.) Calculations (Dry & Liquid)

• Dry Formulation Example: A pesticide label specifies an application rate of $2.0 \text{ lbs of active ingredient (a.i.)}$ per acre. The product is formulated as a 75% Water Dispersible Granule (75 WDG, meaning 0.75 lb a.i. per 1 lb product). How many pounds of commercial product are needed per acre? $\text{Pounds Product/Acre} = \frac{\text{Target a.i. (lbs/acre)}}{\text{Formulation Concentration (fraction)}} = \frac{2.0 \text{ lbs a.i.}}{0.75 \text{ a.i. fraction}} = 2.67 \text{ lbs product/acre}$
• Liquid Formulation Example: A label requires applying $1.5 \text{ lbs a.i.}$ per acre. The formulation is a 4 EC (contains $4.0 \text{ lbs a.i.}$ per gallon). How many quarts of product are needed per acre? $\text{Gallons Product/Acre} = \frac{1.5 \text{ lbs a.i.}}{4.0 \text{ lbs a.i./gal}} = 0.375 \text{ gallons}$ $\text{Quarts Product/Acre} = 0.375 \text{ gal} \times 4 \text{ qts/gal} = 1.5 \text{ quarts product/acre}$

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Essential Unit Conversion Factors Reference