7.4 Dosage, Dilution & Tank Mixing Calculations

7.4 Dosage, Dilution & Tank Mixing Calculations

Precision in mathematical calculations is a fundamental core competency for licensed pesticide applicators in New Hampshire. Calculating dosages incorrectly causes chemical under-dosing that fails to suppress target pests, or chemical over-dosing that inflicts crop phytotoxicity, violates federal tolerance residues, and contaminates sensitive groundwater resources. Applicators must master active ingredient conversions, tank mix capacities, geometric area calculations, and spot dilution formulas.

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Formulated Product vs. Active Ingredient (a.i.)

Pesticide label directions specify application rates in one of two ways:

1. Formulated Commercial Product Rate: The label directly states the volume or weight of product as sold in the jug or bag (e.g., "Apply 2.5 pints of Treflan per acre" or "Apply 3.0 ounces of Barricade 65WG per 1,000 sq ft").
2. Active Ingredient (a.i.) Rate: The label states the weight of the pure toxic chemical to be applied per acre (e.g., "Apply 1.5 pounds of chlorothalonil active ingredient per acre"). The applicator must calculate how much formulated product is required to supply that weight of active ingredient.

Formulation Nomenclature Rules

• Dry Formulations (WP, WDG, DF, SP, G): The number in the trade name represents the percentage of active ingredient by weight.
  • Example: Captan 80 WDG contains 80% active ingredient (0.80 lb a.i. per pound of product).
  • Example: Sevin 50 WP contains 50% active ingredient (0.50 lb a.i. per pound of product).
• Liquid Formulations (EC, SC, F, L): The number in the trade name represents the pounds of active ingredient per gallon of liquid product.
  • Example: Dursban 4 EC contains 4.0 pounds of active ingredient per gallon.
  • Example: Atrazine 4L contains 4.0 pounds of active ingredient per gallon.
  • Example: Roundup PRO (3 lb a.i./gal glyphosate acid equivalent).

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Active Ingredient Conversion Calculations

+-------------------------------------------------------------------------+
|                   ACTIVE INGREDIENT CONVERSION FORMULAS                 |
+-------------------------------------------------------------------------+
| 1. DRY FORMULATIONS (WP, WDG, SP, G):                                   |
|                                                                         |
|    Pounds of Formulated Product = Pounds a.i. Recommended               |
|                                   ------------------------              |
|                                   Percent a.i. (as decimal)             |
|                                                                         |
| 2. LIQUID FORMULATIONS (EC, SC, F, L):                                  |
|                                                                         |
|    Gallons of Formulated Product = Pounds a.i. Recommended              |
|                                    -----------------------              |
|                                    Pounds a.i. per Gallon               |
+-------------------------------------------------------------------------+

Dry Formulation Example

A field corn grower needs to apply 2.0 pounds of active ingredient per acre using Atrazine 80 WP (80% wettable powder). $\text{Pounds of 80 WP per Acre} = \frac{2.0\text{ lb a.i.}}{0.80} = 2.5\text{ pounds of product per acre}$ If the grower is treating a 40-acre field: $\text{Total Product Needed} = 40\text{ acres} \times 2.5\text{ lb/acre} = 100\text{ pounds of Atrazine 80 WP}$

Liquid Formulation Example

An orchardist needs to apply 3.0 pounds of active ingredient per acre to control apple scab using a fungicide formulated as 4 EC (4 pounds a.i. per gallon). $\text{Gallons of 4 EC per Acre} = \frac{3.0\text{ lb a.i.}}{4.0\text{ lb a.i./gal}} = 0.75\text{ gallon per acre}$ Converting 0.75 gallons into quarts (since 1 gallon = 4 quarts = 128 fl oz): $\text{Quarts per Acre} = 0.75 \times 4 = 3.0\text{ quarts per acre (or 96 fl oz/acre)}$ If the orchard block is 12 acres: $\text{Total Product Needed} = 12\text{ acres} \times 0.75\text{ gal/acre} = 9.0\text{ gallons of 4 EC}$

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Tank Mix Loading & Coverage Calculations

When preparing to spray an entire field or turf property, applicators must determine how many acres a single full tank will cover, how much chemical to add per tankload, and how to mix partial loads for remaining slivers of land.

Step 1: Determine Acres Covered per Full Tank

Divide total usable spray tank capacity by the calibrated application rate in Gallons Per Acre:

$\text{Acres Covered per Tank} = \frac{\text{Tank Capacity (Gallons)}}{\text{Calibrated Application Rate (GPA)}}$

• Example: A boom sprayer has a 500-gallon tank and is calibrated to apply 20 GPA. $\text{Acres Covered per Tank} = \frac{500\text{ gallons}}{20\text{ GPA}} = 25\text{ acres per tankload}$

Step 2: Determine Chemical Product per Tankload

Multiply acres covered per tank by the labeled application rate per acre:

$\text{Product to Add per Tank} = \text{Acres per Tank} \times \text{Label Rate per Acre}$

• Example: The label instructs applying 1.5 pints of herbicide per acre. $\text{Product per Tank} = 25\text{ acres} \times 1.5\text{ pints/acre} = 37.5\text{ pints}$ Converting to gallons (8 pints per gallon): $\text{Gallons per Tank} = \frac{37.5}{8} = 4.69\text{ gallons of product}$

Step 3: Partial Tank Calculations (Preventing Waste Disposal)

Applicators should never mix a full tank of pesticide when only a fraction of acreage remains. Mixing excess chemical creates illegal disposal hazards.

• Scenario: An applicator completes several full tanks and has 6.5 acres remaining to spray. The sprayer is calibrated at 20 GPA, and the label rate is 2.0 pounds of 75 WP per acre.
  1. Water Volume Needed: $\text{Water Needed} = 6.5\text{ acres} \times 20\text{ GPA} = 130\text{ gallons of water}$
  2. Product Quantity Needed: $\text{Product Needed} = 6.5\text{ acres} \times 2.0\text{ lb/acre} = 13.0\text{ pounds of 75 WP}$
  3. Mixing Procedure: Fill the spray tank with approximately 70 gallons of clean water, engage agitation, slowly add the 13.0 pounds of 75 WP, then add remaining water to reach the 130-gallon line.

Commercial Turf Calculations (Rate per 1,000 Sq Ft)

Turf and landscape pesticide labels specify rates per 1,000 square feet rather than per acre.

• Example: A lawn care technician has a 200-gallon spray rig calibrated to apply 2.0 gallons per 1,000 sq ft. The label specifies 1.5 fluid ounces of herbicide per 1,000 sq ft.
  1. Units of 1,000 sq ft per Tank: $\text{Tank Capacity in 1,000 sq ft units} = \frac{200\text{ gallons}}{2.0\text{ gal / 1,000 sq ft}} = 100\text{ units (100,000 sq ft)}$
  2. Product per Tank: $\text{Herbicide Needed} = 100\text{ units} \times 1.5\text{ fl oz/unit} = 150\text{ fluid ounces}$ Converting to gallons (128 fl oz per gallon): $\text{Gallons Needed} = \frac{150}{128} = 1.17\text{ gallons (or 1 gallon + 22 fl oz)}$

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Geometric Land Area Calculations

Accurately measuring the surface area of target sites is mandatory. Miscalculating area leads directly to under- or over-application.

+-------------------------------------------------------------------------+
|                    STANDARD GEOMETRIC AREA FORMULAS                     |
+-------------------------------------------------------------------------+
| RECTANGLE / SQUARE:                                                     |
|   Area = Length x Width                                                 |
|                                                                         |
| TRIANGLE:                                                               |
|   Area = 0.5 x Base x Height                                            |
|                                                                         |
| CIRCLE:                                                                 |
|   Area = pi x r^2  (where pi = 3.1416, and r = radius)                  |
|                                                                         |
| TRAPEZOID:                                                              |
|   Area = 0.5 x (Base 1 + Base 2) x Height                               |
|                                                                         |
| CONVERSION TO ACRES:                                                    |
|   Acres = Square Feet / 43,560                                          |
+-------------------------------------------------------------------------+

Worked Scenarios for Applicators

1. Rectangular Field: A hay field measures 660 feet long by 330 feet wide. $\text{Area} = 660 \times 330 = 217,800\text{ sq ft}$ $\text{Acres} = \frac{217,800}{43,560} = 5.0\text{ acres}$
2. Triangular Field Corner: An irregular field corner has a base of 300 feet and a perpendicular height of 150 feet. $\text{Area} = 0.5 \times 300 \times 150 = 22,500\text{ sq ft}$ $\text{Acres} = \frac{22,500}{43,560} = 0.516\text{ acres}$
3. Circular Golf Green: A putting green has a diameter of 80 feet (radius $r = 40\text{ feet}$). $\text{Area} = \pi \times r^2 = 3.1416 \times 40^2 = 3.1416 \times 1,600 = 5,026.5\text{ sq ft}$
4. Highway Right-of-Way Strip: An applicator is treating a roadside brush corridor that extends 4 miles in length with an average treated width of 15 feet.
   • Total length in feet: $4\text{ miles} \times 5,280\text{ ft/mile} = 21,120\text{ feet}$.
   • Area in square feet: $21,120 \times 15 = 316,800\text{ sq ft}$.
   • Total acreage: $316,800 / 43,560 = 7.27\text{ acres}$.

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Percentage Dilution Math for Spot and Hand-Wand Treatments

Backpack sprayers, hand-wand sprayers, greenhouse space sprayers, and woody brush spot treatments frequently specify chemical rates as a percentage concentration of spray solution (e.g., "Mix a 1.5% v/v solution of herbicide in water").

The Percentage Dilution Master Formula

To determine the fluid ounces of pesticide concentrate required:

$\text{Fluid Ounces of Pesticide} = \text{Target Solution (Gallons)} \times 128\text{ fl oz/gal} \times \text{Percent Concentration (decimal)}$

Practical Step-by-Step Spot Mix Scenarios

• Scenario 1 (Backpack Invasive Weed Treatment): An applicator needs to prepare a 3-gallon backpack sprayer with a 2% solution of triclopyr to spot-treat oriental bittersweet in a forest edge. $\text{Fl Oz of Pesticide} = 3\text{ gallons} \times 128\text{ fl oz/gal} \times 0.02$ $\text{Fl Oz of Pesticide} = 384 \times 0.02 = 7.68\text{ fluid ounces}$ Procedure: Add approximately 1.5 gallons of clean water to the backpack, measure and add 7.7 fluid ounces of triclopyr, agitate gently, then fill with clean water to the 3-gallon fill line.

• Scenario 2 (High-Volume Hydraulic Spot Spray): A utility contractor is preparing a 100-gallon skid tank with a 1.5% solution of glyphosate for invasive knotweed along an electric transmission right-of-way. $\text{Gallons of Pesticide} = 100\text{ gallons} \times 0.015 = 1.5\text{ gallons}$ Converting to fluid ounces: $1.5 \times 128 = 192\text{ fluid ounces}$. Procedure: Add 50 gallons of water to the tank, pour in 1.5 gallons of glyphosate concentrate, start mechanical agitation, and bring water level to exactly 100 gallons.

• Scenario 3 (Dormant Fruit Tree Horticultural Oil): An apple grower is preparing 250 gallons of a 3% dormant oil spray to smother overwintering European red mite eggs. $\text{Gallons of Oil} = 250\text{ gallons} \times 0.03 = 7.5\text{ gallons of horticultural oil}$ The grower adds 7.5 gallons of oil to the tank under vigorous mechanical agitation with water to create a stable, uniform emulsion before spraying.