9.4 Tank Mix & Active Ingredient (a.i.) Calculations

Tank Mix & Active Ingredient (a.i.) Calculations

Once an application rig has been calibrated to determine its delivery rate (GPA or gallons per 1,000 sq ft), the applicator must compute the precise quantity of pesticide formulation to add to the spray tank. Adding too little chemical results in sub-lethal underdosing, economic crop loss, and accelerated pest resistance; adding too much causes crop phytotoxicity, illegal chemical residues, and regulatory penalties.

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1. Quick-Reference Unit Conversion & Measurement Equivalencies

Applicators must know these standard liquid, dry weight, and linear conversion constants:

+-----------------------------------------------------------------------------+
|                        MASTER CONVERSION EQUIVALENCIES                      |
|                                                                             |
|   [LIQUID VOLUME]                                                           |
|   1 Gallon = 4 Quarts = 8 Pints = 16 Cups = 128 Fluid Ounces = 3,785 mL     |
|   1 Quart  = 2 Pints  = 4 Cups  = 32 Fluid Ounces  = 946 mL                 |
|   1 Pint   = 2 Cups   = 16 Fluid Ounces = 473 mL                            |
|   1 Cup    = 8 Fluid Ounces = 16 Tablespoons = 237 mL                       |
|   1 Tablespoon = 3 Teaspoons = 0.5 Fluid Ounce = 14.8 mL                    |
|                                                                             |
|   [DRY WEIGHT]                                                              |
|   1 Pound (lb) = 16 Dry Ounces = 453.6 Grams = 0.4536 Kilograms             |
|                                                                             |
|   [AREA & DISTANCE]                                                         |
|   1 Acre = 43,560 Square Feet = 0.4047 Hectare                              |
|   1 Mile = 5,280 Feet = 1,760 Yards = 1.609 Kilometers                      |
+-----------------------------------------------------------------------------+

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2. Determining Tank Coverage (Acres per Tank)

Before adding chemical to a spray tank, determine how many acres a single full tank load will treat based on the calibrated application rate (GPA):

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

Worked Example 1: Full Tank Acreage Coverage

Problem: A commercial boom sprayer has a 400-gallon tank and is calibrated to apply 25 GPA. How many acres will one full tank treat?

Step-by-Step Solution:

1. Identify given values: $\text{Tank Capacity} = 400\text{ gallons}$, $\text{Rate} = 25\text{ GPA}$.
2. Apply formula:

$\text{Acres per Tank} = \frac{400\text{ gallons}}{25\text{ GPA}} = 16.0\text{ Acres}$

1. Answer: One full tank load will cover 16.0 acres.

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3. Formulated Pesticide Product Calculations (Full Tank)

A. Liquid Formulations (EC, SL, F, L)

Liquid product rates are typically given in fluid ounces, pints, quarts, or gallons per acre.

$\mathbf{\text{Liquid Product per Tank} = \text{Acres Covered per Tank} \times \text{Product Rate per Acre}}$

Worked Example 2: Liquid Herbicide Batching

Problem: A 300-gallon sprayer calibrated at 20 GPA treats 15 acres per tank. The herbicide label prescribes an application rate of 1.5 pints per acre. How many total gallons of herbicide product must be added to a full tank?

Step-by-Step Solution:

1. Calculate total pints needed: $15\text{ acres} \times 1.5\text{ pints/acre} = 22.5\text{ pints}$.
2. Convert pints to gallons ($8\text{ pints} = 1\text{ gallon}$):

$\text{Gallons Product} = \frac{22.5\text{ pints}}{8\text{ pints/gal}} = 2.8125\text{ gallons}$

1. Convert remainder to fluid ounces if needed: $0.8125 \times 128\text{ fl oz/gal} = 104\text{ fl oz}$.
2. Answer: Add 2.81 gallons of herbicide to the full tank. Expressed in field units, 22.5 pints is 2 gallons (16 pints) + 3 quarts (6 pints) + 1 cup (0.5 pint).

B. Dry Formulations (WP, WDG, DF, SP, Granules)

Dry pesticide rates are given in ounces or pounds of product per acre.

$\mathbf{\text{Dry Product per Tank} = \text{Acres Covered per Tank} \times \text{Lbs of Product per Acre}}$

Worked Example 3: Wettable Powder Tank Batching

Problem: A 400-gallon sprayer calibrated at 20 GPA treats 20 acres per tank. The insecticide label prescribes 2.5 pounds of 75 WP per acre. How many pounds of 75 WP are needed for a full tank?

Step-by-Step Solution:

1. Calculate total pounds: $20\text{ acres} \times 2.5\text{ lbs/acre} = 50.0\text{ lbs}$.
2. Answer: Add 50.0 pounds of 75 WP to the full spray tank.

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4. Partial Tank Mix Calculations

When the remaining field acreage is less than a full tank load, mix only the exact volume of spray solution required to avoid chemical waste and disposal problems.

$\mathbf{\text{Product for Partial Tank} = \text{Partial Volume (gal)} \times \frac{\text{Product per Full Tank}}{\text{Full Tank Capacity (gal)}}}$

Worked Example 4: Partial Tank Batching

Problem: A full 500-gallon tank requires 15 quarts of fungicide. You only need to mix 200 gallons of spray solution to finish the field. How many quarts of fungicide must be added?

Step-by-Step Solution:

1. Apply proportional formula:

$\text{Quarts Needed} = 200\text{ gal} \times \frac{15\text{ quarts}}{500\text{ gal}} = 200 \times 0.03 = 6.0\text{ Quarts}$

1. Answer: Add 6.0 quarts (or 1.5 gallons) of fungicide to the 200 gallons of water.

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5. Active Ingredient (a.i.) Calculations

University extension recommendations and research bulletins frequently state application rates in terms of pounds of Active Ingredient (lbs a.i.) per acre rather than pounds of formulated product. The applicator must convert the active ingredient rate to the commercial product rate.

+-----------------------------------------------------------------------------+
|                      ACTIVE INGREDIENT (a.i.) CONVERSIONS                   |
|                                                                             |
|   [DRY FORMULATION (% a.i. by weight)]                                      |
|   Lbs Product = Lbs a.i. Recommended / Decimal % of a.i.                    |
|                                                                             |
|   [LIQUID FORMULATION (lbs a.i. per gallon)]                                |
|   Gallons Product = Lbs a.i. Recommended / Lbs a.i. per Gallon              |
+-----------------------------------------------------------------------------+

A. Dry Formulations (% a.i. by Weight)

Dry formulations (such as 80 WP, 60 WDG, or 50 SP) list active ingredient as a percentage by weight.

$\mathbf{\text{Lbs of Commercial Product needed} = \frac{\text{Lbs a.i. recommended per acre}}{\%\text{ a.i. in formulation (as decimal)}}}$

Worked Example 5: Dry Product a.i. Conversion

Problem: An extension recommendation calls for 1.5 lbs a.i. per acre of an herbicide. You purchase an 80% Wettable Powder (80 WP). How many pounds of the commercial 80 WP product must be applied per acre?

Step-by-Step Solution:

1. Convert percentage to decimal: $80% = 0.80$.
2. Apply formula:

$\text{Lbs Product} = \frac{1.5\text{ lbs a.i.}}{0.80} = 1.875\text{ lbs product per acre}$

1. Convert decimal fraction to ounces ($0.875 \times 16\text{ oz/lb} = 14\text{ oz}$).
2. Answer: Apply 1.875 lbs (or 1 lb 14 oz) of 80 WP product per acre.

B. Liquid Formulations (Lbs a.i. per Gallon)

Liquid formulations (such as 4 EC, 2.5 SL, or 6 F) state the concentration in pounds of active ingredient contained in one gallon of concentrate (e.g., 4 EC contains 4.0 lbs a.i. per gallon).

$\mathbf{\text{Gallons of Product needed} = \frac{\text{Lbs a.i. recommended per acre}}{\text{Lbs a.i. per gallon of product}}}$

Worked Example 6: Liquid Product a.i. Conversion

Problem: An agronomist recommends applying 0.75 lb a.i. per acre of an insecticide. You have a 4 EC formulation (which contains 4.0 lbs a.i. per gallon). How many fluid ounces of commercial product are required per acre?

Step-by-Step Solution:

1. Calculate gallons needed:

$\text{Gallons Product} = \frac{0.75\text{ lb a.i.}}{4.0\text{ lbs a.i./gal}} = 0.1875\text{ gallons per acre}$

1. Convert gallons to fluid ounces ($1\text{ gallon} = 128\text{ fluid ounces}$):

$\text{Fluid Ounces} = 0.1875\text{ gal} \times 128\text{ fl oz/gal} = 24.0\text{ Fluid Ounces}$

1. Answer: Apply 24.0 fluid ounces (or 1.5 pints) of 4 EC insecticide per acre.

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6. Turf & Ornamental Landscape Calculations (Per 1,000 Sq Ft)

Landscape and turfgrass pesticide labels commonly state dosage rates per 1,000 square feet.

$\mathbf{\text{Number of 1,000 sq ft Units} = \frac{\text{Total Lawn Area (sq ft)}}{1,000}}$ $\mathbf{\text{Total Product Needed} = \text{Number of Units} \times \text{Product Rate per 1,000 sq ft}}$

Worked Example 7: Turfgrass Herbicide Dosing

Problem: A commercial landscape applicator is contracted to treat a 45,000 square foot commercial turfgrass lawn. The herbicide label prescribes applying 3.0 fluid ounces of product per 1,000 sq ft. How many total gallons of herbicide product are required to treat the property?

Step-by-Step Solution:

1. Calculate the number of 1,000 sq ft units:

$\text{Units} = \frac{45,000\text{ sq ft}}{1,000} = 45.0\text{ units}$

1. Calculate total fluid ounces needed:

$\text{Total Fluid Ounces} = 45.0 \times 3.0\text{ fl oz} = 135.0\text{ Fluid Ounces}$

1. Convert fluid ounces to gallons ($128\text{ fl oz} = 1\text{ gallon}$):

$\text{Total Gallons} = \frac{135.0\text{ fl oz}}{128\text{ fl oz/gal}} = 1.0547\text{ Gallons}$

1. Answer: The applicator needs 135.0 fluid ounces (or 1 gallon and 7 fluid ounces) of herbicide concentrate.