Problem 7
How many gallons of surfactant would you add to a 1,500-gallon tank at the use rate of 0.25%?

Adjuvants are usually mixed or expressed as "percent solution." One percent is equal to a part per hundred and is usually expressed as a decimal equivalent.

In this problem take the ¼% (0.25) and divide it by 100 to get the decimal equivalent (0.0025). Multiply 0.0025 times the gallons in the tank, 1,500 gal, and you will need 3.75 gal of surfactant to add to the tank.

Problem 8
You are preparing for a high volume application of a 1% spray solution with a 400- gallon tank. The instructions are to "spray to wet." How much product should you add to the tank?

Problem 9
A low volume application is to be made with a 5-gallon backpack sprayer applying a 12% spray solution. How much product would be added to the tank?

Five gallons multiplied by 12% (0.12) equal 0.6 gallons (5 X 0.12 = 0.6).

Problem 10
Your label instructs you to use 5 oz of surfactant per 100 gallons of mixed solution. How many quarts of surfactant will be required for a 1,500-gallon tank?

(16 fluid oz = 1 pint; 32 fluid oz = 1 quart; 128 fluid oz = 1 gal)
If you have a 1,500-gallon tank you simply need to divide this number by 100 and then multiply the answer by 5 oz. Thus, 15 multiplied by 5 oz gives 75 oz of surfactant in a 1,500-gallon tank. To convert this into quarts divide 75 oz by 32 oz per quart. The amount of surfactant to add to the 1,500 gallons is 2.3 quarts.

Problem 11
If a sprayer applies 30 gal per acre (gpa), and is traveling at 10 miles per hour, how many gallons would be applied at 15 miles per hour?

Since speed and output are inversely related, increasing one decreases the other, the formula

Old Rate X Old Speed = New Rate X New Speed

can be used to calculate the answer. Use the known variables, such as speed and output (rate), and then divide them by the new variables trying to be found. Multiply the original speed (10 mph) and original output (30 gpa) and divide by the new speed (15 mph) to arrive at the new output rate of 20 gallons per acre. The equation would read:

(10 mph) X (30 gpa) divided by (15 mph) = (20 gpa) the new rate.

If speed increases the gallons applied per acre must decrease.

Problem 12
A sprayer is applying 40 gallons per acre at 12 mph, how many gallons per acre would be applied at a new speed of 10 mph?

When speed is decreased, output must increase. Using the same formula as the previous problem, the original values of 40 gpa at 12mph would be equal to the new values of 10 mph at 48 gpa. The new application rate will increase since the equipment is going slower.

(12 mph) X (40 gpa) divided by (10 mph) = (48 gpa) the new rate.

Problem 13
A sprayer is applying 20 gallons per acre at 12 mph. How fast would it have to travel to apply 30 gallons per acre?

As with the previous problem, the gallons per acre are increasing, consequently the mph must decrease. The product of the original speed and output (20 gpa X 12 mph = 240) divided by the new output (30 gpa) will give the new speed, which is 8 mph.

(12 mph) X (20 gpa) = (8 mph) X (30 gpa)

Problem 14
The sprayer is applying 40 gallons of spray solution per acre and the tank size is 2,200 gal. The mix is Karmex at 6 lb per acre. To refill the sprayer when there is 300 gal of solution left in the tank, how much Karmex do you have to add to refill?

A full tank is 2,200 gallons but there are 300 gallons left. As a result, you only need to add 1,900 gallons to fill the tank (2,200 - 300 = 1,900). The 1,900 gallons of fresh water divided by 40 gal per acre solution rate = 47.5, which is the amount of acres that will each need to have 6 lb of Karmex. 47.5 acres times 6 lb of Karmex for each acre equals 285 lb of Karmex that needs to be added before the tank is completely filled with the 1,900 gal of fresh water.

Problem 15
How many gallons per minute must a pump supply to apply 30 gallons per acre when treating a 16 ft swath at 12 miles per hour?

To calculate gallons per minute first determine how many minutes it takes, traveling at 12 miles per hour, to go one mile. This can be determined by dividing 60 minutes per hour by 12 miles per hour, which is equal to 5 minutes per mile. Next, determine how many acres were covered in one mile of right-of-way treating a 16 ft swath. One mile of right-of-way with an 8-foot swath is equal to one acre. Consequently, one mile with a 16-foot swath is 2 acres. The only other calculation required is to determine how many gallons to apply over the two acres. Since you already know that 30 gallons will be applied on one acre, then 60 gallons will be applied to 2 acres. Because 2 acres are covered per mile and one mile takes 5 minutes to cover, then 60 gallons will be applied every 5 minutes. By dividing 60 gallons by 5 minutes the gallons applied per minute will be 12, the minimum pump capacity that must be used to make a proper application.

To properly calibrate the amount of material to apply to a given area, it is important to know a couple of formulas. For example, a one-mile strip of an 8-foot swath is equal to one acre. In another example, the original output times the speed is equal to the new output times the new speed. With some simple basics on how to properly calibrate equipment and calculate the proper amount of herbicide to add to a tank, potential over or under applications can be avoided and effective weed control can be achieved. Knowing how your equipment is calibrated, how to make adjustments to it for accurate calibration, and how to figure the amounts of acres and chemical requirements are the most important concepts to learn. Minor amounts of application error may seem unimportant when looking at a single nozzle output, but when multiplied by numerous nozzles on a boom and by numerous miles of treatment, the error can be tremendous at the end of a day, week, or job, if not corrected.