Need a worksheet for calibrating a plot sprayer? Well, we just so happen to have one here:
Category: Calculators & Applications
Simple on-line calculators, linked to Agrimetrixapps.com
Sprayer Turn Compensation
Turn compensation is a feature in pulse width modulation (PWM) sprayers in which nozzle output matches the boom’s speed during a turn. When turning, the inside and outside of a boom travel at different speeds, resulting in over-dosing on the inside and under-dosing on the outside. Read about PWM systems here, here, and here.
The degree of the problem depends on the inside turn diameter. Clearly, the tighter the turn, the more severe the over-and under-dosing. The ability of a PWM sprayer to compensate also depends on the turn tightness, as well as the Duty Cycle (DC) the system is operating at during the turn.
In the above example, a 120 ft boom makes a turn around an object with a 60 ft diameter. Assuming a 12 mph speed and an application volume of 10 gpa, the inside of the boom travels at 4 mph and applies 30 gpa, or 3x. On the outermost nozzle, the speed is 20 mph with an application volume of 6 gpa, or 0.6 x. A sprayer operating at 60% DC would be able to correct the application in this turn by operating at 100% DC on the outside and 20% DC on the inside.
But completing the turn at other DCs may be problematic. In this case, lower sprayer DC would require the inside DC to operate below 20%, which may not be possible, depending on the system. Conducting the turn at higher DC would prevent the outer boom from meeting the flow requirements, resulting in under-dosing.
Optimizing the benefit of turn compensation requires the operator to enter the turn at a DC that meets the objectives. Is it more important to prevent under-dosing of the outside perimeter? If so, slow down in the turn (reducing DC) and maximize the extra capacity at the outside of the boom, possibly at the cost of over-dosing the inside.
The agronomic benefit of turn compensation is to provide sufficient pesticide dosage where it’s needed. It’s been reported that repeatedly applying sublethal herbicide dosages at the same site can lead to the development of polygenic resistance in some outcrossing weed species. These areas are likely to occur at the outside boom location of a permanent landscape feature that the sprayer moves around year after year.
Turn compensation is a valuable feature in all agricultural operations where input distribution uniformity is important. Spraying is no exception, and PWM makes it possible.
Diluting 20,000-Fold with a 30 Gallon Remaining Volume in a 1,200 Gallon Tank
(This short article is an addendum to this article)
Our goal in this example is to dilute by a factor of 20,000.
The maximum amount of dilution possible with a 1,200 gallon tank and a 30 gallon remainder is 1200/30=40.
The formulae:
Dilution per Rinse = final dilution ^(1/# of rinses)
Rinse Volume = (dilution per rinse * remaining volume) – remaining volume
- One rinse diluting by 20,000 – impossible with a 1,200 gallon tank (max achievable is 40-fold);
- Two sequential rinses, each diluting by a factor of 20,000^(1/2) = 141. Also impossible with a 1,200 gallon tank;
- Three sequential rinses, each diluting by a factor of 20,000^(1/3) = 27. A volume of 780 gallons can do this (27*30)-30=780 gallons. For three rinses, the total volume is 2,340 gallons.
- Four sequential rinses, each diluting by a factor of 20,000^(1/4) = 12. A volume of 330 gallons can do this, for a total volume of 1,320 gallons;
- Five sequential rinses, each diluting by a factor of 20,000^(1/5) = 7. A volume of 180 gallons can do this, for a total volume of 900 gallons;
- Six sequential rinses, each diluting by a factor of 20,000^(1/6) = 5.2. A volume of 126 gallons can do this, for a total volume of 757 gallons.
Second, let’s assume the operator is prepared to prime the boom where it doesn’t harm soybeans. Now the first new product tank takes care of the last dilution, lowering the cleanout dilution requirement by 1,200/30 = a factor of 40. Now the cleanout dilution requirement is only 20,000/40 = 500.
- One 1,200 gallon tank rinse can only achieve 40-fold dilution.
- Two rinses, each diluting by 500^(1/2) = 22. Rinse volumes of 640 gallons are sufficient, for a total of 1,280 gallons.
- Three sequential rinses, each diluting by a factor of 500^(1/3) = 7.9. A volume of 210 gallons can do this, for a total volume of 630 gallons;
- Four sequential rinses, each diluting by a factor of 500^(1/4) = 4.7. A volume of 112 gallons can do this, for a total volume of 448 gallons.
Tank Rinse Dilution
If you have a limited amount of clean water to rinse your sprayer, this can help you decide how to make the best use of it. Simple enter two values – the amount of pesticide remaining in your tank sump (including the suction line to the pump and the return line to the tank), and the amount of clean water you want to add. The units (gal, L) are not important as long as they are the same for both entries.
The app allows you to calculate the dilution power of up to 5 sequential rinses.
