formula.expert

Sprayer calibration morning

Sprayer calibration · speed, tanks and mix strength

The sprayer was calibrated in the spring at 100 L/ha travelling 6.0 km/h, and the card taped inside the cab door still says so. Today the ground is firm and the operator is running 7.5 km/h with the same nozzles and pressure. The tank holds 800 L, and the product label calls for 2.5 L/ha. Work out the day as it actually is: the rate the faster speed really applies, the area one tank now covers, the product that goes into each fill, and the percent solution in the tank.

Tv₂v₁AV₂

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • V₁ = 100 L/ha — Calibrated rate
  • v₁ = 6 km/h — Calibrated speed
  • v₂ = 7.5 km/h — Today's speed
  • T = 800 L — Tank volume
  • Rᵥ = 2.5 L/ha — Label rate of product
Determine
  1. (a)the application rate at today's speed
  2. (b)the area one tank covers now
  3. (c)the product per tank fill
  4. (d)the percent solution in the tank
Step 1 of 4(a) · solve for Rate at the new speed

Same nozzles, same pressure: the boom puts out the same litres per minute regardless of what the wheels do, so rate falls exactly as speed rises. A quarter more speed is a fifth less water — 80 L/ha, not the 100 on the card.

v1V1v2V2
Rearranged for V₂
V2=V1v1v2V_2 = V_1 \frac{v_1}{v_2}
Your values, in your units
V2=(100 L/ha)(6 km/h)(7.5 km/h)V_2 = \left( 100\ \text{L/ha} \right) \frac{\left( 6\ \text{km/h} \right)}{\left( 7.5\ \text{km/h} \right)}
Answer
V2=80 L/haV_2 = 80\ \text{L/ha}

Carried onward at full precision, not this rounded figure.

Open the Effect of Speed on Application Rate solver →

Step 2 of 4(b) · solve for Area covered per tank

Every downstream number keys off the real rate from part (a), not the card's. At 80 L/ha the 800 L tank covers 10 ha — a full hectare more than the calibration assumed, which is exactly how a block gets finished with a suspiciously empty tank.

TVA
Rearranged for A
A=TVA = \frac{T}{V}
Your values, in your units
A=(800 L)(80 L/ha)A = \frac{\left( 800\ \text{L} \right)}{\left( 80\ \text{L/ha} \right)}
Answer
A=100 dunamA = 100\ \text{dunam}

Carried onward at full precision, not this rounded figure.

Open the Area Covered per Tank solver →

Step 3 of 4(c) · solve for Product per tank

The label rate is per hectare, so the product per fill follows the AREA, not the tank: 2.5 L/ha across the 10 ha this tank now covers is 25 L into each fill. Dose to the card's 9 ha instead and every hectare sprayed runs 10 % light.

VpRvA
Rearranged for Vₚ
Vp=Rv AV_p = R_v \, A
100 dunamcarried from step 2
Your values, in your units
Vp=(2.5 L/ha) (100,000 m2)V_p = \left( 2.5\ \text{L/ha} \right) \, \left( 100{,}000\ \text{m}^{2} \right)
Converted to base units
Vp=(2.5 L/ha) (10 ha)V_p = \left( 2.5\ \text{L/ha} \right) \, \left( 10\ \text{ha} \right)
Answer
Vp=25 LV_p = 25\ \text{L}

Carried onward at full precision, not this rounded figure.

Open the Liquid Product per Tank solver →

Step 4 of 4(d) · solve for Mix strength

The strength check closes the loop: 25 L of product in the 800 L tank is a 3.125 % solution — the figure to compare against the label's maximum concentration before anything goes near the crop.

PTVp
Rearranged for P
P=100 VpTP = 100 \, \frac{V_p}{T}
25 Lcarried from step 3
Your values, in your units
P=100 (0.025 m3)(800 L)P = 100 \, \frac{\left( 0.025\ \text{m}^{3} \right)}{\left( 800\ \text{L} \right)}
Converted to base units
P=100 (25 L)(800 L)P = 100 \, \frac{\left( 25\ \text{L} \right)}{\left( 800\ \text{L} \right)}
Answer
P=31.25 per milleP = 31.25\ \text{per mille}

Carried onward at full precision, not this rounded figure.

Open the Percent Solution in the Tank solver →

Answer

Therefore today's speed applies 80 L/ha, one tank covers 10 ha, each fill takes 25 L of product, and the mix runs at 3.125 % — the water changed with the speed, but the dose per hectare only stays right because the product was figured on the real area, not the card's.

Why this order

The whole morning turns on one fact about hydraulics and one about labels. The hydraulics: nozzles meter litres per minute, wheels meter hectares per hour, and the application rate is just their quotient — so rate moves inversely with speed, exactly and immediately, with no grace period. The label: the dose is written per hectare of ground, never per tank of water. Water is only the carrier; the chemical does the work. That is why the order of operations here matters — real rate first, real area second, and only then the product, so the dose rides on ground actually covered.

The classic mistake is treating the card as the truth after conditions change: dosing 25 L against the card's 9 ha coverage while actually covering 10 puts 10 % less active on every hectare — invisible on the day, visible in the weeds three weeks later, and usually blamed on the product.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.