DRAFT
This is converted to hectares by dividing by 10,000:
Area treated | = | 1000 x S x W = | S x W ha/minute |
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| 60 x 10,000 | 600 |
This gives the standard formula for calculating the coverage of a sprayer:
Area/min = | swath width (m) x speed (km/hr) | ha/min |
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| 600 |
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Example: |
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Speed | : |
| 8km/hour |
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Swath width: |
| 12 m |
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Therefore: |
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Area treated | = | S x W | = | 8 x 12 | = | 0.16 ha/min |
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| 600 |
| 600 |
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Table 2 shows the coverage of the sprayer for various swath widths and spraying speeds.
Table 2 – Coverage of sprayer in ha/min
SWATH WIDTH (metres)
SPEED | 4 | 6 | 8 | 10 | 15 | 20 | 25 | 30 |
km/hr |
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4 | 0.03 | 0.04 | 0.05 | 0.07 | 0.10 | 0.13 | 0.17 | 0.20 |
6 | 0.04 | 0.06 | 0.08 | 0.10 | 0.15 | 0.20 | 0.25 | 0.30 |
8 | 0.05 | 0.08 | 0.11 | 0.13 | 0.20 | 0.27 | 0.33 | 0.40 |
10 | 0.07 | 0.10 | 0.13 | 0.17 | 0.25 | 0.33 | 0.42 | 0.50 |
12 | 0.08 | 0.12 | 0.16 | 0.20 | 0.30 | 0.40 | 0.50 | 0.60 |
14 | 0.09 | 0.14 | 0.19 | 0.23 | 0.35 | 0.47 | 0.58 | 0.70 |
4.Calculate the required output from the sprayer in litres/minute to give the correct application rate for the chemical being used.
The output rate of chemical is given by the area sprayed (in hectares) per minute multiplied by the required application rate in litres per hectare.
Example: | 0.16 ha/min |
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Coverage: |
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Application rate: | 100 l/ha | = | 16 l/min |
Output = | 0.16 x 100 |
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