To solve this problem, we need to determine the speed ratio between the ground wheel and the metering wheel to achieve a specific application rate of fertilizer (200 kg/ha).

Step 1: Understand the given data.

• Row-to-row spacing: 40 cm
• Fertilizer dispensed per revolution of the metering wheel: 38 g
• Diameter of the ground wheel: 60 cm
• Desired application rate: 200 kg/ha

Step 2: Convert units for the application rate.

The application rate is given in kg/ha, so: $200 \text{ kg/ha} = 200,000 \text{ g/ha}$

Step 3: Calculate the area covered per revolution of the ground wheel.

The ground wheel's circumference determines the distance traveled for each revolution. The formula for the circumference of a circle is: $\text{Circumference} = \pi \times \text{diameter}$ Given the diameter of the ground wheel is 60 cm, the circumference is: $\text{Circumference of ground wheel} = 3.14 \times 60 = 188.4 \text{ cm}$ This means for each revolution, the ground wheel covers a distance of 188.4 cm.

Step 4: Calculate the area covered per revolution of the ground wheel.

The area covered per revolution is the product of the distance covered and the row-to-row spacing: $\text{Area covered per revolution} = \text{distance} \times \text{spacing}$ Convert the row-to-row spacing from cm to meters (since we're working with hectares): $\text{Spacing} = 40 \text{ cm} = 0.4 \text{ m}$ Now, calculate the area per revolution: $\text{Area covered per revolution} = 1.884 \text{ m} \times 0.4 \text{ m} = 0.7536 \text{ m}^2$ Each hectare (ha) is 10,000 m², so the number of revolutions required to cover 1 ha is: $\frac{10,000}{0.7536} = 13,270.57 \text{ revolutions per hectare}$

Step 5: Determine the required fertilizer application per revolution of the ground wheel.

To achieve an application rate of 200 kg/ha, we need 200,000 g of fertilizer distributed over 13,270.57 revolutions: $\frac{200,000 \text{ g}}{13,270.57 \text{ revolutions}} = 15.08 \text{ g/revolution of the ground wheel}$

Step 6: Calculate the speed ratio between the ground wheel and the metering wheel.

The metering wheel dispenses 38 g per revolution, but we need only 15.08 g per revolution of the ground wheel. The speed ratio between the ground wheel and the metering wheel is: $\text{Speed ratio} = \frac{38}{15.08} = 2.52$

Thus, the correct speed ratio between the ground wheel and the metering wheel is 2.52:1.

Final Answer:

The speed ratio is 2.52:1.

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Would you like a more detailed explanation of any step?

Related Questions:

1. How would the calculation change if the ground wheel diameter were increased?
2. What effect would slippage of the ground wheel have on the application rate?
3. How would different row-to-row spacing affect the speed ratio calculation?
4. If the metering wheel dispensed a different amount per revolution, how would this affect the speed ratio?
5. How can this calculation be adjusted for different field sizes?

Tip: Always double-check unit conversions to ensure accuracy, especially when dealing with area-based application rates.