Question

Which of the following expression is correct for increased draft due to increase in speed?
[\(D_o = \text{Static component of draft independent of speed}, Ds = \text{Draft at speed } 'S', k = \text{a constant depending on machine type and operating condition.}\)]

• A. \(Ds = D_o + k \, S^2\)
• B. \(Ds = k \, D_o + S^2\)
• C. \(Ds = k \, D_o - k \, S^2\)
• D. \(Ds = k \, D_o + k \, S^2\)

Hint:

ASABE draft model: $D = F_i [A + B(S) + C(S)^2] W \cdot d$. The quadratic term $k S^2$ accounts for soil kinetic acceleration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Soil-tool dynamic interaction models show that soil resistance consists of static cutting/cohesion forces plus dynamic acceleration forces proportional to the square of forward velocity.
Key Formula or Approach:
\[ D(S) = D_0 + k \, S^2 \]
where \(D_0\) is static soil cutting resistance and \(k S^2\) represents kinetic energy imparted to accelerate the furrow slice.

Step 2: Detailed Explanation:
As forward speed \(S\) increases, additional energy is required to accelerate and throw the soil slice.
The dynamic inertial resistance of the soil varies with the square of speed (\(v^2\)).
Therefore, the total draft at forward speed \(S\) is modeled as:
\[ Ds = D_o + k \, S^2 \]

Step 3: Final Answer:
Hence, the correct expression is \(Ds = D_o + k \, S^2\), corresponding to option (A).