Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A tractor-mounted rotavator operates with a rotor diameter of $480\text{ mm}$ at a rotor rotational speed of $N = 210\text{ rpm}$. The tractor travels forward at a speed of $v_f = 2.7\text{ km/h}$. Each rotor flange carries $z = 3$ L-shaped blades cutting successively in the same plane. The tilling pitch (bite length) of the rotavator in $\text{cm}$ is ________ (round off to 2 decimal places).
Answer Key:
7.14
Step-by-Step Solution & Explanation
```
Rotavator Kinematics:
Bite Length (L)
|<---- L ---->|
/\ /\ /\
/ \ / \ / \ (Trochoidal cut paths)
```
**Step-by-step Calculation:**
1. Forward speed of tractor:
$v_f = 2.7\text{ km/h} = \frac{2.7 \times 1000}{3600} = 0.75\text{ m/s}$
2. Rotor frequency in revolutions per second:
$n = \frac{N}{60} = \frac{210}{60} = 3.5\text{ rev/s}$
3. Blade cut frequency (cuts per second in the same plane with $z = 3$ blades):
$f_{\text{cut}} = n \times z = 3.5 \times 3 = 10.5\text{ cuts/s}$
4. Tilling pitch / bite length ($L$):
$L = \frac{v_f}{f_{\text{cut}}} = \frac{0.75\text{ m/s}}{10.5\text{ s}^{-1}} = 0.07143\text{ m} = 7.14\text{ cm}$
**Alternative Formula:**
$L = \frac{60 \times v_f}{N \times z} = \frac{60 \times 0.75}{210 \times 3} = \frac{45}{630} = 0.07143\text{ m} = 7.14\text{ cm}$