Discussion
A vertical disc plough with 5 discs is operated at a depth of 0.15 m. The disc angle and disc diameter are 40° and 0.6 m, respectively. If overlap between two consecutive discs is 0.12 m at 0.15 m depth of cut, the total width of cut at the specified depth in m is
Answer: A
The total width of cut is 1.19 m.
Each disc cuts a chord of the disc circle at the working depth, and the disc angle (between the disc face and the direction of travel) turns that chord across the line of travel:
\[ \begin{aligned} \text{chord} &= 2\sqrt{d(D - d)} = 2\sqrt{0.15 \times 0.45} = 0.520 \\ w &= 0.520 \times \sin 40^{\circ} = 0.334\ \text{m} \end{aligned} \]
Five discs cut 5 w, less the four overlaps between neighbours:
\[ W = 5(0.334) - 4(0.12) = 1.670 - 0.48 = 1.19\ \text{m} \]
Ignoring the overlaps leaves 1.67 m, which is not offered; using the full chord without the sine factor gives 2.60 - 0.48 = 2.12, close to 2.11. The key's method projects the chord with sin 40 degrees and subtracts the overlaps once between each pair of discs.
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