Question

A cylindrical workpiece is turned using two different tools. Tool 1 has zero nose radius; the side and end cutting edge angles are \(20^\circ\) and \(10^\circ\), respectively. Tool 2 has a 0.5 mm nose radius. Both tools machine at a feed of \(0.2\ \text{mm/rev}\). The ratio of the ideal maximum height of unevenness on the surface produced by Tool 1 to that produced by Tool 2 is \(\underline{\hspace{2cm}}\) (round off to one decimal place).

Hint:

For turning with a sharp tool, surface roughness depends strongly on the side and end cutting edge angles; with a rounded nose, it mainly varies with \(f^2/r\). Even a small nose radius drastically reduces the theoretical roughness.

Answer 1

Correct Answer: 2.9 - 3.7

Step 1: Maximum height of unevenness for Tool 1 (sharp tool).
For a sharp tool (zero nose radius) with side cutting edge angle \(C_s\) and end cutting edge angle \(C_e\), the theoretical peak-to-valley surface roughness is \[ h_1 \;=\; \frac{f}{\tan C_s + \cot C_e}, \] where \(f\) is the feed per revolution.
Here, \[ C_s = 20^\circ, C_e = 10^\circ, f = 0.2\ \text{mm/rev}. \] So, \[ h_1 = \frac{0.2}{\tan 20^\circ + \cot 10^\circ}. \] Numerically, \[ \tan 20^\circ \approx 0.3640, \cot 10^\circ = \frac{1}{\tan 10^\circ} \approx 5.6713, \] \[ \tan 20^\circ + \cot 10^\circ \approx 0.3640 + 5.6713 = 6.0353, \] \[ h_1 \approx \frac{0.2}{6.0353} \approx 0.0331\ \text{mm}. \]

Step 2: Maximum height of unevenness for Tool 2 (with nose radius).
For a tool with nose radius \(r\), the theoretical peak-to-valley roughness is \[ h_2 = \frac{f^2}{8r}. \] With \(f = 0.2\ \text{mm/rev}\) and \(r = 0.5\ \text{mm}\): \[ h_2 = \frac{(0.2)^2}{8 \times 0.5} = \frac{0.04}{4} = 0.01\ \text{mm}. \]

Step 3: Ratio of maximum heights (Tool 1 / Tool 2).
\[ \frac{h_1}{h_2} = \frac{0.0331}{0.01} \approx 3.31. \] Rounded off to one decimal place, \[ \frac{h_1}{h_2} \approx 3.3. \]

Final Answer: \(3.3\)