Drilling time and metal removal rate

Given: D = 20 mm, t = 50 mm, v = 25 m/min, f = 0.25 mm/rev, point angle 118°, overtravel = 2 mm.

(a) Spindle speed

$$N = \frac{1000\,v}{\pi D} = \frac{1000 \times 25}{\pi \times 20} = 397.9\ \text{rpm}$$

(b) Drill approach (point allowance)

The drill must travel until its full diameter enters the plate.

$$A = \frac{D}{2}\cot\left(\frac{118^\circ}{2}\right) = \frac{20}{2 \times \tan 59^\circ} = \frac{10}{1.664} = 6.01\ \text{mm}$$

(c) Feed rate

$$f_m = f \times N = 0.25 \times 397.9 = 99.5\ \text{mm/min}$$

(d) Time

Total tool travel:

$$L = t + A + \text{overtravel} = 50 + 6.01 + 2 = 58.01\ \text{mm}$$ $$T = \frac{L}{f_m} = \frac{58.01}{99.5} = 0.583\ \text{min} \approx 35\ \text{s}$$

(e) Metal removal rate

$$\text{MRR} = \frac{\pi}{4}D^2 \times f_m = \frac{\pi}{4}(20)^2 \times 99.5 = 31{,}250\ \text{mm}^3/\text{min}$$

Answer: N = 398 rpm, approach = 6.01 mm, feed rate = 99.5 mm/min, time = 0.583 min (35 s), MRR = 31.25 cm³/min.