---
title: "Drilling time and metal removal rate"
subject: "Workshop Technology"
chapter: 5
key: "semester_2/workshop_technology:drilling-time-numerical"
source: https://www.bctengineer.com/q/semester_2/workshop_technology/drilling-time-numerical
site: BCT Engineering
---

# Drilling time and metal removal rate

Workshop Technology · Chapter 5: Drills and Drilling Processes

- Asked in: year not given, 8 marks
- Video: [Introduction to machining time calculation](https://www.youtube.com/watch?v=MojfK66plik) — Dr. Rahamathullah I
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/workshop_technology/drilling-time-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_2/workshop_technology/chapter-5

## Question

A 20 mm diameter hole is to be drilled through a 50 mm thick mild steel plate with a twist drill of 118° point angle at a cutting speed of 25 m/min and feed of 0.25 mm/rev. Allow 2 mm for breakout (overtravel). Find (a) spindle speed, (b) drill approach, (c) feed rate, (d) time to drill the hole and (e) metal removal rate.

## Short answer

Spindle speed N=1000×25/(π×20)=398 rpm; approach A=(20/2) cot(59°)=6.01 mm; feed rate=0.25×398=99.5 mm/min; total travel 58.01 mm gives time 0.583 min (≈35 s); MRR=(π/4)×20²×99.5=31.25 cm³/min.

## Answer

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

| Quantity | Value |
|---|---|
| N | 397.9 rpm |
| Approach | 6.01 mm |
| Feed rate | 99.5 mm/min |
| Time | 0.583 min |
| MRR | 31.25 cm³/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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For D = 20 mm and v = 25 m/min, the spindle speed N = 1000v/(pi D) is about
   A. 79.6 rpm
   B. 398 rpm
   C. 1250 rpm
   D. 39.8 rpm
   Answer: B. N = 1000 x 25 / (pi x 20) = 397.9 rpm. (Key point: Spindle speed formula)

2. The drill approach (point allowance) for a 118° drill of 20 mm diameter is given by (D/2) cot(59°), which equals about
   A. 6.01 mm
   B. 10 mm
   C. 16.6 mm
   D. 3.0 mm
   Answer: A. A = 10 / tan 59° = 10 / 1.664 = 6.01 mm. (Key point: Drill approach)

3. What is the total tool travel L in this problem?
   A. 50 mm
   B. 52 mm
   C. 58.01 mm
   D. 56.01 mm
   Answer: C. L = plate thickness + approach + overtravel = 50 + 6.01 + 2 = 58.01 mm. (Key point: Total tool travel)

4. With feed rate 99.5 mm/min, the drilling time and MRR (pi/4 D^2 x feed rate) are about
   A. 0.583 min and 31,250 mm^3/min
   B. 5.83 min and 3,125 mm^3/min
   C. 0.583 min and 99.5 mm^3/min
   D. 0.502 min and 31,250 mm^3/min
   Answer: A. T = 58.01/99.5 = 0.583 min; MRR = (pi/4)(400)(99.5) = 31,250 mm^3/min. (Key point: Time and MRR)

## Related in Workshop Technology

- [Turning time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Alathe-turning-time-mrr-numerical) — Workshop Technology, chapter 6
- [Geometry of a twist drill](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Atwist-drill-geometry) — Workshop Technology, chapter 5
- [Sheet metal cutting and forming operations](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Asheet-metal-operations-explained) — Workshop Technology, chapter 8

## Related in other subjects

- [EDM principle and ECM metal removal rate](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Aedm-ecm-principle-mrr-numerical) — Manufacturing and Production Processes, chapter 8
- [Milling, shaper and drilling machine operations](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Amilling-shaper-drilling-machines-operations) — Manufacturing and Production Processes, chapter 7
- [Settling tank 30 m x 3 m, 0.03 mm at 25°C; 0.025 mm at 20°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-k72) — Water Supply Engineering, chapter 6, asked 3×
- [Steel shaft 200 hp: torque, shear, twist](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-shaft-200hp-150rpm) — Strength of Materials, chapter 6