Turning, milling, drilling and grinding; machining time

Turning, drilling, milling and grinding: kinematics, spindle speed, approach and over-travel, machining time and material removal rate, with turning, slab-milling and drilling numericals.

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Why it matters

Turning, milling, drilling and grinding make most of the precise parts in a mechatronic system – shafts, housings, brackets, bearing seats. Estimating machining time and material removal rate (MRR) is how a process planner quotes a job, balances a CNC line and writes the feed and speed values into a part program.

Key ideas

Cutting speed, spindle speed and feed. The cutting speed V is the surface speed of the work (turning) or of the tool's cutting edge (milling, drilling, grinding). The machine is set in spindle speed N, so N = 1000·V/(π·D), with D the diameter that is rotating (work diameter in turning, tool diameter in milling and drilling). Feed is the advance per revolution (turning, drilling), per tooth (milling) or per unit time (table feed).

Turning (lathe): the work rotates, a single-point tool moves parallel to the axis (straight turning), perpendicular to it (facing) or at an angle (taper turning). Depth of cut d is the radial depth, so one pass reduces the diameter by 2d. Cylindrical surfaces are turned because the rotating work generates a true circle with one simple tool.

Drilling: a rotating twist drill (two cutting lips, usual point angle 118°) feeds axially into the work. The conical tip must travel an extra distance (D/2)·cot(θ/2) ≈ 0.3D before the full diameter cuts, and for a through hole it must emerge completely. Drilled holes are not very accurate; reaming, boring or honing follows when tolerance or finish matters.

Milling: a multi-tooth rotating cutter removes material while the table feeds the work past it. Slab (plain/peripheral) milling uses teeth on the periphery, with the axis parallel to the surface; face milling uses teeth on the end face, with the axis perpendicular. Up (conventional) milling has the tooth moving against the feed and chip thickness growing from zero; down (climb) milling has them in the same direction, giving better finish and tool life but needing a backlash-free feed drive (standard on CNC ball-screw machines). Milling chips are discontinuous and the cut is interrupted, so tools must be tough.

Grinding: a bonded abrasive wheel with many randomly oriented grains, each with a large negative rake, removes tiny chips at very high speed (typically 20–45 m/s). Specific energy is high and most of it becomes heat, so coolant is essential to avoid burn and residual stress. Grinding gives the best tolerance and finish (Ra around 0.2–0.8 µm) and can cut hardened steel. Wheels are specified by abrasive (Al₂O₃ for steels, SiC for cast iron and non-ferrous, cBN and diamond for very hard materials), grit size, grade (bond strength), structure and bond type. A soft-grade wheel is used on hard work, so that dulled grains are shed and fresh sharp grains are exposed (self-sharpening); a hard-grade wheel suits soft work.

Machining time is the length the tool must actually travel divided by the feed velocity: always add approach (A) and over-travel (O) to the work length. MRR is the volume removed per unit time and links to power through specific cutting energy: P = u·MRR.

Formulas

N = 1000·V / (π·D)

  • N = spindle speed (rev/min); V = cutting speed (m/min); D = rotating diameter (mm). The 1000 converts m to mm.

t = (L + A + O) / (f·N) (turning, drilling)

  • t = time per pass (min); L = length of cut (mm); A, O = approach and over-travel (mm); f = feed (mm/rev).

MRR (turning) = π·D_avg·d·f·N ≈ 1000·V·f·d

  • MRR in mm³/min; d = depth of cut (mm); D_avg = (D₀ + D₁)/2. The approximate form uses V in m/min.

Facing time t = (D/2) / (f·N)

  • For facing a full face from the outside to the centre at constant N.

Drill tip allowance = (D/2)·cot(θ/2)

  • θ = point angle (118° gives about 0.3D).

MRR (drilling) = (π·D²/4)·f·N

f_m = f_z·Z·N

  • f_m = table feed (mm/min); f_z = feed per tooth (mm/tooth); Z = number of teeth.

t = (L + A + O) / f_m (milling)

A = √(d·(D − d)) (slab milling approach)

  • d = depth of cut (mm); D = cutter diameter (mm).

MRR (milling) = w·d·f_m

  • w = width of cut (mm).

P = u·MRR

  • P = cutting power (W) with u = specific cutting energy (J/mm³) and MRR in mm³/s. Take u from a data book for the work material.

Worked examples

Example 1 (turning, standard). A 60 mm diameter bar is turned over a length of 250 mm at V = 90 m/min, f = 0.25 mm/rev, d = 2 mm. Approach plus over-travel = 5 mm. Find N, the time for one pass and the MRR.

  1. N = 1000·V/(π·D) = 1000 × 90 / (π × 60) = 477.5 rev/min
  2. t = (L + A + O)/(f·N) = 255 / (0.25 × 477.5) = 255 / 119.4 = 2.14 min
  3. D_avg = (60 + 56)/2 = 58 mm. MRR = π·D_avg·d·f·N = π × 58 × 2 × 0.25 × 477.5 = 43 500 mm³/min (the approximation 1000·V·f·d gives 45 000 mm³/min). Answer: N = 477.5 rev/min, t = 2.14 min, MRR ≈ 43 500 mm³/min.

Example 2 (slab milling, GATE level). A 300 mm long, 60 mm wide block is slab milled with an 80 mm diameter, 8-tooth HSS cutter at V = 25 m/min, f_z = 0.1 mm/tooth, d = 4 mm. Neglect over-travel. Find the machining time and MRR.

  1. N = 1000 × 25 / (π × 80) = 99.47 rev/min
  2. f_m = f_z·Z·N = 0.1 × 8 × 99.47 = 79.58 mm/min
  3. A = √(d(D − d)) = √(4 × 76) = 17.44 mm
  4. t = (L + A)/f_m = (300 + 17.44) / 79.58 = 3.99 min
  5. MRR = w·d·f_m = 60 × 4 × 79.58 = 19 100 mm³/min Answer: t ≈ 3.99 min, MRR ≈ 19 100 mm³/min.

Example 3 (drilling). A 20 mm through hole is drilled in a 40 mm plate, point angle 118°, V = 25 m/min, f = 0.2 mm/rev. Neglect approach clearance.

  1. N = 1000 × 25/(π × 20) = 397.9 rev/min
  2. Tip allowance = 10·cot 59° = 10 × 0.6009 = 6.01 mm
  3. t = (40 + 6.01)/(0.2 × 397.9) = 46.01/79.58 = 0.578 min
  4. MRR = (π × 20²/4) × 0.2 × 397.9 = 25 000 mm³/min Answer: t ≈ 0.58 min (about 35 s), MRR = 25 000 mm³/min.

Common mistakes

  • Using the cutting speed V in m/min directly with D in mm without the factor 1000.
  • Using feed per tooth as if it were table feed in milling – multiply by Z and N.
  • Ignoring approach and over-travel (especially the slab-milling approach √(d(D − d)) and the drill-tip allowance).
  • Taking the depth of cut in turning as the diameter reduction; diameter reduces by 2d.
  • Using the tool diameter in turning or the work diameter in milling for N.
  • Thinking a harder grinding wheel is needed for harder work; it is the opposite.
  • Forgetting that grinding is also a rotating-tool process when classifying operations.

For GATE ME

  • Machining time for turning, facing, drilling (with tip allowance) and slab or face milling (with approach).
  • MRR and cutting power from specific energy; time for several passes.
  • Up versus down milling, grinding wheel specification and selection, and the effect of feed and nose radius on surface roughness (covered with cutting mechanics). Practise unit handling (m/min, mm/rev, mm/min) until it is automatic.

Quick check

  1. A 40 mm bar is turned at 75 m/min. What is the spindle speed?
  2. A 6-tooth cutter runs at 200 rev/min with 0.05 mm/tooth. What is the table feed?
  3. Why is down milling preferred on CNC machines but avoided on old manual mills?
  4. Which grade of wheel is used to grind hardened steel?
  5. In one turning pass with depth of cut 1.5 mm, by how much does the diameter reduce?

Answers: 1. N = 1000 × 75/(π × 40) ≈ 597 rev/min. 2. 0.05 × 6 × 200 = 60 mm/min. 3. It gives better finish and tool life but pulls the work into the cutter, so it needs a backlash-free (ball-screw) feed drive. 4. A soft grade, so worn grains are shed. 5. 3 mm.

Interview questions

All Engineering Materials and Manufacturing Processes interview questions

Try answering each one aloud before you open it.

  1. 1.What is turning in the context of machining processes?Concept

    Turning is a machining process where a cutting tool, typically a non-rotary tool bit, moves linearly while the workpiece rotates. This process is used to remove material from the outer diameter of a rotating cylindrical workpiece, creating a desired shape and size. It is commonly performed on a lathe.

  2. 2.Explain the milling process and its applications.Concept

    Milling removes material with a rotating multi-tooth cutter while the work, clamped on the table, is fed past it. Each tooth takes an intermittent chip, so the table feed is f_m = f_z·Z·N. Slab (peripheral) milling uses teeth on the cutter periphery and face milling uses teeth on the end face. It is used for flat faces, steps, slots, keyways, pockets, gear teeth and, on CNC machining centres, complex 3D contours.

  3. 3.Describe the drilling process and its primary purpose.Concept

    Drilling is a machining process used to create round holes in a workpiece. It involves the use of a drill bit, which is a rotary cutting tool, to cut into the material. The primary purpose of drilling is to create holes of various sizes and depths, which can be used for fastening, assembly, or further machining operations.

  4. 4.What is grinding, and how does it differ from other machining processes?Concept

    Grinding is a machining process that uses an abrasive wheel as the cutting tool to remove material from a workpiece. Unlike other machining processes that use sharp cutting tools, grinding relies on the abrasive action of the wheel to achieve a smooth surface finish and precise dimensions. It is often used for finishing operations and to achieve tight tolerances.

  5. 5.Why is turning preferred over milling for cylindrical parts?Application

    Turning is preferred over milling for cylindrical parts because it is specifically designed for machining round shapes. The rotating workpiece allows for efficient material removal along the outer diameter, resulting in high precision and smooth surface finishes. Additionally, turning can be more cost-effective for producing cylindrical parts due to its simplicity and speed.

  6. 6.What happens if the feed rate is too high during a milling operation?Application

    If the feed rate is too high during a milling operation, it can lead to several issues such as poor surface finish, increased tool wear, and potential damage to the workpiece. The excessive feed rate can cause the cutting tool to remove too much material too quickly, resulting in rough surfaces and increased heat generation, which can degrade the tool and affect the quality of the machined part.

  7. 7.How does the choice of grinding wheel affect the grinding process?Application

    The wheel specification – abrasive, grit size, grade, structure and bond – sets removal rate, finish and wheel wear. Al₂O₃ suits steels, SiC suits cast iron and non-ferrous metals, and cBN or diamond suit very hard materials. A coarse grit removes stock faster while a fine grit gives a better finish. For hard workpieces a soft-grade wheel is chosen so that dulled grains are released and fresh sharp grains are exposed (self-sharpening); soft work uses a harder grade, and an open structure helps when chips are large or heat must be limited.

  8. 8.Calculate the machining time for a turning operation with a cutting speed of 100 m/min, a workpiece diameter of 50 mm, and a length of 200 mm. Assume a feed rate of 0.2 mm/rev.Numerical

    To calculate the machining time, first determine the spindle speed (N) using the formula: N = (1000 × cutting speed) / (π × diameter). N = (1000 × 100) / (π × 50) ≈ 636.62 rev/min. Then, calculate the machining time (T) using: T = length / (feed rate × N). T = 200 / (0.2 × 636.62) ≈ 1.57 minutes.

  9. 9.A milling operation requires a surface finish of 0.8 µm. What factors should be considered to achieve this finish?Application

    To achieve a surface finish of 0.8 µm in a milling operation, factors such as cutting tool material and geometry, cutting speed, feed rate, and depth of cut should be considered. Using a sharp tool with the appropriate rake angle, selecting a suitable cutting speed, and maintaining a low feed rate can help achieve the desired finish. Additionally, ensuring machine stability and minimizing vibrations are crucial for achieving a smooth surface.

  10. 10.Determine the time required to drill a hole with a depth of 50 mm and a diameter of 10 mm using a drill bit with a feed rate of 0.1 mm/rev and a spindle speed of 800 rev/min.Numerical

    Feed velocity = f·N = 0.1 × 800 = 80 mm/min. Ignoring the tip allowance, t = 50/80 = 0.625 min (37.5 s). Strictly, the drill travels an extra (D/2)·cot(θ/2) ≈ 5 × 0.601 = 3.0 mm for a 118° point, so t ≈ 53/80 ≈ 0.66 min to obtain the full diameter over the whole depth or break through.

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