Appearance
Material removal rate is the single number that determines whether a deep hole drilling operation is profitable or marginal. But maximising MRR is not a simple matter of increasing feed until something breaks — it must be balanced against tool life, chip evacuation capacity, spindle power, and the fundamental differences between gun drilling and BTA.
MRR Fundamentals
The Drilling MRR Formula
For any drilling operation, the material removal rate is:
Q = (vf × π × D²) / 4,000
Where:
- Q = material removal rate (cm³/min)
- vf = feed rate (mm/min) = fn × n
- D = drill diameter (mm)
- fn = feed per revolution (mm/rev)
- n = spindle speed (RPM)
Since n = (Vc × 1,000) / (π × D), the formula simplifies to:
Q = (fn × D × Vc) / 4
Where Vc = cutting speed (m/min).
Imperial version:
Q = (vf × π × D²) / 4
Where Q = in³/min, D = inches, vf = in/min.
Key Insight
The simplified formula Q = (fn × D × Vc) / 4 shows that MRR scales linearly with feed, diameter, and cutting speed. This means:
- Doubling feed doubles MRR
- Doubling speed doubles MRR
- Doubling diameter doubles MRR (at the same feed and speed)
However, each variable has practical limits imposed by the process and tooling, and these limits are different for gun drilling vs BTA.
MRR Calculation Examples
BTA Drilling
| Parameter | Value |
|---|---|
| Diameter | 40 mm |
| Material | Alloy steel (200 HB) |
| Cutting speed | 80 m/min |
| Feed per rev | 0.18 mm/rev |
| Spindle speed | n = (80 × 1,000) / (π × 40) = 637 rpm |
| Feed rate | vf = 0.18 × 637 = 115 mm/min |
| MRR | Q = (115 × π × 40²) / 4,000 = 144 cm³/min |
Alternatively: Q = (0.18 × 40 × 80) / 4 = 144 cm³/min.
Gun Drilling
| Parameter | Value |
|---|---|
| Diameter | 10 mm |
| Material | Alloy steel (200 HB) |
| Cutting speed | 80 m/min |
| Feed per rev | 0.03 mm/rev |
| Spindle speed | n = (80 × 1,000) / (π × 10) = 2,546 rpm |
| Feed rate | vf = 0.03 × 2,546 = 76 mm/min |
| MRR | Q = (76 × π × 10²) / 4,000 = 6.0 cm³/min |
Alternatively: Q = (0.03 × 10 × 80) / 4 = 6.0 cm³/min.
Tip: Notice that the 40 mm BTA operation removes 144 cm³/min of material while the 10 mm gun drill removes 6 cm³/min — a 24× difference. However, the BTA head uses multiple cutting edges while the gun drill uses one. The material removal per cutting edge tells a different story about tool load.
Cutting Speed Selection
Recommended Cutting Speeds
| Material | Cutting Speed Range (m/min) | Optimal for Tool Life |
|---|---|---|
| Brass | 80–150 | 100–120 |
| Aluminium | 80–160 | 120–140 |
| Carbon steel (< 150 HB) | 80–120 | 90–100 |
| Alloy steel (150–250 HB) | 70–100 | 75–85 |
| Alloy steel (250–350 HB) | 50–80 | 60–70 |
| Grey cast iron | 70–100 | 80–90 |
| Ductile cast iron | 60–90 | 70–80 |
| Stainless steel (austenitic) | 30–60 | 40–50 |
| Titanium alloy | 20–40 | 25–35 |
| Aluminium (wrought) | 100–200 | 120–150 |
Speed Effect on MRR
At a fixed feed and diameter, MRR is directly proportional to cutting speed. Increasing Vc from 70 to 100 m/min (43% increase) increases MRR by 43% — but reduces tool life by approximately 50–60% in most materials. The optimum cutting speed for maximum productivity is not the maximum speed; it is the speed that balances MRR against tool change frequency.
Feed Rate Selection
Feed by Diameter and Material
| Diameter (mm) | Carbon/Alloy Steel | Cast Iron | Aluminium | Brass |
|---|---|---|---|---|
| 3 | 0.007–0.013 | 0.009–0.038 | 0.006–0.037 | 0.006–0.030 |
| 5 | 0.016–0.026 | 0.018–0.068 | 0.010–0.109 | 0.010–0.069 |
| 10 | 0.030–0.049 | 0.050–0.120 | 0.025–0.174 | 0.020–0.139 |
| 15 | 0.042–0.071 | 0.070–0.154 | 0.035–0.194 | 0.028–0.179 |
| 20 | 0.060–0.107 | 0.106–0.207 | 0.060–0.254 | 0.040–0.249 |
| 25 | 0.069–0.117 | 0.120–0.221 | 0.072–0.295 | 0.048–0.291 |
| 30 | 0.079–0.134 | 0.140–0.237 | 0.084–0.360 | 0.056–0.327 |
| 40 | 0.091–0.169 | 0.180–0.254 | 0.105–0.488 | 0.072–0.399 |
| 50 | 0.091–0.169 | 0.180–0.254 | 0.105–0.488 | 0.072–0.399 |
Feed Limits by Process
| Process | Typical Feed Range | Limiting Factor |
|---|---|---|
| Gun drilling (single-lip) | 0.01–0.05 mm/rev | Chip evacuation through V-groove |
| BTA drilling (indexable) | 0.08–0.25 mm/rev | Chip breaking and evacuation through tube |
| BTA drilling (brazed) | 0.06–0.18 mm/rev | Heat generation at cutting edge |
| Ejector drilling | 0.08–0.20 mm/rev | Chip evacuation through inner tube |
Warning: The feed ranges in the table above are for stable, continuous cutting. Reduce feed by 20–30% when: drilling through cross-holes or interrupted cuts, operating near the maximum L/D ratio for the tool, using a machine with less than recommended power, or drilling materials at the upper end of their hardness range.
Comparing MRR: BTA vs Gun Drilling
Direct Comparison at Equal Diameter
For a 20 mm hole in alloy steel:
| Parameter | Gun Drill | BTA Drill | Ratio |
|---|---|---|---|
| Cutting speed (m/min) | 80 | 80 | 1:1 |
| Feed (mm/rev) | 0.04 | 0.12 | 1:3 |
| MRR (cm³/min) | 16 | 48 | 1:3 |
| Cutting edges | 1 | 3–4 | — |
| MRR per edge (cm³/min) | 16 | 12–16 | ~1:1 |
The apparent 3:1 MRR advantage of BTA over gun drilling at the same diameter is explained by the additional cutting edges. The MRR per cutting edge is similar — but BTA's multiple inserts are arranged around a rigid round tube, while the gun drill's single edge is limited by the V-groove chip evacuation constraint.
Practical Application Range
| Diameter Range | Preferred Process | Typical MRR Range |
|---|---|---|
| 3–10 mm | Gun drilling | 1–20 cm³/min |
| 10–20 mm | Both (MRR favours BTA) | 5–80 cm³/min |
| 20–50 mm | BTA drilling | 30–300 cm³/min |
| 50–100 mm | BTA drilling | 100–800 cm³/min |
| 100–200 mm | BTA drilling | 300–2,000 cm³/min |
Optimising MRR
The MRR Optimisation Equation
The objective is not maximum MRR — it is maximum productive MRR, defined as:
Productive MRR = Q × (tool life) / (tool change time + cycle time)
Increasing Q reduces tool life, which increases tool change frequency. The optimum is where the marginal gain in MRR equals the marginal cost of additional tool changes.
Constraints on MRR Increase
| Constraint | Symptom of Exceeding Limit | Mitigation |
|---|---|---|
| Spindle power | Spindle stalls or trips overload | Reduce feed or speed; verify adequate motor power |
| Chip evacuation | Chips pack in drill tube or flute | Reduce feed; increase coolant pressure or flow |
| Tool life | Edge breaks before completing hole | Reduce speed (affects tool life exponentially) |
| Machine stability | Chatter, vibration marks on bore | Reduce feed; adjust speed to avoid resonance |
| Coolant system | Coolant temperature exceeds 50°C | Reduce MRR; increase coolant tank capacity |
| Guide pad wear | Bore taper exceeds tolerance | Reduce feed; check pad condition |
Practical Optimisation Steps
- Start at recommended parameters for the material and diameter combination
- Increase feed first (feed affects tool life less than speed does)
- Increase speed second (speed has an exponential effect on tool life)
- Monitor chip form — chips should be C-type or small broken segments
- Monitor spindle power — if power exceeds 80% of rated, reduce feed
- Monitor surface finish — if finish degrades, check tool wear before adjusting parameters
- Document the optimum — the best parameters for each material–diameter combination
Tip: In deep hole drilling, feed rate should be the first variable adjusted when optimising MRR. Increasing feed increases chip thickness, which improves chip breaking and has a relatively mild effect on tool life. Increasing speed increases MRR but reduces tool life by an exponent of approximately 3–5 (depending on the material). As a rule: raise feed to the chip-breaking limit first, then adjust speed to achieve the target tool life.
MRR in Multi-Insert BTA Heads
Load Distribution
In a typical three-insert BTA head, the MRR is distributed unevenly:
| Insert Position | Share of Total MRR | Cutting Speed |
|---|---|---|
| Peripheral (outer) | 50–60% | 100% of Vc |
| Intermediate | 25–30% | 60–80% of Vc |
| Centre (inner) | 15–20% | 10–30% of Vc |
The peripheral insert does the majority of the work and operates at the highest cutting speed. It is typically the first insert to require replacement. The centre insert, operating nearest to zero radius, has the lowest chip load and the lowest speed.
Per-Insert MRR Calculation
For a 40 mm BTA head with three inserts:
- Total MRR = 144 cm³/min (from earlier example)
- Peripheral insert: ~75 cm³/min
- Intermediate insert: ~40 cm³/min
- Centre insert: ~29 cm³/min
This distribution explains why different chipbreaker types are often used at different positions — the peripheral insert needs aggressive chip breaking at high chip load, while the centre insert needs a geometry that can break chips at low feed per tooth.
MRR and Power Relationship
Direct Calculation
The relationship between MRR and spindle power:
Pc = (Q × kc) / (60,000 × η)
Where Pc = net power (kW), kc = specific cutting force (N/mm²), η = efficiency.
Example: For the 40 mm BTA operation at 144 cm³/min in alloy steel (kc = 2,500 N/mm²):
Pc = (144 × 2,500) / (60,000 × 0.8) = 7.5 kW
This is the net cutting power. The machine's rated motor power should be 20–30% higher (~10 kW minimum).
Maximum MRR from Available Power
If machine power is the limiting factor:
Qmax = (Pc_available × 60,000 × η) / kc
Example: Machine with 50 kW spindle, kc = 2,500:
Qmax = (50 × 60,000 × 0.8) / 2,500 = 960 cm³/min
At a feed of 0.18 mm/rev and Vc = 80 m/min, this corresponds to a maximum diameter of:
D = (Qmax × 4) / (fn × Vc) = (960 × 4) / (0.18 × 80) = 267 mm
Coolant and MRR Relationship
Coolant system capacity also limits MRR. The coolant must remove the heat generated by material removal:
| Heat Source | Percentage of Total Heat | Removed By |
|---|---|---|
| Chip formation (shear zone) | ~75% | Coolant + chips |
| Friction (tool-chip interface) | ~15% | Coolant |
| Friction (guide pad-bore) | ~10% | Coolant |
Higher MRR generates proportionally more heat. The coolant flow must be sufficient to maintain the cutting zone below the temperature limit for the tool material and workpiece.
The standard coolant flow formula for BTA:
Qcoolant = 4.5 × D (L/min)
For a 40 mm hole: Qcoolant = 180 L/min. At MRR = 144 cm³/min, this provides approximately 1.25 L of coolant per cm³ of material removed — sufficient for steel but potentially marginal for titanium, which requires more coolant per unit of material removed.
FAQ
What is the material removal rate formula for deep hole drilling?
Q = (vf × π × D²) / 4,000 (metric, cm³/min) which simplifies to Q = (fn × D × Vc) / 4. For imperial: Q = (vf × π × D²) / 4 (in³/min), where D is in inches and vf is in/min.
How much higher is BTA MRR compared to gun drilling?
BTA achieves 3–7× higher MRR than gun drilling at equivalent diameters. This is due to multiple cutting edges and the rigid round-section drill tube that permits higher feed rates. The MRR per cutting edge, however, is similar between the two processes.
What feed rate should I use for deep hole drilling in steel?
For carbon and alloy steels in gun drilling (3–20 mm): 0.007–0.10 mm/rev depending on diameter. For BTA drilling in steel (20–50 mm): 0.06–0.17 mm/rev. Use the lower end of the range for harder materials and higher L/D ratios.
What cutting speed is recommended for BTA drilling in alloy steel?
70–100 m/min for alloy steel at 150–250 HB. Use the lower end (70–80 m/min) when tool life is the priority, the upper end (90–100 m/min) when MRR is the priority. For hardened alloy steel above 250 HB, reduce to 50–80 m/min.
What limits MRR in BTA drilling?
The four primary constraints are: spindle power (Pc = Q × kc / 60,000), chip evacuation capacity (limited by drill tube internal diameter), tool life (cutting edge temperature increases with MRR), and coolant system capacity (both pressure and flow). The weakest link determines the practical MRR limit.
How does diameter affect MRR?
MRR scales linearly with diameter at constant feed and speed. A 40 mm drill removes 2× the material of a 20 mm drill at the same feed and speed. However, power scales with D² and torque scales with D² × fn — so the machine requirements increase faster than MRR.
What is the MRR per cutting edge in BTA vs gun drilling?
In a typical BTA head with three inserts, MRR per edge is approximately 12–16 cm³/min per cutting edge in alloy steel at 20 mm diameter. For a gun drill in the same material and diameter, MRR per edge is approximately 16 cm³/min — similar, but spread across one edge instead of three.
Should I increase feed or speed to improve MRR?
Increase feed first. Feed has a milder effect on tool life than speed does. Speed affects tool life exponentially (typically n^3 to n^5 depending on material), while feed affects tool life approximately linearly. Raise feed to the chip-breaking limit, then adjust speed for target tool life.
How do I calculate MRR from available spindle power?
Use Qmax = (Pc × 60,000 × η) / kc, where Pc is available spindle power (kW), η is machine efficiency (0.7–0.85), and kc is specific cutting force (2,000–3,000 N/mm² for steels). Then back-calculate achievable feed and speed from the diameter.
What MRR can I expect for 40 mm BTA drilling in steel?
At Vc = 80 m/min and fn = 0.18 mm/rev: approximately 144 cm³/min. At more conservative parameters (Vc = 70 m/min, fn = 0.12 mm/rev): approximately 84 cm³/min. The actual practical MRR depends on machine power, coolant capacity, and tool life requirements.
Conclusion
Material removal rate is the productivity metric for deep hole drilling, calculated simply as Q = (fn × D × Vc) / 4 but constrained by spindle power, chip evacuation, tool life, and coolant capacity. BTA drilling achieves 3–7× higher MRR than gun drilling at equivalent diameters, driven by multi-insert design and rigid round-section drill tubes. The optimisation strategy is consistent across both processes: increase feed to the chip-breaking limit first (milder effect on tool life), then adjust cutting speed to achieve the target balance between MRR and tool change frequency. The specific cutting force of the workpiece material (kc) links MRR to power consumption and is the essential input for determining whether a machine has the capacity to deliver the target MRR for a given diameter and material combination.