Deep Hole Drilling Speed, Feed, Power, Cycle Time Calculator

Estimating cutting parameters for deep hole drilling requires accounting for factors that general drilling formulas ignore — depth-dependent pressure drops, chip evacuation constraints, and tool deflection. The formulas in this article are adapted from industry standards (Sandvik Coromant, Guhring, ISCAR) with corrections specific to deep hole drilling operations.

This article provides formulas and step-by-step calculation methods for the key parameters in deep hole drilling: cutting speed, feed rate, power, torque, thrust force, and cycle time. Each formula includes correction factors for depth, coolant conditions, and machine rigidity.

Cutting Speed and Spindle Speed ​

Surface Cutting Speed ​

Formula:

Metric: Vc = (π × D × n) / 1000
Imperial: SFM = (π × D × RPM) / 12

Where:

• Vc = cutting speed (m/min)
• SFM = cutting speed (surface feet per minute)
• D = drill diameter (mm for metric, inches for imperial)
• n / RPM = spindle speed (rev/min)

Spindle Speed from Cutting Speed ​

Formula:

Metric: n = (Vc × 1000) / (π × D)
Imperial: RPM = (SFM × 3.82) / D

Quick constants:

• Metric: 1000 / π = 318.3, so n = (Vc × 318.3) / D
• Imperial: 3.82 / π = 1.22, so RPM = SFM / (D × 0.262)

Depth correction: For L/D ratios above 10:1, multiply calculated RPM by a depth factor (0.9 at 10:1, 0.8 at 20:1, 0.7 at 50:1, 0.6 at 100:1). The reduced speed compensates for increased friction and reduced coolant effectiveness at depth.

Calculation Example ​

Problem: Calculate RPM for a 12 mm gun drill in low-carbon steel (Vc = 180 m/min), depth 240 mm.

Step 1 — Base RPM: n = (180 × 1000) / (π × 12) = 180,000 / 37.70 = 4,775 RPM

Step 2 — Depth correction: L/D = 240 / 12 = 20:1. Depth factor = 0.8. Adjusted n = 4,775 × 0.8 = 3,820 RPM

Result: Start at 3,800 RPM.

Feed Rate Formulas ​

Feed per Revolution ​

Feed per revolution (fn or f) is the axial advance of the tool per spindle revolution.

Formula for feed rate (linear):

Metric: Vf = fn × n (mm/min)
Imperial: IPM = IPR × RPM

Where:

• Vf / IPM = linear feed rate
• fn / IPR = feed per revolution
• n / RPM = spindle speed

Feed per Revolution by Method ​

Gun drilling — by insert size:

fn (mm/rev) = insert size code × 0.01 (typical starting value)

Example: Insert size 10 → 0.10 mm/rev starting feed.

BTA drilling — by diameter:

fn (mm/rev) = 0.002 × D + 0.04 (for D in mm, approximate starting value)

Example: 50 mm BTA → 0.002 × 50 + 0.04 = 0.14 mm/rev starting feed.

Material-Specific Feed Factors ​

Calculation Example ​

Problem: Calculate feed rate for a 12 mm gun drill at 3,800 RPM with 0.10 mm/rev feed.

Vf = 0.10 × 3,800 = 380 mm/min

Result: Feed rate = 380 mm/min.

Material Removal Rate ​

Formula ​

Metric: MRR = (D × fn × Vc) / 4 (cm³/min)
Imperial: MRR = (D × IPR × SFM × 12) / π (in³/min)

Where MRR is material removal rate and D is drill diameter.

Simplified imperial:

MRR (in³/min) = 0.26 × D × IPR × SFM

Typical MRR Values ​

Power Consumption ​

Cutting Power ​

Formula:

Metric: Pc = (MRR × Kc) / 60 (kW)
Imperial: HP = (MRR × K) / 396,000

Where:

• Pc = cutting power (kW)
• HP = cutting power (horsepower)
• MRR = material removal rate (cm³/min metric, in³/min imperial)
• Kc = specific cutting force (N/mm²)
• K = specific cutting force (psi)

Specific Cutting Force Values ​

Total Power Required ​

Ptotal = Pc / η

Where η is machine efficiency (typically 0.75–0.90 for deep hole drilling machines).

Machine efficiency values:

• New, well-maintained machine: η = 0.85–0.90
• Average machine condition: η = 0.75–0.85
• Older machine with wear: η = 0.65–0.75

Power Calculation Example ​

Problem: Calculate power required for BTA drilling a 50 mm hole in medium-carbon steel at Vc = 120 m/min, fn = 0.14 mm/rev. Machine efficiency = 0.80.

Step 1 — Material removal rate: MRR = (50 × 0.14 × 120) / 4 = 210 cm³/min

Step 2 — Cutting power: Kc = 2,200 N/mm² (medium-carbon steel) Pc = (210 × 2,200) / 60 = 7,700 W = 7.7 kW

Step 3 — Total power: Ptotal = 7.7 / 0.80 = 9.6 kW

Result: A minimum 10 kW (13.4 HP) spindle motor is required.

Quick Power Rules of Thumb ​

Torque and Thrust Force ​

Torque ​

Formula:

Metric: Mc = (Kc × D² × fn) / 8,000 (N·m)
Imperial: T = (K × D² × IPR) / 8,000 (lbf·ft)

Where D is drill diameter.

Alternative from power:

Metric: Mc = (Pc × 9,550) / n (N·m)
Imperial: T = (HP × 5,252) / RPM (lbf·ft)

Thrust Force ​

Formula (approximate):

Metric: Ff = Kc × D × fn × 0.7 (N)
Imperial: Ff = K × D × IPR × 0.7 (lbf)

Typical Torque and Thrust Values ​

Note: These are approximate ranges. Actual values depend on material, feed rate, tool geometry, and coolant conditions.

Torque Calculation Example ​

Problem: Estimate torque for BTA drilling a 50 mm hole in medium-carbon steel at fn = 0.14 mm/rev.

Mc = (2,200 × 50² × 0.14) / 8,000 = (2,200 × 2,500 × 0.14) / 8,000 Mc = 770,000 / 8,000 = 96 N·m

Result: Approximately 96 N·m cutting torque.

Cycle Time Estimation ​

Drilling Time Formula ​

Metric: Tc = L / (fn × n) (min)
Imperial: Tc = L / (IPR × RPM) (min)

Where L is the total axial travel (hole depth + approach + overrun).

Approach and Overrun ​

Approach formula:

A = (D / 2) × tan(90° − point angle / 2)

Or for common angles:

A (120° point) = 0.289 × D
A (118° point) = 0.300 × D
A (90° point) = 0.500 × D

Total Cycle Time ​

Ttotal = Tc + Taux

Where Taux includes:

• Tool change time (if multiple tools): 0.5–3 min per change
• Part loading/unloading: 0.5–5 min
• Indexing time (multi-diameter): 0.1–0.5 min per index
• Coolant on/off delay: 0.1–0.2 min

Cycle Time Example ​

Problem: Calculate cycle time for gun drilling a 10 mm hole, 300 mm deep, in low-carbon steel. Point angle = 120°. Parameters: n = 3,800 RPM, fn = 0.10 mm/rev.

Step 1 — Approach distance: A = 0.289 × 10 = 2.89 mm

Step 2 — Total travel: Ltotal = 300 + 2.89 = 303 mm (assuming through-hole)

Step 3 — Drilling time: Tc = 303 / (0.10 × 3,800) = 303 / 380 = 0.80 min

Step 4 — Total cycle time (with 1.5 min auxiliary): Ttotal = 0.80 + 1.50 = 2.30 min

Result: Cycle time ≈ 2.3 min per hole.

Worked Examples ​

Example 1: Complete Gun Drilling Calculation ​

Given:

• Material: Low-carbon steel (Kc = 1,800 N/mm²)
• Drill diameter: 8 mm
• Hole depth: 200 mm (L/D = 25:1)
• Insert size: 8
• Machine efficiency: 0.80

Step 1 — Cutting speed: Base Vc = 180 m/min. Depth factor at 25:1 = 0.75. Adjusted Vc = 180 × 0.75 = 135 m/min

Step 2 — Spindle speed: n = (135 × 1,000) / (π × 8) = 5,370 RPM

Step 3 — Feed: Base fn = 0.10 mm/rev (insert size 8). Depth factor = 0.75. Adjusted fn = 0.10 × 0.75 = 0.075 mm/rev

Step 4 — Feed rate: Vf = 0.075 × 5,370 = 403 mm/min

Step 5 — MRR: MRR = (8 × 0.075 × 135) / 4 = 20.25 cm³/min

Step 6 — Power: Pc = (20.25 × 1,800) / 60 = 608 W = 0.61 kW Ptotal = 0.61 / 0.80 = 0.76 kW

Step 7 — Torque: Mc = (1,800 × 8² × 0.075) / 8,000 = 1.08 N·m

Step 8 — Cycle time: A = 0.289 × 8 = 2.3 mm Ltotal = 200 + 2.3 = 202.3 mm Tc = 202.3 / 403 = 0.50 min Ttotal = 0.50 + 1.5 = 2.0 min

Results summary:

• Speed: 5,370 RPM
• Feed: 0.075 mm/rev (403 mm/min)
• Power: 0.76 kW required
• Torque: 1.1 N·m
• Cycle time: 2.0 min per hole

Example 2: Complete BTA Drilling Calculation ​

Given:

• Material: Medium-carbon steel (Kc = 2,200 N/mm²)
• Drill diameter: 60 mm
• Hole depth: 900 mm (L/D = 15:1)
• Machine efficiency: 0.85

Step 1 — Cutting speed: Base Vc = 130 m/min. Depth factor at 15:1 = 0.85. Adjusted Vc = 130 × 0.85 = 110 m/min

Step 2 — Spindle speed: n = (110 × 1,000) / (π × 60) = 583 RPM

Step 3 — Feed: Base fn = 0.002 × 60 + 0.04 = 0.16 mm/rev. Depth factor = 0.85. Adjusted fn = 0.16 × 0.85 = 0.136 mm/rev

Step 4 — Feed rate: Vf = 0.136 × 583 = 79 mm/min

Step 5 — MRR: MRR = (60 × 0.136 × 110) / 4 = 224 cm³/min

Step 6 — Power: Pc = (224 × 2,200) / 60 = 8,213 W = 8.2 kW Ptotal = 8.2 / 0.85 = 9.6 kW

Step 7 — Torque: Mc = (2,200 × 60² × 0.136) / 8,000 = 134.6 N·m

Step 8 — Cycle time: A = 0.289 × 60 = 17.3 mm Ltotal = 900 + 17.3 = 917.3 mm Tc = 917.3 / 79 = 11.6 min Ttotal = 11.6 + 3.0 = 14.6 min

Results summary:

• Speed: 583 RPM
• Feed: 0.136 mm/rev (79 mm/min)
• Power: 9.6 kW required
• Torque: 135 N·m
• Cycle time: 14.6 min per hole

Quick-Reference Formula Sheet ​

Mobile Calculator Tips ​

For quick calculations on the shop floor:

Spindle speed (mental math): Divide 318,000 by drill diameter in mm, then multiply by Vc in m/min, and divide by 1,000. Example: D = 12 mm, Vc = 180 → (318,000 / 12) × 180 / 1,000 = 26,500 × 0.18 = 4,770 RPM

Feed rate (mental math): Multiply RPM by feed per revolution, then divide by 1,000 to get m/min. Example: 3,800 RPM × 0.10 mm/rev = 380 mm/min = 0.38 m/min

Cycle time (mental math): Double the depth in meters, double again, then divide by feed rate in m/min. Example: 0.2 m depth, 0.38 m/min → 0.2 / 0.38 × 60 = 31.6 seconds

Note: For precise cycle time estimates, always use a calculator or spreadsheet. Mental math is useful for ballpark estimates (±20%) but not for production quoting.

Summary Table ​

FAQ ​

How do I calculate the spindle speed for a deep hole drilling operation? ​

Use the formula n = (Vc × 1000) / (π × D) where Vc is cutting speed in m/min from the material table and D is drill diameter in mm. Then apply a depth correction factor based on the L/D ratio. For example, a 10 mm gun drill in low-carbon steel (Vc = 180 m/min) at L/D = 20:1: base RPM = (180 × 1000) / (π × 10) = 5,730 RPM, corrected with depth factor 0.8 = 4,584 RPM final.

What is the formula for power consumption in deep hole drilling? ​

Cutting power is Pc = (MRR × Kc) / 60, where MRR is material removal rate in cm³/min and Kc is specific cutting force in N/mm². MRR = (D × fn × Vc) / 4 for drilling. So the complete formula becomes Pc = (D × fn × Vc × Kc) / 240 in kW. For example, BTA drilling a 60 mm hole at 0.14 mm/rev and 110 m/min in steel (Kc = 2,200): Pc = (60 × 0.14 × 110 × 2,200) / 240 = 8.2 kW.

How do I estimate cycle time for a deep hole drilling operation? ​

Cycle time is calculated as Tc = L / (fn × n), where L is total axial travel (hole depth + approach distance). For a 120° point drill, approach = 0.289 × D. Add auxiliary time (part loading, coolant on/off, tool changes) for total cycle time. For example, a 300 mm deep hole with a 10 mm drill at 0.10 mm/rev and 3,800 RPM: drilling time = 303 / 380 = 0.80 min, plus 1.5 min auxiliary = 2.3 min total.

How much power do I need for BTA drilling a 100 mm hole in steel? ​

Using the power rule of thumb: BTA drilling in steel requires approximately 0.20–0.40 kW per mm of diameter. For 100 mm: 20–40 kW. Using the formula with typical parameters (Vc = 100 m/min, fn = 0.20 mm/rev, Kc = 2,200): MRR = (100 × 0.20 × 100) / 4 = 500 cm³/min, Pc = (500 × 2,200) / 60 = 18.3 kW, Ptotal = 18.3 / 0.85 = 21.5 kW. So a 22–25 kW (30–35 HP) spindle motor is appropriate.

What is the relationship between feed rate and surface finish in deep hole drilling? ​

Surface finish in deep hole drilling is primarily affected by feed rate, tool geometry, and vibration. As a general rule, surface roughness Ra increases with feed rate: Ra ≈ fn² / (32 × re), where fn is feed per revolution and re is the corner radius of the cutting edge. However, in deep hole drilling, coolant conditions and chip evacuation often have a larger effect on surface finish than the theoretical Ra calculation. A feed rate that produces reliable chip evacuation should be selected first, then surface finish optimized through tool geometry and cutting speed adjustment.

How do I calculate torque for a deep hole drilling operation? ​

Use the formula Mc = (Kc × D² × fn) / 8,000, where Kc is specific cutting force in N/mm², D is drill diameter in mm, and fn is feed per revolution in mm/rev. For example, BTA drilling a 60 mm hole at 0.136 mm/rev in steel (Kc = 2,200): Mc = (2,200 × 60² × 0.136) / 8,000 = 134.6 N·m. Verify torque against the machine spindle's rated capacity, especially at low RPM where maximum torque is typically available.