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Speeds and Feeds worksheet

Grades 6–7 · CCSS 7.G.B.4, 6.RP.A.3.d

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Units: Metric (m/min, mm)
- Workpiece material: Aluminum and its alloys: 250 SFM (HSS)
- Tool material: Carbide (× 2.5)
- Tool diameter (or workpiece diameter when turning): 25 mm
- Number of teeth (flutes): 6
- Chip load (feed per tooth): 0.03 mm/tooth
Find: Spindle speed
Answer
2.
Problem 2.
- Units: Metric (m/min, mm)
- Workpiece material: Stainless steel (free machining): 40 SFM (HSS)
- Tool material: High-speed steel
- Tool diameter (or workpiece diameter when turning): 25 mm
- Number of teeth (flutes): 3
- Chip load (feed per tooth): 0.05 mm/tooth
Find: Spindle speed
Answer
3.
Problem 3.
- Units: Inch (SFM, in)
- Workpiece material: Cast iron (medium hard): 80 SFM (HSS)
- Tool material: High-speed steel
- Tool diameter (or workpiece diameter when turning): 0.125 in
- Number of teeth (flutes): 3
- Chip load (feed per tooth): 0.001 in/tooth
Find: Spindle speed
Answer
4.
Problem 4.
- Units: Inch (SFM, in)
- Workpiece material: Stainless steel (free machining): 40 SFM (HSS)
- Tool material: High-speed steel
- Tool diameter (or workpiece diameter when turning): 0.25 in
- Number of teeth (flutes): 1
- Chip load (feed per tooth): 0.003 in/tooth
Find: Spindle speed
Answer
5.
Problem 5.
- Units: Metric (m/min, mm)
- Workpiece material: Stainless steel (work hardening): 20 SFM (HSS)
- Tool material: High-speed steel
- Tool diameter (or workpiece diameter when turning): 6 mm
- Number of teeth (flutes): 4
- Chip load (feed per tooth): 0.08 mm/tooth
Find: Spindle speed
Answer
6.
Problem 6.
- Units: Metric (m/min, mm)
- Workpiece material: Magnesium and its alloys: 300 SFM (HSS)
- Tool material: Carbide (× 2.5)
- Tool diameter (or workpiece diameter when turning): 10 mm
- Number of teeth (flutes): 4
- Chip load (feed per tooth): 0.15 mm/tooth
Find: Spindle speed
Answer
7.
Problem 7.
- Units: Inch (SFM, in)
- Workpiece material: Stainless steel (free machining): 40 SFM (HSS)
- Tool material: High-speed steel
- Tool diameter (or workpiece diameter when turning): 0.25 in
- Number of teeth (flutes): 3
- Chip load (feed per tooth): 0.002 in/tooth
Find: Spindle speed
Answer
8.
Problem 8.
- Units: Inch (SFM, in)
- Workpiece material: Stainless steel (free machining): 40 SFM (HSS)
- Tool material: Carbide (× 2.5)
- Tool diameter (or workpiece diameter when turning): 0.125 in
- Number of teeth (flutes): 2
- Chip load (feed per tooth): 0.001 in/tooth
Find: Spindle speed
Answer
9.
Problem 9.
- Units: Metric (m/min, mm)
- Workpiece material: Titanium alloys: 20 SFM (HSS)
- Tool material: High-speed steel
- Tool diameter (or workpiece diameter when turning): 25 mm
- Number of teeth (flutes): 4
- Chip load (feed per tooth): 0.05 mm/tooth
Find: Spindle speed
Answer
10.
Problem 10.
- Units: Inch (SFM, in)
- Workpiece material: Alloy steel (300–400 Brinell): 30 SFM (HSS)
- Tool material: Carbide (× 2.5)
- Tool diameter (or workpiece diameter when turning): 1 in
- Number of teeth (flutes): 4
- Chip load (feed per tooth): 0.008 in/tooth
Find: Spindle speed
Answer

Answer key: Speeds and Feeds worksheet

1. Spindle speed: 2,426 rpm
1. Carbide runs about 2.5 times the high-speed steel starting speed for aluminum and its alloys
  $250 SFM \times 2.5$ $= 625 SFM$
2. Convert to meters per minute, since you chose metric (1 ft = 0.3048 m)
  $625 SFM \times 0.3048$ $= 190.5 m/min$
3. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 1,000 turns meters into millimeters
  $\frac{1,000 \times 190.5}{π \times 25}$ $= 2,425.521333 rpm$
4. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $2,425.521333 \times 6 \times 0.03$ $= 436.59384 mm/min$
2. Spindle speed: 155 rpm
1. Starting speed for stainless steel (free machining) with a high-speed steel tool
  $V$ $= 40 SFM$
2. Convert to meters per minute, since you chose metric (1 ft = 0.3048 m)
  $40 SFM \times 0.3048$ $= 12.192 m/min$
3. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 1,000 turns meters into millimeters
  $\frac{1,000 \times 12.192}{π \times 25}$ $= 155.233365 rpm$
4. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $155.233365 \times 3 \times 0.05$ $= 23.285005 mm/min$
3. Spindle speed: 2,445 rpm
1. Starting speed for cast iron (medium hard) with a high-speed steel tool
  $V$ $= 80 SFM$
2. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 12 turns feet into inches
  $\frac{12 \times 80}{π \times 0.125}$ $= 2,444.619926 rpm$
3. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $2,444.619926 \times 3 \times 0.001$ $= 7.33386 in/min$
4. Spindle speed: 611 rpm
1. Starting speed for stainless steel (free machining) with a high-speed steel tool
  $V$ $= 40 SFM$
2. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 12 turns feet into inches
  $\frac{12 \times 40}{π \times 0.25}$ $= 611.154981 rpm$
3. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $611.154981 \times 1 \times 0.003$ $= 1.833465 in/min$
5. Spindle speed: 323 rpm
1. Starting speed for stainless steel (work hardening) with a high-speed steel tool
  $V$ $= 20 SFM$
2. Convert to meters per minute, since you chose metric (1 ft = 0.3048 m)
  $20 SFM \times 0.3048$ $= 6.096 m/min$
3. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 1,000 turns meters into millimeters
  $\frac{1,000 \times 6.096}{π \times 6}$ $= 323.402844 rpm$
4. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $323.402844 \times 4 \times 0.08$ $= 103.48891 mm/min$
6. Spindle speed: 7,277 rpm
1. Carbide runs about 2.5 times the high-speed steel starting speed for magnesium and its alloys
  $300 SFM \times 2.5$ $= 750 SFM$
2. Convert to meters per minute, since you chose metric (1 ft = 0.3048 m)
  $750 SFM \times 0.3048$ $= 228.6 m/min$
3. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 1,000 turns meters into millimeters
  $\frac{1,000 \times 228.6}{π \times 10}$ $= 7,276.563998 rpm$
4. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $7,276.563998 \times 4 \times 0.15$ $= 4,365.938399 mm/min$
7. Spindle speed: 611 rpm
1. Starting speed for stainless steel (free machining) with a high-speed steel tool
  $V$ $= 40 SFM$
2. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 12 turns feet into inches
  $\frac{12 \times 40}{π \times 0.25}$ $= 611.154981 rpm$
3. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $611.154981 \times 3 \times 0.002$ $= 3.66693 in/min$
8. Spindle speed: 3,056 rpm
1. Carbide runs about 2.5 times the high-speed steel starting speed for stainless steel (free machining)
  $40 SFM \times 2.5$ $= 100 SFM$
2. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 12 turns feet into inches
  $\frac{12 \times 100}{π \times 0.125}$ $= 3,055.774907 rpm$
3. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $3,055.774907 \times 2 \times 0.001$ $= 6.11155 in/min$
9. Spindle speed: 78 rpm
1. Starting speed for titanium alloys with a high-speed steel tool
  $V$ $= 20 SFM$
2. Convert to meters per minute, since you chose metric (1 ft = 0.3048 m)
  $20 SFM \times 0.3048$ $= 6.096 m/min$
3. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 1,000 turns meters into millimeters
  $\frac{1,000 \times 6.096}{π \times 25}$ $= 77.616683 rpm$
4. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $77.616683 \times 4 \times 0.05$ $= 15.523337 mm/min$
10. Spindle speed: 286 rpm
1. Carbide runs about 2.5 times the high-speed steel starting speed for alloy steel (300–400 brinell)
  $30 SFM \times 2.5$ $= 75 SFM$
2. Find the spindle speed: divide the cutting speed (V) by the distance around the tool (π × D). The 12 turns feet into inches
  $\frac{12 \times 75}{π \times 1}$ $= 286.478898 rpm$
3. Find the feed rate: each tooth takes one bite per turn, so multiply turns per minute by teeth and bite size
  $286.478898 \times 4 \times 0.008$ $= 9.167325 in/min$