18 Turning & Milling Setup Practice Questions & Answers

1. 1. You are drilling a .750 in. diameter hole in the end of a 3.50 in. diameter bar gripped in a lathe chuck. The recommended cutting speed for the material is 100 SFM. Using the NIMS turning formula RPM = (CS x 3.82) / D, what spindle speed should you set?

   • A.509 RPM✓ Answer
   • B.109 RPM, because the 3.50 in. bar is the diameter that is actually revolving in the chuck
   • C.533 RPM
   • D.382 RPM, taken as 3.82 x 100 divided by a 1.00 in. nominal tool size

   In the RPM formula, D is the diameter where the cut actually occurs, not the diameter of the stock, so drilling on a lathe uses the drill's diameter: (100 x 3.82) / .750 = 509 RPM. The 3.50 stock figure is a planted distractor, and 533 RPM comes from the 4 x CS / D constant used on the milling side rather than the 3.82 keyed for turning.

   Source: NIMS Machining Level I Preparation Guide, Turning — Turning Speeds and Feeds (RPM = CS x 3.82 / D; D = diameter being cut)Report a problem with this question

2. 2. A lathe is running at 300 RPM, the feed is set to .010, and the cut is 3.000 in. long. How long will the cut take?

   • A.About 300 minutes, because a feed of .010 means the tool travels .010 in. per minute no matter what the spindle speed is
   • B.About 1 minute, because lathe feed is stated in inches per revolution✓ Answer
   • C.About 3 minutes
   • D.About 0.3 minute

   Cut time is T = L / (f x N), and on a lathe f is inches per revolution, so T = 3.000 / (.010 x 300) = 1 minute. This is the contrast NIMS tests: lathe feed is per revolution while milling feed is inches per minute, and treating .010 as IPM gives the 300-minute trap answer.

   Source: NIMS Machining Level I Preparation Guide, Turning — Turning Speeds and Feeds (feed in IPR; cut time T = L / (f x N))Report a problem with this question

3. 3. A 3.00 in. diameter x 25 in. long shaft must have a hole drilled and bored in one end on the lathe. Which setup is correct?

   • A.Mount the shaft between centers with a dog and drive plate, then feed the drill in from the tailstock
   • B.Grip one end in a three-jaw chuck and carry the overhang on a follower rest bolted to the carriage so the support travels along with the cutting tool
   • C.Grip one end in a three-jaw chuck and support the shaft near the free end with a steady rest bolted to the bed✓ Answer
   • D.Grip the shaft in a four-jaw chuck with the full 25 in. hanging out and take very light cuts so that deflection stays inside the tolerance

   A steady rest bolts to the bed and stays in one place, its soft brass or bronze jaws set finger-tight around the rotating shaft so the end being machined runs true without whipping. A follower rest bolts to the carriage and travels with the tool, which supports a long slender shaft while turning but leaves the end unsupported for drilling; and between centers is impossible here because the tailstock center occupies the very end you must drill.

   Source: NIMS Machining Level I Preparation Guide, Turning — Work Holding Devices and Basic Setup (steady rest mounts to the bed with soft jaws; follower rest mounts to the carriage)Report a problem with this question

4. 4. Which bar cannot be gripped and centered by a three-jaw universal scroll chuck?

   • A.A 1.000 in. round bar
   • B.A 1.000 in. across-flats hexagon bar
   • C.A round bar that already has a .750 in. diameter step turned on one end
   • D.A 1.000 in. across-flats octagon bar✓ Answer

   All three jaws of a universal chuck are driven by one scroll plate and sit 120 degrees apart, so only shapes whose number of sides divides evenly by three - round, triangular, hexagonal - seat on all three jaws at once. An octagon has eight sides and a square has four, so both require a four-jaw independent chuck where each jaw is set separately.

   Source: NIMS Machining Level I Preparation Guide, Turning — Work Holding Devices and Basic Setup (three-jaw scroll chuck holds shapes divisible by three)Report a problem with this question

5. 5. What is the correct way to true a rough square blank in a four-jaw independent chuck?

   • A.Set the jaws roughly on the concentric rings scribed in the chuck face, then sweep the work with a dial indicator✓ Answer
   • B.Tighten the four jaws in a criss-cross pattern and let the scroll pull the blank to center the way it does in a three-jaw chuck
   • C.Hold a piece of chalk against the revolving blank and keep tightening whichever jaw is nearest the chalk mark until the mark disappears
   • D.Set each jaw by measuring out from the chuck body with a steel rule, which is close enough because the blank will be faced and turned anyway

   A four-jaw independent chuck has no scroll: each jaw is driven by its own screw, which is exactly why it holds square and irregular work but must be trued deliberately. The scribed rings give rough centering quickly, and final concentricity comes from a dial indicator, loosening one jaw while tightening the one opposite it so the work shifts without ever losing its grip.

   Source: NIMS Machining Level I Preparation Guide, Turning — Work Holding Devices and Basic Setup (four-jaw independent chuck: rings for rough setting, dial indicator for final truing)Report a problem with this question

6. 6. The Between Centers turning project requires the part to be turned end for end during machining. What makes that setup hold the coaxiality called out on the print?

   • A.The lathe dog grips the work tightly enough that nothing can shift while the cut is being taken
   • B.The part always rotates on the same two 60 degree center holes, so reversing it does not change the axis✓ Answer
   • C.The tailstock center revolves with the work, so it cannot introduce any runout of its own
   • D.The compound rest is set parallel to the centerline, which forces every diameter to come out concentric

   Between centers, the axis of rotation is defined by the two center holes in the part rather than by chuck jaws, so the identical axis is re-established every time the part is remounted. That is why NIMS names swapping the part end for end without losing concentricity as the advantage of between-centers work, and why chucking projects instead demand deliberate second-op indicating.

   Source: NIMS Machining Level I Standards, Duty 2.3 Between Centers Turning (part must be turned end for end; coaxiality .002 TIR)Report a problem with this question

7. 7. Trial cuts on a shaft turned between centers measure 2.240 in. at the headstock end and 2.220 in. at the tailstock end. What correction is required?

   • A.Move the tailstock .020 in. away from the operator
   • B.Move the tailstock .010 in. toward the operator
   • C.Move the tailstock .010 in. away from the operator✓ Answer
   • D.Move the tailstock .020 in. toward the operator

   An offset tailstock tilts the work axis while the tool still travels parallel to the ways, and because the tilt changes the radius on one side the diameter changes by twice the offset - so the correction is always half the measured difference, here .020 / 2 = .010. The tailstock end is the small end, meaning it is currently offset toward the operator and feeding into the tool, so the tailstock must be moved away from the operator, the direction that makes the tailstock end larger.

   Source: NIMS Machining Level I Preparation Guide, Turning — Process Improvement and Troubleshooting (tailstock offset = half the difference; offset away from operator enlarges the tailstock end)Report a problem with this question

8. 8. A part is faced in the chuck and a small pip or nub is left standing at the center of the face. What caused it?

   • A.The tool point was set slightly above the workpiece centerline
   • B.The feed was too heavy for the depth of cut being taken
   • C.The compound rest was left swiveled at 29.5 degrees instead of 30 degrees
   • D.The tool point was set slightly below the workpiece centerline✓ Answer

   As the tool feeds toward the axis, a cutting edge sitting below center passes underneath the last of the stock and never reaches the centerline, so a small cone of material is left standing. The fix is to set the tool point exactly on center, checked against a center held in the tailstock or against a rule held between the tool and the work.

   Source: NIMS Machining Level I Preparation Guide, Turning — Process Improvement and Troubleshooting (nub after facing = tool set below center)Report a problem with this question

9. 9. Starting with the compound rest perpendicular to the lathe centerline, how is it set for single-pointing an external right-hand 60 degree Unified thread, and what squares the threading tool to the work?

   • A.Swivel 29.5 degrees to the right and square the tool with a center (fishtail) gage✓ Answer
   • B.Swivel 29.5 degrees to the left and square the tool with a center (fishtail) gage
   • C.Swivel 29.5 degrees to the right and square the tool with a screw pitch gage
   • D.Swivel 30 degrees to the left and square the tool with a screw pitch gage

   Swiveling just under half of the 60 degree included angle makes the compound feed the tool almost entirely along the leading flank, so one flank cuts and chip control stays manageable; the swivel goes to the right for an external right-hand thread and to the left for an internal one, and direction is the discriminator NIMS tests. A center gage has the 60 degree vee needed to set the tool square to the work axis, while a screw pitch gage only counts threads per inch.

   Source: NIMS Machining Level I Preparation Guide, Turning — Single Point Threading (compound 29.5 degrees right for external RH threads; center gage squares the tool)Report a problem with this question

10. 10. A 2.500 in. diameter clearance-fit hole must be produced in the chucking part. Which process sequence should the plan call for?

    • A.Center drill, drill, step the drill up in size, then ream to the finished diameter
    • B.Center drill, drill, step the drill up in size, then bore to the finished diameter✓ Answer
    • C.Center drill and then drill straight to 2.500 in. with a single drill
    • D.Lay out and prick punch the hole, then bore it from the solid without drilling

    A reamer follows whatever hole it is given, so it can improve size and finish but cannot correct location, roundness or straightness - and a 2.500 in. reamer is not a practical shop tool anyway. Boring is the operation that precisely enlarges an existing hole, and it is the only step that can bring the bore concentric to the datum diameter the print requires.

    Source: NIMS Machining Level I Preparation Guide, Turning — Tapping, Fits and Allowances / Turning Operations (large clearance holes are center drilled, drilled and bored, not reamed)Report a problem with this question

11. 11. A parting tool starts to chatter and squeal as it feeds into the cut. Which corrective action attacks the actual cause?

    • A.Set the tool a little above center so that the cutting edge digs in and clears the chip
    • B.Back the tool out and let it dwell at the bottom of the cut until the chatter damps out
    • C.Extend the blade only as far past the holder as the depth of cut requires and lock the carriage to the ways✓ Answer
    • D.Raise the spindle speed to the RPM you would use to drill a hole of the same diameter

    Parting chatter is a rigidity problem: a long unsupported blade plus a carriage free to creep on the ways lets the tool spring in and out of the cut at its natural frequency. Minimizing overhang, locking the carriage, keeping the tool exactly on center and square, and feeding without dwelling keeps the edge loaded and stops the vibration - and note that parting speed is not the same as drilling speed for the same diameter.

    Source: NIMS Machining Level I Preparation Guide, Turning — Turning Operations (parting: tool on center and square, carriage locked, keep it cutting)Report a problem with this question

12. 12. You pick up the side of a workpiece on a vertical mill with a .200 in. diameter cylindrical edge finder and zero the axis dial the instant the tip kicks over. How far must the table then move to put the spindle centerline on that edge?

    • A..200 in.
    • B..050 in.
    • C.No move at all - the dial was zeroed on the edge
    • D..100 in.✓ Answer

    At kick-over the edge finder tip is running concentric with the spindle while its outside surface touches the work, so the spindle centerline is standing off the edge by exactly one tip radius. Moving half the tip diameter, .100 in., brings the spindle axis onto the edge, and this half-diameter compensation is the same logic used for touching off with any cutter or pin.

    Source: NIMS Machining Level I Preparation Guide, Milling — Basic Milling Operations (edge finder compensation = half the tip diameter)Report a problem with this question

13. 13. Before machining, the vise must be aligned on the mill table. Which method is correct?

    • A.Sweep a dial indicator along the solid (fixed) jaw and tap the vise until the reading does not change from end to end✓ Answer
    • B.Sweep a dial indicator along the movable jaw and tap the vise until the reading does not change from end to end
    • C.Push the vise body up against the machined edge of the table and clamp it there, since that edge is parallel to the X axis
    • D.Clamp a parallel lightly between the jaws and run the indicator along the parallel instead of the vise itself

    Only the solid jaw is rigidly fixed to the vise body and machined square to its base, so it is the one surface that can serve as a reference plane. The movable jaw rides on clearance and tends to lift and tip as it is tightened, which is also why the finished face of a part is always seated against the solid jaw.

    Source: NIMS Machining Level I Preparation Guide, Milling — Clamping and Vise Applications (indicate the solid jaw)Report a problem with this question

14. 14. While squaring up a block, face 1 is milled flat, the block is then turned and face 2 is milled with face 1 held flat against the solid vise jaw. Why is face 1 placed against the solid jaw?

    • A.Because the movable jaw would scratch the face that has already been finished
    • B.Because the solid jaw is fixed square to the table, which squares face 2 to face 1✓ Answer
    • C.Because it allows the parallels to be pulled out from under the block
    • D.Because the largest surface of a block must always be machined last

    Squaring works by transferring one good surface into the setup as a reference: face 1 becomes the primary datum, and the solid jaw is the only vise surface trustworthy enough to carry that reference. Seat the block down on parallels with a soft-faced hammer and put a round rod between the part and the movable jaw so the jaw's tipping action clamps without twisting the block out of square.

    Source: NIMS Machining Level I Standards, Duty 2.5 Square Up a Block (six surfaces square within .002 over 4.5 in.); Milling guide — Clamping and Vise ApplicationsReport a problem with this question

15. 15. What does it mean for a vertical mill head to be in tram, and why does it matter to the squaring and hole work on the Level I milling project?

    • A.The spindle is perpendicular to the X axis only, because movement in Y comes from the knee and cannot affect the cut
    • B.The quill is locked at zero so that all vertical travel comes from the knee, which keeps the cutter from drifting downward during a heavy cut
    • C.The spindle is perpendicular to both the X and the Y axis, so face-milled surfaces come out flat✓ Answer
    • D.The table is level with the floor, checked by laying a machinist's level across the table slots so that coolant drains evenly

    Tram is checked by sweeping an indicator in the spindle through a full circle, front to back and side to side, and adjusting the head until both readings match. If the spindle leans in either plane, a face mill cuts a shallow dish or leaves a step between passes and every drilled, reamed or bored hole leans off perpendicular, which destroys the squareness and true-position callouts on the project print.

    Source: NIMS Machining Level I Preparation Guide, Milling — Milling Operations Setup (head trammed perpendicular to both X and Y)Report a problem with this question

16. 16. What is the correct sequence for removing an R-8 collet from a vertical mill spindle?

    • A.Unscrew the drawbar completely and then pry the collet down out of the spindle taper with a screwdriver, using a rag to protect the spindle nose from marring
    • B.Tighten the drawbar first to seat the taper fully, strike the top of the drawbar, and then unscrew it while holding on to the collet
    • C.Loosen the drawbar a few turns and pull down on the tool while running the spindle in reverse at its lowest speed until the taper breaks free
    • D.Loosen the drawbar a few turns, tap the top of the drawbar with a dead-blow hammer to break the taper loose, then unscrew the drawbar the rest of the way✓ Answer

    The R-8 collet is held by a self-holding taper that a drawbar pulls up tight, so it will not drop out on its own once the drawbar is released. Leaving several threads engaged means the hammer blow drives the collet down out of the taper instead of damaging the drawbar threads or letting the tool fall on the table.

    Source: NIMS Machining Level I Preparation Guide, Milling — Tool Holding Applications (R-8 spindle taper; loosen drawbar, strike, then remove)Report a problem with this question

17. 17. A .875 in. diameter 4-flute end mill is run at 55 SFM with a chip load of .008 in. per tooth. Using RPM = (4 x CS) / D, the constant used in the NIMS milling guide, what table feed should be set?

    • A.8.05 IPM✓ Answer
    • B.2.01 IPM
    • C.7.68 IPM
    • D..032 IPM

    RPM = (4 x 55) / .875 = 251.4, and milling feed is inches per minute: IPM = RPM x chip load per tooth x number of teeth = 251.4 x .008 x 4 = 8.05. Dropping the tooth count gives 2.01, using the turning constant 3.82 gives 7.68, and .032 is only the feed per revolution, which is why the constant and the tooth count must both be stated and used.

    Source: NIMS Machining Level I Preparation Guide, Milling — Feeds and Speeds (RPM = 4 x CS / D; IPM = RPM x chip load x number of teeth)Report a problem with this question

18. 18. A 2.000 in. diameter shaft is clamped in V-blocks on the mill table. You touch a .250 in. diameter end mill against the side of the shaft and zero the dial. How far must the table move to bring the cutter's centerline over the shaft's centerline?

    • A.1.000 in.
    • B.1.125 in.✓ Answer
    • C.1.250 in.
    • D..875 in.

    At the touch-off point the cutter's centerline is offset from the shaft's surface by the cutter radius, and the shaft's centerline lies another shaft radius beyond that surface. The move is therefore shaft radius plus cutter radius, 1.000 + .125 = 1.125 in.; using the shaft radius alone leaves the cutter off center by the amount of its own radius.

    Source: NIMS Machining Level I Preparation Guide, Milling — Milling Operations Setup (centering a cutter over round stock: workpiece radius + cutter radius)Report a problem with this question

Practice questions based on NIMS/ANSI 101-2001, Duties and Standards for Machining Skills Level I, and the published content of the NIMS Machining Level I theory exams, together with standard precision-machining practice. NIMS is a mark of the National Institute for Metalworking Skills; this site is not affiliated with or endorsed by NIMS. Machining Level I is a set of separate credentials, and most of them also require a hands-on performance test that this bank does not cover. The NIMS theory exams are open-reference, but questions here never depend on recalling a handbook table value, a citation number or a machine rating — always work from the print in front of you, your employer's written procedures, and the machine's own documentation, and confirm current requirements before testing. About the NIMS machining credentials →