Machinist milling, indexing and gear cutting practice questions (429A).

Five free practice questions on milling, indexing and gear cutting: block F of the Red Seal 429A Machinist exam, about 28 of its 135 questions. They are in the exam's format: four options, one correct answer. Answer each one before opening the explanation. No account, no email.

01The questions

1. Procedural · Sets up work holding devices for conventional milling machines

   A vise must be set with its jaws at 90° to the table's lengthwise travel. How is it aligned?

   1. Set the swivel base to its 90° graduation and clamp it.
   2. Indicate the solid jaw while moving the table lengthwise.
   3. Square the jaw to the column face with a machinist square.
   4. Indicate the solid jaw while moving the saddle in and out.
   Show the answer

   D. Correct. With the jaws across the table, the jaw face runs in the cross (saddle) direction, so the indicator must travel that way along it. Tap the vise until the reading holds steady over the jaw's length, then tighten it down and check again.

   Why not the others

   A. Swivel-base graduations are a coarse setting, good for a rough start but not for precision alignment. Indicate the jaw to finish the job.

   B. Traversing lengthwise checks a jaw that runs parallel to the table travel. A jaw set at 90° to that travel lies along the saddle's direction, so the saddle movement is the one to use.

   C. The column face is not a precision reference for the table's travel, and a square against it is checked only by eye or feeler. An indicator moved along the jaw with the saddle checks it against the actual travel.

2. Procedural · Sets up milling accessories

   How is the axis of a rotary table set directly under the vertical-mill spindle?

   1. Edge-find the rim on one side and move half the table diameter.
   2. Sweep its centre bore with an indicator turned by the spindle.
   3. Line it up on the middle T-slot and the table's ruled scale.
   4. Bring a pointed centre in the spindle down into the bore by eye.
   Show the answer

   B. Correct. An indicator held in the spindle, turned by hand, sweeps the bore at the centre of the rotary table. When the reading is the same all the way round, the spindle axis is on the table's axis; lock the table and saddle there and note the dial or readout positions.

   Why not the others

   A. One edge-finder touch plus half the nominal diameter assumes the rim is exactly its stated size and true to the table's axis. Sweeping the centre bore uses the axis itself.

   C. Lining up with a T-slot and the machine table's ruled scale is coarse; neither locates the rotary table's axis to the accuracy needed. Indicate the centre bore.

   D. A centre point dropped into the bore lines things up only by eye and feel. It gets close, but an indicator sweep gives the precision the job needs.

3. Procedural · Selects conventional milling machine speeds and feeds

   A 0.625 in. HSS end mill is to run at the chart's 100 sfpm for this steel. What spindle speed is set, to the nearest rpm?

   1. 1 222 rpm
   2. 611 rpm
   3. 1 920 rpm
   4. 3 840 rpm
   Show the answer

   B. Correct. rpm = (12 × CS) ÷ (π × D) = (12 × 100) ÷ (3.1416 × 0.625) = 1 200 ÷ 1.9635 = 611.2, so 611 rpm. The 12 turns feet per minute into inches per minute to match a diameter in inches.

   Why not the others

   A. This uses the radius, 0.3125 in., in place of the diameter, which doubles the answer. The formula takes the cutter's full diameter.

   C. This leaves π out: 1 200 ÷ 0.625 = 1 920. Without π the bottom line is not the cutter's circumference, and the speed comes out about three times too high.

   D. This leaves π out and uses the radius as well: 1 200 ÷ 0.3125 = 3 840. Both mistakes push the speed up.

4. Procedural · Mills slots, grooves and keyways using a conventional milling machine

   A T-slot is to be milled into a solid block. What is the correct order of cuts?

   1. Mill the throat to full depth, then cut the head with the T-slot cutter.
   2. Cut the head with the T-slot cutter, then mill the throat down to meet it.
   3. Plunge the T-slot cutter to full depth, then feed it along the slot.
   4. Mill the throat halfway, cut the head, then finish the throat to depth.
   Show the answer

   A. Correct. The T-slot cutter has a narrow neck that must run in an open slot, so the throat is milled first to the full depth of the T-slot. The T-slot cutter is then fed in from the end along the throat and cuts the wide head at the bottom; the top edges are often chamfered last.

   Why not the others

   B. The T-slot cutter's neck needs the throat to pass through; it cannot reach the bottom of solid metal. The throat is cut first to full depth, then the T-slot cutter cuts the head.

   C. A T-slot cutter is not a plunging tool, and its neck needs a slot to run in. The throat is milled to full depth first and the T-slot cutter fed in from the end of the block along it.

   D. With the throat only half deep, the T-slot cutter's neck and head would be buried in solid metal below it. The throat goes to the full depth of the slot first so the cutter has a path to follow.

5. Procedural · Cuts gears and splines using a conventional milling machine

   Two spur gears of module 2 mesh with each other, one with 20 teeth and one with 40 teeth. What is the centre distance between their shafts?

   1. 60 mm
   2. 64 mm
   3. 20 mm
   4. 120 mm
   Show the answer

   A. Correct. Pitch diameter = module × N: 2 × 20 = 40 mm and 2 × 40 = 80 mm. Meshing gears roll on their pitch circles, so the centre distance is the sum of the pitch radii: (40 + 80) ÷ 2 = 60 mm.

   Why not the others

   B. This uses the outside diameters (44 mm and 84 mm). Gears mesh on their pitch circles; the outside diameters overlap by the working depth of the teeth. Centre distance = (40 + 80) ÷ 2 = 60 mm.

   C. This uses the difference of the tooth counts, 2 × (40 − 20) ÷ 2, which is the centre distance for an internal gear pair. For two external spur gears the pitch radii add: (40 + 80) ÷ 2 = 60 mm.

   D. This adds the two pitch diameters (40 + 80) without halving. The distance between shaft centres is the sum of the pitch radii: (40 + 80) ÷ 2 = 60 mm.

02Answer key

03What block F covers

Block F is the conventional milling machine, vertical and horizontal. Set-up covers vises, clamps and angle plates, arbors, collets and cutters, dividing heads and rotary tables, locating the workpiece, and speeds and feeds. Operation covers face and end milling, pockets and profiles, slots, keyways, T-slots and dovetails, gear and spline cutting with indexing, and hole work on the mill.

The Red Seal Occupational Standard names this block “Machines using conventional milling machines”. It carries 20.7% of the exam, about 28 of the 135 questions, split across two tasks. The share column is each task's weight within the block, as the standard publishes it.

04What the questions turn on

Table feed needs the tooth count

Feed rate per minute is feed per tooth times the number of teeth times rpm. The cutter diameter sets the spindle speed, not the feed, once the rpm is known.

Take up backlash when indexing

A dividing head's worm drive has backlash. If the crank overshoots the hole, back it off well past the hole and come forward into it again, so the backlash is taken up in the same direction as every other division.

Some pockets need a start hole

An end mill whose end teeth stop short of the centre cannot cut the core beneath it. Drill a start hole to pocket depth, a little smaller than the end mill, then mill outward to the walls.

Gang milling runs at one rpm

Cutters on one arbor all turn at the same rpm, and the largest has the highest surface speed. Calculate the spindle speed from the largest diameter so no cutter runs above the chart cutting speed.

05The other blocks

The 429A exam has eight blocks. Each has its own page of free questions:

The 429A Machinist practice exam page describes the full question bank. How these questions are written and checked: how our questions are made. Found a mistake? Tell us and we will fix it.