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Gear manufacturing methods and indexing

Short answer

Manufacturing methods: casting, form milling, generating (hobbing, shaping), finishing; hobbing is accurate and fast. Indexing: 40/24 = 1 + 10/15; 40/37 = 1 + 3/37. Module 3, 24 teeth: PCD 72 mm, OD 78 mm, depth 6.75 mm, circular pitch 9.42 mm.

(a) Classify the methods of manufacturing gears and explain gear hobbing with its advantages. (b) A spur gear with 24 teeth and module 3 mm is to be cut on a milling machine by form milling using a dividing head (worm ratio 40:1). Calculate the indexing, the pitch circle diameter, outside diameter, whole depth and circular pitch. Also state the indexing for 37 teeth.

(a) Methods of gear manufacture

  1. Forming/shaping from material: casting (sand, die, investment), powder metallurgy, forging, cold rolling, extrusion and stamping (thin gears) - low cost for large quantity but lower accuracy.
  2. Machining (cutting):
    • Form milling: a form cutter (one of a set of 8 cutters for each module, numbered by tooth number range) cuts one tooth space at a time; the work is indexed in a dividing head. Slow and less accurate.
    • Generating: a tool and work move relative to each other as if a pair of gears (hobbing, gear shaping by pinion or rack cutter) - accurate, with one cutter for all tooth counts of one module.
  3. Finishing: shaving, grinding, lapping, burnishing to improve accuracy and finish.

Gear hobbing: the hob is a worm-like cutter with gashes forming cutting teeth. Hob and gear blank rotate in a fixed speed ratio (as worm and wheel), and the hob is fed across the face of the blank. For a single-start hob and zz teeth, the blank turns 1/z1/z revolution for each revolution of the hob. Tooth form is generated by the successive cuts.

Advantages: continuous cutting and high production rate; one hob cuts all gears of the same module and pressure angle; good accuracy; can cut spur, helical, worm wheels (not internal gears or gears near shoulders).

(b) Indexing and gear dimensions

Simple indexing with 40:1 dividing head: the crank turns per division

n=40zn = \frac{40}{z}

For z=24z = 24:

n=4024=123=1+1015n = \frac{40}{24} = 1\frac{2}{3} = 1 + \frac{10}{15}

One full turn and 10 holes in a 15-hole circle (or 20 holes in a 30-hole circle).

For z=37z = 37:

n=4037=1337n = \frac{40}{37} = 1\frac{3}{37}

One full turn and 3 holes in a 37-hole circle.

Gear dimensions (module m=3m = 3 mm, z=24z = 24):

ItemFormulaValue
Pitch circle diametermzmz72 mm
Outside diameterm(z+2)m(z+2)78 mm
Whole depth2.25m2.25m6.75 mm
Circular pitchπm\pi m9.42 mm

Answer: 1 turn + 10 holes on the 15-hole circle (24 teeth); 1 turn + 3 holes on the 37-hole circle (37 teeth); PCD 72 mm, OD 78 mm, depth 6.75 mm, circular pitch 9.42 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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