A parallel helical gearset uses a 20-tooth pinion driving a 36-tooth gear. The pinion has a right hand helix angle of 30-degrees, a normal pressure angle of 25-degrees, and a normal diametral pitch of 6 teeth/in. Find:

This question is incomplete, the complete question is;

A parallel helical gearset uses a 20-tooth pinion driving a 36-tooth gear. The pinion has a right hand helix angle of 30-degrees, a normal pressure angle of 25-degrees, and a normal diametral pitch of 6 teeth/in. Find:

a) the normal, transversa and axial circular pitches

b) the normal base circular pitch

a)

- normal circular pitch  Pn is 0.5236 in

- Transverse circular pitch Pt is 0.6046 in

- Axial circular Pitch Px is 1.0472 in

b)

normal base circular pitch ( Pnb ) is 0.4745 in

Explanation:

Given that;

Number of teeth on pinion Np = 20 teeth

Number of teeth on gear Ng = 36 teeth

Helix angle W = 30°

Normal pressure angle ∅ = 25°

diametral pitch p = 6 teeth/in

a)

-The normal circular pitch  Pn

Pn = π / p

we substitute

Pn = π / 6

Pn = 0.52359877 ≈ 0.5236 in

The normal circular pitch  Pn is 0.5236 in

- Transverse circular pitch Pt

Pt = Pn / cosW

we substitute

Pt = 0.52359877 / cos30°

Pt = 0.60459978 ≈ 0.6046 in

Transverse circular pitch Pt is 0.6046 in

- The Axial circular Pitch Px

Px = Pt / tanW

we substitute

Px = 0.60459978 / tan30°

Px = 1.047197 ≈ 1.0472 in

Axial circular Pitch Px is 1.0472 in

b)

The normal base circular pitch ( Pnb )

Pnb = Pn cos∅

Pnb = 0.52359877 × ( cos25°)

Pnb = 0.4745 in

normal base circular pitch ( Pnb ) is 0.4745 in

answer:

what

explanation: