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
Answer:
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
Step-by-step 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