---
title: "Helical gears on parallel shafts numerical"
subject: "Theory of Mechanism and Machine I"
chapter: 5
key: "semester_6/theory_of_mechanism_and_machine_i:helical-gear-parallel-shafts-geometry"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/helical-gear-parallel-shafts-geometry
site: BCT Engineering
---

# Helical gears on parallel shafts numerical

Theory of Mechanism and Machine I · Chapter 5: Bevel, Helical and Worm Gears

- Asked in: year not given, 8 marks
- Video: [Helical Gear Numerical#Machine Design](https://www.youtube.com/watch?v=lMcwtR1g_X8) — Vreenda Gurukul Foundation 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/helical-gear-parallel-shafts-geometry
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-5

## Question

Define normal and transverse planes for a helical gear and obtain the relations between normal and transverse module, pitch and pressure angle. A pair of parallel-shaft helical gears has 20 and 50 teeth, normal module 4 mm, helix angle 25 degrees and normal pressure angle 20 degrees. Determine the transverse module, pitch circle diameters, centre distance, transverse pressure angle, axial pitch and the equivalent (formative) numbers of teeth.

## Short answer

Normal plane is perpendicular to tooth helix, transverse plane is perpendicular to axis; m_n = m_t cosψ, tan φ_t = tan φ_n / cosψ. Given m_n=4 mm, ψ=25°, φ_n=20°, z1=20, z2=50: m_t=4.4135 mm, d1=88.27 mm, d2=220.68 mm, C=154.47 mm, φ_t=21.88°, axial pitch p_a=29.73 mm, equivalent teeth z_v1=26.9, z_v2=67.2.

## Answer

#### Planes and relations
- **Transverse plane**: the plane perpendicular to the gear axis (the end face).
- **Normal plane**: the plane perpendicular to the tooth (helix) at the pitch point.

```text
      helix angle psi
   p_t (transverse pitch)
   |<------->|
   ///////////  <- teeth at angle psi to axis
   |<-p_n->|    p_n = p_t cos(psi)
```

From the geometry of the pitch plane (development):

$$
p_n = p_t\cos\psi \Rightarrow m_n = m_t\cos\psi
$$

$$
\tan\phi_t = \frac{\tan\phi_n}{\cos\psi}, \qquad p_a = \frac{p_t}{\tan\psi}
$$

The pitch diameter is $d = m_tz = \dfrac{m_nz}{\cos\psi}$. For parallel shafts, the two gears must have equal helix angles of opposite hand. An axial thrust is produced, and for continuous overlap of the teeth the face width must be $b \ge p_a$ (preferably $b \ge 1.15 p_a$).

#### Calculation
Data: $m_n = 4$ mm, $\psi = 25^\circ$ ($\cos\psi = 0.9063$), $\phi_n = 20^\circ$, $z_1 = 20$, $z_2 = 50$.

Transverse module:

$$
m_t = \frac{m_n}{\cos\psi} = \frac{4}{0.9063} = 4.4135\ \text{mm}
$$

Pitch circle diameters:

$$
d_1 = m_tz_1 = 4.4135 \times 20 = 88.27\ \text{mm}, \qquad d_2 = m_tz_2 = 4.4135 \times 50 = 220.68\ \text{mm}
$$

Centre distance:

$$
C = \frac{d_1 + d_2}{2} = 154.47\ \text{mm}
$$

Transverse pressure angle:

$$
\phi_t = \tan^{-1}\frac{\tan 20^\circ}{\cos 25^\circ} = \tan^{-1}\frac{0.3640}{0.9063} = 21.88^\circ
$$

Pitches:

$$
p_t = \pi m_t = 13.865\ \text{mm}, \quad p_n = \pi m_n = 12.566\ \text{mm}, \quad p_a = \frac{p_t}{\tan 25^\circ} = 29.73\ \text{mm}
$$

Formative numbers of teeth (equivalent spur gear in the normal plane):

$$
z_v = \frac{z}{\cos^3\psi}: \quad z_{v1} = 26.9, \quad z_{v2} = 67.2
$$

The pinion's $z_{v1} = 26.9$ is well above 17, so there is no risk of interference.

**Answer: $m_t = 4.4135$ mm, $d_1 = 88.27$ mm, $d_2 = 220.68$ mm, $C = 154.47$ mm, $\phi_t = 21.88^\circ$, $p_a = 29.73$ mm, $z_{v1} = 26.9$, $z_{v2} = 67.2$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The transverse plane of a helical gear is the plane:
   A. Perpendicular to the tooth at the pitch point
   B. Containing the gear axis
   C. Perpendicular to the gear axis (the end face)
   D. Tangent to the pitch cylinder
   Answer: C. The transverse plane is perpendicular to the axis; the normal plane is perpendicular to the tooth (helix). (Key point: Normal and transverse planes)

2. The relation between normal module m_n and transverse module m_t is:
   A. m_n = m_t / cos(psi)
   B. m_n = m_t sin(psi)
   C. m_t = m_n cos(psi)
   D. m_n = m_t cos(psi)
   Answer: D. From p_n = p_t cos(psi), m_n = m_t cos(psi), so the pitch diameter is d = m_n z/cos(psi). (Key point: Module relation)

3. For m_n = 4 mm and psi = 25°, the transverse module m_t is:
   A. 4.4135 mm
   B. 3.625 mm
   C. 4.000 mm
   D. 9.46 mm
   Answer: A. m_t = 4/cos25° = 4/0.9063 = 4.4135 mm. (Key point: Transverse module value)

4. Formative number of teeth of a helical gear is z/cos^3(psi). For the 20-tooth pinion this is 26.9. What does this mean, and what hand do parallel-shaft mating gears need?
   A. Interference is likely; same hand
   B. No interference problem as 26.9 > 17; equal helix angle but opposite hand
   C. No interference; same helix angle and same hand
   D. Interference; opposite helix angles
   Answer: B. z_v1 = 26.9 exceeds 17 so there is no interference. Parallel-shaft helical gears need equal helix angles of opposite hand. (Key point: Equivalent teeth and conditions)

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