Helical gears on parallel shafts numerical

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.

      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, $z_{v1} = 26.9$, $z_{v2} = 67.2$.