Bevel, Helical and Worm Gears

Definitions

• Pitch cone angle ($\gamma$): the angle between the gear axis and the pitch cone element.
• Cone distance ($A_0$): length of a pitch cone element from the apex to the large-end pitch circle.
• Back cone: a cone whose elements are perpendicular to the pitch cone elements at the large end of the tooth. Its development is a spur gear.
• Formative (virtual) number of teeth: number of teeth on the spur gear whose pitch radius equals the back cone distance.

Tredgold's approximation

The tooth profile of a bevel gear is taken as that of a spur gear lying on the back cone surface. The back cone radius is $r_b = \dfrac{d}{2\cos\gamma}$, so the virtual spur gear has pitch diameter $d/\cos\gamma$ and

$$z' = \frac{z}{\cos\gamma}$$

This $z'$ is used for tooth strength design and to check interference ($z'$ need not be an integer).

       apex
        /|\
       / | \   A0 = cone distance
      /  |  \
     /   |   \
    /_(gamma)_\
    gear pitch dia d

Shafts at 90 degrees

For shaft angle $\Sigma = 90^\circ$:

$$\tan\gamma_1 = \frac{z_1}{z_2}, \qquad \gamma_2 = 90^\circ - \gamma_1$$

Calculation

$$\gamma_1 = \tan^{-1}\frac{20}{40} = 26.57^\circ, \qquad \gamma_2 = 90^\circ - 26.57^\circ = 63.43^\circ$$

Pitch diameters (large end):

$$d_1 = mz_1 = 6 \times 20 = 120\ \text{mm}, \qquad d_2 = mz_2 = 6 \times 40 = 240\ \text{mm}$$

Cone distance:

$$A_0 = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2} = \sqrt{60^2 + 120^2} = 134.16\ \text{mm}$$

Face width limit (to avoid weak tooth tips and cutting problems): $b \le A_0/3$ (or $10m$, whichever is less):

$$b_{max} = \frac{134.16}{3} = 44.72\ \text{mm}$$

Formative numbers of teeth:

$$z_1' = \frac{20}{\cos 26.57^\circ} = 22.36, \qquad z_2' = \frac{40}{\cos 63.43^\circ} = 89.44$$

Tooth proportions (large end): addendum $= m = 6$ mm, dedendum $= 1.2m = 7.2$ mm.

Answer: $\gamma_1 = 26.57^\circ$, $\gamma_2 = 63.43^\circ$, $d_1 = 120$ mm, $d_2 = 240$ mm, $A_0 = 134.16$ mm, $b_{max} = 44.72$ mm, $z_1' = 22.36$, $z_2' = 89.44$.

Planes and relations

• Transverse plane: plane perpendicular to the gear axis (the end face).
• Normal plane: 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 angle but of opposite hand. Axial thrust results, and overlap of teeth requires face width $b \ge p_a$ (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 (spur gear equivalent in the normal plane):

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

The pinion's $z_{v1} > 17$, so there is no interference problem although it has only 20 teeth.

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

Worm drive relations

For a worm with $z_1$ starts (threads), axial pitch $p_x = \pi m$, and diameter $d_w = qm$:

• Lead $L = z_1p_x$.
• Lead angle $\tan\lambda = \dfrac{L}{\pi d_w} = \dfrac{z_1}{q}$.
• Wheel pitch diameter $d_g = mz_2$; the wheel helix angle equals $\lambda$ (for 90 degree shafts, with the same hand).
• Centre distance $C = \dfrac{d_w + d_g}{2}$; velocity ratio $= z_2/z_1$.

Calculation

Axial pitch: $p_x = \pi \times 6 = 18.850$ mm.

Lead: $L = 3 \times 18.850 = 56.55$ mm.

$$\lambda = \tan^{-1}\frac{z_1}{q} = \tan^{-1}\frac{3}{10} = 16.70^\circ$$

Worm diameter: $d_w = qm = 10 \times 6 = 60$ mm. Wheel diameter: $d_g = 6 \times 45 = 270$ mm.

$$C = \frac{60 + 270}{2} = 165.0\ \text{mm}$$ $$\text{Velocity ratio} = \frac{z_2}{z_1} = \frac{45}{3} = 15$$

Efficiency

$$\eta = \frac{\cos\phi_n - \mu\tan\lambda}{\cos\phi_n + \mu\cot\lambda} = \frac{\cos 20^\circ - 0.05 \times 0.3000}{\cos 20^\circ + 0.05/0.3000} = \frac{0.9247}{1.1064}$$ $$\eta = 83.6\ \%$$

Advantages and disadvantages

Answer: $L = 56.55$ mm, $\lambda = 16.70^\circ$, $d_w = 60$ mm, $d_g = 270$ mm, $C = 165.0$ mm, VR $= 15$, $\eta \approx 83.6\ \%$.

(a) Spiral bevel gears

• Bevel gears with curved, oblique teeth; the spiral angle is about 35 degrees.
• Teeth engage gradually, so contact is smooth, quiet and the contact ratio is higher than straight bevel gears.
• Stronger and can run at high speeds.
• Produce axial thrust, whose direction depends on spiral hand and rotation.
• Shafts intersect. Application: final drive of rear-wheel drive vehicles (older cars), machine tool gearboxes.

(b) Hypoid gears

• Similar to spiral bevel gears but the shaft axes are offset (do not intersect); pitch surfaces are hyperboloids.
• The pinion is larger and stronger than in a spiral bevel set of the same ratio, and can be mounted with bearings on both sides.
• There is sliding along the tooth length, so a special hypoid (EP) lubricant is needed.
• Application: automobile differential (propeller shaft is lowered, giving a lower floor).

(c) Crossed helical gears

• Two helical gears of any helix angle (and hand) on non-parallel, non-intersecting shafts. Shaft angle $\Sigma = \psi_1 + \psi_2$ (for like hand).
• Theoretically point contact, so only light loads can be carried; sliding velocity along the teeth is high.
• Velocity ratio $\dfrac{N_1}{N_2} = \dfrac{z_2}{z_1} = \dfrac{d_2\cos\psi_2}{d_1\cos\psi_1}$, so it depends on the pitch diameters too.
• Application: oil pumps, distributor drives, light instrument drives.