Spur Gears

Law of gearing

For a constant angular velocity ratio, the common normal at the point of contact of the two teeth must always pass through a fixed point on the line of centres, called the pitch point $P$. Then

$$\frac{\omega_1}{\omega_2} = \frac{O_2P}{O_1P} = \text{constant}$$

(Proof: along the common normal the velocities of the contact point on both bodies must have equal components. This gives $\omega_1 \cdot O_1N_1 = \omega_2 \cdot O_2N_2$, where $O_1N_1$, $O_2N_2$ are perpendiculars from the centres to the common normal. Similar triangles give $O_1N_1/O_2N_2 = O_1P/O_2P$.)

Generation of involute

An involute is the curve traced by a point on a taut string as it is unwound from a circle, called the base circle. A point on a straight line that rolls without slipping on the base circle also traces the involute.

        T      string (tangent)
        |\
        | \
        |  * P  (point on involute)
        |  /
   -----|-/-----
  /     T0      \
 |   base circle  |
  \              /
   --------------

The string length $TP$ equals the arc $T_0T$ on the base circle that has been unwound.

Involute satisfies the law of gearing

Take two base circles of radii $r_{b1}$, $r_{b2}$ and a common tangent $N_1N_2$ to both (the string). When one circle drives the other, the contact point of the two involutes always lies on this tangent. So the common normal at the contact point is always $N_1N_2$, a fixed straight line crossing the line of centres at the same pitch point. Hence the velocity ratio is constant:

$$\frac{\omega_1}{\omega_2} = \frac{r_{b2}}{r_{b1}}$$

This line $N_1N_2$ is the line of action; the angle it makes with the common tangent of the pitch circles is the pressure angle $\phi$.

Properties of involute

1. The normal to an involute at any point is tangent to the base circle.
2. The radius of curvature at any point equals the length of the tangent from that point to the base circle.
3. The involute starts on the base circle and exists only outside it.
4. Path of contact is a straight line, so the pressure angle and the direction of the tooth force remain constant.
5. $r_b = r\cos\phi$ (base circle radius from pitch circle radius).

Advantages

• Constant velocity ratio is kept even if the centre distance changes slightly (the base circles do not change).
• Constant pressure angle gives smooth transmission and steady bearing loads.
• Simple to manufacture: a rack cutter has straight sides, so one cutter serves all gears of the same module and pressure angle.
• Interchangeability of gears of the same module and pressure angle.

Standard 20 degree full-depth system: addendum $= m$, dedendum $= 1.25m$, clearance $= 0.25m$, tooth thickness $= p_c/2$ (no backlash).

(a) Pitch circle diameters

$$d_1 = mz_1 = 5 \times 24 = 120\ \text{mm}, \qquad d_2 = mz_2 = 5 \times 60 = 300\ \text{mm}$$

(b) Centre distance

$$C = \frac{d_1 + d_2}{2} = \frac{120 + 300}{2} = 210\ \text{mm}$$

(c) Speed of gear

$$N_2 = N_1\frac{z_1}{z_2} = 1200 \times \frac{24}{60} = 480\ \text{rpm}$$

Velocity ratio $= z_2/z_1 = 2.5$.

(d) Circular pitch and tooth thickness

$$p_c = \pi m = \pi \times 5 = 15.708\ \text{mm}, \qquad t = \frac{p_c}{2} = 7.854\ \text{mm}$$

(e) Tooth proportions

(f) Outside and root diameters

(g) Base circle (radius $r_b = r\cos\phi$, $\cos 20^\circ = 0.9397$)

$$r_{b1} = 60\cos 20^\circ = 56.38\ \text{mm}, \quad r_{b2} = 150\cos 20^\circ = 140.95\ \text{mm}$$

So the base circle diameters are about $2 \times 56.38$ and $2 \times 140.95$ mm.

Answer: $d_1 = 120$ mm, $d_2 = 300$ mm, $C = 210$ mm, $N_2 = 480$ rpm, $p_c = 15.708$ mm, $t = 7.854$ mm.

Derivation

Let $R_a$, $r_a$ be the addendum circle radii of the wheel and pinion, $R$, $r$ the pitch circle radii, and $R_b$, $r_b$ the base circle radii. The line of action is $N_1N_2$ (tangent to both base circles) through pitch point $P$. Contact starts where the wheel addendum circle cuts the line of action (point $K$) and ends where the pinion addendum circle cuts it (point $L$).

   N1 ..........K....P....L......... N2
        approach | recess

$$\text{Path of approach } KP = \sqrt{R_a^2 - R_b^2} - R\sin\phi$$ $$\text{Path of recess } PL = \sqrt{r_a^2 - r_b^2} - r\sin\phi$$ $$\text{Path of contact } KL = KP + PL$$

Arc of contact $=\dfrac{\text{path of contact}}{\cos\phi}$; contact ratio $=\dfrac{\text{arc of contact}}{p_c}$.

Data

$r = 40$ mm, $R = 80$ mm, $r_a = 44$ mm, $R_a = 84$ mm, $\phi = 20^\circ$.

$$r_b = 40\cos 20^\circ = 37.588\ \text{mm}, \quad R_b = 80\cos 20^\circ = 75.175\ \text{mm}$$

(a) Path of approach and recess

$$KP = \sqrt{84^2 - 75.175^2} - 80\sin 20^\circ = 37.479 - 27.362 = 10.117\ \text{mm}$$ $$PL = \sqrt{44^2 - 37.588^2} - 40\sin 20^\circ = 22.873 - 13.681 = 9.192\ \text{mm}$$

(b) Path of contact

$$KL = 10.117 + 9.192 = 19.31\ \text{mm}$$

(c) Arc of contact

$$\frac{KL}{\cos 20^\circ} = \frac{19.31}{0.9397} = 20.55\ \text{mm}$$

(d) Contact ratio

Circular pitch $p_c = \pi m = 12.566$ mm.

$$\text{Contact ratio} = \frac{20.55}{12.566} = 1.64$$

A contact ratio of about 1.64 means that on average one pair is in contact for a time and two pairs for the rest, so the drive is continuous (it must exceed 1).

Answer: approach $= 10.117$ mm, recess $= 9.192$ mm, path of contact $= 19.31$ mm, arc of contact $= 20.55$ mm, contact ratio $= 1.64$.

Interference

Involute action exists only outside the base circle. If the tip of the driven gear tooth goes below the base circle of the pinion (that is, beyond the tangent point $N$ of the line of action), the tip digs into the flank of the pinion below its base circle. This mismatch is called interference. It causes undercutting, weakens the tooth and gives noise and wear.

   pinion base circle
         ____
        /    \----- N (tangent point)
       |  O   |   \
        \____/     \  line of action
                    \ K  <- gear tip beyond N:
                           interference

Minimum teeth with a rack

Let pinion pitch radius be $R = mz/2$ and the rack addendum $= a_w m$ ($a_w = 1$ for full depth). The rack tip should not go beyond the point $N$ (the tangent point on the pinion base circle). The distance from the pitch point $P$ to the rack's tip along the line of action is $\dfrac{a_w m}{\sin\phi}$. The distance $PN = R\sin\phi$. For no interference:

$$\frac{a_w m}{\sin\phi} \le R\sin\phi = \frac{mz}{2}\sin\phi$$ $$z_{min} = \frac{2a_w}{\sin^2\phi}$$

Values for full depth ($a_w = 1$), $\sin^2 20^\circ = 0.1170$

Pinion meshing with a gear (ratio $G = 3$)

Interference is avoided when the gear addendum circle does not pass the pinion's base-circle tangent point $N_1$:

$$(R_G + a_wm)^2 \le R_G^2\cos^2\phi + (R + R_G)^2\sin^2\phi$$

with $R_G = GR$. Solving the resulting quadratic for the pinion pitch radius $R$:

$$z_{min} = \frac{2a_w\left[G + \sqrt{G^2 + (2G+1)\sin^2\phi}\right]}{(2G+1)\sin^2\phi}$$

For $G = 3$, $\phi = 20^\circ$, $a_w = 1$: $z_{min} = 14.98$, so a pinion of at least 15 teeth is needed (the value rises to 17.1 as $G \to \infty$, which is the rack case, and drops to 12.32 for $G = 1$).

Methods to avoid interference

• Use more teeth on the pinion than $z_{min}$.
• Use stub teeth (shorter addendum) or larger pressure angle.
• Use profile (addendum) correction, that is, a long-addendum pinion and short-addendum gear (X-gears).
• Increase the centre distance (gives a larger working pressure angle).

Answer: $z_{min} = 2a_w/\sin^2\phi = 31.90$ (14.5 deg), 17.10 (20 deg), 11.20 (25 deg); for $G = 3$, 20 deg: 14.98, take 15 teeth.

(a) Module and diametral pitch

• Module $m = d/z$ (mm): pitch circle diameter divided by number of teeth. It is the metric size of the tooth. Gears must have the same module to mesh. Standard modules (ISO/IS) include 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12 mm.
• Diametral pitch $P = z/d$ (teeth per inch) in the inch system, with $P = 25.4/m$.
• Circular pitch $p_c = \pi m = \pi d/z$.

(b) Standard tooth systems

• Gears from the same system and module are interchangeable.
• The 20 degree system has a stronger tooth (thicker at the root) and a smaller $z_{min}$ (18) than the 14.5 degree system (32).
• Stub teeth are stronger and avoid undercutting but have a smaller contact ratio.

(c) Undercutting

When a gear is cut with a hobbing cutter or rack cutter and the number of teeth is below $z_{min}$, the cutter tip removes part of the tooth root below the base circle. The tooth becomes thin at the root and weak, and the useful involute is shortened.

Remedies: use a larger pressure angle (20 or 25 degrees), stub teeth, or profile shifting (the cutter is moved outward by $xm$ so that the tooth gets a thicker root).

The minimum number of teeth to avoid undercutting for a rack cutter is $z_{min} = 2/\sin^2\phi$.