N. P. Shubin.

Determination of the Contact Ratio in a Twenty-Degree Involute Gear Mesh

In involute gearing the contact ratio \(\varepsilon\) is determined by the ratio of the length of the line of action \(AB\) (Fig. 1) to the base pitch \(t_e\). The points of intersection of the addendum circles of the gears with the generating straight line \(A\) and \(B\) define the beginning and end of meshing and bound that segment of the generating straight line on which meshing of both gears takes place.

\[ \varepsilon = \frac{AB}{t_e} \]

Fig. 1.

Let us denote:

\(R_1\) and \(R_2\)—the radii of the pitch circles of the gears,
\(r_1\) and \(r_2\)—the radii of the base circles of the gears,
\(R_{a_1}\)—the radius of the addendum circle of gear \(O_1\),
\(R_{a_2}\)—the same for gear \(O_2\),
\(\alpha\)—the angle of inclination of the generating straight line to the horizontal,

\[ x_1=\frac{h_1}{m} \]

—the ratio of the tooth addendum height to the module for gear \(O_1\),

\[ x_2=\frac{h_2}{m} \]

—the same for gear \(O_2\),

\[ t=\pi m \]

—the pitch along the pitch circle,

\[ t_e=\pi m\cos\alpha \]

—the pitch along the base circle,

\(Z_1\) and \(Z_2\)—the numbers of teeth of the gears.

From drawing 1,

\[ AB=AC+CB, \]

\[ \varepsilon=\frac{AC}{t_e}+\frac{CB}{t_e}=\varepsilon_1+\varepsilon_2, \]

where:

\(\varepsilon_1\)—the part of the contact ratio corresponding to segment \(AC\) of the line of action,
\(\varepsilon_2\)—the same, corresponding to segment \(CB\).

We determine the value \(\varepsilon_1\) by considering the triangles \(O_1AN\) and \(O_1CN\).

\[ AC=\sqrt{R_{a_1}^{\,2}-R_1^{\,2}\cos^2\alpha}-R_1\sin\alpha. \]

Expressing the values \(R_{a_1}\) and \(R_1\) in terms of the number of teeth and the module, we obtain

\[ AC=\frac{Z_1m}{2} \left[ \sqrt{\left(1+\frac{2x_1}{Z_1}\right)^2-\cos^2\alpha} -\sin\alpha \right]. \]

Substituting the value of \(t_e\), we shall have

\[ \varepsilon_1= \frac{Z_1}{2\pi\cos\alpha} \left[ \sqrt{\left(1+\frac{2x_1}{Z_1}\right)^2-\cos^2\alpha} -\sin\alpha \right]. \tag{1} \]

Analogously, the value \(\varepsilon_2\) is determined from the triangles \(O_2BM\) and \(O_2CN\).

\[ \varepsilon_2= \frac{Z_2}{2\pi\cos\alpha} \left[ \sqrt{\left(1+\frac{2x_2}{Z_2}\right)^2-\cos^2\alpha} -\sin\alpha \right]. \tag{2} \]

Thus the total contact ratio will be:

\[ \varepsilon=\varepsilon_1+\varepsilon_2= \frac{Z_1}{2\pi\cos\alpha} \left[ \sqrt{\left(1+\frac{2x_1}{Z_1}\right)^2-\cos^2\alpha} + \frac{Z_2}{Z_1} \sqrt{\left(1+\frac{2x_2}{Z_2}\right)^2-\cos^2\alpha} - \left(1+\frac{Z_2}{Z_1}\right)\sin\alpha \right]. \]

Replacing \(\dfrac{Z_2}{Z_1}\) by the transmission ratio \(i\), we obtain

\[ \varepsilon= \frac{Z_1}{2\pi\cos\alpha} \left[ \sqrt{\left(1+\frac{2x_1}{Z_1}\right)^2-\cos^2\alpha} + \sqrt{\left(i+\frac{2x_2}{Z_1}\right)^2-(i\cos\alpha)^2} - (1+i)\sin\alpha \right]. \tag{3} \]

Using this equation, it is possible to determine the contact ratio for any pair of gears at different angles \(\alpha\), gear ratios, and addendum heights.

However, determining the contact ratio from Eq. (3) is associated with considerable calculations and presents great difficulties in transmission analyses when many variants have to be dealt with.

The calculations can be greatly simplified if the teeth of both gears have the same addendum height.

At the end a table is given for determining the contact ratio for gears having an angle of inclination of the generating straight line of \(20^\circ\) and an addendum height equal to the module.

For

\[ x_1 = x_2 = 1 \]

equations (1) and (2) take the form:

\[ \varepsilon_1 = \frac{Z_1}{2\pi \cos\alpha} \left[ \sqrt{\left(1+\frac{2}{Z_1}\right)^2-\cos^2\alpha} -\sin\alpha \right]. \tag{4} \]

\[ \varepsilon_2 = \frac{Z_2}{2\pi \cos\alpha} \left[ \sqrt{\left(1+\frac{2}{Z_2}\right)^2-\cos^2\alpha} -\sin\alpha \right]. \tag{5} \]

It follows from these equations that the part of the contact ratio \(\varepsilon_1\) corresponding to the segment \(AC\) depends, at a constant angle \(\alpha\), only on the number of teeth \(Z_1\) of the first gear and does not depend on the number of teeth of the second gear. Similarly, the part of the contact ratio \(\varepsilon_2\) for the segment \(CB\) depends only on the number of teeth \(Z_2\) of the second gear and does not depend on the number of teeth of the first gear.

Both equations have exactly the same structure and differ from each other only in the different values of \(Z\).

This circumstance makes it possible to carry out separately the calculations of the partial contact ratios \(\varepsilon_1\) for gears with different numbers of teeth. To obtain the total contact ratio for any number of teeth of the meshing gears, the entire calculation operation is reduced to the simple addition of these two partial values \(\varepsilon_1\).

Table 1 gives data for determining the contact ratio for gears with a number of teeth up to 400 at a pressure angle \(\alpha = 20^\circ\) and addendum height \(h = 1m\).

The second column gives the values of \(\varepsilon_1\) calculated from equation (4).

The third column gives the value of the contact ratio when two gears with equal numbers of teeth are meshed, at a gear ratio equal to unity. The data of the third column are obtained by multiplying the values of \(\varepsilon_1\) by 2.

The fourth column gives the maximum values of the contact ratio in external gearing for gears with the given number of teeth \(Z\).

As the gear ratio increases, the contact ratio increases and reaches its maximum at a gear ratio equal to infinity (meshing with a rack).

This case is shown in Fig. 2

\[ \varepsilon = \frac{AB}{t_e} = \frac{AC}{t_e} + \frac{CB}{t_e} = \varepsilon_1+\varepsilon_2. \]

Here:

\(\varepsilon_1\)—the part of the contact ratio depending only on the dimensions of gear \(O_1\).

The values of \(\varepsilon_1\) are determined from the second column of Table 1.

![Drawing 2]

Fig. 2.

\(\varepsilon_2\)—the part of the contact ratio corresponding to the segment \(CB\) and depending only on the addendum height of the rack tooth \(h_2\).

From triangle \(CBD\),

\[ \varepsilon_2=\frac{CB}{t_e}=\frac{BD}{\sin\alpha \cdot t_e}=\frac{h_2}{t_e \cdot \sin\alpha}. \]

Substituting the values of \(h_2\) and \(t_e\), we obtain

\[ \varepsilon_2=\frac{x_2 m}{\pi m \sin\alpha \cos\alpha}=\frac{2x_2}{\pi \sin 2\alpha}. \tag{6} \]

For \(x_2=1\),

\[ \varepsilon_2=\frac{2}{\pi \sin 2\alpha}. \tag{6a} \]

It follows from equation (6a) that \(\varepsilon_2\) does not depend on the dimensions of the wheel meshing with the rack, and is a constant quantity.

For \(\alpha=20^\circ\),

\[ \varepsilon_2=\frac{2}{\pi \sin 40^\circ}=0.9904. \]

Thus, the maximum value of the contact ratio (the fourth column of Table 1) can be obtained by adding to the previously found value \(\varepsilon_1\) the constant quantity 0.9904,

\[ \varepsilon \text{ at } i=\sim=\varepsilon_1+0.9904. \]

In the fifth column of Table 1, the values of the relative length of the normal to the involute at the extreme point of the tooth are calculated. Using this quantity, it is possible to draw a conclusion about undercutting of the tooth.

From Fig. 1,

\[ \frac{AN}{t_e}=\frac{1}{t_e}\sqrt{R_{a_1}^{\,2}-R_1^{\,2}\cos^2\alpha}. \]

Substituting the values \(R_a\), \(R_1\), and \(t_e\), we obtain

\[ \frac{AN}{t_e}=\frac{Z_1}{2\pi\cos\alpha} \sqrt{\left(1+\frac{2x_1}{Z_1}\right)^2-\cos^2\alpha}. \tag{7} \]

For \(x=1\),

\[ \frac{AN}{t_e}=\frac{Z_1}{2\pi\cos\alpha} \sqrt{\left(1+\frac{2}{Z_1}\right)^2-\cos^2\alpha}. \tag{7a} \]

Examples of the use of Tables 1.

a) Determine the contact ratio for gears with numbers of teeth:

\[ Z_1=24 \quad \text{and} \quad Z_2=47. \]

From the table, for

\[ \begin{aligned} Z_1&=24; & \varepsilon_{24}&=0.801,\\ Z_2&=47; & \varepsilon_{47}&=0.872, \end{aligned} \]

\[ \varepsilon=\varepsilon_{24}+\varepsilon_{47}=0.801+0.872=1.673. \]

There will be no undercutting of the teeth, since in the present case

\[ \varepsilon<\frac{AN}{t_e}; \qquad 1.673<2.191. \]

b) The same for:

\[ Z_1=122, \qquad Z_2=191 \]

\[ \varepsilon=\varepsilon_{122}+\varepsilon_{191}=0.936+0.954=1.890. \]

There will be no undercutting, since

\[ 1.890<8.004 \]

c) Will there be undercutting of the teeth for \(Z_1=15,\ Z_2=182\)?

The theoretical contact ratio (without taking undercutting into account)

\[ \varepsilon=\varepsilon_{15}+\varepsilon_{182}=0.740+0.953=1.693. \]

There will be undercutting of the teeth, since

\[ \varepsilon>\frac{AN}{t_e}; \qquad 1.693>1.609. \]

d) Determine the greatest number of teeth of a gear which would work with a gear \(Z_1=13\) without undercutting.

From the table: for \(Z_1=13\),

\[ \varepsilon_{13}=0.721, \qquad \frac{AN}{t_e}=1.474. \]

In the absence of undercutting, the condition must be satisfied

\[ \varepsilon=\varepsilon_{13}+\varepsilon_2\leq \frac{AN}{t_e}. \]

Hence

\[ \varepsilon_2\leq \frac{AN}{t_e}-\varepsilon_{13}; \qquad \varepsilon_2\leq 1.474-0.721=0.753. \]

From the second column of the table we find that this condition is satisfied by

\[ Z_2=16; \qquad \varepsilon_{16}=0.749. \]

Thus a gear with number of teeth \(Z_1=13\) will work without undercutting with gears whose numbers of teeth are: 13, 14, 15, 16; all other gears with different numbers of teeth will produce undercutting.

d) Establish the limits of the numbers of teeth of gears with which gears having number of teeth \(Z_1=14, 15, 16, 17, 18\) can work without undercutting, and determine for these gears the greatest and the least contact ratio.

By analogy with example (c) we obtain:

e) Establish the limits of the numbers of teeth of gears with which a gear having number of teeth \(Z_1=87\) can work without undercutting, and determine \(\varepsilon_{\min}\) and \(\varepsilon_{\max}\).

From the table of example (d) it follows:

the smallest number of teeth—\(Z_{2\min}=16\)

the greatest number of teeth—\(Z_{2\max}=\infty\) (rack).

From Table 1:

\[ \text{for } Z_1=87, \qquad \varepsilon_{87}=0.918, \]

\[ \text{for } Z_2=16, \qquad \varepsilon_{16}=0.749, \]

\[ \varepsilon_{\min}=\varepsilon_{87}+\varepsilon_{16}=0.918+0.749=1.667. \]

The maximum possible contact ratio for a gear with \(Z_1=87\) will occur when the gear meshes with a rack and is determined directly from Table 1, column 4:

\[ \varepsilon_{\max}=1.908. \]

To determine the contact ratio for a twenty-degree involute mesh with various addendum heights, Table 2 has been compiled.

The table gives the values of the relative length of the normal to the end point of the involute for various addendum heights according to equation (7).

The table is compiled for gears with numbers of teeth from 10 to 100 and with \(x\) varying from 1 to 0.

When the addendum height of the tooth is decreased, point \(A\) (Fig. 1) approaches the pitch point \(C\) and, at \(x=0\), coincides with the pitch point. The length of the normal in this case will be equal to the segment \(CN\). The values of the relative length of the normal \(CN\) are given in the last column of the table.

To obtain the partial contact ratio \(\varepsilon_1\) corresponding to the segment \(AC\), it is necessary to subtract from the relative length of the normal \(AN\) the relative length of the normal \(CN\).

\[ \varepsilon_1=\frac{AN}{t_e}-\frac{CN}{t_e}. \tag{8} \]

Examples of using Tables 2.

a) Determine the contact ratio for gears: \(Z_1=20\) and \(Z_2=49\), with tooth addendum height \(h=0.9m\).

\[ \varepsilon_{20}=\frac{AN_{20}}{t_e}-\frac{CN_{20}}{t_e}=1.871-1.159=0.712, \]

\[ \varepsilon_{49}=\frac{AN_{49}}{t_e}-\frac{CN_{49}}{t_e}=3.634-2.838=0.796, \]

\[ \varepsilon=\varepsilon_{20}+\varepsilon_{49}=0.712+0.796=1.508. \]

There will be no undercutting of the teeth, since

\[ \varepsilon_{20}<\frac{CN_{49}}{t_e};\qquad 0.712<2.838 \quad\text{and}\quad \varepsilon_{49}<\frac{CN_{20}}{t_e};\qquad 0.796<1.159. \]

b) Likewise: for

\[ \begin{aligned} Z_1&=15; & h_1&=0.7m\\ Z_2&=29; & h_2&=0.9m \end{aligned} \]

\[ \varepsilon_{15}=1.420-0.869=0.551 \]

\[ \varepsilon_{29}=2.431-1.680=0.741 \]

\[ \varepsilon=\varepsilon_{15}+\varepsilon_{29}=0.551+0.741=1.292. \]

There will be no undercutting, since

\[ \varepsilon_{15}<\frac{CN_{29}}{t_e} \quad\text{and}\quad \varepsilon_{29}<\frac{CN_{15}}{t_e}. \]

c) Determine the greatest tooth addendum height for gears with numbers of teeth \(Z_1=10\) and \(Z_2=10\), at which undercutting is absent.

In the absence of undercutting the following condition must be satisfied:

\[ \varepsilon_{10}\leq \frac{CN_{10}}{t_e}. \tag{a} \]

From equation (8)

\[ \varepsilon_{10}=\frac{AN_{10}}{t_e}-\frac{CN_{10}}{t_e}. \tag{b} \]

Solving equations (a) and (b) jointly, we obtain

\[ 2\frac{CN_{10}}{t_e}\geq \frac{AN_{10}}{t_e}. \tag{c} \]

From the first row of Table 1, last column,

\[ \frac{CN_{10}}{t_e}=0.579;\qquad 2\frac{CN_{10}}{t_e}=1.158. \]

Condition (c) is satisfied by the value

\[ \frac{AN_{10}}{t_e}=1.152, \]

at which the tooth addendum height is

\[ h=0.8m. \]

The contact ratio in this case will be:

\[ \varepsilon_{10}=1.152-0.579=0.573;\qquad \varepsilon=\varepsilon_{10}+\varepsilon_{10}=1.146. \]

d) A gear with number of teeth \(Z_1=12\) and addendum height \(h_1=1m\) is in mesh with a gear \(Z_2=45\).

Determine the greatest addendum height \(h_2\) for gear \(Z_2\), under the condition of absence of undercutting.

Condition for absence of undercutting:

\[ \varepsilon_{45}\leq \frac{CN_{12}}{t_e} \tag{a} \]

According to equation (8)

\[ \varepsilon_{45}=\frac{AN_{45}}{t_e}-\frac{CN_{45}}{t_e}. \tag{b} \]

From (a) and (b) we obtain

\[ \frac{CN_{12}}{t_e}+\frac{CN_{45}}{t_e}\leq \frac{AN_{45}}{t_e}; \qquad 0.695+2.607\leq \frac{AN_{45}}{t_e}; \qquad \frac{AN_{45}}{t_e}\leq 3.302. \]

This condition is satisfied for the wheel \(Z_2=45\) by an addendum height between \(0.8m\) and \(0.7m\).

Interpolating, we find

\[ x_2=0.7+0.1\,\frac{3.302-3.235}{3.317-3.235}=0.78; \qquad h_2=0.78m. \]

With this addendum height, the relative length of the normal will be:

\[ \frac{AN_{45\,(0.78)}}{t_e} = 3.235+(3.317-3.235)\frac{0.8}{10} = 3.301. \]

Contact ratio:

\[ \varepsilon_{12}=1.405-0.695=0.710; \qquad \varepsilon_{45}=3.301-2.607=0.694; \]

\[ \varepsilon=\varepsilon_{12}+\varepsilon_{45}=0.710+0.694=1.404. \]

d) Two wheels \(Z_1=20\) and \(Z_2=69\), with equal tooth addendum heights, are in mesh. Determine the tooth addendum height for a contact ratio \(\varepsilon=1.1\).

According to equation (8)

\[ \varepsilon_{20}=\frac{AN_{20}}{t_e}-\frac{CN_{20}}{t_e}, \tag{a} \]

\[ \varepsilon_{60}=\frac{AN_{60}}{t_e}-\frac{CN_{60}}{t_e}. \tag{b} \]

Substituting the values \(\varepsilon_{20}\) and \(\varepsilon_{60}\) into the equation

\[ \varepsilon=\varepsilon_{20}+\varepsilon_{60}, \]

we obtain:

\[ \varepsilon+\frac{CN_{20}}{t_e}+\frac{CN_{60}}{t_e} = \frac{AN_{20}}{t_e}+\frac{AN_{60}}{t_e} \]

\[ 1.1+1.159+3.476 = \frac{AN_{20}}{t_e}+\frac{AN_{60}}{t_e}; \qquad 5.735= \frac{AN_{20}}{t_e}+\frac{AN_{60}}{t_e}. \]

The sum of the relative normals satisfying the last equality is located, for wheels \(Z_1=20\) and \(Z_2=60\), between \(h=0.6m\) and \(h=0.7m\).

For

\[ h=0.6m; \qquad \frac{AN_{20}}{t_e}+\frac{AN_{60}}{t_e} = 1.662+4.034=5.696 \]

\[ h=0.7m; \qquad \frac{AN_{20}}{t_e}+\frac{AN_{60}}{t_e} = 1.734+4.118=5.852. \]

For the sum \(5.735\)

\[ x_2=0.6+\frac{5.735-5.696}{5.852-5.696}\,0.1=0.625; \qquad h=0.625m. \]

The relative normals for this addendum value are:

\[ \frac{AN_{20}}{t_e} = 1.662+(1.734-1.662)\frac{0.25}{10} = 1.680, \]

\[ \frac{AN_{60}}{t_e} = 4.034+(4.118-4.034)\frac{0.25}{10} = 4.055. \]

Contact ratio:

\[ \varepsilon_{20}=1.680-1.159=0.521; \qquad \varepsilon_{60}=4.055-3.476=0.579; \]

\[ \varepsilon=\varepsilon_{20}+\varepsilon_{60}=0.521+0.579=1.100. \]

Table 1

Determination of the contact ratio \(\varepsilon\) for a twenty-degree involute mesh with addendum height equal to the module. \(\alpha = 20,\ h = 1m.\)

External gearing.

Table 2.

Relative length of the normal to the end point of the involute for a twenty-degree external involute mesh with different tooth addendum heights.