Contact Ratio of Meshing Gear and Pinion A gearset has N1 tooth pinion, N2 tooth gear, pressure angle ϕ, and diametral pitch P (Figure 11.7). Find: a. The contact ratio b. The pressure angle and contact ratio, if the center distance is increased by 0.2 in. Given: N1 = 15, N2 = 45, ϕ = 20°,

Question

Contact Ratio of Meshing Gear and Pinion

A gearset has [latex]N_{1}[/latex] tooth pinion, [latex]N_{2}[/latex] tooth gear, pressure angle [latex]\phi [/latex], and diametral pitch [latex]P[/latex] (Figure 11.7).

Find:

a. The contact ratio

b. The pressure angle and contactratio, if the center distance is increased by 0.2 in.

Given: [latex]N_{1}[/latex] = 15, [latex]N_{2}[/latex] = 45, [latex]\phi [/latex] = 20°, [latex]P[/latex] = 2.5 in.[latex]^{−1}[/latex]

[latex] a=\frac{1}{P}=\frac{1}{2.5}=0.4 ext { in. (by Table } 11.1 ext { ) } [/latex]

Assumption: Standard gear sizes are considered.

Table 11.1 Commonly Used Standard Tooth Systems for Spur Gears
Item	20° Full Depth	20° Stub	25° Full Depth
Addendum [latex]a[/latex]	[latex]1/P[/latex]	[latex]0.8/P[/latex]	[latex]1/P[/latex]
Dedendum [latex]b_{d}[/latex]	[latex]1.25/P[/latex]	[latex]l/P[/latex]	[latex]1.25/P[/latex]
Clearance [latex]f[/latex]	[latex]0.25/P[/latex]	[latex]0.2/P[/latex]	[latex]0.25/P[/latex]
Working depth [latex]h_{k}[/latex]	[latex]2/P[/latex]	[latex]1.6/P[/latex]	[latex]2/P[/latex]
Whole depth [latex]h[/latex]	[latex]2.25/P[/latex]	[latex]1.8/P[/latex]	[latex]2.25/P[/latex]
Tooth thickness [latex]t[/latex]	[latex]1.571/P[/latex]	[latex]1.571/P[/latex]	[latex]1.571/P[/latex]

Accepted Answer

Applying Equation 11.2, the pitch diameter for the pinion and gear are found to be

[latex]P = \frac{N}{d}[/latex] (11.2)

[latex] d_1=\frac{15}{2.5}=6 in .=152.4 mm \quad ext { and } \quad d_2=\frac{45}{2.5}=18 in .=457.2 mm [/latex]

Hence, the gear pitch radii are

[latex] r_1=3 in .=76.2 mm ext { and } r_2=9 in .=228.6 mm [/latex]

a. The center distance [latex]c[/latex] is the sum of the pitch radii. So

[latex]c[/latex] = 3 + 9 = 12 in. = 304.8 mm

The radii of the base circles, using Equation 11.9, are

[latex] r_b=r \cos \phi [/latex] (11.9)

[latex] \begin{array}{l} r_{b 1}=3 \cos 20^{\circ}=2.819 in .=71.6 mm \ r _{b 2}=9 \cos 20^{\circ}=8.457 in .=214.8 mm \end{array} [/latex]

Substitution of the numerical values into Equation 11.14 gives the contactratio as

[latex] C_r=\frac{1}{p \cos \phi}\left[\sqrt{\left(r_p+a_p ight)^2-\left(r_p \cos \phi ight)^2}+\sqrt{\left(r_g+a_g ight)^2-\left(r_g \cos \phi ight)^2} ight]-\frac{c an \phi}{p} [/latex] (11.14)

[latex] \begin{aligned} c_r &=\frac{2.5}{\pi \cos 20^{\circ}}\left[\sqrt{(3+0.4)^2-(2.819)^2}+\sqrt{(9+0.4)^2-(8.457)^2} ight]-\frac{12 an 20^{\circ}}{\pi / 2.5} \ &=1.61 \end{aligned} [/latex]

Comment: The result, about 1.6, represents a suitable value.

b. For the case in which the center distance is increased by 0.2 in., we have [latex]c[/latex] = 12.2 in. It follows that

[latex] c=\frac{1}{2}\left(d_1+d_2 ight), \quad d_1+d_2=2(12.2)=24.4 in . [/latex] (b)

By Equation 11.2,

[latex] \frac{N_1}{d_1}=\frac{N_2}{d_2}, \frac{15}{d_1}=\frac{45}{d_2} [/latex] (c)

Solving Equations (b) and (c), we have [latex]d_{1}[/latex] = 6.1 in. and [latex]d_{2}[/latex] = 18.3 in. or [latex]r_{1}[/latex] = 3.05 in. and [latex]r_{2}[/latex] = 9.15 in.

The diametral pitch becomes [latex]P = N_{1}/d_{1}[/latex] = 15/6.1 = 2.459 in·[latex]^{−1}[/latex]. The addendum is therefore [latex]a = a_{1} = a_{2}[/latex] = 1/2.459 = 0.407 in. Base radii of the gears will remain the same. The new pressure angle can now be obtained from Equation 11.9:

[latex] \phi_{ ext {new }}=\cos ^{-1}\left(\frac{r_{b 1}}{r_1} ight)=\cos ^{-1}\left(\frac{2.819}{3.05} ight)=22.44^{\circ} [/latex]

Through the use of Equation 11.14, the new contact ratio is then

[latex] \begin{aligned} c_{r, ext { new }} &=\frac{2.459}{\pi \cos 22.44^{\circ}}\left[\sqrt{(3.05+0.407)^2-(2.819)^2}+\sqrt{(9.15+0.407)^2-(8.457)^2} ight]-\frac{12.2 an 22.44^{\circ}}{\pi / 2.459} \ &=1.52 \end{aligned} [/latex]

Comment: Results show that increasing the center distance leads to an increase in pressure angle but decrease in the contact ratio.

Table 11.1 Commonly Used Standard Tooth Systems for Spur Gears
Item	20° Full Depth	20° Stub	25° Full Depth
Addendum $a$	$1/P$	$0.8/P$	$1/P$
Dedendum $b_{d}$	$1.25/P$	$l/P$	$1.25/P$
Clearance $f$	$0.25/P$	$0.2/P$	$0.25/P$
Working depth $h_{k}$	$2/P$	$1.6/P$	$2/P$
Whole depth $h$	$2.25/P$	$1.8/P$	$2.25/P$
Tooth thickness $t$	$1.571/P$	$1.571/P$	$1.571/P$

Question 11.3: Contact Ratio of Meshing Gear and Pinion A gearset has N1 to...

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