Determine:

1. sliding velocities at engagement at an disengagement of pair of a teeth, and

2. Contact ratio.

Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: Medium

Given $\phi= 20^{\circ}$; t= 30; T= 50; m=4; N= 1000 rpm; or $\omega_{1}=2\pi\times(1000/60)=104.7 rad/s$.

1.Sliding velocities at engagement and at disengagement of pair of a teeth

First of all, let us find the Radius of the addendum circles of the smaller gear and the larger gear.

We know that,

Addendum of the smaller gear,

$=\frac{mt}{2}[\sqrt{1+\frac{T}{t}(\frac{T}{t}+2)}\sin^{2}\phi-1]$

$=\frac{4\times 30}{2}[\sqrt{1+\frac{50}{30}(\frac{50}{30}+2)}\sin^{2}20^{\circ}-1]$

$=60(1.31-1)=18.6 mm$

And Addendum of the larger gear,

$=\frac{mt}{2}[\sqrt{1+\frac{t}{T}(\frac{t}{T}+2)}\sin^{2}\phi-1]$

$=\frac{4\times 50}{2}[\sqrt{1+\frac{30}{50}(\frac{30}{50}+2)}\sin^{2}20^{\circ}-1]$

$=100(1.09-1)=9$ mm

Pitch circle Radius of the smaller gear,

r= mt/2=4(30/2)=60 mm

Radius of the Addendum circle of the smaller gear,

$r_{A}$=r+Addendum of the smaller gear= 60+18.6=78.6 mm

Pitch circle Radius of the larger gear,

R=mt/2=4(50/2)=100 mm

Radius of addendum circle of the larger gear,

$R_{A}$=R+Addendum of the larger gear = 100+9=109 mm

We know that the path of approach(i.e path of contact when engagement occurs),

$KP=\sqrt{(R_{A})^{2}-R^{2}\cos^{2}\phi}-R\sin\phi$

$=\sqrt{(109)^{2}-(100)^{2}\cos^{2}20^{\circ}}-100\sin20^{\circ}=55.2-34.2=21mm$

and the path of recess(i.e path of contact when disengagement occurs),

$PL=\sqrt{(r_{A})^{2}-r^{2}\cos^{2}\phi}-r\sin\phi$

$=\sqrt{(78.6)^{2}-(60)^{2}\cos^{2}20^{\circ}}-60\sin20^{\circ}=54.76-20.52=34.24mm$

Let $\omega_{2}$ = Angular speed of the larger gear in rad/s

We know that $ \frac{\omega_{1}}{\omega_{2}}=\frac{T}{t}$ or $ \omega_{2}=\frac{\omega_{1}\times t}{T} = \frac{10.47\times30}{50} = 62.82 rad/s $

Sliding velocity at engagement of a pair of teeth

​ $=(\omega_{1}+\omega_{2})KP=(104.7+62.82)21=3518$ rad/s

=3.518 m/s

And Sliding velocity at Disengagement of a pair of teeth

​ $=(\omega_{1}+\omega_{2})PL=(104.7+62.82)34.24=5736$ rad/s

=5.736 m/s

2. Contact Ratio: We know that the length of the arc of contact

=Length of the path of contact/ $\cos \phi$

$= \frac {KP+PL}{\cos \phi}=\frac{21+34.24}{\cos 20^{\circ}}=58.78$ mm

And circular pitch=$\pi\times m=3.142\times 4=12.568 mm$

Contact Ratio= Length of arc of contact / circular pitch= 58.78/12.568=4.67 say 5