Analyzing Ferguson’s Paradox by the Formula Method. Consider the same Ferguson paradox train as in Example 9-6 which has the following tooth numbers and initial conditions (see Figure 9-37): Sun gear #2 N2 = 100-tooth external gear Sun gear #3 N3 = 99-tooth external gear Sun gear #4 N4 = 101-tooth

Question

Analyzing Ferguson’s Paradox by the Formula Method.
Consider the same Ferguson paradox train as in Example 9‑6 which has the following tooth numbers and initial conditions (see Figure 9‑37):
Sun gear #2 [latex] N _{2} [/latex]= 100-tooth external gear
Sun gear #3 [latex] N _{3} [/latex]= 99-tooth external gear
Sun gear #4 [latex] N _{4} [/latex]= 101-tooth external gear
Planet gear [latex] N _{5} [/latex]= 20-tooth external gear
Input to sun # 2 0 rpm
Input to arm 100 rpm counterclockwise
Sun gear 2 is fixed to the frame, providing one input (zero velocity) to the system.
The arm is driven at 100 rpm CCW as the second input. Find the angular velocities of the two outputs that are available from this compound train, one from gear 3 and one from gear 4, both of which are free to rotate on the main shaft.

Accepted Answer

1 We will have to apply equation 9.14 twice, once for each output gear. Taking gear 3 as the last gear in the train with gear 2 as the first, we have:

[latex] R=\pm \frac{ ext { product of number of teeth on driver gears }}{ ext { product of number of teeth on driven gears }}=\frac{\omega_{L}-\omega_{ ext {arm }}}{\omega_{F}-\omega_{ ext {arm }}} [/latex] (9.14)

[latex]\begin{aligned}N_{2} &=100 & N_{3} &=99 & N_{5} &=20 \\omega_{ ext {arm }} &=+100 & \omega_{F} &=0 & \omega_{L} &=?\end{aligned}[/latex] (a)

2 Substituting in equation 9.14 we get:

[latex]\begin{aligned}\left(-\frac{N_{2}}{N_{5}} ight)\left(-\frac{N_{5}}{N_{3}} ight) &=\frac{\omega_{L}-\omega_{ ext {arm }}}{\omega_{F}-\omega_{ ext {arm }}} \\left(-\frac{100}{20} ight)\left(-\frac{20}{99} ight) &=\frac{\omega_{3}-100}{0-100} \\omega_{3} &=-1.01\end{aligned}[/latex] (b)

3 Now taking gear 4 as the last gear in the train with gear 2 as the first, we have:

[latex]\begin{aligned}N_{2} &=100 & N_{4} &=101 & N_{5} &=20 \\omega_{ ext {arm }} &=+100 & \omega_{F} &=0 & \omega_{L} &=?\end{aligned}[/latex] (c)

4 Substituting in equation 9.14, we get:

[latex]\begin{aligned}\left(-\frac{N_{2}}{N_{5}} ight)\left(-\frac{N_{5}}{N_{4}} ight) &=\frac{\omega_{L}-\omega_{ ext {arm }}}{\omega_{F}-\omega_{ ext {arm }}} \\left(-\frac{100}{20} ight)\left(-\frac{20}{101} ight) &=\frac{\omega_{4}-100}{0-100} \\omega_{4} &=+0.99\end{aligned}[/latex] (d)

Question 9.7: Analyzing Ferguson’s Paradox by the Formula Method. Consider...

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