Editing Virtual work (section)

== Gear train ==
A gear train is formed by mounting gears on a frame so that the teeth of the gears engage.  Gear teeth are designed to ensure the pitch circles of engaging gears roll on each other without slipping, this provides a smooth transmission of rotation from one gear to the next.  For this analysis, we consider a gear train that has one degree-of-freedom, which means the angular rotation of all the gears in the gear train are defined by the angle of the input gear.   
[[Image:Transmission of motion by compund gear train (Army Service Corps Training, Mechanical Transport, 1911).jpg|thumb|right|300px|Illustration from Army Service Corps Training on Mechanical Transport, (1911), Fig. 112 Transmission of motion and force by gear wheels, compound train]]

The size of the gears and the sequence in which they engage define the ratio of the angular velocity ''ω<sub>A</sub>'' of the input gear to the angular velocity ''ω<sub>B</sub>'' of the output gear, known as the speed ratio, or [[gear ratio]], of the gear train.  Let ''R'' be the speed ratio, then
<math display="block"> \frac{\omega_A}{\omega_B} = R.</math>

The input torque ''T''<sub>''A''</sub> acting on the input gear ''G''<sub>''A''</sub> is transformed by the gear train into the output torque ''T''<sub>''B''</sub> exerted by the output gear ''G''<sub>''B''</sub>.  If we assume, that the gears are rigid and that there are no losses in the engagement of the gear teeth, then the principle of virtual work can be used to analyze the static equilibrium of the gear train.

Let the angle ''θ'' of the input gear be the generalized coordinate of the gear train, then the speed ratio ''R'' of the gear train defines the angular velocity of the output gear in terms of the input gear, that is
<math display="block"> \omega_A = \omega, \quad \omega_B = \omega/R.</math>

The formula above for the principle of virtual work with applied torques yields the generalized force
<math display="block"> Q = T_A \frac{\partial\omega_A}{\partial\omega} - T_B \frac{\partial \omega_B}{\partial\omega} = T_A - T_B/R = 0.</math>

The [[mechanical advantage]] of the gear train is the ratio of the output torque ''T''<sub>''B''</sub> to the input torque ''T''<sub>''A''</sub>, and the above equation yields
<math display="block"> MA = \frac{T_B}{T_A} = R.</math>

Thus, the speed ratio of a gear train also defines its mechanical advantage.  This shows that if the input gear rotates faster than the output gear, then the gear train amplifies the input torque.  And, if the input gear rotates slower than the output gear, then the gear train reduces the input torque.