A Differential relates the angular positions and forces of three gears on a common axis of rotation. They are commonly used in car steering systems to make sure that when a car rounds a corner, the inside wheel spins less than the outside wheel. In our erg, we use the differential to split an input force from the handle to the two output forces that act on flywheel and the tires.
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Subscripts
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\large L, R, M |
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the
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left
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half-shaft,
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right
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half-shaft,
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and
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middle
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gear,
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respectively.Three
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observations:
- When we hold the right axle fixed and turn the left axle one revolution, the middle sprocket makes half a revolution in the same angular direction.
- Holding the middle sprocket in place and turning the left axle one revolution causes the right axle to make one revolution in the opposite direction.
- The above observations are the same if the left and right axle are switched.
The results can be summarized as such:
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* When we hold the right axle fixed and turn the left axle one revolution, the middle sprocket makes half a revolution in the same angular direction. * Holding the middle sprocket in place and turning the left axle one revolution causes the right axle to make one revolution in the opposite direction. * The above observations are the same if the left and right axle are switched. The results can be summarized as such: {latex} \large\begin{equationalign*} \theta_{\text{M}} = \frac{1}{2}\theta_{\text{L}} + \frac{1}{2}\theta_{\text{R}} \end{equationalign*} {latex} where {latex} |
where
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\large $\theta$ |
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angular
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position.
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In
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other
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words,
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the
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position
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of
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the
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middle
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gear
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is
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half
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the
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sum
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of
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the
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left
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and
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right
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gears
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.
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Say
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that
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\large $\omega$ |
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angular
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velocity,
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\large $\alpha$ |
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angular
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acceleration,
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and
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\large $\tau$ |
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torque.
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\large $r$ |
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the
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radius
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of
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the
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gear,
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and
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\large $v$ |
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the
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linear
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velocity
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at
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the
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edge
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of
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a
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gear.
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From
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the
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first
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equation,
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we
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can
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find
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the
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relationships
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between
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the
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angular
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and
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linear
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velocities
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of
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the
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gears.
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{latex} \large \begin{align*} \omega_{\text{M}} =& \frac{1}{2} \left( \omega_{\text{L}} + \omega_{\text{R}} \right) \\ \omega_{\text{M}} r_{\text{M}} =& \frac{1}{2} ( \omega_{\text{L}} r_{\text{M}} + \omega_{\text{R}} r_{\text{M}}) \\ v_{\text{M}} =& \frac{1}{2} \left( \omega_{\text{L}} r_{\text{M}} \frac{r_{\text{L}}}{r_{\text{L}}} + \omega_{\text{R}} r_{\text{M}} \frac{r_{\text{R}}}{r_{\text{R}}} \right) \\ v_{\text{M}} =& \frac{1}{2} \left( v_{\text{L}} \frac{r_{\text{M}}}{r_{\text{L}}} + v_{\text{R}} \frac{r_{\text{M}}}{r_{\text{R}}} \right) \end{align*} {latex} |
We
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assume
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that
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these
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ideal
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gears
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are
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and
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massless.
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Therefore,
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we
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can
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use
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conservation
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of
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energy
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to
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say
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that
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input
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power
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equals
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output
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power.
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Say
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\large $P$ |
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power
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as
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a
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function
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of
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time.
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} \large \begin{align*} &P_{\text{M}} = P_{\text{L}} + P_{\text{R}} \\ &\tau_{\text{M}} \cdot \omega_{\text{M}} = \tau_{\text{L}} \cdot \omega_{\text{L}} + \tau_{\text{R}} \cdot \omega_{\text{R}} \\ &\tau_{\text{M}} \cdot \left[ \frac{1}{2} \left( \omega_{\text{L}} + \omega_{\text{R}} \right) \right] = \tau_{\text{L}} \cdot \omega_{\text{L}} + \tau_{\text{R}} \cdot \omega_{\text{R}} \end{align*} {latex} |
If
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we
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consider
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\large $\omega_{\text{L}}$ |
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$\omega_{\text{R}}$ |
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we
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find
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that
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} \large \begin{align*} \frac{1}{2} \tau_{\text{M}} \omega_{\text{L}} &= \tau_{\text{L}} \omega_{\text{L}}, \;\;\; \frac{1}{2} \tau_{\text{M}} \omega_{\text{R}} = \tau_{\text{R}} \omega_{\text{R}} \\ \frac{1}{2} \tau_{\text{M}} &= \tau_{\text{L}}, \;\;\;\;\;\;\; \;\; \; \;\; \frac{1}{2} \tau_{\text{M}} = \tau_{\text{R}} \end{align*} {latex} |
Therefore,
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we
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can
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conclude
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that
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the
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relationships
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between
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torques
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is
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the
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same
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as
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the
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relationships
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between
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angular
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position;
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a
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torque
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on
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the
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middle
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gear
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is
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evenly
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divided
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between
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the
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torque
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on
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the
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left
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gear
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and
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the
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torque
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on
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the
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right
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gear.
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\large
\begin{align*}
\tau_{\text{M}} &= \tau_{\text{L}} + \tau_{\text{R}} \\
\tau_{\text{L}} &= \tau_{\text{R}} = \frac{1}{2} \tau_{\text{M}}
\end{align*}
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| Widget Connector | ||
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