!differentialwithangles.png|border=1!
 
Subscripts {latex} \large L, R, M{latex} correspond the left half-shaft, right half-shaft, and middle gear, respectively.Three 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: 
{latex}
\large\begin{equation*}
\theta_{\text{M}} = \frac{1}{2}\theta_{\text{L}} + \frac{1}{2}\theta_{\text{R}}
\end{equation*} 
{latex}
where {latex} \large $\theta$ {latex} represents angular position. In other words, the *position of the middle gear is half the sum of the left and right gears*. Say that{latex} \large $\omega$ {latex} represents angular velocity, {latex} \large $\alpha${latex} represents angular acceleration, and {latex} \large $\tau$ {latex} represents torque. {latex} \large $r$ {latex} is the radius of the gear, and {latex} \large $v$ {latex} is the linear velocity at the edge of a gear. From the first equation, we can find the relationships between the angular and linear velocities of the gears. 
{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 assume that these ideal gears are [frictionless|http://xkcd.com/669/] and massless. Therefore, we can use conservation of energy to say that input power equals output power. Say {latex} \large $P$ {latex} represents power as a function of time.
{latex}
\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 we consider {latex}\large $\omega_{\text{L}}$ {latex} and {latex} $\omega_{\text{R}}$ {latex} separately, we find that
{latex}
\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, we can conclude that the relationships between torques is the same as the relationships between angular position; *a torque on the middle gear is evenly divided between the torque on the left gear and the torque on the right gear.*
{latex}
\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*}
{latex}