I am somewhat stymied on what appears to be a simple formula. Here is the problem statement:
Assume that a rigid body is traveling with constant velocity $v$, and is rotating with a constant yaw rate of $\dot{\theta}$. Find the distance travelled in one time step, $\Delta t$.
My professor then goes on to say that:
The instantaneous yaw-angle at a given time step is: $\theta = \dot{\theta} \Delta t$. The distance travelled in one time step is therefore given by:
$$ d = \frac{2 v \Delta t \ \text{sin}(\frac{\theta}{2})}{\theta}$$
I am completely unclear as to how this was possibly derived. In order to convince myself that this at least made sense, I set $\theta = 0$ and used l'opitals rule to derive $d$ in the case of no yaw or yaw-rate, and we get the trivial solution. However I am still unclear as to how this formula was derived in the first place. Many thanks.
Okay, let the constant yaw-rate be $\omega$. As the body has a constant linear forward velocity of $v$, it must follow a perfectly circular path.
Let us consider the path of the body from $t=0$ to $t=\Delta t$.
We see from the figure that: $$\phi_{G} + R = \frac{\pi}{2}$$ $$\theta + 2R = \pi$$ $$\Rightarrow \theta = 2\phi_{G}$$
From the sine rule in the triangle we get: $$\frac{sin\theta}{r_{G}} = \frac{sinR}{r_{C}}$$ $$\Rightarrow \frac{sin2\phi_{G}}{r_{G}} = \frac{sin(\frac{\pi}{2} - \phi_{G})}{r_{C}}$$ $$\Rightarrow r_{G} = 2r_{C}sin \phi_{G}$$ $$\Rightarrow r_{G} = 2r_{C}sin \frac{\theta}{2}$$
Now, we know that: $$\theta = \omega \Delta t$$ And, (from rotational mechanics) $$v = r_{C} \omega$$ $$\Rightarrow r_{C} = \frac{v}{\omega} = \frac{v \Delta t}{\theta}$$ Therefore, from the previous equation we see that: $$r_{G} = \frac{2 v \Delta t sin(\frac{\theta}{2})}{\theta}$$
Same as your equation for $d$.