estimate the perimeter of the island

I'm assigned a task involving solving a problem that can be described as follows: Suppose I'm driving a car around a lake. In the lake there is an island of irregular shape. I have a GPS with me in the car so I know how far I've driven and every turns I've made. Now suppose I also have a camera that takes picture of the island 30 times a second, so I know how long sidewise the island appears to me all the time. Also assume I know the straight line distance between me and the island all the time. Given these conditions, if I drive around the lake for one full circle, will I be able to estimate the perimeter of the island? If yes, how? Thanks.

Assume the island $\Omega$ is convex. Parametrize the observed width $w$ of $\Omega$ with the direction from which it is observed. So we have a function $\phi\to w(\phi)$ of period $\pi$. The average observed width is $$\bar w:={1\over 2\pi}\int_0^{2\pi }w(\phi)\ d\phi\ .$$ Consider a line element $ds$ of $\partial\Omega$. Its contribution to $\bar w$ is half its average projection length over all directions $\phi$ (here we use the convexity of $\partial\Omega$), so it is $\ \lambda\thinspace\thinspace ds\ $ for some universal constant $\lambda>0$ yet to be determined. In any case we have a formula of the following kind:

$$\bar w = \lambda \int_{\partial\Omega} ds =\lambda\ L(\partial\Omega)\ .$$

When $\Omega$ is the unit disk then $\bar w=2$ and $L(\partial\Omega)=2\pi$. It follows that $\lambda={1\over\pi}$, and solving for $L(\partial\Omega)$ we get

$$L(\partial\Omega)=\pi\ \bar w={1\over 2}\int_0^{2\pi} w(\phi)\ d\phi\ .$$