Numerical example

The following example show the first iterations of the algorithm for a very flat ellipse:

    $\displaystyle \left\{\begin{aligned}a_e &= 100\\ f &= 0.9 \end{aligned}\right.$

We consider a test point $ A$ having the following coordinates:

$\displaystyle T\left\{\begin{aligned}\varphi &= 75^\circ\\ h &= 0.1 \end{aligne...
...\begin{aligned}r &= 93.7139699\\ z &= \phantom{0}3.5930796 \end{aligned}\right.$

We assume we know only $ r$ and $ z$ and want to compute the unknown $ \varphi$ and $ h$.

Equations (6) give us the tangent of the half slope:

    $\displaystyle t_0=0.0191634$

Applying equation (7), we get the signed distance between $ A$ and the intersection of the first line and the ellipse:

    $\displaystyle k_0= 0.3419763 \Rightarrow \varphi_0=75.3818944^\circ$

>From this value of $ k$, we deduce the first coefficients $ \alpha$, $ \beta$, $ \gamma$ and $ \delta$ of the quartic (we don't compute $ \varepsilon$). We reduce it to a cubic using the already known root $ t_0$:

$\displaystyle \tau^3 + a_1 \tau^2 + a_2 \tau + a_3=0$   where$\displaystyle \quad \left\{\begin{aligned}a_1 &= -3.5542558\\ a_2 &= \phantom{-}1.3365805\\ a_3 &= -3.5478057\\ \end{aligned}\right.$

This polynomial has only one real root:

    $\displaystyle \tilde{t}_0=3.4640705 \Rightarrow \tilde{\zeta}_0=147.7954987^\circ \Rightarrow \tilde{\varphi}_0=74.5916410^\circ$

The middle point between $ \varphi_0$ and $ \tilde{\varphi}_0$ is the best approximation we have at the end of this first iteration:

    $\displaystyle \varphi_1=74.9867677^\circ$

We know an upper bound for the error of this approximation is the half-width of the interval, which is $ 0.3951267^\circ$ (in fact the error is about 30 times smaller). As we consider this is too much, we compute the starting values for a second iteration:

    $\displaystyle k_1=0.1005980 \quad t_1=0.8577741$

The steps of the second iteration are similar to the steps of the first one. They lead to the following result:

    $\displaystyle \tilde{\varphi}_1=75.0132026^\circ$

So at the end of the second iteration, we have an estimate $ \varphi_2=74.9999851^\circ$ and an upper bound of its error of $ 0.0132174^\circ$ (the real error being almost 900 times smaller).

Luc Maisonobe 2006-02-04