yields to which have a maximum at θ = 1 / 2 arctan(4 / 3) ≈ 0 . 464. A graph of this results appears in figure 4(red line). Note that this value is different from that we obtain if we eval- uate θ ( α = π/ 4) = 1 / 2, the launch angle of maximum range. The maximum curvature happens at a smaller angle than the angle of maximun range. It is interesting to note that the angle 2 θ = arctan(4 / 3) corresponds to a triangle which sides fulfill the relation 3 2 + 4 2 = 5 $^{2}$, a Pythagoras’ triple. Using the numerical results for α ( θ ) from the previous section, it is possible to carry on the calculation of K (see figure 4) performing the derivatives of ˜ r from Eq. (13) in a direct form and evaluating numerically the re- quired values of α and its derivatives. For the required derivatives of W ( z ) we used the expressions [2] and Using this method, we obtained good results for values of ε up to ≈ 1 but we require to calculate arguments of the Lambert W function near the limit z = − 1 /e for larger values of ε . The reliability of our numerical result was done comparing the first and second derivatives of r ( θ ) with those corresponding to the ellipse. As can be seen in figure 4, K present a maximum in all the cases which can be calculated as well. We left to an ulterior work the analysis of the maxima distribution as a function of the pertubative parameter, that is not the case for the curvature with α parameterization as we shall see in the next section. For the launch angle parameterization of C m we shall use expression [10] for calculate the curvature from the rectangular form of equations (7) and (8), where the primes denote derivative respect the parameterization variable, α in this case. A direct calculation yields which have a maximum at α = π/ 4 in the interval α ∈ [0 , π/ 2], as expected. A plot of ρκ ( α ) for several values of ε beginning at zero and ending at ε = 10 appears in Figure 5(a). In red appears the drag-free case. Note that both extremal values increase for increasing ε value as ρκ (0) ≈ (16 + 6 $^{ε}$− 6 ε $^{2}$) and ρκ ( π/ 2) ≈ ε . Notice that for small ε , the κ crosses the drag-free curvature κ$_{0}$ . The angles α ∗ at which κ attain their maxima are ob- tained in the usual way and requires to solve, numerical or graphically, the equation with In the limit ε → 0, we recover the drag-free curvature and With FIG. 4: (color online) Curvature of C m ( ε ) using the polar angle as the parameter from Eq. (19). In red line appears the corresponding result for the ellipse, Eq. (20). The maximum happens at θ = 1 / 2 arctan(4 / 3), different from the value of maximum range α = π/ 4. B. Launch angle parameterization. 4 K$_{0}$ = 1 $_{2}$√$_{2}$$_{ρ}$ (5 − 3 cos 2 θ ) 6 (1 + 3 sin 2 θ ) $^{3}$(47 60 cos 2 θ + 21 cos 4 θ ) 3 / $_{2}$, (20) d dx W ( x ) = W ( x ) x (1 + W ( x )) , (21) d $^{2}$W ( x ) dx 2 = − $^{exp(}$−2 W ( x ))( W ( x ) + 2) (1 + W ( x )) 3 . (22) P $_{1}$( α ) = 16 + 6 ε 2 − 8 ε $^{2}$cos 2 α + 2 ε $^{2}$cos 4 α +30 ε sin α − ε sin 3 α + ε sin 5 α, (25) P $_{2}$( α ) = 5 + 3 cos 4 α + 3 ε 2 − 4 ε $^{2}$cos 2 α + ε $^{2}$cos 4 α + 10 ε sin α − 5 ε sin 3 α + ε sin 5 α. (26) κ$_{0}$ = $_{16}$√$_{2}$ ρ 1 (5 + 3 cos 4 α ) 3 / $_{2}$, (27) κ = x ′ y ′′ − y ′ x ′′ ( x ′ 2 + y ′ $^{2}$)3 / $_{2}$, (23) 3(1 + ε sin α ∗ ) sin 2 α ∗ × Q $_{1}$( α ∗ $^{)}$Q $_{2}$( α ∗ )+ ε cos α $^{∗}$Q $_{3}$( α $^{)}$Q $_{4}$( α ) = 0 , (28) κ = √$_{2}$ ρ P 1 ( α ) ( √$_{P}$ 2 ( α )) 3 × (1 + ε sin α ) $^{2}$. (24) Q 1 = 4 (3 + ε $^{2}$) cos 2 α ∗ + ε $^{(}$−4 ε − 15 sin α ∗ + 5 sin 3 α ∗ ); (29) Q 2 = 16 + 6 ε 2 − 8 ε $^{2}$cos 2 α ∗ + 2 ε $^{2}$cos 4 α ∗ + 30 ε sin α ∗ ε sin 3 α ∗ + ε sin 5 α ∗ ; (30) 0 0.5 1 1.5 0 20 40 60 80