Asymmetrical Pythagorean-hodograph spline-based ${{\mathrm{C}}}^{4}$ continuous local corner smoothing method with jerk-continuous feedrate scheduling along linear toolpath

Abstract

In computer numerical control systems, linear segments generated by computer-aided manufacturing software are the most widely used toolpath format. Since the linear toolpath is discontinuous at the junction of two adjacent segments, the fluctuations on velocity, acceleration and jerk are inevitable. Local corner smoothing is widely used to address this problem. However, most existing methods use symmetrical splines to smooth the corners. When any one of the linear segments at the corner is short, the inserted spline will be micro to avoid overlap. This will increase the curvature extreme of the spline and reduce the feedrate on it. In this article, the corners are smoothed by a ${C}^{4}$ continuous asymmetric Pythagorean-hodograph (PH) spline. The curvature extreme of the proposed spline is investigated first, and $K=2.5$ is determined as the threshold to constrain the asymmetry of the spline. Then, a two-step strategy is used to generate a blended toolpath composed of asymmetric PH splines and linear segments. In the first step, the PH splines at the corners are generated under the condition that the transition lengths do not exceed half of the length of the linear segments. In the second step, the splines at the corners are re-planned to reduce the curvature extremes, if the transition error does not reach the given threshold and there are extra linear trajectories on both sides of the spline trajectory. Finally, the bilinear interpolation method is applied to determine the critical points of the smoothed toolpath, and a jerk-continuous feedrate scheduling scheme is presented to interpolate the smoothed toolpath. Simulations show that, under the condition of not affecting the machining quality, the proposed method can improve the machining efficiency by $7.27$ to $75.11\%$ compared to ${G}^{3}$ and ${G}^{4}$ methods.

Appendices

Appendix 1. Proof process of Theorem 1

When trigonometric functions are used to express $\overrightarrow{{n}_{1}}$ and $\overrightarrow{{n}_{2}}$, $\overrightarrow{{n}_{1}}={l}_{1}\overrightarrow{{T}_{1}}={l}_{1}{\left(cos\alpha ,sin\alpha \right)}^{T}={\left({u}_{0}^{2}-{v}_{0}^{2},2{u}_{0}{v}_{0}\right)}^{T}/15$ and $\overrightarrow{{n}_{2}}={l}_{2}\overrightarrow{{T}_{2}}={l}_{2}{\left(cos\beta ,sin\beta \right)}^{T}={\left({u}_{7}^{2}-{v}_{7}^{2},2{u}_{7}{v}_{7}\right)}^{T}/15$. According to the half angle formula of trigonometric function, ${u}_{0}$, ${v}_{0}$, ${u}_{7}$ and ${v}_{7}$ can be written as ${u}_{0}=\pm \sqrt{15{l}_{1}}cos\frac{\alpha }{2}$, ${v}_{0}=\pm \sqrt{15{l}_{1}}sin\frac{\alpha }{2}$, ${u}_{7}=\pm \sqrt{15{l}_{2}}cos\frac{\beta }{2}$ and ${v}_{7}=\pm \sqrt{15{l}_{2}}sin\frac{\beta }{2}$, where ${u}_{0}$ and ${v}_{0}$, ${u}_{7}$ and ${v}_{7}$ take the same sign. Substituting the above four parameters into $\overrightarrow{m}=\frac{{\left({u}_{0}{u}_{7}-{v}_{0}{v}_{7},{u}_{0}{v}_{7}+{u}_{7}{v}_{0}\right)}^{T}}{15}$, $\overrightarrow{{B}_{7}{B}_{8}}$ can be written as $\overrightarrow{{B}_{7}{B}_{8}}=\overrightarrow{m}=\pm \sqrt{{l}_{1}{l}_{2}}{\left(cos\frac{\alpha +\beta }{2},sin\frac{\alpha +\beta }{2}\right)}^{T}$. It is obvious that the length of $\overrightarrow{{B}_{7}{B}_{8}}$ equals to $\sqrt{{l}_{1}{l}_{2}}$. In addition, according to the sum-to-product trigonometric identities, $\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}={\left(2cos\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2},2sin\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}\right)}^{T}=2cos\frac{\alpha -\beta }{2}{\left(cos\frac{\alpha +\beta }{2},sin\frac{\alpha +\beta }{2}\right)}^{T}$. Since ${P}_{0}{P}_{1}$ and ${P}_{1}{P}_{2}$ are two different linear segments, and they are not on the same line, $cos\frac{\alpha -\beta }{2}\ne 0$ can be deduced. If $cos\frac{\alpha -\beta }{2}>0$, $\frac{\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}}{\left|\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}\right|}={\left(cos\frac{\alpha +\beta }{2},sin\frac{\alpha +\beta }{2}\right)}^{T}$, let ${u}_{0}$ and ${v}_{0}$ take the same sign, then $\frac{\overrightarrow{{B}_{7}{B}_{8}}}{\left|\overrightarrow{{B}_{7}{B}_{8}}\right|}{=\left(cos\frac{\alpha +\beta }{2},sin\frac{\alpha +\beta }{2}\right)}^{T}=\frac{\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}}{\left|\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}\right|}$. On the contrary, let ${u}_{0}$ and ${v}_{0}$ take different signs, then $\frac{\overrightarrow{{B}_{7}{B}_{8}}}{\left|\overrightarrow{{B}_{7}{B}_{8}}\right|}{=-\left(cos\frac{\alpha +\beta }{2},sin\frac{\alpha +\beta }{2}\right)}^{T}=\frac{\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}}{\left|\overrightarrow{{T}_{1}}+\overrightarrow{{T}_{2}}\right|}$.

Appendix 2. Proof process of Theorem 2

According to Eq. (9), $\overrightarrow{{B}_{4}{B}_{7}}=\overrightarrow{{B}_{4}{B}_{5}}+\overrightarrow{{B}_{5}{B}_{6}}+\overrightarrow{{B}_{6}{B}_{7}}=\frac{398}{429}\overrightarrow{m}+\frac{889}{429}\overrightarrow{{n}_{1}}$. At the same time, $\overrightarrow{{B}_{4}{B}_{7}}=\overrightarrow{{B}_{4}{Q}_{1}}+\overrightarrow{{Q}_{1}{B}_{7}}$. Since $\overrightarrow{m}$ and $\overrightarrow{{n}_{1}}$ are two different vectors, $\overrightarrow{{B}_{4}{B}_{7}}$ has only unique expression when expressed by base {$\overrightarrow{m}, \overrightarrow{{n}_{1}}\}$. Considering $\overrightarrow{{B}_{4}{Q}_{1}}$ and $\overrightarrow{{n}_{1}}$, $\overrightarrow{{Q}_{1}{B}_{7}}$ and $\overrightarrow{m}$ have the same directions, it can be deduced that $\overrightarrow{{B}_{4}{Q}_{1}}=\frac{889}{429}\overrightarrow{{n}_{1}}$ and $\overrightarrow{{Q}_{1}{B}_{7}}=\frac{398}{429}\overrightarrow{m}$, i.e. $\left|\overrightarrow{{B}_{4}{Q}_{1}}\right|=\frac{889}{429}{l}_{1}$. Similarly, $\left|\overrightarrow{{B}_{11}{Q}_{2}}\right|=\frac{889}{429}{l}_{2}$. From Theorem 1, it is known that ${B}_{7}{B}_{8}$ and the angle bisector of $\mathrm{\angle }{{P}_{0}P}_{1}{P}_{2}$ are perpendicular. As shown in Fig. 1, the angle bisector of $\mathrm{\angle }{{P}_{0}P}_{1}{P}_{2}$ and ${B}_{7}{B}_{8}$ intersect at point ${B}_{m}$; then, ${B}_{m}$ is the midpoint of ${B}_{7}{B}_{8}$. In right triangle ∆${Q}_{1}{P}_{1}{B}_{m}$, $\left|\overrightarrow{{B}_{m}{Q}_{1}}\right|=\left|\overrightarrow{{Q}_{1}{B}_{7}}\right|+\frac{\left|\overrightarrow{{B}_{7}{B}_{8}}\right|}{2}=\frac{1225}{858}\sqrt{{{l}_{1}l}_{2}}$, $\angle {{Q}_{1}P}_{1}{B}_{m}=\frac{\pi -\theta }{2}$, then $\left|\overrightarrow{{Q}_{1}{P}_{1}}\right|=\frac{1225\sqrt{{{l}_{1}l}_{2}}}{858\mathrm{cos}\frac{\theta }{2}}$. Similarly, $\left|\overrightarrow{{Q}_{2}{P}_{1}}\right|=\frac{1225\sqrt{{{l}_{1}l}_{2}}}{858cos\frac{\theta }{2}}$.

Appendix 3. Proof process of Theorem 3

When ${l}_{1}={l}_{2}$, the proposed PH spline is symmetrical about the angle bisector of $\angle {{P}_{0}P}_{1}{P}_{2}$; then, the maximum deviation error occurs between the mid-point of the proposed PH spline and corner point ${P}_{1}$, which can be expressed as

$$\begin{aligned}{e}_{max}&=\left|r\left(0.5\right)-{P}_{1}\right|=\frac{1}{{2}^{15}}|{(B}_{0}-{P}_{1})+15{(B}_{1}-{P}_{1})\\&+105{(B}_{2}-{P}_{1})+455{(B}_{3}-{P}_{1})+1365{(B}_{4}-{P}_{1})\\&+3003{(B}_{5}-{P}_{1})+5005{(B}_{6}-{P}_{1})+6435{(B}_{7}-{P}_{1})\\&+6435{(B}_{8}-{P}_{1})+5005{(B}_{9}-{P}_{1})+3003{(B}_{10}-{P}_{1})\\&+1365{(B}_{11}-{P}_{1})+455{(B}_{12}-{P}_{1})+105{(B}_{13}-{P}_{1})\\&+15{(B}_{14}-{P}_{1})+{(B}_{15}-{P}_{1})|=\frac{1}{{2}^{15}}|{(B}_{0}+{B}_{15}-{2P}_{1})\\&+15{(B}_{1}+{B}_{14}-{2P}_{1})+105{(B}_{2}+{B}_{13}-{2P}_{1})\\&+455{(B}_{3}+{B}_{12}-{2P}_{1})+1365{(B}_{4}+{B}_{11}-{2P}_{1})\\&+3003{(B}_{5}+{B}_{10}-{2P}_{1})+5005{(B}_{6}+{B}_{9}-{2P}_{1})\\&+6435{(B}_{7}+{B}_{8}-{2P}_{1})|.\end{aligned}$$

(24)

From Eq. (10), it can be deduced that

$$\left\{\begin{array}{c}\begin{array}{c}{B}_{0}+{B}_{15}-{2P}_{1}=(4{l}_{1}+\frac{889}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\\ {B}_{1}+{B}_{14}-{2P}_{1}=(3{l}_{1}+\frac{889}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\\ {B}_{2}+{B}_{13}-{2P}_{1}=(2{l}_{1}+\frac{889}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\end{array}\\ \begin{array}{c}{B}_{3}+{B}_{12}-{2P}_{1}=({l}_{1}+\frac{889}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\\ {B}_{4}+{B}_{11}-{2P}_{1}=(\frac{889}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\\ \begin{array}{c}{B}_{5}+{B}_{10}-{2P}_{1}=(\frac{490}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\\ {B}_{6}+{B}_{9}-{2P}_{1}=(\frac{175}{429}{l}_{1}+l)(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\\ {B}_{7}+{B}_{8}-{2P}_{1}=l(\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}})\end{array}\end{array}\end{array}.\right.$$

(25)

Combining Eqs. (24) and (25), the maximum deviation error ${e}_{\mathrm{max}}$ is expressed as

$${e}_{\mathrm{max}}=\frac{1}{{2}^{15}}\left|\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}}\right|\left(\left(\frac{397}{429}+10207\right){l}_{1}+16384l\right).$$

(26)

Considering $\left|\overrightarrow{{T}_{2}}-\overrightarrow{{T}_{1}}\right|=2\mathit{sin}\left(\frac{\theta }{2}\right)$ and $l=\frac{16384\times 1225}{858cos(\frac{\theta }{2})}\sqrt{{{l}_{1}l}_{2}}=\frac{16384\times 1225}{858cos(\frac{\theta }{2})}{l}_{1}$, Eq. (13) can be obtained and Theorem 3 is proved.

Appendix 4. Proof process of Theorem 4

If ${l}_{1}={l}_{2}$, $R(t)$ is the same as $r(t)$ and ${e}_{r}={e}_{R}$. Without loss of generality, translate, rotate and, if necessary, reflect transition PH spline such that ${P}_{1}$ is at the origin, ${P}_{0}$ is on the negative x-axis, ${P}_{2}$ is above the x-axis and ${l}_{1}>{l}_{2}$. Replace ${l}_{1}$ and ${l}_{2}$ in Eq. (10) with ${l}_{0}=\mathrm{max}\left({l}_{1},{l}_{2}\right)={l}_{1}$ to get the control points of $R(t)$, which is denoted by ${\overline{B} }_{i} (i=\mathrm{0,1},2,\dots , 15)$. Subtracting ${B}_{i}$ from ${\overline{B} }_{i}$, we arrive at

$$\left\{\begin{array}{c}\begin{array}{c}\begin{array}{c}{\overline{B} }_{0}-{B}_{0}=-\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}(\sqrt{{l}_{1}}-\sqrt{{l}_{2}})\overrightarrow{{T}_{1}}\\ {\overline{B} }_{1}-{B}_{1}=-\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}(\sqrt{{l}_{1}}-\sqrt{{l}_{2}})\overrightarrow{{T}_{1}}\\ {\overline{B} }_{2}-{B}_{2}=-\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}(\sqrt{{l}_{1}}-\sqrt{{l}_{2}})\overrightarrow{{T}_{1}}\end{array}\\ \begin{array}{c}{\overline{B} }_{3}-{B}_{3}=-\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}(\sqrt{{l}_{1}}-\sqrt{{l}_{2}})\overrightarrow{{T}_{1}}\\ {\overline{B} }_{4}-{B}_{4}=-\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}(\sqrt{{l}_{1}}-\sqrt{{l}_{2}})\overrightarrow{{T}_{1}}\\ {\overline{B} }_{5}-{B}_{5}=-\frac{1195}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{1}}+\frac{10}{286cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{2}}\end{array}\\ \begin{array}{c}{\overline{B} }_{6}-{B}_{6}=-\frac{1081}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{1}}+\frac{48}{286cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{2}}\\ {\overline{B} }_{7}-{B}_{7}=-\frac{827}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{1}}+\frac{398}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{2}}\\ {\overline{B} }_{8}-{B}_{8}=-\frac{398}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{1}}+\frac{827}{286cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{2}}\end{array}\end{array}\\ \begin{array}{c}{\overline{B} }_{9}-{B}_{9}=-\frac{48}{286cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{1}}+\left(\frac{175}{429}\left({l}_{1}-{l}_{2}\right)+\frac{1081}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\\ {\overline{B} }_{10}-{B}_{10}=-\frac{10}{286cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\overrightarrow{{T}_{1}}+\left(\frac{490}{429}\left({l}_{1}-{l}_{2}\right)+\frac{1195}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\\ {\overline{B} }_{11}-{B}_{11}=\left(\frac{889}{429}\left({l}_{1}-{l}_{2}\right)+\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\end{array}\\ \begin{array}{c}{\overline{B} }_{12}-{B}_{12}=\left(\left(\frac{889}{429}+1\right)\left({l}_{1}-{l}_{2}\right)+\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\\ {\overline{B} }_{13}-{B}_{13}=\left(\left(\frac{889}{429}+2\right)\left({l}_{1}-{l}_{2}\right)+\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\\ \begin{array}{c}{\overline{B} }_{14}-{B}_{14}=\left(\left(\frac{889}{429}+3\right)\left({l}_{1}-{l}_{2}\right)+\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\\ {\overline{B} }_{15}-{B}_{15}=\left(\left(\frac{889}{429}+4\right)\left({l}_{1}-{l}_{2}\right)+\frac{1225}{858cos\frac{\theta }{2}}\sqrt{{l}_{1}}\left(\sqrt{{l}_{1}}-\sqrt{{l}_{2}}\right)\right)\overrightarrow{{T}_{2}}\end{array}\end{array}\end{array}\right.$$

(27)

It is obvious that ${\overline{B} }_{i}-{B}_{i}={a}_{i}\overrightarrow{{T}_{1}}+{b}_{i}\overrightarrow{{T}_{2}}$, ${a}_{i}\le 0$ and ${b}_{i}\ge 0$ for $i=\mathrm{0,1},\dots ,15$. Considering $r\left(t\right)={\sum }_{i=0}^{15}{B}_{i}{b}_{i}^{15}\left(t\right)$ and $R\left(t\right)={\sum }_{i=0}^{15}{\overline{B} }_{i}{b}_{i}^{15}\left(t\right)$, we have

$$\overrightarrow{r\left(t\right)R\left(t\right)}={\sum }_{i=0}^{15}\left({\overline{B} }_{i}-{B}_{i}\right){b}_{i}^{15}\left(t\right)={f}_{1}(t)\overrightarrow{{T}_{1}}+{f}_{2}(t)\overrightarrow{{T}_{2}}$$

(28)

where ${f}_{1}(t)\le 0$ and ${f}_{2}(t)\ge 0$ for arbitrary $t\in [0, 1]$. As shown in Fig. 24, the region bounded by the transition PH spline $r(t)$ and the linear trajectory ${P}_{0}{-P}_{1}-{P}_{2}$ is denoted as $\mathrm{S}$. Then, it can be deduced from Eq. (27) that $R\left(t\right)$ is outside of $\mathrm{S}$ for arbitrary $t\in [0, 1]$. From the convex hull property of Bezier spine, the transition PH spline $R(t)$ is inside of ∆${P}_{0}{P}_{1}{P}_{2}$. Then, for arbitrary ${t}_{1}\in [0, 1]$, there is an intersection point between the line segment $\overline{{P }_{1}R({t}_{1})}$ and the transition PH spline $r(t)$; in other words, there is a ${t}_{2}\in [0, 1]$ such that $\left|\overrightarrow{{P}_{1}r({t}_{2})}\right|\le \left|\overrightarrow{{P}_{1}R({t}_{1})}\right|$, which gives ${e}_{r}\le {e}_{R}$ naturally.

Fig. 24

The schematic diagram of two transition PH spline $r(t)$ and $R(t)$

Appendix 5. Parameters of butterfly-shaped curve

The degree: p = 3.

The control point (mm): P = [(54.493, 52.139), (55.507,52.139), (56.082, 49.615), (56.780, 44.971), (69.575,51.358), (77.786, 58.573), (90.526, 67.081), (105.973,63.801), (100.400, 47.326), (94.567, 39.913), (92.369,30.485), (83.440, 33.757), (91.892, 28.509), (89.444,20.393), (83.218, 15.446), (87.621, 4.830), (80.945, 9.267),(79.834, 14.535), (76.074, 8.522), (70.183, 12.550), (64.171,16.865), (59.993, 22.122), (55.680, 36.359), (56.925, 24.995),(59.765, 19.828), (54.493, 14.940), (49.220, 19.828), (52.060,24.994), (53.305, 36.359), (48.992, 22.122), (44.814, 16.865),(38.802, 12.551), (32.911, 8.521), (29.152, 14.535), (28.040,9.267), (21.364, 4.830), (25.768, 15.447), (19.539, 20.391),(17.097, 28.512), (25.537, 33.750), (16.602, 30.496), (14.199,39.803), (8.668, 47.408), (3.000, 63.794), (18.465, 67.084),(31.197, 58.572), (39.411, 51.358), (52.204, 44.971), (52.904,49.614), (53.478, 52.139), (54.492, 52.139)].

The knot vector: U = [0, 0, 0, 0, 0.0083, 0.015, 0.0361, 0.0855, 0.1293, 0.1509, 0.1931, 0.2273, 0.2435, 0.2561, 0.2692, 0.2889, 0.3170, 0.3316, 0.3482, 0.3553, 0.3649, 0.3837, 0.4005, 0.4269, 0.4510, 0.4660, 0.4891, 0.5000, 0.5109, 0.5340, 0.5489, 0.5731, 0.5994, 0.6163, 0.6351, 0.6447, 0.6518, 0.6683, 0.6830, 0.7111, 0.7307, 0.7439, 0.7565, 0.7729, 0.8069, 0.8491, 0.8707, 0.9145, 0.9639, 0.9850, 0.9917, 1.0, 1.0, 1.0, 1.0].

The weight vector: W = [1.0, 1.0, 1.0, 1.2, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1, 2, 1.0, 1.0, 5.0, 3.0, 1.0, 1.1, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.1, 1.0, 3.0, 5.0, 1.0, 1.0, 2.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.2, 1.0, 1.0, 1.0].