A cubic Bézier curve is defined by four points: $P_0,P_1,P_2,$ and $P_3$.

The curve is constructed as follows:

On the segments $P_0{P}_1$, $P_1{P}_2$, and $P_2{P}_3$ the points $Q_0,Q_1,$ and $Q_2$ are drawn such that ${\displaystyle \frac{P_0{Q}_0}{P_0{P}_1}}={\displaystyle \frac{P_1{Q}_1}{P_1{P}_2}}={\displaystyle \frac{P_2{Q}_2}{P_2{P}_3}}=t$, with $t$ in $[0,1]$.

On the segments $Q_0{Q}_1$ and $Q_1{Q}_2$ the points $R_0$ and $R_1$ are drawn such that
${\displaystyle \frac{Q_0{R}_0}{Q_0{Q}_1}}={\displaystyle \frac{Q_1{R}_1}{Q_1{Q}_2}}=t$ for the same value of $t$.

On the segment $R_0{R}_1$ the point $B$ is drawn such that ${\displaystyle \frac{R_{0}B}{R_0{R}_1}}=t$ for the same value of $t$.

The Bézier curve defined by the points $P_0,P_1,P_2,P_3$ is the locus of $B$ as $Q_0$ takes all possible positions on the segment $P_0{P}_1$.
(Please note that for all points the value of $t$ is the same.)

From the construction it is clear that the Bézier curve will be tangent to the segments $P_0{P}_1$ in $P_0$ and $P_2{P}_3$ in $P_3$.

A cubic Bézier curve with $P_0=(1,0),P_1=(1,v),P_2=(v,1),$ and $P_3=(0,1)$ is used to approximate a quarter circle.
The value $v>0$ is chosen such that the area enclosed by the lines $O{P}_0,O{P}_3$ and the curve is equal to ${\displaystyle \frac{\pi }{4}}$ (the area of the quarter circle).

By how many percent does the length of the curve differ from the length of the quarter circle?
That is, if $L$ is the length of the curve, calculate $100\times {\displaystyle \frac{L-\frac{\pi }{2}}{\frac{\pi }{2}}}$
Give your answer rounded to 10 digits behind the decimal point.