Escape velocity from a rotating black hole

2 hours left to award the bounty, but no answear so far. At least I found out that the local 3-velocity $$\rm v=\sqrt{v_r^2+v_{\theta}^2-v_{\phi}^2}=\sqrt{v_x^2+v_y^2-v_z^2}$$ is related to the coordinate derivatives $$\rm \dot r, \ \dot \theta, \ \dot \phi$$ by $$\rm \frac{v_{r}}{\sqrt{1-v^2}} = \dot r \ \sqrt{\frac{\Sigma}{\Delta}}$$ for the radial component, $$\rm \frac{v_{\theta} \ \sqrt{\Sigma}}{\sqrt{1-v^2}} = \dot \theta \ \Sigma$$ with the radius of gyration $$\bar R = \sqrt{\frac{\left(a^2+r^2\right)^2-a^2 \Delta \sin ^2 \theta }{a^2 \cos ^2 \theta +r^2}}$$ for the motion along the latitude and $$\rm \frac{v_{\phi} \ \bar{R}}{\sqrt{1-v^2}} = \dot \phi \ \Sigma$$ for the motion along the axis of symmetry, with the terms $$\rm \Sigma = r^2 + a^2 \cos^{2} \theta, \quad \Delta = r^2 - 2 \ r + a^2$$ and $\rm x, \ y, \ z$ as the cartesian transformation of the Boyer-Lindquist coordinates.

With the gravitational time dilation

$$\rm \varsigma =\sqrt{\frac{\left(a^2+r^2\right)^2-a^2 \left(a^2+(r-2) r\right) \sin ^2(\theta )}{\left(a^2+(r-2) r\right) \left(a^2 \cos ^2(\theta )+r^2\right)}}$$

and

$$\rm \varsigma = \frac{1}{\sqrt{1-v_{esc}^2 }}$$

the radial escape velocity should be

$$\rm v_{esc}=\frac{\sqrt{\varsigma ^2-1}}{\varsigma }$$

which, in the limit of $a\to 0$ gives the same result as Schwarzschild.

Example of a testparticle being thrown radially upwards with the local escape velocity by a corotating ZAMO sitting close above the horizon at θ=45°:

(The Spin parameter in the animation is a=Jc/G/M²=0.998)