Lingpoche, 凌波车

Consider this, every turn of time $t_g$ around the circular path, $t_g$ advances by $t_{\lambda\,c}$ along the axis of rotation.

$2\pi r=kt_{\lambda}$

$k$ represents the resistance along the time axis, for every $2\pi$ around the circular path, it advances by one unit along the axis.  If $k$ is measured in unit per radian, it is a constant.

In general, for a radial path of $r_v$,

$2\pi r_v=k_vt_{\lambda\,c}$

$k_v=2\pi\cfrac{ r_v}{t_{\lambda\,c}}$

$v_t=\cfrac{c}{k_v}=\cfrac{ct_{\lambda\,c}}{2\pi r_v}$

where

$\cfrac{\partial\,t_{g\,t}}{\partial\,t}=v_t$

But when $v_c$ is normal time speed.

$T_c=\cfrac{2\pi r_{c }}{c}=\cfrac{t_{\lambda\,c}}{v_c}$ --- (1)

$ct_{\lambda\,c}=2\pi r_{c }v_c$

Therefore,

$v_{ t }=\cfrac { r_{ c }v_{ c } }{ r_{ v } }$

When $r_v$ is varied in general, $f_v$, the frequency varies, $k_v$ meausred in unit per perimeter changes such that for every cycle around the circular path, time advances by $t_{\lambda\,c}$ still.  $k$ when measured in unit per radian (per $2\pi$) however, remains a constant.

Alternatively,

$T_v=\cfrac{2\pi r_{v }}{c}=\cfrac{t_{\lambda\,c}}{v_t}$ --- (2)

We formulate (1) divided by (2),

$\cfrac{T_c}{T_v}=\cfrac{r_{c }}{r_{v }}=\cfrac{v_t}{v_c}$

Since $r_c$ is small it seems that we can only go backwards in time with this.

$r_v\gt r_c$  

$v_t\lt v_c$

$T_c\lt T_v$

For every $T_v$, many $T_c$ would have passed, $v_c$ rushes forward and we travel back in time.

Maybe $v_t$ can be reduced further by counter rotating, in which case,

$T_v=\cfrac{2\pi r_{v }}{v}=\cfrac{t_{\lambda\,c}}{v_t}$

And,

$\cfrac{T_c}{T_v}=\cfrac{r_{c }}{r_{v }}\cfrac{v}{c}=\cfrac{v_t}{v_c}$

When,

${r_v}\gt{r_c}$

${r_v}c\gt\gt{r_c}v$

such that,

$v_t\lt\lt v_c$

$T_c\lt\lt T_v$

If it is possible for,

${r_v}\lt{r_c}$

then,

$v_t\gt v_c$

Normal time recedes, as we travel forward in time.

Time travels in a helix at constant speed.  Time advances after every cycle. $k$ measured in unit per radian is a constant. When the radius of the time circle is increased, at constant speed around the circular path, time advances is slowed.  Normal time is in constant flux, as such normal time goes forward as we travel back in time.  If we rotate the sonic cone anti-clockwise, $v_t$ slows down further.

What exactly is "lingpo"?  A sonic cone at 7.489 Hz.  Two of such cones to form an enclosure.

And off you go, in time.