Derivation of path equation for automobile steering so I have this equation related to the picture. My problem is, I can’t see how the equation geometrically/mathematically makes any sense. The equation is true and returns the correct result when being used.

$R_{sp,3}$ is the radius that approximately goes where I have marked it with red on the picture. $d_{ki}$ and $b_v$ are both distances.
I don’t know if you need the equations for radius $R_{ki,1}$ og $R_{ki,2}$ but here they are ($V_p$ is just a number related to the specific case):
$$R_{ki,1} = \frac{1}{\dfrac{1}{R_{sp,3}}+\dfrac{108}{V_p^2}} \quad\text{and}\quad R_{ki,2} = \frac{1}{\dfrac{1}{R_{sp,3}\pm b_v}+\dfrac{108}{V_p^2}}$$
The equation I need help with is:
$$d_{ki} = R_{sp,3} + \sqrt{R_{sp,3} \cdot (R_{sp,3} \pm b_v)}$$
Can anyone tell me how that equation is derived?
 A: Here's a reimagining of OP's figure:

(In OP's figure, the "red arc" extends from $B$ to $G$. The portion to the left of $C$ is irrelevant here; so, for the purposes of this discussion, the "red arc" simply connects $C$ to $G$.)
To simplify notation, define
$$r := R_{sp,3} \qquad s := R_{ki,1} \qquad t := R_{ki,2} \qquad\quad c := \frac12b_v \quad m := \frac14 L_{ki} \qquad\qquad d := d_{ki} $$
Also, let $R$, $S$, $T$ be the centers of the arcs of respective radii $r$, $s$, $t$. Let the measures of the central angles be $2\theta$ at $R$, $2\phi$ at $T$, and $2\theta+2\phi$ at $S$.
Taking OP's $L_{ki}/2$ measurements to indicate horizontal distances between landmark points on the roadway ($C$, $D$, $E$ in the figure), we readily determine
$$\begin{align}
2m &= s(\sin2\theta+\sin2\phi) \tag1 \\
2m &= t \sin2\phi \tag2
\end{align}$$
Vertically, we find
$$\begin{align}
2c &= r - (r-s)\cos2\theta-(s+t)\cos2\phi+t \\
\to\qquad c &= r\sin^2\theta+t\sin^2\phi + s\sin(\phi+\theta)\sin(\phi-\theta) 
\tag3
\end{align}$$
These equations impose dependencies that reduce our collection of parameters from seven $(r, s, t; c, m; \theta, \phi)$ to some choice of four. For instance, we can solve for the radii in terms of the angles and horizontal/vertical offsets:
$$\begin{align}
r &= \sec\phi\csc^2\theta\sec(\phi-\theta) (
c \cos\phi\cos(\phi-\theta) - m\sin(2\phi-\theta)) \\
s &= m \csc(\phi+\theta)\sec(\phi-\theta)\\ 
t &= m \csc\phi\sec\phi
\end{align} \tag4$$
From here ... It's not quite clear how to proceed.
OP has additional relations we can write as
$$\frac1s - \frac1r \;= \frac1v =\; \frac1t - \frac1{r+2c} \tag5$$
where $v := V_p^2/108$ (and where we deduce that OP's "$\pm$" must be "$+$", since $t$ must be larger than $s$, as the $s$-arc makes a sharper turn than the $t$-arc). However, OP has commented "To make the equations for [$s$ and $t$], it is assumed that [$d=c$]. But that is not $100\%$ accurate." But applying $d=c$ to the target relation ($(\star)$ below) implies $d=c=0$, so the assumption is inconsistent with the target, which suggests that there are other layers of approximation going on. So, $(5)$ doesn't seem directly applicable here. Even if it were, by introducing an additional parameter $v$, the two equations in $(5)$ only reduce the number of free parameters to $3$; and it's not clear (to me) which might be subject to approximation.
As for the target relation itself, OP's rendering of it is inconsistent with the figure given, as it makes $d_{ki}$ larger than $R_{sp,3}$, which is one of the largest radii. Presumably, the relation should be a subtraction. Moreover, to ensure $d_{ki}$ is positive, we must take the "$\pm$" to be "$-$". Thus, it would appear that the target "should be"
$$d = r - \sqrt{r(r-2c)} \qquad\to\qquad (d-r)^2 = r(r-2c) \tag{$\star$}$$
The right-hand version is tantalizing in that it suggests a power of a point situation, in which $D$ lives on a circle orthogonal to a family of circles whose line-of-centers is the perpendicular bisector of the $2c$ segment:

In this situation, the $d$ segment (OP's $d_{ki}$) is an extension of $\overline{RD}$ that meets $\bigcirc R$, so that $(\star)$ asserts $|RD|^2=r(r-2c)$. One can show (I used coordinates) that this is equivalent to the condition
$$c r = 2 (r - s) s \sin^2(\phi+\theta) =2m(r-s)\tan(\phi+\theta) \tag6$$
where $r$ and $s$ can be rewritten via $(4)$, if desired.
Unfortunately, OP has commented that the $d$ segment is not orthogonal to the red arc (ie, is not an extension of $\overline{RD}$), but rather the vertical distance from $D$ to that arc. The messiness of the counterpart of $(6)$ for that situation exceeds my patience to TeX-ify it, so for now I'll leave its derivation as an exercise to the reader.
I'm inclined to wonder if maybe part of the engineer's genius here is to approximate the vertical distance by the orthogonal-to-arc one, leveraging the power of a point formula to simplify calculations. After all, OP has commented that $r$ tends to be on the order of $500$m, while $d$ is in the neighborhood of $1.5$m; at those scales, the computational error from the different interpretations of $d$ may be "acceptable".
Not that it really matters. I'm at a loss for how to arrive at even the nicer relation $(6)$. (Fiddling in GeoGebra shows that $D$ is not on the green circle for arbitrary values of the free parameters; more conditions are needed to lock this in.) Nor do I have the intuition to see why it would be an appropriate approximator. (I guess that's why I'm a mathematician, not an engineer.)
Without more information or insight, this is about as far as I can go.

Some additional thoughts ...
The flip side of my saying $(\star)$ implies "$c=d=0$" when $c=d$ is that $(\star)$ could alternatively imply $r=\infty$ when $c=d$. That is, the $c=d$ condition behind the approximations in $(5)$ effectively assumes "$r$ is large" (compared to other parameters). I suspect that this assumption also drives the target relation $(\star)$ ... which is to say: I don't believe $(\star)$ is a derivable geometric fact, but a contrived computational convenience.
Even so, let me cover one more bit of geometry:

Let the horizontal from my point $E$ and the vertical from $D$ meet at $D'$. Then
$$|D'D|=2m\tan\phi \tag{A.1}$$
Now, for the contrivance(s). Beware of vigorously waving hands!
For "large" $r$ ...

*

*The red arc is "flat", so that path $CDE$ becomes a "symmetric" $S$-curve that places $D$ at the vertical half-way mark of the $2c$ segment. This gives us
$$c \approx |D'D| = 2m\tan\phi \tag{A.2}$$

*The central angle at $R$ is "small", so that $\theta\approx 0$ and thus
$$c\approx 2m\tan(\phi+\theta) \tag{A.3}$$

*Now, if $r$ is large compared to $s$ then we can write $1-\dfrac{s}{r}\approx 1$, so that
$$c\approx 2m\left(1-\frac{s}{r}\right)\tan(\phi+\theta) \tag{A.4}$$
But this is simply (though only approximately) $(6)$, which is equivalent to the target $(\star)$; I'll write this as
$$|RD|^2 = r(r-2c) \tag{A.5}$$

*Finally, the target $d$ segment —defined by OP as the vertical distance from $D$ to the "red arc"— is, as described before, approximated by the extension of $|DR|$ that meets that arc, so that $|DR|\approx r-d$, giving
$$(r-d)^2 \approx r(r-2c) \tag{A.6}$$
as (approximately) desired.

This loosey-goosey approach almost-certainly doesn't pass muster with numerical analysts (of which I am not one), but it might at least serve as a scaffolding for a rigorous treatment.
A: The way I understand it at present time.
The radius of curvature of a curve with cartesian equation $y=f(x)$ is known to be:
$$R(x)=\dfrac{(1+f'(x)^2)^{3/2}}{f''(x)}\tag{1}$$
In our case where
$$f(x)=\sqrt{x(x+b)}\tag{2}$$
(whose curves are hyperbolas), the radius of curvature is
$$R_b(x)=\frac{1}{2b^2}\left(8x^2+8bx+b^2\right)^{3/2}\tag{3}$$
The corresponding curves grow like $kx^3$ and (almost) cover the first quadrant.

Fig. 1: (abscissas: distances ; ordinates: radii of curvature) Determination of an adequate value of $b$ for the connecting arc, being known the initial (resp. final) radius of curvature $R'=100$ (resp. $R''=300$). Among all candidates (the arc of curves situated between the two red lines), only one will correspond to a given length ; for example if we desire a curve length $2$ meters (to be read on the $x$ axis), we should take $b \approx 4$ (probably unrealistic values, I agree ; our aim here is understanding/agreeing on the principle).
Otherwise said, Fig. 1 is a kind of nomogram allowing, through the determination of a certain curve to get the good value of parameter $b$, and then use equation (2).
What do you think about all this interpretation ?
A: The path model is clear so far.. ok for me to start understanding/interpreting. There could be typos with given labels:
At the end of interpretation, we should be able to state what the distances and acceleration are given and resulting equation is for path going above center line.
Basically, $d_k$ depends on R which again depends on acceleration due to $V_p$
$R_{k1}, R_{k2}$ are radii or $y$ coordinate on either side of designed inflection point at center.In other words with this point, an odd function describes the curve.
Curve $R_{sp,1}$ transitions to $R_{sp,3}$ through a point of inflection. Likewise $R_{sp,2}$ transitions to $R_{sp,4}$ through a point of inflection. The indicated vertical line is not labelled.
$L_{bu}/2$ is length designed between inflection and horizontal tangent point of merge completion where the vehicle resumes pure x- velocity without acceleration.
At distance $dki$ there is an inflection point maneuver designed on the center line of the road on x-axis.
To find $d_k$ following quadratic equation has been solved :
The first and second roots  are
$$d_{ki1} = R_{sp,3} + \sqrt{R_{sp,3} \cdot (R_{sp,3} \pm b_v)};\;d_{ki2} = R_{sp,3} - \sqrt{R_{sp,3} \cdot (R_{sp,3} \pm b_v)} \tag 1$$
we can find sum and product and form quadratic equation that contains both these roots $d_{ki}$:
$$ d_k^2 - 2 R_{sp,3} \; d_k \mp R_{sp,3}= 0 \tag 2 $$
Radial correction/compensation is due to influence of a lateral acceleration, follows a parallelly connected electric resistances model or pattern ( effective less than least radius etc). 108 could be some permissible limit acceleration value. Radius $ R_{ki,1} $ has no influence of acceleration. $R_{ki,2}$ takes acceleration into account.
$$ \dfrac{1}{R_{ki,2}}=\dfrac{1}{R_{sp,3 \pm b_v}}+ \dfrac{108}{V_p^2} \tag3 $$
We can continue through comments if you wish.
