Calculate curvature necessary for line segment to cover given linear distance I have to apologize in advance for probably not knowing all the proper terms, and not being properly skilled in reading equations. I hope whoever is kind enough to answer this is also patient enough to explain it in relatively simple terms.
For illustration purposes, consider the below line segments. Assume they are all of the exact same length (I don't have the drawing tools to draw them exactly). I start with a straight segment, then curve it more and more, but keep the starting point of the segment the same.

The rate of curvature in a given segment is constant (i.e. each curved segment, if continued, would form a perfect circle).
Observe that the vertical distance the segment covers decreases as the angle of the curve increases. "Bending over" makes it "shorter" without decreasing the actual length of the segment.
Now consider that there is a specific vertical distance I am trying to get the line segment to cover. I want to curve the line so that no part of it goes above the yellow line.

Note that no part of the line can be above the yellow, not just the end point of the line.
My question is: given a line segment of known length, and a known limit for vertical distance, by what formula can I calculate the angle of curvature required to make the line segment fit below said limit. Or, to put it more simply: how can I figure out how much I need to curve the black line before it fits below the yellow line?
 A: The animation below illustrates how to obtain the bending line.

Suppose the length of the arc is $L$, and the maximum altitude of the arc above the starting point is $H$.  Since this is an arc, it has a radius and an angle $\theta$.  Then
$ L = R \theta $
Now for $0 \le t \le \theta$ , the altitude of a point on the line is
$ y(t) = R \sin(t)  = \dfrac{L \sin(t)}{\theta} $
The maximum of this expression has two cases
Case I:  $0 \le \theta \le \dfrac{\pi}{2} $
In this case,
$ \dfrac{H}{L} = \dfrac{\sin(\theta) }{\theta } $
Case II:  $ \dfrac{\pi}{2} \le \theta \le \pi $
In this case,
$ \dfrac{H}{L} = \dfrac{1}{\theta} $
So from the required ratio of $\dfrac{H}{L}$ we can solve for $\theta$, and then plot the arc.
Note that
$ \dfrac{2}{\pi} \le \dfrac{\sin(\theta)}{\theta} \le 1 $ for  $0 \le \theta \le \dfrac{\pi}{2} $
and
$ \dfrac{1}{\pi} \le \dfrac{1}{\theta} \le \dfrac{2}{\pi} $ for $ \dfrac{\pi}{2} \le \theta \le \pi  $
Therefore, based on the value of $\dfrac{H}{L}$ over the interval
$[ \dfrac{1}{\pi} , 1 ] $ we can select which function to invert to find $\theta$, as follows
$  \theta = \left\{ \begin{aligned} 
&\text{sinc}^{-1}(\dfrac{H}{L}) \\ &\dfrac{1}{(H/L)} \end{aligned} \right. \qquad 
\begin{aligned} &\dfrac{2}{\pi} \le \dfrac{H}{L} \le 1 \\ &\dfrac{1}{\pi} \le \dfrac{H}{L} \le \dfrac{2}{\pi} \end{aligned}  $
where $\text{sinc}(\theta) = \dfrac{\sin(\theta)}{\theta} $

A: Here's the code I used to find $\theta = \text{sinc}^{-1}\bigg(\dfrac{H}{L}\bigg) $
In the code, the variable $\text{h2l}$ is used to represent $\dfrac{H}{L}$.
Public Function sincinv(ByVal h2l As Double) As Double
Dim f, fp As Double

' Find t such that sin(t)/t - h2l = 0

' function:    f = sin t - h2l t = 0
' derivative:  fp = cos t - h2l

p = WorksheetFunction.Pi()

lconverged = 0

t = h2l * p / 2

For i = 1 To 100

f = Sin(t) - h2l * t

fp = Cos(t) - h2l

If Abs(f) < 0.00000001 Then
   lconverged = 1
   Exit For 
End If

dt = -f / fp

If Abs(dt) < 0.00000001 Then
   lconverged = 1
   Exit For
End If

t = t +  dt

Next i

If Not (lconverged = 1) Then
   MsgBox ("did not converge")
   Exit Function
End If

sincinv = t

End Function

