A line starts at $A=(x_s, y_s)$ and finishes at $B=(x_e,y_e)$
Some things we know about this line:
- It is made up of $n$ circular arcs that are tangential to one another where they join together.
- Every arc is upwards facing (like a smile)
- The arcs have radii $R_1, R_2, ...$ etc. that we know the value of
- We can assume $x_s < x_e$ and $y_s < y_e$
- The gradient of the line at the start point $A$ of the line is zero, at the end point $B$ the gradient can be whatever
My aim is to be able to sketch this line, and to do that I expect this problem reduces to finding the $n$ center points $(x_{c1},y_{c1}), (x_{c2},y_{c2}), ...$etc. and the $n-1$ tangential points $(x_{t1},y_{t1}), (x_{t2},y_{t2}), ...$etc. but I keep getting stuck when I try to do that myself.
If finding the general solution to this is too tricky, any answer for $n=2$ or $n=3$ would still be very helpful. My end goal is to write some code that sketches the required line given (start point, end point, radii) as an input.
We can set WLOG $A=(0,0)$. Hence $C_1=(0,R_1)$ and (see figure below): $$ \begin{align} C_2 =\ & C_1+(R_1-R_2)\big(\sin\theta_1,-\cos\theta_1\big)\\ C_3 =\ & C_2+(R_2-R_3)\big(\sin(\theta_1+\theta_2),-\cos(\theta_1+\theta_2)\big)\\ \dots & \\ \end{align} $$ $$ \begin{align} T_1 =\ & C_1+R_1\big(\sin\theta_1,-\cos\theta_1\big)\\ T_2 =\ & C_2+R_2\big(\sin(\theta_1+\theta_2),-\cos(\theta_1+\theta_2)\big)\\ T_3 =\ & C_3+R_3\big(\sin(\theta_1+\theta_2+\theta_3), -\cos(\theta_1+\theta_2+\theta_3)\big)\\ \dots & \\ \end{align} $$ where $\theta_i$ is the angle insisting on the $i$th arc.
As $T_n= B$ we must then satisfy two equations: $T_{nx}=x_e$ and $T_{ny}=y_e$ for the $n$ unknowns $\theta_1,\dots\theta_n$.
For $n=2$ Mathematica finds an explicit (but very lengthy) solution. For $n>2$ one could in principle find a solution depending on $n-2$ angles among $\theta_i$.