This is a problem from the design of roadways. I know someone has the answer, but I haven't been able to work it out myself. The sketch below shows a single curve made up of two asymmetrical reversing circular arcs. The arcs are tangent to each-other at the point they meet and there is no tangent segment between them. Moving from left to right, the curve begins at an angle of zero and ends in a direction that is not equal to zero.
If the radius of the first curve ($R_1$), the orthogonal distance between the start of the first arc and the end of the second arc ($y_T$), and the angle from zero at which the second arc ends ($\theta_1$) are known, what is the subtended angle ($\Delta_1$) of the first arc that will result in two arcs of equal length?
Since, $L_i=R_i\Delta_i$ and $L_{1}=L_2$, it is fairly simple to see that $R_2=R_1\frac{\Delta_1}{\Delta_2}$ and that $\Delta_1=R_2\frac{\Delta_2}{\Delta_1}$. But, I cannot use this relationship unless I have some other independent means of determining $R_2$ or $\Delta_1$. The only variable that is not dependent upon $\Delta_1$ or $R_2$ is the given distance between the start and end of the curve (measured in a direction orthogonal to the start direction of the first arc).
I can easily use the Solver function in Excel to determine the unique angle that will make everything work, but I am trying to write an algorithm that will draw the reversing curve in cad. That means I can't use Excel's solver function and I need to figure out how to arrange the equations to find $\Delta_1$ without using $R_2$ or $\Delta_2$.
Assume that any of the typical formulas for circular arcs are available, such as the tangent length, the chord length, mid-ordinate length, etc. I also know calculus and linear algebra so feel free to use whatever mathematics you think best answers the question. Thank you.
There is another easy relation to exploit: $$ \theta=\Delta_1-\Delta_2, $$ where $\theta$ has to be taken as negative if it goes "downwards" in the figure.
Another equality can be found for $y_T$: $$ y_T=R_1(1-\cos\Delta_1)+2R_2 \sin{\Delta_2\over2} \sin{2\Delta_1-\Delta_2\over2}. $$ Inserting here $\Delta_2=\Delta_1-\theta$ and using some trig identities one gets: $$ y_T=R_1(1-\cos\Delta_1)+R_2(\cos\theta-\cos\Delta_1). $$ Use now $R_2=R_1(\Delta_1/\Delta_2)=R_1\Delta_1/(\Delta_1-\theta)$ to get the final equation for $\Delta_1$. Unfortunately, that equation cannot be solved in terms of elementary functions.