I am more or less new to Mathematica, but have never dealt with more complex problems requiring more advanced coding and would be grateful for some guidance. I have got two parameterized surfaces, more explicitly a tube surface
$S(s,t)=\vec{c}(s)+sin(t)\,\vec{B}(s)+cos(t)\,\vec{N}(s)$
and a simple plane
$A(w,v)=P_1+w\,\vec{V_1}+v\,\vec{V_2}$
with all parameters and vectors known apart from the parameterization parameters $s$, $t$ , $w$, $v$.
$P_1$ is one point on the intersection curve of $S$ and $E$ and I need to find a second point $P_2$ along the intersection curve with a certain distance $d$. The distance is measured by the arc length of the intersection curve.
I do not need an explicit formula for the intersection curve, just the next Point with a distance $d$ along the curve. I tried to solve it explicitly but I have no idea how to solve this non trivial system.
Solve[S==E,{s,t,w}]
does not find a solution. I arrived by hand at an equation $w=f(w,v)$. But this equation won't be solved by Solve[ ] either. As my solution will be dependent on a parameter e.g. $v$, I cannot use a numeric tool like NSolve[ ].
The two surfaces are
S={25464.8 - 25464.8 Cos[0.0000392699 s] - 50. Cos[0.0000392699 s] Cos[2\[Pi] t], 0. + 50 Sin[2 \[Pi] t], 25464.8 Sin[0.0000392699 s] + 50. Cos[2 \[Pi] t] Sin[0.0000392699 s]}
A={1606.98 + 0.387712 u + 0.910297 w, 9.74064*10^-15 - 4.38749*10^-16 u + 5.00745*10^-15 w, 9044.85 + 0.921781 u - 0.413957 w}
Eis a protected symbol, function arguments need[...]andS/Eare not functions, so you don't need the[...]for them. – Lukas Lang Sep 07 '17 at 09:46