Given a curve $c:[a,b] \to \mathbb{R}^3$ and defining a surface $S:[a,b] \times [0,2 \pi]\to \mathbb{R}^3$ where
$S(s,t)=c(s)+R\cdot sin(t)\, N(s)+R\cdot cos(t)\, B(s)$
with being $N(s)$ and $B(s)$ the normal and binormal vectors of the curve $c$. And $r$ is the radius of the tube along $c$.
Now, I want to find a change of basis from the Cartesian to the orthonormal basis $F(s)$ (defined by the frenet frame) at every point of the curve $c(s)$ consisting of its tangent vector $T(s)$, the normal vector $N(s)$ and binormal vector $B(s)$.
If I construct the matrix $M_F^C$ for the change of basis from the Cartesian to the orthonormal basis of the curve $F(s)$ and apply this to the Surface, I will obtain the resulting vector $X(s,t)=(X_1,X_2,X_3):=M_F^C S(s,t)$. This vector should describe the surface of the tube for all $s\in[a,b]$ and satisfy ${X_1}^2+{X_2}^2={R}^2$ $(*)$ (This is the actual result I am looking for).
When I calculate this for an exemplary curve I fail to meet $(*)$. I assume I am wrong somewhere about the change of basis. I have never dealt with these topics before and would appreciate some help.
I'll make an example:
Let's take the well known helix with Radius $r$ and being translated along the x-axis by $a$:
$c(s)=(r \,sin(\frac{s}{k})+a, r\, cos(\frac{s}{k}), h\, \frac{s}{k})$
with $k=\sqrt{h^2+r^2}$ the curve $c(s)$ is defined by its arc length $s$. The frenet frame is given by
$T(s)= (\frac{r}{k}\, cos(\frac{s}{k}),-\frac{r}{k}\, sin(\frac{s}{k}),\frac{h}{k})$
$B(s)=(-\frac{h}{k}\, cos(\frac{s}{k}),\frac{h}{k}\, sin(\frac{s}{k}),\frac{r}{k})$
$N(s)=(-sin(\frac{s}{k}),-cos(\frac{s}{k}),0)$
Then the surface of my tube with radius $R$ will be defined as
$S(s,t)=c(s)+R\, cos(2 \pi t) \,B(s)+R\, sin(2\pi t)\,N(s)$
That's right this far, isn't is? Now, I define the change of basis $M=(T(s), B(s), N(s))$ with $T(s)$, $N(s)$, $B(s)$ as column vectors and define its inverse for the transformation. Simplified it yields
$M^{-1}= \begin{bmatrix} \frac{r}{k}\, cos(\frac{s}{k})&-\frac{r}{k}\, sin(\frac{s}{k})&\frac{h}{k}\\ -\frac{h}{k}\, cos(\frac{s}{k})&\frac{h}{k}\, sin(\frac{s}{k})&\frac{r}{k})\\ -sin(\frac{s}{k})&-cos(\frac{s}{k})&0 \end{bmatrix}$
Which turns out to be the rows of the frenet frame $T(s)$, $N(s)$, $B(s)$. The transformed curve results in
$M^{-1} S(s,t)=(\frac{h^2s+a\, r\,k \,cos(\frac{s}{k})}{k^2},\frac{h\,r\,s - a\, h\,k\, cos(\frac{s}{k})}{k^2}+R\,cos(2\pi t),-r+ R\, sin(2\pi t)-a\,sin(\frac{s}{k}))$
With $R$ being the radius of the tube, the root of the sum of the squared first and second argument should result in $R$. At least this is what I imagine should happen when I change my basis to the frenet frame of the curve. But when I do the math, it does not turn out to be right:
$\sqrt{{\left(\frac{h^2s+a\, r\,k \,cos(\frac{s}{k})}{k^2}\right)}^2+{\left(\frac{h\,r\,s - a\, h\,k\, cos(\frac{s}{k})}{k^2}+R\,cos(2\pi t)\right)}^2}\neq R$