A space curve consisting of a spiral wound around a helix. (Slinky curve.) I'm trying to use the parametric equations for the sinky curve to construct a meshable model for 3D magneto-static FEA of a transformer coil. I have the modellling and meshing covered but, I need to intertwine two conductors (twisted pair) and whilst I have  gotton close by trial and error, I've yet to get the two conductors to intertwine closely without intersecting.
(The red areas in the image below show the slight intersection between them.)

So my question is, can a mathematician or two, help me avoid that intersection?
The equations used are:
x = ([r]+[a]Cos([w][t]))*Cos([t])
y = ([r]+[a]Cos([w][t]))*Sin([t])
z = [h]*[t]+[a]Sin([w][t])
Where:
a = diameter of the spiral = 0.25
h = the slope angle of the helix = 0.08 (1 translates to 2pi in the z per revolution)
r = the radius of the helix - 3.95
w = the number of turns of the spiral per revolution of the helix = 4
t runs from 0 to 14 pi.
These parameters give the curve below, and second copy rotated 45° gives the intertwine:
The software that generates the model and detects the interferance is using math to do so; but it is closed source.
Can a mathematician show me how to adjust the parameters I am using to avoid the interferance?
I realise that this is couched in 'engineering terms', but I am an engineer. I've provided as much of the math as I understand; can you nudge me the rest of the way?
Buk
Update: Thanks to Jyrki's assistance below I've now produced my model of the transformer coil:

 A: I think the problem can be solved by using the tube around the helix from an old answer and then placing the two wires on diametrically opposite sides of that tube to prevent contact.
Here's what it looks like. First from the side:

Then a view from above:

On with the explanation. There is a helix curve in between the red and blue wires in my image. It has the usual parametrization
$$
\vec{r}=(R\cos t, R \sin t, ht),
$$
where $R$ is the radius of the tube the helix sits on, and the curve will rise $2\pi h$ per revolution (using a negative $h$ reverses the handedness of the screw).
Differentiation w.r.t. $t$ gives the tangent vector along the central helix
$$
\vec{t}=(-R\sin t, R\cos t, h).
$$
We record that this has constant length $\sqrt{R^2+h^2}$. We see that the derivative of the tangent, after normalizing to unit length, gives
$$
\vec{n}=(-\cos t,-\sin t,0)
$$
as the normal (perpendicular to $\vec{t}$).
The binormal (again normalized) is the cross product
$$
\vec{b}=\frac1{||\vec{t}||}\vec{t}\times\vec{n}=\frac1{\sqrt{R^2+h^2}}(h \sin t,-h\cos t,R).
$$
We see that $\vec{n}$ and $\vec{b}$ are perpendicular to each other and to $\vec{t}$. Therefore the points given by vectors like
$$\vec{r}+ a\cos u\vec{n}+a\sin u\vec{b}\qquad(*)$$
$0\le u\le 2\pi$ form a circle centerd at $\vec{r}$ and perpendicular to the helix. By also varying $u$ they thus parametrize that tubular surface.
We also see that by replacing $u$ with $u+\pi$ we get two points on the opposite sides of the helix. We see that the same effect is gotten by replacing $a$ with $-a$.
To get the wires intertwined around the central helix you want $u=k t$ for some parameter $k$. This is thus my suggested answer:
wire number 1:
$$
\vec{r}+a\cos(kt)\vec{n}+a\sin(kt)\vec{b},
$$
wire number 2:
$$
\vec{r}-a\cos(kt)\vec{n}-a\sin(kt)\vec{b}.
$$
This way they are separated from each other by a constant distance $2a$ (assuming that $2a$ is smaller than the rise per revolution $2\pi h$).
When $R$ is much bigger than $h$ you may get away with using
$\vec{b}'=(0,0,1)$ instead of the 'technically correct' binormal $\vec{b}$ I used. The distortion won't probably be much. It seems to me that the formulas you showed do just that.
In my images, $R=10$, $h=1/\pi$, $a=0.6$, $k=4$.
Below there is a small section of the above image with a partially transparent (Opacity[0.6] in Mathematica) tube between the red and blue wires.

