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Given two wires each of diameter D and length L, what will be the new length after you twist the wires around each other 1 Turn?

I mean that when you start with the pair of straight wires and put them against a scale, the length is L. When you twist them around each other tightly, and place the straightened twisted wire against the scale, its new length will be reduced.

And what I mean by 1 Turn is that looking at the cross-section, if you begin with the Red wire on top, Black at bottom then after half Turn the Red wire is at bottom and after another half Turn (or Total 1 Turn) the Red is on top again. And both wires are twisting in same direction, say clockwise, when looking at the cross-section. Successive cross-sections will look like the number 8 rotating.

I suppose we should also give the pitch? But when you play with an actual mouse wire, twisting it around itself, you will realize that you can not reduce the pitch beyond a certain extent. So what is that minimum pitch when twisting the wires? What is causing that restriction? Can it be quantified?

Perhaps the question is not too clear, so please feel free to edit/clarify.

The axis of each wire will take the shape of a helix of radius $D/2$ and height $H$. The length of each helix is (for one turn)
$$
L=sqrtH^2+pi^2 D^2
$$
and we can invert this to obtain the desired new length:
$$
H=sqrtL^2-pi^2 D^2.
$$
This is the vertical distance between the centers of the end sections of a wire, total length will be $H+Dcostheta$, where the slope $theta$ of a wire (called "pitch" in the question) is given by $tantheta=Hoverpi D$ (see figure).

As for the minimum slope, my previous answer was not correct. Experimenting with GeoGebra, it seems that $theta_min=45°$, which is the slope in the figure.

EDIT.

The equations of the two helices, if $a=D/2$ and $b=H/(2pi)$, can be written as:
$$
(acos t, asin t, bt),quad (-acos t, -asin t, bt),quad 0le tle 2pi.
$$
Let's take any point on the first helix, e.g. $P=(a,0,0)$ (corresponding to $t=0$). Its distance $s$ from any point on the second helix is:
$$
s(t)=sqrt2a^2(1+cos t)+b^2t^2,quad 0le tle 2pi.
$$
The wires don't intersect if $sge2a$, but it's easy to check that the minimum of $s(t)$ is $s(0)=2a$ only if $bge a$. On the contrary, if $b<a$ the minimum of $s(t)$ occurs when $t$ is the root of $sin tover t=b^2over a^2$, and this minimum is strictly less than $2a$.

For the wires not to intersect, we must therefore require $bge a$, which translates in a bound on the slope:
$$
tantheta=bover age1,quadtexti.e.quad thetage45°.
$$

EDIT.

Here's the GeoGebra code to draw the two surfaces in the figure above:

Surface((a cos(t), a sin(t), b t) + r cos(u) (cos(t), sin(t), 0) + r sin(u) (b sin(t), -b cos(t), a) / sqrt(a² + b²), t, 0, 2π, u, 0, 2π)

Surface((-a cos(t), -a sin(t), b t) + r cos(u) (-cos(t), -sin(t), 0) + r sin(u) (-b sin(t), b cos(t), a) / sqrt(a² + b²), t, 0, 2π, u, 0, 2π)

Where $a$ is the distance of each helix from the axis of symmetry, $b$ is the height of each helix, and $r$ is the radius of the wire (which is the same as $a$ in figure above).