How can I calculate the position of a wire-stripper's set screw based on the wire diameter?

Question

I have a wire stripper with set screw that looks pretty much like this

As you can see, position of set screw determines distance between jaws when fully closed, therefore gauge of wire to be stripped. It occurred to me that it was possible to derive a precise relationship between screw position and jaw opening. Any suggestions on how to go about this? It is obvious there are similar triangles formed at the fixed central pin, but I can't see how that helps me.

Answer 1

The abstract version might look like this:

Answer 2

Below is a sketch of the tool in a very open position with a few points defined:

The blue and red bits are the two arms of the tool and the green bit is the screw. Aside from the presence of the screw in one arm, it is assumed that the two arms are mirror images of each other. Points $B$ and $C$ are the bottoms of the wire stripping notches in each arm. Thus, when $B$ and $C$ coincide, the tool is completely closed and the diameter of the gap $\overline{BC}$ is $0$.

To work out the various relationships, it is better to have the tool completely closed, as in the figure below:

Let’s assume that $\angle BAD = \frac{\pi}{2} + \alpha$, where $\alpha \gt 0$. As $\angle BAD + \angle DAK = \pi$ and $\angle ADK = \frac{\pi}{2}$ we see that $\angle DKA = \alpha$ and therefore $\angle LKM = 2 \alpha$. So when the tool is completely closed, the angle between the arms is $2 \alpha$. As the arms open, $\angle LKM = 2 \alpha + \theta$, where $\theta = \angle BAC$. I.e. the angle between the arms, minus a constant ($2 \alpha$), equals the angle between the bottom of the notches.

To solve the puzzle we therefore only need to know how the position of the screw determines the angle between the arms. Zooming in on the area of interest we have the following figure:

We first note that $\angle GNF = \angle LKM$, i.e $\angle GNF$ is the angle between the tools's arms. We also see that $\triangle AEN$ is similar to $\triangle FGN$. This means $$\frac{r_1}{r_2} = \frac{\overline {AN}}{\overline {FN}}$$

Let's call the distance between the center of the fixed central pin and the center of the screw $d = \overline{AF}$. We can then write $$\overline{FN} = \frac{r_2}{r_1 + r_2}d$$

And since $$sin(\angle GNF) = \frac{r_2}{\overline {FN}}$$

we find that $$sin(\angle LKM) = \frac{r_1+r_2}{d}$$

or $$d = \frac{r_1+r_2}{sin(\theta + 2 \alpha)}$$

We have now found the relationship between the angle of the arms and the position of the screw. The diameter of the gap $g = \overline{BC}$, which is a chord of the circle with radius $\overline{AB}$ and angle $\theta$, is $$g=2\overline{AB}sin(\frac{\theta}{2})$$

which means $$\theta =2 \, arcsin(\frac{g}{2\overline{AB}})$$

And we are done. Given the diameter of the wire $g$ we can calculate the necessary angle $\theta$ from the equation above, and the necessary position of the screw from the equation for $d$.