## Task 36/71

On a lathe with a centrally clamping three-jaw chuck, a circular cross-section eccentric feature is to be turned on a circular part with diameter $d$.

The axis of the feature is to be offset from the axis of the part by $e$ (see figure).

To eccentrically clamp the part, a metal shim of thickness $z$ is to be placed under one of the jaws. Determine $z$ as a function of $d$ and $e: f=f(d ; e)$ !

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-215.jpg?height=531&width=702&top_left_y=314&top_left_x=680)
We represent the situation in a right-angled Cartesian coordinate system, where the axis of rotation pierces the drawing plane at the origin of the coordinate system; the clamping jaw, which holds the intermediate piece, lies (without loss of generality) on the negative x-axis (see figure). Then the center $M_{1}$ of the entire part lies on the positive x-axis at a distance $e$ from the origin. From the drawing, it can be seen that

$$
z=r_{2}-r_{1}+e
$$

where $r_{2}$ is the distance from the center to the clamping jaw, and $r_{1}=\frac{d}{2}$. Therefore, it is necessary to determine $r_{2}$. The following equations are available for this purpose:

$$
\begin{aligned}
& K_{1}: \quad\left(x_{1}-e\right)^{2}-y_{1}^{2}=r_{1}^{2} \\
& K_{2}: \quad x_{1}^{2}+y_{1}^{2}=r_{2}^{2} \\
& \frac{y_{1}}{x_{1}}=\tan 60^{\circ}
\end{aligned}
$$

From (3) it follows that $y_{1}=x_{1} \sqrt{3}$. Substituting this into (1) and (2), we obtain after a short calculation

$$
\begin{array}{r}
4 x_{1}^{2}-2 x_{1} e+e^{2}-r_{1}^{2}=0 \\
r_{2}=2 x_{1}
\end{array}
$$

From (1a) it follows that

$$
x_{1}=\frac{e}{4} \pm \frac{1}{4} \sqrt{4 r_{1}^{2}-3 e^{2}}
$$

Thus, we have

$$
r_{2}=\frac{1}{2}\left(e+\sqrt{d^{2}-3 e^{2}}\right)
$$

(since $r_{2}>\frac{1}{2} e$, the negative root value is discarded). Consequently, $z=\frac{1}{2}\left(3 e-d+\sqrt{d^{2}-3 e^{2}}\right)$. Additional note: For a part to be manufactured in this way, $r_{1}=\frac{d}{2}>e$ must hold.

### 2.12 Problems and Solutions 1972