## Task 12/61

In a design office, the construction documents for the special manufacture of a laboratory centrifuge are to be prepared. A stepped gear is to be installed between the drive motor and the centrifuge, which must meet the following requirements:

1. The first stage is a direct coupling of the centrifuge to the motor, which has a speed of $a_{1}=$ 6400 min $^{-1}$.
2. The gear should enable five different speeds.
3. The speed $a_{2}$ should be $75 \%$ of the speed $a_{1}$, similarly, the speed $a_{3}$ should be $75 \%$ of $a_{2}$, and so on.

The gear receives the schematic arrangement shown in the figure. ( $a$ drive, $b$ driven, $S$ switching position)

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-015.jpg?height=262&width=1487&top_left_y=303&top_left_x=296)

a) The sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ of the speeds is to be established.

b) What transmissions must the gear pairs $(3,4),(5,6),(7,8)$ receive if the gear pair $(1,2)$ has a ratio of $1: 1$?

c) How large must the radii of the gears $1,2,3,4,5,6,7,8$ be chosen? The distance of the layshaft from the drive and driven shaft is $175 \mathrm{~mm}$ (measured from shaft center to shaft center).
a) It is $a_{1}=6400 \mathrm{~min}^{-1}$ (Direct Drive) the first stage (Condition 1).

Since $75 \%=0.75=\frac{3}{4}$, the following stages result from multiplying the respective preceding stage by $\frac{3}{4}$:

$a_{2}=4800 \mathrm{~min}^{-1} ; a_{3}=3600 \mathrm{~min}^{-1} ; a_{4}=2700 \min ^{-1} ; a_{5}=2025 \mathrm{~min}^{-1}$

A sequence in which each subsequent term is calculated from the preceding one by multiplying with the same factor is called a geometric sequence.

b) In gear position 2, the gear pairs $(1,2)$ and $(5,6)$ are engaged. If $a_{v}$ denotes the rotational speed of the input shaft, then:

$$
\frac{a_{1}}{a_{2}}=\frac{a_{1}}{a_{v}} \cdot \frac{a_{v}}{a_{2}}=\frac{6400}{4800}=\frac{4}{3}=4: 3
$$

In gear position 2: $\frac{a_{1}}{a_{v}}=\ddot{\mathrm{U}}(1,2)=\frac{1}{1}=1$, where $\ddot{\mathrm{U}}(1,2)$ denotes the transmission ratio of the gear pair $(1,2)$. Then $\frac{a_{v}}{a_{2}}=\ddot{\mathrm{U}}(5,6)=4: 3$. Therefore, the gear pair $(5,6)$ must have a transmission ratio of Ü $(5,6)=4: 3$.

In gear position 3, the gear pairs (1,2) and (7,8) are engaged. Similarly, we get $\frac{a_{1}}{a_{3}}=\frac{6400}{3600}=16: 9$, so ÜU $(7,8)=16: 9$.

In gear position 4, the gear pairs (3,4) and (5,6) are engaged. Therefore, $\frac{a_{1}}{a_{4}}=\frac{6400}{2700}=64: 27$. Now, $\frac{a_{v}}{a_{4}}=4: 3$, so $\frac{a_{1}}{a_{4}}=\frac{a_{1}}{a_{v}} \cdot \frac{4}{3}=\frac{64}{27}$. This results in $\ddot{\mathrm{U}}(3,4)=16: 9$.

c) The circumferences of a gear pair are inversely proportional to the transmission ratio. For example, if the gear pair $(3,4)$ has a transmission ratio of Ü $(3,4)=\frac{16}{9}$, then (with $U_{k}$ denoting the circumference and $r_{k}$ the radius of gear $k$):

$$
\frac{U_{4}}{U_{3}}=\frac{16}{9}, \quad \frac{2 \pi r_{4}}{2 \pi r_{3}}=\frac{16}{9}, \quad \frac{r_{4}}{r_{3}}=\frac{16}{9}
$$

Furthermore, $r_{3}+r_{4}=175 \mathrm{~mm}$. This results in a system of equations with two unknowns. The solution is: $r_{3}=63 \mathrm{~mm}, r_{4}=112 \mathrm{~mm}$. Similarly, we get $r_{1}=87.5 \mathrm{~mm}, r_{2}=87.5 \mathrm{~mm}, r_{5}=75 \mathrm{~mm}$, $r_{6}=100 \mathrm{~mm}, r_{7}=63 \mathrm{~mm}, r_{8}=112 \mathrm{~mm}$.