A **multiplier** in mechanics is a gearbox with a gear ratio of less than 1 - designed to transmit torque with increasing rotation at the output shaft, while reducing torque. An example of a multiplier is an ordinary bicycle, where one revolution of the sprocket with pedals corresponds to several turns of the driven wheel.

According to the number of nodes, the multipliers are divided into one-stage and multi-stage. Torque transmission is by chain, belt or gear. An important characteristic of the multiplier is the multiplication factor.

**Multiplication factor, **$\mathit{i}$ – is the ratio of the angular velocity of rotation of the input (primary) shaft ${\mathit{\omega}}_{\mathit{1}}$ to the angular velocity of rotation of the output (secondary) shaft ${\mathit{\omega}}_{\mathit{2}}$:

$\mathit{i}\mathit{=}\frac{{\mathit{\omega}}_{\mathit{1}}}{{\mathit{\omega}}_{\mathit{2}}}$

All ratios in the multiplier are determined by this factor, taking into account, of course, the losses of the multiplier.

Another not unimportant multiplier parameter is torque $\mathit{M}$, which he is able to convey and not break. It should not be forgotten that as the speed rises, the torque drops. So in order to save the required torque on the load, the primary shaft must be fed torque to $\mathit{i}$ times more than you need to get. Knowing the multiplication factor and the efficiency of the selected transmission type, you can calculate all the necessary parameters of a given device.

For example, we have a source of mechanical energy of sufficient power, which develops 500 rpm, and an electric generator with a power of 500 W (Data abstract). The task is to determine the multiplier parameters.

Choose the type of transmission belt, with a V-belt, its efficiency is 0.96 (the efficiency of the bearings used will not be considered). Next, let's define the generator. In most cases, the power generators are designed in such a way that the rated power is reached at 1500 rpm.

Determine the required multiplication factor. Because the angular velocity $\mathit{\omega}$, is defined as

$\mathit{\omega}\mathit{=}\mathit{n}\mathit{\xb7}\frac{\mathit{2}\mathit{\pi}}{\mathit{60}}$

where $\mathit{n}$ – the number of revolutions per minute, the coefficient of recovery can be defined as the ratio of revolutions of the leading and driven pulleys:

$\mathit{i}\mathit{=}\frac{{\mathit{n}}_{\mathit{1}}}{{\mathit{n}}_{\mathit{2}}}\mathit{=}\frac{\mathit{500}}{\mathit{1500}}\mathit{=}\frac{\mathit{1}}{\mathit{3}}\mathit{=}\mathit{0}\mathit{.}\mathit{3}$

Further, it is logical that the speed of movement of the belt on both pulleys is the same, because of its linear speed

$\mathit{v}\mathit{=}\mathit{\omega}\mathit{\xb7}\mathit{R}$

determine the ratio of the radii of the pulleys:

${\mathit{R}}_{\mathit{1}}\mathit{=}\frac{{\mathit{R}}_{\mathit{2}}}{\mathit{i}}$

We find the power consumed by the generator to drive the generator from the ratio.

${\mathit{N}}_{\mathit{1}}\mathit{=}\frac{{\mathit{N}}_{\mathit{2}}}{\mathit{\eta}}\mathit{=}\frac{\mathit{500}}{\mathit{0}\mathit{.}\mathit{96}}\mathit{=}\mathit{521}\mathit{}\mathit{W}$

The torque on the primary shaft of the multiplier will be found through the power:

$\mathit{N}\mathit{=}\frac{\mathit{M}\mathit{\xb7}\mathit{n}}{\mathit{9}\mathit{.}\mathit{55}}$

where 9,55 – some coefficient, conversion from rad/s to rpm. From here

${\mathit{M}}_{\mathit{1}}\mathit{=}\frac{\mathit{9}\mathit{.}\mathit{55}\mathit{\xb7}{\mathit{N}}_{\mathit{1}}}{{\mathit{n}}_{\mathit{1}}}\mathit{=}\frac{\mathit{9}\mathit{.}\mathit{55}\mathit{\xb7}\mathit{521}}{\mathit{500}}\mathit{=}\mathit{10}\mathit{}\mathit{\u041d}\mathit{\xb7}\mathit{\u043c}$

With that, the torque on the generator shaft will be:

${\mathit{M}}_{\mathit{2}}\mathit{=}\frac{\mathit{9}\mathit{.}\mathit{55}\mathit{\xb7}{\mathit{N}}_{\mathit{2}}}{{\mathit{n}}_{\mathit{2}}}\mathit{=}\frac{\mathit{9}\mathit{.}\mathit{55}\mathit{\xb7}\mathit{500}}{\mathit{1500}}\mathit{=}\mathit{3}\mathit{\u041d}\mathit{\xb7}\mathit{\u043c}$

As we can see, the moment on the drive shaft is better known as many times as it is less than its speed. If we express through the coefficient of recovery then we get:

${\mathit{M}}_{\mathit{1}}\mathit{=}\frac{{\mathit{M}}_{\mathit{2}}}{\mathit{i}}\mathit{=}\frac{\mathit{3}}{\mathit{0}\mathit{.}\mathit{3}}\mathit{=}\mathit{10}\mathit{}\mathit{\u041d}\mathit{\xb7}\mathit{\u043c}$