```Hello Mechatronitians,
Welcome to a new edition of the blog on the subject:

July 2021

```I am sure that the title of this article has alerted you. Yes, the gearbox is one of the biggest
enemies of machinery manufacturers. Why?```

## 1. There is always an exeption: slow applications with gearbox

```We go by parts, as there are always exceptions and one of them are applications with slow movement,
we are talking about speeds below 100 mm / s (which is the same as 6 m / min), where the speed of any electric motor is too high.
Door closing, position adjustment, high force movements, vertical lift applications, etc.
In this case, the gearbox offers a huge advantage, it reduces speed and increases torque, allowing large loads to be moved with a relatively small motor.
Remember that the output torque of a gearbox is:```

Ts = Te x i

```where
Ts – is the gearbox output torque, Nm
Te – is the gearbox input torque, Nm
i – is the reduction ratio
It is like this for slow applications, but today we are going to talk about dynamic, fast applications and especially with many changes of direction (sense of movement).
In this type of application, where the motor is constantly accelerating and braking, the gearbox no longer offers any advantage, on the contrary,
it introduces errors, performance losses and increased costs. Even the increase in torque is almost negligible. To show it, we are going to see the power calculation formulas.
We're going to talk about power first, focusing only on mechanical power. Power is the amount of energy you need to accelerate, move, or stop a mass in motion or at rest.
Power is always the function of two variables: speed and force.```

Formula 1

P = ʃ (V x F)

```where
V – is the movement of speed
F – is the necessary load/force

The higher the speed and force, the higher the power. If with the same available power we reduce the speed, we can obtain greater force.```

## 2. Real case of a dynamic application with gearbox

```We are going to simulate a dynamic application. In our case we cannot reduce the speed, the application needs it, so we can only use a small reduction ratio reducer.
To show it graphically, we are going to calculate 5 different cases with reduction ratios from 1:1 to 5:1.
Example application:

Transmission: toothed belt
Mounting position: horizontal
Pulley development: 130 mm/rev
Pulley width: 30 mm (aluminum material)
Mass to move: 10 kg
Stroke: 1000 mm
Positioning time:0.5s
Acceleration / braking time:0.15s
V = 2.8 m/s
A = 19 m/s^2.
3.8 Nm servomotor inertia:5kgcm²
Planetary gear inertia:1.5kgcm²

In table 1 you can find the torque values ​​necessary to carry out the example application for 5 different options. We use the option with gearbox (i=1:1)
and the options with gearbox from i=2:1 to i=5:1.```

Table 1

 `Reduction ratio` i=1:1 i=2:1 i=3:1 i=4:1 i=5:1 `Required motor torque, Nm` 4,6 3,4 3,12 3,24 3,51 `Motor RPM (gearbox input)` 1318 2637 3956 5275 6594 `Angular acceleration ɛ, on the motor shaft` 920 1841 2761 3682 4602

Image 1: Graph of torque required according to the reduction ratio Image 2: RPM of the servomotor according to the reduction ratio ```In image 1 and 2 you can see graphically the data in table 1.

Sum up:

In image 1 it can be seen that from the 3: 1 ratio there is no longer any use in increasing the reduction ratio because it no longer offers any mechanical advantage,
on the contrary, every time we need a larger servomotor.

How can this phenomenon be explained? Why increasing the reduction ratio the required motor torque is increased instead of reduced?

To carry out the simulation we use the following formula:
```

Formula 2

T = ɛ x j

```where
T – is the drive torque, Nm
ɛ – is the angular acceleration, rad / s
j – is the inertia of the elements to move, kgcm ^ 2```
```With an increase in the reduction ratio of the gearbox,
the revolutions of the servomotor (image 2) and the angular acceleration ɛ in the motor shaft increase in the same proportion (see formula 3).
This fact generates a greater need for torque on the motor shaft and in this way even if we use a gearbox,
a large part of the torque is spent on acceleration of the motor shaft and not on acceleration of the load.
```

Formula 3

T = ɛ x i x j

```where
i – is the reduction ratio

In the example we have not considered the losses produced by the gearbox performance.
Although planetary gearbox manufacturers speak of an approximate efficiency of 97%, this data corresponds to a work under the nominal torque.
```

If the torque transmitted by the gearbox is less than nominal, the efficiency drops by up to 85%. That is, we lose up to 15% of power.

## 3. Use a gearbox or not?

```There is no clear rule, it depends on each case.
In the example calculated between the ratio i = 1: 1 and i = 2: 1 we observe that there is a decrease in the necessary torque of 27%.
Depending on the application this may be a minor advantage. On the other hand, if we simply select a larger motor we can avoid the use of a gearbox.
The option without gearbox is more interesting from the point of view of costs. The cost of a gearbox in this example is 300-350 euros.
The cost of a larger engine can be as high as 50-80 euros.

There are cases where the use of a gearbox can be justified by the inertia ratio issue (the inertia of the load versus the inertia of the servomotor shaft)
which has not been considered in this example. Efficiency losses due to the heating of the gearbox oil in applications where the input revolutions are higher than
3000 RPM have not been considered either.

```

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See you in the next chapter. Until then, all the best!