Introduction

Transmission are available in many different types, depending on the application, e.g. as gear drive, belt drive, friction drive, worm drive, planetary drive, etc. However, the physical principle of transforming rotational speed (angular velocity) and torque or velocity and force are identical for all transmission types. However, before we go into more detail on how transmissions work, we will briefly explain the most important types. More detailed information can be found in the main articles.

Gear drive

In gear drives, gear wheels, wich are also referred to as cogwheels or toothed wheels or simply gears, engage with each other and thus form-locking convert the revolution speed and torque of the drive shaft to the desired value on the output shaft. The gearbox input shaft is referred to as the drive shaft. This corresponds to the shaft which is connected to the motor and whose speed or torque is to be changed by the gear unit. The output shaft therefore corresponds to the gearbox output.

Form-locking (or positive locking) transmissions transfer the power through interlocking geometric forms!

The animation below schematically shows a three-stage gear transmission. If you look at it, you will notice that the direction of rotation of the gears changes with each gear pair! If, for example, the driving gear rotates counterclockwise, the driven gear will rotate clockwise. This reversal of the direction of rotation must be taken into account when designing gear drives.

The sense of rotation changes with each gear pair (gear stage)!

Belt drive and chain drive

In belt drives or chain drives, the speed and torque are converted by wheels which drive each other via belts (belt drives) or chains (chain drives). The wheels of belt drives are also referred to as belt pulleys, and the those of a chain drives are referred to as chain wheels or sprocket wheels.

While the power transmission of chain drives is also form-locked, the power transmission of belt drives is not effected by interlocking forms but by frictional forces between belt and pulley. In such a case one speaks of a friction-locking transmission or somewhat imprecisely of a force-locking transmission.

Friction-locking (or force-locking) transmissions transfer the power through frictional forces!

The advantage of friction-locking power transmissions is the integrated overload protection. While with gear drives the teeth could break in case of overload or the chains could break in case of chain drives, with belt drives the belt is only pulled over the belt pulley in case of overload. Belt drives are therefore frequently used where many load peaks are to be expected, e.g. in cone crushers or jaw crushers for crushing stones.

The animation above schematically shows a three-stage belt drive. In contrast to gear drives or friction wheel drives (explained below), the direction of rotation of the individual pulleys does not change in the manner shown. However, this does not necessarily have to be the case with belt drives. To reverse the sense of rotation, the belts can also be crossed (crossed belt drive).

Friction wheel drive

With special friction-locking drives, the toothless wheels can also roll directly onto each other. This is then referred to as a friction wheel drive, shown in the animation below.

The advantage of a friction drive compared to a gear drive is that in case of overload the friction wheels simply slip onto each other and thus protect the transmission from major damage. The disadvantage, however, is the lower efficiency, as relative movements occur due to non-optimal adhesion conditions between the friction wheels. Such a minimal slipping of the wheels will always be present with friction-locking power transmission. In technical terminology, this is also referred to as slippage and reduces the efficiency. Slippage also occurs between belt and pulley in belt drives.

Slippage is the relative movement between a driving element and a driven element in friction-locking power transmissions!

The elastic deformations of the friction wheels or belts at the contact points also lead to efficiency losses, since permanent “flexing” is associated with high forces. The flexing work becomes noticeable in a warming of the wheels or the belt.

Where are gearboxes used?

In mechanical engineering there are many technical systems, which are driven either by muscle power or by motors. For example, the rear wheel of a bicycle is driven either by the muscles of the cyclist or by an electric motor. Electric motors are also used in drilling machines, while in cars internal combustion engines are used. These motors supply the energy needed to drive the respective components, e.g. to the drill chuck of a drilling machine or to the rear wheel of a bicycle.

However, all these different examples have one thing in common. The mechanical power of the motors is generally not directly used. Rather, the mechanical power must be provided in different ways, depending on the situation. When starting off with a car or a bicycle, the “force” behind the drive power should be as large as possible in order to be able to set the respective vehicle in motion.

Later on, the “speed” is more important in order to be able to cover a large in a short time. The mechanical power can therefore be used either for a high force or for a high velocity. Such a control between force and velocity, or more precisely between torque and rotational speed, is taken over by transmissions. Transmissions are sometimes simply referred to as gearings or gears, although not every transmission consists of gears. Transmissions are important elements in mechanical engineering.

Transmissions control the power supplied in favour of a high velocity (rotational speed) or in favour of a high force (torque)!

In addition, transmissions also have the task of influencing the direction of rotation. Think, for example, of the reverse gear of a car. Gear units therefore basically fulfil the following tasks:

• power transfer
• influencing the direction of rotation
• control of speed and torque

Mechanical power

The mechanical power results with translatory motion by the multiplication of force F and velocity v and with rotary motion by the multiplication of torque M and rotational speed n:

\begin{align}
&\boxed{P = F \cdot v} ~~~~~~~\text{translational power} \\[5px]
\label{p}
&\boxed{P = 2 \pi \cdot M \cdot n} ~~~~~~~\text{rotational power} \\[5px]
\end{align}

The article mechanical power deals with the derivation of these formulas in more detail.

Since a transmission can only ever change one of the two variables (velocity or force, or rather rotational speed or torque) in favour of or at the expense of the other variable, the mechanical power always remains constant. This fact is ultimately a direct consequence of the principle of energy conservation, because if both influencing variables could be reduced or increased simultaneously, then a transmission unit would destroy energy or generate it out of nothing (see next section).

So, a transmission does not change the mechanical power supplied but only the ratio of velocity and force or the ratio of revolution speed and torque, which is behind the power! Of course, this only applies as long as friction losses are not taken into account. Taking friction effects into account, the transmission output shaft will actually have slightly less power than the transmission input shaft. In no case, however, a transmission can increase the mechanical power. The term power is meant in the physical sense, as energy transmitted per unit time! This is actually the reason why it is called mechanical power transmission and not mechanical power transformation!

The mechanical power is not changed by a transmission (apart from friction effects which reduce the power)!

Relationship between torque and speed

The mechanical power P of a rotating shaft depends on the torque M and the speed n according to equation (\ref{p}). If friction losses are not taken into account, the mechanical power supplied to the driving shaft must be the same as the power taken from the driven shaft due to the law of energy conservation. After all, the energy transferred within a certain time is ideally transmitted completely from the driving shaft to the driven shaft.

Thus, if the mechanical power of the driving shaft P₁ and the driven shaft P₂ are equated on the basis of the principle of energy conservation, this directly leads to the following relationship between torque and speed ratio:

\begin{align}
&P_1 = P_2 \\[5px]
&2 \pi \cdot M_1 \cdot n_1 = 2 \pi \cdot M_2 \cdot n_2 \\[5px]
&\boxed{M_1 \cdot n_1 = M_2 \cdot n_2} \\[5px]
\end{align}

For a given torque and speed at the gearbox input, the left side of the equation is constant. Obviously, the product on the right side of the equation, i.e. the product of torque and speed at the gearbox output, must also correspond to this constant value. Thus, an increase in torque through the gearbox must inevitably be accompanied by a decrease in speed to the same extent. Conversely, an increase in speed leads to a reduction in torque to the same extent.

The analogous relationships also apply to a translational motion, which a gearbox may only ever change in favor of or at the expense of one of the two quantities (force or speed):

\begin{align}
&P_1 = P_2 \\[5px]
&\boxed{F_1 \cdot v_1 =F_2 \cdot v_2} \\[5px]
\end{align}

How torque and revolution speed (or force and velocity) are converted in a transmission is explained in more detail in the article How does a gearbox (transmission) work?.

Introduction

Gear wheels of transmissions are subject to wear due to sliding processes on the flanks. Therefore, the teeth must generally have a hard and therefore wear-resistant surface. For this purpose, the surface may have to be specifically treated. With steel gears, a hard and wear-resistant tooth flank can be achieved by surface hardening (case hardening) by means of induction hardening.

Another way to reduce wear is to evaporate special layers onto the surface of the gear. For example, certain steels (nitriding steels) can form a hard and wear-resistant nitride layer on the surface by a heat treatment with nitrogen (called nitriding). For example, worms for worm drives are often made of nitriding steel.

However, such heat treatments are not suitable for every gear material (e.g. plastic gears) or make economic sense or they alone are not sufficient to reduce wear to the desired level. In general, gear drives must therefore be additionally lubricated to keep wear on the teeth as low as possible. The measures presented below to reduce wear are becoming more and more effective from case to case.,

Types of lubricants

Non-liquid lubricants such as graphite are used for so-called dry lubrication. Such dry lubricants reduce wear generally more effective than merely heat treating the surface.

In dry lubrication, the force-transmitting elements touch each other via friction-reducing solids!

Compared to dry lubrication, the use of liquid lubricants (lubricating oils) or paste-like lubricants (greases) can significantly reduce wear on the tooth flanks (liquid lubrication).

In liquid lubrication, the force-transmitting elements touch each other via a friction-reducing liquid film!

If the liquid film does not allow any direct contact between the components, pure liquid friction is present. The prerequisite for this is that the lubricant is under high pressure in order to be able to apply the necessary forces to keep the components apart from direct contact. Wear for pure fluid friction is virtually zero. In general  however, the tooth flanks still have direct contact with each other. This is then referred to as mixed friction between liquid friction and solid friction.

Maintenance

The lubricating oils and greases used in liquid lubrication are generally subject to ageing. Dirt particles or abrasion from the rubbing contact points are introduced into the oils or greases over time. For this reason, lubricating oils and greases must be regularly replaced or refilled. This is also the reason for regular oil changes in cars.

Lubricants are subject to aging and must be checked regularly and replaced if necessary!

Ensuring lubrication

While lubricating greases adhere to the components to be lubricated due to their viscous behaviour, however, the liquid lubricating oils will settle off the gear wheels over time. To ensure permanent lubrication, it is sufficient, for example, if one of the gears is immersed in an oil pan and the lubricating oil is distributed throughout the entire gearbox due to splashing effects. However, the resistance when immersed in the oil pan causes additional efficiency losses.

However, this principle of immersion can only be used in closed gearboxes  in which the housing is sealed. For technical equipment which is used in different positions (e.g. a demolition hammer), it must also be ensured that lubrication is guaranteed in every conceivable position and that at least one gear wheel dips into the oil.

In addition, when designing closed gear units, care should be taken to ensure that it is possible to change oil or fill new oil at a later date without having to disassemble the transmission completely. For this reason, such gears often have an oil drain plug that allows an uncomplicated oil change without having to disassemble the gear housing.

It must also be considered that the gearbox heats up during operation. This leads to overpressure in the sealed housing. Excessive pressure may cause the gear oil to be forced out of the housing despite the seals. For this reason, gear units that are exposed to high temperatures are equipped with a air bleed valve that can be used to reduce the overpressure.

To ensure that there is always enough oil in the gearbox, it is often equipped with an oil level indicator. This indicator must be checked at regular intervals and, if necessary, oil refilled. The figure below shows the oil level indicator of a demolition hammer. In this case, the oil gauge also serves as an oil drain plug through which the oil is directly refilled.

Definition

The figure below schematically shows a toothed gear whose basic function was explained in more detail in the article Operating principle. As this animation shows, a transmission usually consists not only of one pair of gears but of several, each mounted on different shafts. Each pair of gears that meshes with each other represents a so-called gear stage.

A gear stage is characterized by a change in speed and torque between the driving wheel and the driven wheel. The gearbox shown above consists of three gear stages, whereby each gear stage can be assigned a specific transmission ratio. The transmission ratio is defined as the ratio of the rotational speeds between the driving wheel (n₁) and the driven wheel (n₂):

\begin{align}
\label{def_uebersetzungsverhaeltnis}
&\boxed{i = \frac{n_1}{n_2}} \\[5px]
\end{align}

A gear stage is a wheel pairing within a gearbox at which the speed and torque changes!

The first gear stage is represented by the green gear (z₁ = 15) and the orange gear (z₂ = 30). The second gear stage results from the gear pairing of the orange gear wheel (z₂) and the blue gear wheel (z₃ = 90). The green gear (z₄ = 15) and the red gear (z₅ = 60) form the third gear stage. For the different gear stages this results in the following transmission ratios i:

\begin{align}
\label{1}
&\text{1. gear stage:}~~~ \underline{i_1} = \frac{z_2}{z_1} = \frac{30}{15} = \underline{2} \\[5px]
\label{2}
&\text{2. gear stage:}~~~ \underline{i_2} = \frac{z_3}{z_2} = \frac{90}{30} = \underline{3} \\[5px]
\label{3}
&\text{3. gear stage:}~~~ \underline{i_3} = \frac{z_5}{z_4} = \frac{60}{15} = \underline{4} \\[5px]
\end{align}

Note that gearwheels 3 and 4 do not represent a gear stage, since the gears are on a common shaft. The speed of the two gears is therefore identical. Thus neither a change of the speed nor a change of the torque takes place. Therefore, this gear pair is not a gear stage.

Overall transmission ratio

The torque increases as the speed decreases from stage to stage according to the transmission ratios calculated above. In the first gear stage, the torque of the gearbox input shaft is doubled. In the second gear stage, this doubled torque is now tripled. As a result, the torque after the second stage is 6 times higher than at the input shaft. Finally, in the third gear stage this 6 times greater torque is now quadrupled. The total torque at the output shaft of the gearbox is 24 times higher than the torque applied to the gearbox input shaft!

The reverse relationship applies to the speeds. This means, that the rotational speed between input and output of the gearbox is reduced by a factor of 24. Consequently, the input shaft must rotate 24 times for one revolution of the output shaft.

If only the gearbox input and the gearbox output are compared, then the entire transmission behaves like a single gear stage with a transmission ratio of 24. As this example shows, the total transmission ratio i_t of the entire gearbox can therefore be determined by multiplying the individual transmission ratios of the respective gear stages:

\begin{align}
\label{4}
&\boxed{i_t = i_1 \cdot i_2 \cdot i_3 \cdot \dots}  \\[5px]
\end{align}

The overall transmission ratio of a gearbox results from the multiplication of the individual transmission ratios of the respective gear stages!

Since the transmission ratio of a gear stage depends on the ratio of the number of teeth involved, a change in the number of teeth on one of the gears generally results in a change of the overall transmission ratio.

If, for example, gear 5 (z₅ = 60 teeth) is replaced by a gear wheel twice as large with twice the number of teeth (z₅‘= 120 teeth), the gear ratio is doubled in this gear stage to i₃‘= 8. This doubling also doubles the overall transmission ratio from 24 to i_t‘= 48.

Note that by multiplying the individual transmission ratios, each gear stage influences the total transmission ratio linearly according to equation (\ref{4}). Doubling or tripling a single gear stage therefore also means doubling or tripling the overall transmission ratio.

Pro and cons

The total transmission ratio of 24 obtained in this example could also be achieved with only one gear stage. In such a case, the gear wheel on the output shaft would have to be 24 times the size of the gear wheel on the input shaft. However, the dimensions of the gearbox would be very large.

The figure below shows the true-to-scale dimension of a single-stage gearbox with the same overall transmission ratio as the 3-stage gearbox.

Multi-stage gearboxes offer the advantage of dividing the desired transmission ratio into several smaller gear stages, thus keeping the overall dimensions of the gearbox small.

However, it must be noted that the friction increases with each gear stage. This is partly due to the fact that a total of more teeth mesh with each other, which generally slide against each other and thus also generate more friction. On the other hand, the respective shafts of the gear stages have to be mounted and thus cause increased bearing friction.

Influence of the number of teeth

It has already been explained that changing the number of teeth of a gear has a direct effect on the transmission ratio of the corresponding gear stage and generally also affects the overall transmission ratio.

In the case of gear wheel 2, however, a change in the number of teeth has no effect on the overall transmission ratio! This can be seen immediately if you take a closer look at the formula for determining the overall transmission ratio. For this, equation (\ref{1}), (\ref{2}) and (\ref{3}) are used directly in equation (\ref{4}). It becomes apparent that the number of teeth z₂ cancel each other out. Thus the overall transmission ratio is obviously not dependent on the number of teeth z₂:

\begin{align}
&\underline{i_{ges}} = i_1 \cdot i_2 \cdot i_3 = \frac{z_2}{z_1} \cdot \frac{z_3}{z_2} \cdot \frac{z_5}{z_4} = \underline{\frac{z_3 \cdot z_5}{z_1 \cdot z_4}}   \\[5px]
\end{align}

But if the total transmission ratio does not depend on this gearwheel, what is its purpose? In fact, this gearwheel 2 can be dispensed with completely and thus gear 1 can mesh directly with gear 3 without any change in the overall transmission ratio of 24! A recalculation of the now 2-stage gearbox proofs it:

\begin{align}
\text{1. gear stage:}~~~ \underline{i_1} &= \frac{z_3}{z_1} = \frac{90}{15} = \underline{6} \\[5px]
\text{2. gear stage:}~~~ \underline{i_2} &= \frac{z_5}{z_4} = \frac{60}{15} = \underline{4} \\[5px]
\text{overall transmission ratio:}~~~ \underline{i_{ges}} &= i_1 \cdot i_2 = 6 \cdot 4  = \underline{24} \\[5px]
\end{align}

The fact that the total transmission ratio is independent of the number of teeth z₂ can also be descriptively explained. If the gear wheel 1 of the drive shaft moves one tooth further, the gear wheel 2 also moves one tooth further. This movement of one tooth is now directly transferred from gearwheel 2 to gearwheel 3. In this respect, gearwheel 2 merely serves as a intermediate gear for pushing on the following gear. Thus gear 1 could also push the gear 3 directly one tooth further without changing anything in the basic process.

Depending on the function of the gearbox, however, gearwheel 2 is by no means superfluous. Because without it the output shaft rotates against its original direction of rotation! The intermediate gearwheel 2 serves as a so-called idler gear and thus has the function of adjusting the sense of rotation of the output shaft (reversal of the direction of rotation). In order to fulfill this task, the idler gear 2 could in principle also be placed between gearwheel 4 and 5.

An idler gear reverses the direction of rotation without affecting the overall transmission ratio!

Arrangement of the gears

Under what condition does the number of teeth of a gear wheel influence the overall transmission ratio and when not? The figure below shows a gear train in which gears are mounted on separate shafts. In such a case, the total transmission ratio depends only on the number of teeth of the first gear (z₁) and the last gear (z₆):

\begin{align}
\require{cancel}
&\underline{i_{t}} = i_1 \cdot i_2 \cdot i_3 \cdot i_4 \cdot i_5 = \frac{\bcancel{z_2}}{z_1} \cdot \frac{\bcancel{z_3}}{\bcancel{z_2}} \cdot \frac{\bcancel{z_4}}{\bcancel{z_3}} \cdot \frac{\bcancel{z_5}}{\bcancel{z_4}} \cdot \frac{z_6}{\bcancel{z_5}} = \underline{\frac{z_6}{z_1}} \\[5px]
\end{align}

In order to design transmissions effectively, two different gears must always be arranged on a common shaft, which in turn drives a shaft with another pair of gears. In this case, the number of teeth of all gears is then included in the calculation of the overall transmission ratio!

\begin{align}
\require{cancel}
&\underline{i_{t}} = i_1 \cdot i_2 \cdot i_3 = \underline{ \frac{z_2}{z_1} \cdot \frac{z_4}{z_3} \cdot \frac{z_6}{z_5}} \\[5px]
\end{align}

Introduction

The term power (in the physical sense) plays an important role in transmissions, since they have to be dimensioned according to the motor performance. For this reason, the following article discuss the term power in more detail. In connection with gears, the question of mechanical power at the output shaft (e.g. at the drill chuck of a drill or at the rear wheel of a bicycle) always seems to be an interaction between two decisive parameters. Namely the

• translational speed (velocity) and force at translational motions; respectively the
• rotary speed and torque at rotary motions.

Everyday experience shows that for a given drive power (motor power) the gearbox can only ever change one parameter at the expense of the other parameter. For example, one will not be able to ride up a steep hill with a bicycle or a car (when a great force is required) at such a high speed compared to a flat road. The speed of the driven wheels must be reduced accordingly in favor of the force. You have to literally “shift down a gear”. Only after overcoming the ascent, when no great force is required anymore, one can pick up speed again with a higher gear.

Even a drill in the highest gear may not be able to develop the torque required to drill a large bore. At this point, too, a lower gear must be used in favour of the torque and at the expense of the speed.

Definition of mechanical power

So while an increase of force can only take place at the expense of velocity, an increase of speed results in an inevitable reduction of force. This situation is ultimately a direct consequence of the principle of energy conservation (more details on this see article operating principle).

To understand this, a basic knowledge of the terms power and work is required. For this reason, the following sections describe these terms in more detail.

Physically, the mechanical power P of a moving body is defined by the work or energy W transferred within a certain time Δt:

\begin{align}
\label{def_leistung}
&P = \frac{W}{\Delta t} \\[5px]
\end{align}

The more work is done within a certain period of time, the greater the power. The work W, by definition, results from the product of force F and distance Δs through which the force acts (provided force and distance are rectified):

\begin{align}
\label{def_arbeit}
&W = {F} \cdot {\Delta s} \\[5px]
\end{align}

These basic terms are used in the following sections to look more closely at mechanical power during translational and rotational motions.

Power during translational motions (linear motions)

For the power of translational motions (linear motions) there is a certain connection between force and velocity. This will be shown in the following sections using a winch driven by a motor as an example.

Derivation

The winch pulls a crate upwards with constant velocity v and constant force F. The crate is raised by the displacement Δs within the time Δt. The work W of the winch performed during this time Δt results by definition from the product of applied force F and travelled distance Δs:

\begin{align}
\label{arbeit}
&W = F \cdot \Delta s \\[5px]
\end{align}

This work was obviously done during lifting and thus within the lifting time Δt. From the definition of the power, the converted mechanical power of the winch P can then be determined as follows:

\begin{align}
\label{leistung_trans}
&P =  \frac{W}{\Delta t} = \frac{F \cdot \Delta s}{\Delta t} = F \cdot \underbrace{ \frac{ \Delta s}{\Delta t}}_{=v} = F \cdot v \\[5px]
\end{align}

During this mathematical transformation, the fact was used, that the quotient of the distance covered Δs and the time required Δt corresponds to the velocity v of the raised box.

The required mechanical power P to drive a component with constant velocity v by the force F results from the product of both quantities:

\begin{align}
\label{translationsleistung}
&\boxed{P = F \cdot v} \\[5px]
\end{align}

Influencing the motion by a transmission

In this example, the required mechanical power is supplied by the motor and transmitted directly to the winch. In principle, motors cannot provide any amount of power. Rather, the performance is limited depending on the engine design. If only a certain motor power P is available, then after transforming the equation (\ref{translationsleistung}) it becomes immediately apparent that a higher force F can obviously only be achieved with a correspondingly lower speed v.

\begin{align}
\label{kraft}
&F = \frac{P}{v} \\[5px]
\end{align}

A heavier crate can therefore only be lifted with a higher force if the speed is reduced accordingly. On the other hand, at given engine power P a lighter box (when only a lower force F is necessary for pulling up) can be lifted with higher speed v.

\begin{align}
\label{geschwindigkeit}
&v = \frac{P}{F} \\[5px]
\end{align}

This is exactly where transmissions come into play. They take control of the power in favour of a greater force or in favour of a greater speed. A transmission cannot increase both quantities at the same time, as this would require an increase in power. However, the power is fixed by the motor and cannot be changed even by a gearbox.

Transmissions do not change the mechanical power but only the speed-force ratio, which is behind a certain power!  This means either a higher force at lower speed or a higher speed at lower force.

In an ideal case, the input power P_i supplied by the motor is completely transferred by the gear to the gearbox output (P_o). In reality, however, power losses P_l occur in the gear unit due to friction. These are expressed by the transmission efficiency η_g (≤ 1):

\begin{align}
\label{getriebewirkungsgrad_trans}
&\boxed{P_{o} = P_{i} \cdot \eta_g} \\[5px]
\end{align}

Power during rotary motions (circular motions)

The knowledge of the relationship between force and velocity for translational motions can be transferred to rotary motions. For this purpose, the winch described in the previous section is considered again. This time, however, the circular motion of the rope winch and the acting force is considered in more detail.

Derivation

The winch pulls the rope with the force F along the arc length Δs during winding. The distance covered Δs, however, no longer describes a straight line but an arc of a circle. However, the force is always parallel to the curved path at every point. This means that the formula for the work W=F⋅Δs can be applied again.

\begin{align}
\label{arbeit_01}
&W = F \cdot \Delta s \\[5px]
\end{align}

During a whole revolution, the winch wound the rope once completely around its circumference, i.e. the force F acted along the circular path Δs=2π⋅r. The work W done by the winch during this rotation is then finally determined as follows:

\begin{align}
\label{arbeit_02}
&W = F \cdot 2 \pi r \\[5px]
\end{align}

The time Δt for this single turn is also called period T (period = “time duration per revolution”). Therefore, within the time T the work W=F⋅2π⋅r was done, leading to the following power P:

\begin{align}
\label{leistung_rot}
&P = \frac{W}{\Delta t} = \frac{F \cdot 2 \pi r}{T} = 2 \pi \underbrace{\frac{1}{T}}_{=n} \cdot \underbrace{ F \cdot r}_{=M} = 2 \pi \cdot n \cdot M \\[5px]
\end{align}

The product of acting force F and perpendicular aligned lever arm r ultimately corresponds to the effective torque M at the winch drum with which the rotary motion is carried out. Therefore both variables can be combined to the torque M.

The formula can be interpreted even further, considering the meaning of the expression 1/T. While the period T indicates “time per revolution”, the inverse of the period 1/T therefore indicates “revolutions per time”. This corresponds to the rotational speed n (or frequency f) of the rotational motion! The following relationship between the rotational speed n (or frequency f) and the period T applies:

\begin{align}
\label{drehzahl}
&n = \frac{1}{T} \\[5px]
\end{align}

Note: In principle, the term rotational speed (denoted by n) is equivalent to the term rotational frequency or frequency for short (denoted by f). However, while the term rotational speed is often used in connection with the technical unit “revolutions per minute”, the rotational frequency f is usually used in connection with the physical unit “revolutions per second”. Note that even if the symbol n is used in equations, the unit 1/s must always be used!

The mechanical power P of a component driven at constant rotational speed n by torque M results from the product of both quantities multiplied by the constant factor 2π:
\begin{align}
\label{rotationsleistung}
&\boxed{P = 2 \pi \cdot M \cdot n} \\[5px]
\end{align}

Influencing the motion by a transmission

In the present case, the power required for the rotation of the winch drum is supplied directly by the motor. Since motor performance is always limited, a motor cannot generate any desired torque. However, higher torques are required when lifting larger loads. In such a case, a gearbox must be interposed, which increases the torque M at a given motor power P. After transforming the equation (\ref{rotationsleistung)) it becomes immediately obvious that a higher torque inevitably results in a lower speed n. The crate can no longer be lifted so quickly.

\begin{align}
\label{getriebe}
&n = \frac{P}{2 \pi \cdot M} \\[5px]
\end{align}

If, on the other hand, a low load is to be lifted with a correspondingly reduced torque M, the torque can be reduced by the gear unit in favor of the speed n. In this case, the crate can be lifted faster.

Again, note that the power P is fixed by the motor and cannot be changed by the transmission! A gearbox can only control the ratio of torque and speed! The power supplied by the motor is therefore converted by the gearbox only in favor of a higher torque and thus at the expense of the speed (or vice versa).

Transmissions do not change the mechanical power but only the torque-speed ratio, which is behind a certain power! This means either a high torque at lower speed or a higher speed at lower torque.

In principle, both the torque and the speed can of course be increased at the same time by a higher motor power. But after all, the engine will not be able to deliver unlimited power. Finally, the power limit is reached and a further increase in torque can only be achieved by a gearbox that inevitably reduces the speed. Moreover, for economic reasons, increased engine power does not always make sense, as such motors are usually more expensive than engines with lower power values.

Connection between translational and rotational motion

At this point, the formula (\ref{leistung_rot}) for calculating the power of rotational motions is to be interpreted differently:

\begin{align}
&P = \underbrace{\frac{2\pi}{T}}_{= \omega} \cdot M = \omega \cdot M \\[5px]
&\boxed{P = M \cdot \omega} \\[5px]
\end{align}

In this equation, the term 2π/T ultimately indicates the angle travelled in radians per time unit:

• one full rotation = angle 2π
• time required = period T

Thus the term 2π/T can be interpreted as angular velocity ω. The angular velocity ω is connected to the rotational speed n as follows:

\begin{align}
\label{winkelgeschwindigkeit}
&\boxed{\omega = \frac{2\pi}{T}} = 2\pi \underbrace{\frac{1}{T}}_{= n} = 2 \pi n \\[5px]
&\boxed{\omega = 2 \pi n} \\[5px]
\end{align}

When comparing the power formulas for a translatory motion and a rotatory motion, a direct analogy becomes apparent. The analog quantity to the force for translational motions corresponds to the torque for rotary motions and the quantity of translational speed corresponds to the angular speed. The product of the respective quantities then corresponds to the translational power or rotational power.

The lifting of a crate using a winch is considered again. In this case, the winch drum rotates and the crate translates. However, both motions are obviously not independent of each other. If, for example, the rotational speed of the winch drum is increased, the translational speed of the box also increases. There is obviously a certain relationship between angular velocity and translational velocity.

This relationship can be established through the mechanical power. The rotational power P_rot of the winch drum is completely transferred to translational power P_tra of the crate. If the respective formulas are equated, the following relationship between the angular velocity ω and the translational velocity v applies:

\begin{align}
\label{zusammenhang}
&P_{tra}= P_{rot} \\[5px]
&F \cdot v = M \cdot \omega ~~~~~\text{mit }~~~ M = F \cdot r ~~~\text{folgt:}\\[5px]
&F \cdot v = F \cdot r \cdot \omega \\[5px]
& \boxed{v = \omega \cdot r} \\[5px]
\end{align}

The translational velocity v can be understood as the speed at which a rotating point moves at a distance r from the axis of rotation. A point on the winding rope of the winch drum will rotate at the same speed as the crate is pulled up. Torque and angular velocity of rotational motions are thus directly related to the force and translational velocity of translational motions through the radius (see table above).

Rotational and translational motions are related to each other through the radius!

Conversion of speed

As explained in the article What is a transmission (gearbox) and what is it used for?, transmissions are used, among other things, to set the speed to a desired value. Such a conversion of the speed becomes obvious when looking at the animated gear transmission below. In this case, the speed is reduced from gear to gear.

The reduction in speed can be attributed to the different number of teeth between the respective pairs of gears. For example, the first driving gear (green) on the drive shaft has a total of 15 teeth. As a result, these 15 teeth rotate completely once when the gear wheel is turned. They push the following driven gear (orange) by 15 teeth further.

However, this driven gear has more teeth due to its larger diameter. As a result, it no longer moves by a full turn. In the present case, the driven gearwheel has a total of 30 teeth. Thus, during one rotation of the driving gearwheel, the driven gearwheel is pushed on by only half a rotation. This ultimately means a halving of the speed.

Note that the individual teeth of the larger gears also have the same dimensions as the teeth of the smaller gears, as the respective teeth must fit togehter. Such an interlocking of gears is also called meshing.

Transmission ratio

The change in speed from a driving to a driven wheel is described by the so-called transmission ratio i. It is defined as follows:

\begin{align}
\label{def_uebersetzungsverhaeltnis}
&\boxed{i = \frac{n_1}{n_2}} \\[5px]
\end{align}

In this equation, n₁ denotes the rotational speed of the driving wheel and n₂ the rotational speed of the driven wheel. If the direction of rotation is reversed with a gearstage, this is usually indicated by a negative sign. For reasons of simplicity, however, this convention will not be applied in the following.

In the case described above, the transmission ratio between the two gears is i = 2, which means that the driving wheel rotates twice as fast as the driven wheel or the driven wheel moves only half as fast as the driving wheel. Frequently, transmission ratios are also given in the form 2:1 (“two to one”).

The transmission ratio is defined as the ratio of the rotational speeds of the driving wheel to the driven wheel. It descriptively shows how often the driving wheel has to turn for one rotation of the driven wheel!

Gear drive

For two paired gears, the transmission ratio is determined by the (inverse) ratio of the number of teeth z or the corresponding pitch circle diameter d:

\begin{align}
\label{zaehne_uebersetzungsverhaeltnis}
&\boxed{i = \frac{z_2}{z_1} = \frac{d_2}{d_1}} \\[5px]
\end{align}

The operating pitch circle diameter is the diameter of imaginary pitch cylinders that roll onto each other without sliding (somewhat a bit imprecise just referred to as pitch circle diameter). Consequently, the circumferential speeds on the operating pitch circle of both gears are identical. The pitch circle diameter of a toothed wheel is ultimately the equivalent of the pulley diameter of belt drives.

The operating pitch circle diameter is the diameter of imaginary cylinders that roll onto each other without slipping!

Belt drive and chain drive

In the case of a friction wheel drive or belt drive (or chain drive), the transmission ratio can be determined by the (inverse) ratio of the respective wheel diameters d:

\begin{align}
\label{rad_uebersetzungsverhaeltnis}
&\boxed{i = \frac{d_2}{d_1}} \\[5px]
\end{align}

If, for example, the driven wheel is twice as large as the driving wheel, this also applies to the corresponding wheel circumferences. While the driving wheel rotates once, the double-sized wheel rotates only half a turn (either by rolling onto each other in the case of friction wheels or by chains or belts in the case of belt drives or chain drives). The speed is therefore halved and a transmission ratio of i = 2 is again present.

Gear stages

In principle, a certain transmission ratio can be assigned to each wheel pair within a transmission at which the speed changes. The animations above of the gear transmission and the belt transmission show that a transmission usually does not consist of one pair of wheels but of several, each mounted on a different shaft.

Each pair of wheels that mesh with one another represents a so-called gear stage and is characterized by a certain transmission ratio. In general, a gearbox has several gear stages, each with different transmission ratios.

A gear stage is a wheel pairing within a gearbox at which the speed or torque changes!

So when we talk about the transmission ratio of the entire gearbox, we mean the overall transmission ratio, i.e. the transmission ratio between input shaft and output shaft of the whole gear unit! The total transmission ratio i_t can be calculated by multiplying the individual transmission ratios of the gear stages:

\begin{align}
&\boxed{i_{t} = i_1 \cdot i_2 \cdot i_3 \cdot \dots} \\[5px]
\end{align}

The overall transmission ratio of a gear unit results from the multiplication of the individual transmission ratios of the respective gear stages!

More detailed information on the function and structure of gear stages can be found in the article What is a gear stage?.

Forms of transmission ratios

Gearboxes do not always have to be designed to reduce the speed as is the case in the animations above. In many technical applications, an increase in speed is also desired. This is the case, for example, when driving on highways. In order to move forward as fast as possible, the wheels have to turn as fast as possible. Therefore, it is necessary to increase the speed of the motor shaft by means of a transmission. Then a large gear wheel must then drive a smaller wheel.

In such cases the transmission ratios are smaller than one and one also speaks of a speed ratio. At transmission ratios greater than one, the driven wheel rotates slower than the driving wheel and one speaks somewhat imprecisely of a power ratio. Note, that power in the physical sense ist not transformed but remains constant. Only the torque ist increased at a power ratio. Since the speed is reduced according to the increase of torque, the transmission is often called gear reducer or speed reducer.

A transmission ratio that leads to an increase in speed is called speed ratio. A transmission ratio that leads to an increase in torque is called power ratio.

For example, when starting off with a car in first gear, there is a power ratio with a maximum transmission ratio of about i_max = 3.6. Accordingly, the speed is reduced by a factor of 3.6 compared to the motor speed. In top gear, on the other hand, the shiftable motor gearbox has a speed ratio with a minimum transmission ratio of approx. i_min= 0.8. The speed is therefore increased by a factor of 1.25 (=1/0.8).

Gearboxes that can change their transmission ratio are also referred to as shiftable transmissions or manual transmissions or, for short, gearshifts. An important characteristic of shiftable transmissions is the increase in the transmission ratio from the minimum to the maximum. The greater this increase is, the larger the speed ranges can be shifted. This increase is also referred to as transmission spread S and is calculated as follows:

\begin{align}
&\boxed{S = \frac{i_{max}}{i_{min}}} = \frac{3.6}{0.8}=4.5 \\[5px]
\end{align}

For the described gearbox, the spread is S = 4.5, which means that the gear ratio can be increased by a factor of 4.5 starting from the minimum value.

The ratio of maximum to minimum gear ratio of a shiftable gearbox is called transmission spread!

Conversion of torque

In the previous section, the conversion of the speeds of two gears was described. Due to the energy conservation, a change in torque is always associated with this speed change (see also article What is a gearbox and what is it used for?)! This is discussed in more detail in the following sections.

Gear drive

The change in torque within a pair of gears becomes clear when one looks more closely at the forces occurring. In the following it is assumed that the driving gear wheel has the torque M₁. The adjacent gear wheel is driven by this torque.

Depending on the diameter d₁ of the driving gear, a certain force F is connected to the torque M₁. With this force, the tooth flanks on the pitch circle of the driving gear now press against the tooth flanks of the driven gear (also acting on the pitch circle).

The acting force F can be determined from the definition of torque (“torque = force applied x lever arm”). Thus, at a given torque M₁, the corresponding force F at the tooth flanks can be determined using the respective pitch circle diameter d₁:

\begin{align}
&M_1 = F \cdot r_1 = F \cdot \frac{d_1}{2} \\[5px]
\label{m_t}
&\underline{F = 2 \cdot \frac{M_1}{d_1}} \\[5px]
\end{align}

Note: To simplify matters, it was assumed that the force acts tangentially to the pitch circle, so that force and lever arm (= half pitch circle diameter) are perpendicular to each other. More detailed information on the actual direction of force of involute gears can be found in the corresponding article.

The calculated force F of the driving gear from equation (\ref{m_t}) also acts on the driven gear. However, since the driven gear has a different pitch circle diameter, the force now acts on a changed lever arm (d₂/2). Consequently, this is also associated with a change in torque:

\begin{align}
&M_2 = F \cdot r_2 = F \cdot \frac{d_2}{2} ~~~\text{with equation (2)}~~~F = 2 \cdot \frac{M_1}{d_1} ~~~\text{:} \\[5px]
&M_2 = \underbrace{2 \cdot \frac{M_1}{d_1}}_{= F} \cdot \frac{d_2}{2} \\[5px]
\label{m_1}
&\underline{M_2 = M_1 \cdot \frac{d_2}{d_1}} \\[5px]
\end{align}

It is shown by equation (\ref{m_1}) that the torque M₂ on the driven gear is proportional to the ratio of the respective pitch circle diameters d₂/d₁. The larger the driven gear in relation to the driving gear, the greater the increase in torque will be.

For gears, the pitch circle diameter is directly proportional to the number of teeth. Because with a double (pitch circle) diameter, the gear wheel circumference is twice as large and thus also offers space for twice the number of teeth.

If the driven gear has twice as many teeth as the driving gear, the associated double lever arm ultimately doubles the torque. In this respect, the torque increase can also be expressed by the ratio of the number of teeth:

\begin{align}
\label{m_2}
&\underline{M_2 = M_1 \cdot \frac{z_2}{z_1}} \\[5px]
\end{align}

The ratio of pitch circle diameters in equation (\ref{m_1}) or ratio of the number of teeth in equation (\ref{m_2}) corresponds to the transmission ratio i in equation (\ref{zaehne_uebersetzungsverhaeltnis}). This means that the change in torque can also be expressed directly by the transmission ratio:

\begin{align}
\label{1}
&\boxed{M_2 = M_1 \cdot i }~~~\text{with}~~~\underline{i = \frac{z_2}{z_1}= \frac{d_2}{d_1}=\frac{n_1}{n_2}} \\[5px]
\end{align}

Note that the transmission ratio is defined as the ratio of the rotational speeds of driving gear to driven gears. Thus, for the speed n₂ of the driven gear at a certain transmission ratio i the following relationship to the original speed n₁ applies:

\begin{align}
\label{2}
&\boxed{n_2 = \frac{n_1}{i} } \\[5px]
\end{align}

As the torque increases according to equation (\ref{1}) at a certain transmission ratio, the speed decreases to the same extent according to equation (\ref{2}) and vice versa. This is ultimately a direct consequence of the law of energy conservation. In the section “Energetic approach”, this relationship is explicitly derived using the law of energy conservation.

As the speed is increased by a gearbox, the torque is reduced to the same extent and vice versa!

Note that the equations above apply only to the ideal case of a non-dissipative gearbox. In general, friction cause a reduction in power and thus a reduction in the theoretically calculated torque for the driven shaft. These power losses are taken into account by a gear efficiency \(\eta_g\):

\begin{align}
&\boxed{M_2 = M_1 \cdot i \cdot \eta_g } \\[5px]
\end{align}

For the calculation of the speed, however, the gear efficiency does not play a role, since the speed conversion results from the number of teeth (the teeth cannot penetrate each other and thus produce a lower speed than is preset by the ratio of the number of teeth).

Belt drive

Also with traction mechanism, the change of torque takes place in a similar manner as in gear transmissions. Depending on the diameter d₁, the driving wheel with the torque M₁ pulls on the belt or the chain with a certain force F according to equation (\ref{m_t}).

The same force F also acts on the driven wheel through the belt or the chain. Since the diameter d₂ of the driven wheel differs from the driving wheel, a change in torque M₂ results. The changed torque M₂ at the driven wheel again results from equation (\ref{m_1}). The exact forces acting on belt drives are described in more detail in separate articles.