• How is the relationship between the individual rotational speeds of the components of planetary gears?

## Willis equation

The speed conversion of planetary gearboxes are no longer as easy to understand as those of stationary transmissions. This is due to the fact that the motion of the rotating planet gears is ultimately a superposition of three different motions. The motion no longer consists of a simple rotation around its own axis, but the axis itself performs an additional circular motion around the axis of the sun gear, while the planet gear also performs an additional circular motion because of the rotation of the sun gear.

Thus, the motion of a rotating planet gear can be traced back to the superposition of three separately observable motions:

1. rotation of the carrier around the sun gear
2. rotation of the planet gear around its own center of gravity
3. rotation of the sun gear
 1 2 3 Motion of the carrier Motion of the planet gear Total motion

However, the motions are not independent from each other, because the planet gear rotates on the sun gear. Thus the diameter ratio between the sun gear and the planet gear determines how often the planet gear rotates around its own axis while it moves once around the sun gear.

In order to derive the relationship of the rotational speeds between the sun gear, the planet gear and the carrier, the above-mentioned motions are first described separately and then superposed. For the sake of clarity, the gears are assumed to be (pitch) cylinders.

### Rotation of the carrier around the sun gear

If the sun gear stands still and the planet gear is locked firm on the carrier, then the swept angle of the carrier $$\varphi_c$$ corresponds to the angular position of the planet gear $$\varphi_{p1}$$.

\begin{align}
\label{P1}
&\underline{\varphi_{p1} = \varphi_c}  \\[5px]
\end{align}

### Rotation of the planet gear around its own center of gravity

In fact, the planet gear will roll on the sun gear when mounted rotatably on the carrier and thus rotate around its own center of gravity. The planet gear will thus rotate by an additional angle $$\varphi_{p2}$$.

If one considers a mere rolling motion, then the arc length $$b_c$$, which the carrier has covered on the sun gear, corresponds exactly to the arc length $$b_{p2}$$, by which the planet gear has moved on its circumference. The additional angle $$\varphi_{p2}$$ can be determined by the radian measure as follows:

\begin{align}
&b_{p2} = b_c \\[5px]
&\tfrac{d_p}{2} \cdot \varphi_{p2} = \tfrac{d_s}{2} \cdot \varphi_c  \\[5px]
\label{P2}
&\underline{\varphi_{p2} = \frac{d_s}{d_p} \cdot \varphi_c}  \\[5px]
\end{align}

### Rotation of the sun gear

The carrier is now held in position and the sun gear is rotated clockwise by an angle $$\varphi_s$$. In this case, the planet gear will turn counterclockwise by an angle $$\varphi_{p3}$$. Analogous to the case before, the following statement applies: The arc length $$b_s$$ at the circumference of the sun gear corresponds to the arc length $$b_{p3}$$, by which the planet gear has moved on its circumference:

\begin{align}
&b_{p3} = – b_s \\[5px]
&\tfrac{d_p}{2} \cdot \varphi_{p3} = – \tfrac{d_s}{2} \cdot \varphi_s  \\[5px]
\label{P3}
&\underline{\varphi_{p3} = – \frac{d_s}{d_p} \cdot \varphi_s}  \\[5px]
\end{align}

The negative sign indicates that the motion of the planet gear is in the opposite direction to the motion of the sun gear.

### Superposition of the different motions

The motions of the planet gear according to the equations (\ref{P1}), (\ref{P2}) and (\ref{P3}), which have been considered separately so far, can now be superposed to the total motion:

\begin{align}
&\varphi_p = \varphi_{p1} +\varphi_{p2} + \varphi_{p3}\\[5px]
\label{P}
&\underline{\varphi_{p} = \cdot \varphi_c + \frac{d_s}{d_p} \cdot \varphi_c     –    \frac{d_s}{d_p} \cdot \varphi_s   }  \\[5px]
\end{align}

The angular positions $$\varphi$$ contained in this equation result from the respective angular velocity $$\omega$$ and the elapsed time $$t$$ ($$\varphi = \omega \cdot t$$), whereby the angular velocity is directly related to the rotational speed $$n$$ by $$\omega = 2 \pi \cdot n$$:

\begin{align}
&\varphi = \omega \cdot t  ~~~ \text{with} ~~~ \omega = 2 \pi \cdot n ~~~\text{applies:}  \\[5px]
\label{varp}
&\underline{\varphi = 2 \pi \cdot n  \cdot t}  \\[5px]
\end{align}

If equation (\ref{varp}) is used in equation (\ref{P}), the following relationship ultimately results between the rotational speed of the planet gear $$n_p$$ and the rotational speeds of the sun gear $$n_s$$ and the carrier $$n_c$$:

\begin{align}
&2 \pi \cdot n_p  \cdot t = 2 \pi \cdot n_c  \cdot t + \frac{d_s}{d_p} \cdot 2 \pi \cdot n_c  \cdot t – \frac{d_s}{d_p} \cdot 2 \pi \cdot n_s  \cdot t \\[5px]
&n_p =n_c + \frac{d_s}{d_p} \cdot n_c – \frac{d_s}{d_p} \cdot n_s ~~~~~~~~\text{|} \cdot d_p \\[5px]
&n_p \cdot d_p =n_c \cdot d_p + d_s \cdot n_c – d_s \cdot n_s  \\[5px]
\label{g}
&\boxed{n_p \cdot d_p = n_c \cdot \left(d_p + d_s \right) – n_s \cdot d_s} \\[5px]
\end{align}

Since the pitch circle diameter $$d$$ of a gear is directly proportional to the number of teeth $$z$$, the equation above can also be expressed by the respective number of teeth:

\begin{align}
\label{pln}
&\boxed{n_p \cdot z_p = n_c \cdot \left(z_p + z_s \right) – n_s \cdot z_s} \\[5px]
\end{align}

This equation is called the fundamental formula of planetary gears  (also called Willis equation). The Willis equation is used to determine the different transmission ratios depending on the mode of operation, which will be explained in more detail in the next section.

## Willis equation for planetary gears

The Willis equation (\ref{pln}) generally applies to all planetary gears. Although the planet gears of a classic planetary gearbox are enclosed by a ring gear, this does not change the derived relationships between sun gear, planet gear and carrier. The only question that arises is how the motion of the planet gears is transferred to the ring gear.

Since a mere rolling motion without sliding between the ring gear and the planet gear takes place (considered as pitch cylinders), the velocity at the contact point must be equal. If one knows the speed $$v_{po}$$ with which the outermost point of the planet gear moves, then this corresponds to the velocity $$v_r$$ of the ring gear. Otherwise, a relative motion would come up, which of course can not be the case with toothed wheels. The pitch circle radius $$r$$ (or pitch circle diameter $$d$$) of the ring gear can then be used to determine its rotational speed $$n$$, since the following relationship applies between these parameters:

\begin{align}\;\;\;\;\
\label{o}
&v = \omega \cdot r = \omega \cdot \tfrac{d}{2} ~~~ \text{with} ~~~ \omega = 2 \pi \cdot n ~~~\text{applies}: \\[5px]
\label{v}
&\underline{v = \pi \cdot n \cdot d} \\[5px]
\end{align}

The same situation applies to the rotational speeds at the contact point between the planet gear and the sun gear. At the innermost point, the speed of the planet gear $$v_{pi}$$ must be equal to the speed of the sun gear $$v_s$$. The center of gravity of the planet gear moves with the velocity $$v_c$$ of the carrier. There is a linear relationship between these velocities (see black dotted line in the figure above), so that the circumferential speed of the ring gear $$v_r$$ can be determined for a given circumferential speed of the sun gear $$v_s$$ and a given circumferential speed of the carrier $$v_c$$.

### Why is there such a linear relationship?

Why there is such a relative simple, linear relationship of the speeds will be shown in the following. For the sake of simplicity, the gears are assumed to be pitch cylinders.

The motion of a point on the planet gear can be understood as the superposition of two motions. On the one hand, the planet gear first rotates around its own center of gravity. In this case the typical symmetrical and linear increase of the velocity according to the equation (\ref{o}) is obtained, starting from the axis of rotation of the planet gear. The maximum speeds $$v_p$$ are obtained  at the pitch circle of the planet gear.

In the center of rotation, the speed is zero as long as the planet gear axis does not move. However, the axis of rotation now moves at the speed of the carrier $$v_c$$. Both motions can now be superposed to the total motion.

The velocity of the planet gear at the outmost contact point with the ring gear points in the same direction as the velocity of the carrier. At the innermost point of contact with the sun gear, however, in the opposite direction. Due to the symmetrical speed distribution, the resulting speed of the planet gear at the outermost point of contact with the ring gear is therefore higher ($$v_{po}=v_c+v_p$$) to the same extent as it is lower at the innermost point of contact with the sun gear ($$(v_{pi}=v_c-v_p$$).

In other words, the velocity of a point on the planet gear increases linearly, starting from the point of contact with the sun gear. Since the speed of the carrier $$v_c$$ is assumed to be given, only the speed of the planet gear at the point of contact to the sun gear $$v_s$$ must be known in order to determine the circumferential speed at the opposite point of contact to the ring gear $$v_r$$.

As already explained, for a mere rolling process without relative motions, the speed of the planet gear at the contact point to the ring gear ($$v_{po}=v_c+v_p$$) must be equal to the circumferential speed of the ring gear $$v_r$$:

\begin{align}
&v_r\overset{!}{=}v_{po} \\[5px]
\label{v_r}
&\underline{v_r=v_c+v_p} \\[5px]
\end{align}

The same applies to the contact point between the planet gear and the sun gear. There, the speed of the planet gear ($$v_{pi}=v_c-v_p$$) must be equal to the circumferential speed of the sun gear $$v_s$$:

\begin{align}
&v_s\overset{!}{=}v_{pi} \\[5px]
\label{v_s}
&\underline{v_s=v_c-v_p} \\[5px]
\end{align}

If we subtract equation (\ref{v_s}) from equation (\ref{v_r}), we obtain the following relationship between the circumferential speeds of the sun gear $$v_s$$, the planet gear $$v_p$$ and the ring gear $$v_r$$:

\begin{align}
&v_r – v_s = v_c+v_p-v_c+v_p \\[5px]
&v_r = 2 \cdot v_p + v_s \\[5px]
\label{vvv}
&\underline{ v_p = \frac{v_r}{2} – \frac{v_s}{2} } \\[5px]
\end{align}

If the relationship of equation (\ref{v}) ist used in equation (\ref{vvv}), then the relationship between the corresponding rotational speeds is obtained:

\begin{align}
&v_p = \frac{v_r}{2} – \frac{v_s}{2} \\[5px]
&\pi \cdot n_p \cdot d_p = \frac{\pi \cdot n_r \cdot d_r}{2} – \frac{\pi \cdot n_s \cdot d_s}{2} \\[5px]
\label{nn}
&\boxed{n_p \cdot d_p = n_r \cdot \frac{d_r}{2} – n_s \cdot \frac{d_s}{2}} \\[5px]
\end{align}

The relation resulting from equation (\ref{nn}) can now be equated directly with the fundamental equation (\ref{g}) and one finally gets the following relation between the rotational speeds of the sun gear ($$S$$), the carrier ($$T$$) and the ring gear ($$H$$):

\begin{align}
&n_r \cdot \frac{d_r}{2} – n_s \cdot \frac{d_s}{2} = n_c \cdot \left(d_p + d_s \right)  – n_s \cdot d_s \\[5px]
&n_r \cdot d_r – n_s \cdot d_s  = 2 \cdot n_c \cdot \left(d_p + d_s \right)  – 2 \cdot n_s \cdot d_s \\[5px]
\label{f}
&\underline{n_r \cdot d_r  = 2 \cdot n_c \cdot \left(d_p + d_s \right)  – d_s \cdot n_s} \\[5px]
\end{align}

Additionally, it can be used that the diameters of the ring gear, the planet gear and the sun gear are not independent from each other. The ring gear diameter $$d_r$$ corresponds to the sum of the sun gear diameter $$d_s$$ and twice the planet gear diameter $$d_p$$:

\begin{align}
&d_r = d_s + 2 \cdot d_p \\[5px]
&\underline{d_p = \frac{d_r-d_s}{2}} \\[5px]
\end{align}

This results in the planetary gear equation for classic single-stage planetary gears (independent of the properties of the planet gears!):

\begin{align}
&n_r \cdot d_r = 2 \cdot n_c \cdot \left(\frac{d_r-d_s}{2} + d_s \right) – d_s \cdot n_s \\[5px]
&n_r \cdot d_r =n_c \cdot \left(d_r – d_s + 2 \cdot d_s \right) – d_s \cdot n_s \\[5px]
&\underline{n_r \cdot d_r = n_c \cdot \left(d_r + d_s \right) – d_s \cdot n_s} \\[5px]
\end{align}

Since for toothed wheels the pitch circle diameters $$d$$ are proportional to the number of teeth $$z$$, the equation above can also be expressed by the number of teeth of the ring gear ($$z_r$$) and the number of teeth of the sun gear ($$z_s$$):

\begin{align}
\label{pl}
&\boxed{n_r \cdot z_r = n_c \cdot \left(z_r + z_s \right) – z_s \cdot n_s} \\[5px]
\end{align}

## Transmission ratios of planetary gears

With a single planetary gear set one will obtain three different modes of operation, depending on which component (sun gear, carrier or ring gear) is fixed. Input and output are then carried out by the other two components. Which transmission ratios result in each case, is shown in the next section.

### Fixed sun gear

If the sun gear is fixed ($$n_s=0$$) and the gearbox input is carried out by the ring gear and the output by the carrier, the following transmission ratio $$i_s=\tfrac{n_r}{n_c}$$ results according to the equation (\ref{pln}):

\begin{align}
&n_r \cdot z_r = n_c \cdot \left(z_r + z_s \right) – z_s \cdot \underbrace{n_s}_{=0} \\[5px]
&n_r \cdot z_r = n_c \cdot \left(z_r + z_s \right)  \\[5px]
&\frac{n_r}{n_c} = i_s = \frac{z_r+z_s}{z_r}     \\[5px]
\label{i_s}
&\boxed{i_s = 1+\frac{z_s}{z_r}} ~~~1<i_s<2 \\[5px]
\end{align}

Equation (\ref{i_s}) shows that the transmission ratio is always greater than 1, i.e. the rotational speed is decreased by the planetary gearbox. But the transmission ratio is also limited upwards, since the number of teeth of the sun gear must always be smaller than that of the ring gear (otherwise the sun gear would be larger than the surrounding ring gear). In the theoretical limiting case, if the sun gear is as large as the ring gear and therefore both have identical numbers of teeth, the teeth ratio becomes $$\frac{z_s}{z_r}$$=1 and the transmission ratio 2 at most.

If input and output are reversed, i.e. the gearbox input is carried out by the carrier and the output by the ring gear, then the transmission ratio range lies between 1 and 0.5.

### Fixed ring gear

A further possibility for speed conversion is obtained, when the ring gear is fixed ($$n_r=0$$) and the gearbox input is carried out by the sun gear and the output by the carrier. This results in the following transmission ratio $$i_r=\tfrac{n_s}{n_c}$$:

\begin{align}
&\underbrace{n_r}_{=0} \cdot z_r = n_c \cdot \left(z_r + z_s \right) – z_s \cdot n_s \\[5px]
&0 = n_c \cdot \left(z_r + z_s \right) – z_s \cdot n_s \\[5px]
&\frac{n_s}{n_c} = i_r = \frac{z_r+z_s}{z_s} \\[5px]
\label{i_r}
&\boxed{i_r = 1+\frac{z_r}{z_s}} ~~~2<i_r<\infty \\[5px]
\end{align}

In the present case one also obtains a reduced rotational speed, because the transmission ratio will be greater than 2 in any case, since the number of teeth of the ring gear is always greater than that of the sun gear [the teeth ratio is thus greater than 1 ($$\frac{z_r}{z_s}>1$$)]. The transmission ratio is not limited upwards, since the ring gear and thus its number of teeth can in principle be chosen as large as desired and the transmission ratio then strives towards infinity.

If, in the opposite case, the gearbox input is no longer carried out by the carrier but by the ring gear, then the reciprocal transmission ratios with a range between 0 and 0.5 are obtained.

### Fixed carrier

A last possibility for the transmission ratio is obtained when the carrier ist fixed and the gearbox input is carried out by the sun gear and the output by the ring gear. In this case the following transmission ratio $$i_0=\tfrac{n_s}{n_r}$$ results:

\begin{align}
&n_r \cdot z_r = \underbrace{n_c}_{=0} \cdot \left(z_r + z_s \right) – z_s \cdot n_s \\[5px]
&n_r \cdot z_r = – z_s \cdot n_s \\[5px]
&\frac{n_s}{n_r} = i_0 = -\frac{z_r}{z_s} \\[5px]
\label{i_0}
&\boxed{i_0 = -\frac{z_r}{z_s}} ~~~\text{“stationary transmission ratio”}~~~-\infty<i_0<-1 \\[5px]
\end{align}

First of all, the negative sign is noticeable in the transmission ratio of equation (\ref{i_0}). It indicates that the direction of rotation between input and output shaft changes (“reverse gear”). In the present case, the transmission ratio ranges between $$-\infty$$ and -1 and in the opposite case (when input and output are reversed) between -1 and 0.

Note, that in this case the planetary gear works like a stationary gearbox without  moving rotational axes. For this reason, the transmission ratio in the case of a fixed carrier also called fixed carrier transmission ratio or stationary transmission ratio $$i_0$$!

### Direct drive

A planetary gear can also be used as a so-called direct drive. The carrier and the sun gear are firmly fixed to the ring gear. In this case, the rotary motion is transmitted directly from the input shaft to the output shaft (transmission ratio 1:1).  Such a direct drive is used, for example, in three-speed gear hubs as the “2nd gear”.

### Stationary transmission ratio (fixed carrier transmission ratio)

If one looks at the equations (\ref{i_s}), (\ref{i_r}) and (\ref{i_0}), then obviously all transmission ratios can also be expressed by the fixed carrier transmission ratio $$i_0=-\frac{z_r}{z_s}$$. For a fixed sun gear, the transmission ratio $$i_s$$ then becomes:

\begin{align}
&\boxed{i_s = 1-\frac{1}{i_0}}  \\[5px]
\end{align}

For a fixed ring gear, the transmission ratio $$i_r$$ can be calculated as follows using the fixed carrier transmission ratio $$i_0$$:

\begin{align}
&\boxed{i_r = 1-i_0}\\[5px]
\end{align}

Even the fundamental equation for planetary gears (\ref{pln}) can be expressed by the fixed carrier transmission ratio $$i_0$$:

\begin{align}
&n_r \cdot z_r = n_c \cdot \left(z_r + z_s \right) – z_s \cdot n_s \\[5px]
&n_r \cdot \frac{z_r}{z_s} = n_c \cdot \left( \frac{z_r}{z_s} + 1 \right) – n_s \\[5px]
& – n_r \cdot i_0 = n_c \cdot \left(1-i_0 \right) – n_s \\[5px]
&\boxed{ n_s = n_c \cdot \left(1-i_0 \right) + n_r \cdot i_0 }~~~\text{with}~~~\boxed{i_0=-\frac{z_r}{z_s}}~~~\text{fixed carrier transmission ratio} \\[5px]
\end{align}