Gear Trains
Key Facts
Gear Trains on Fixed Axes
Velocity ratio = $\displaystyle\frac{n_A}{n_B} = \pm \;\displaystyle\frac{t_B}{t_A}$
where,
- Positive sign is taken for Internal gearing,and
- Negative sign is taken for External gearing.
For Compound Chains
If A gears with B, which is then compounded with C, which finally drives D:
An analysis of Gear Trains with particular reference to Epicyclic Gears.
Introduction
One of the most common uses of Gear Trains is in the gear boxes of cars. In the simplest form, the crash gear box, a collection of spur gears were arranged to give different ratios of input to output speed and in the case of the reverse gear, change the direction of the output rotation.
The modern synchromesh gearbox is a development working on very similar principles.
Another, and in many ways more interesting type of train uses epicyclic gears. Many modern cars use a small epicyclic in the starter motor but much larger and more complicated versions have been used as the main gearbox. Before the 1939 - 1945 war the Daimler Car Company used a "pre-selector" Wilson box.
In more recent years these were fitted to London Transport Buses and they were also to be found in some British Army armoured wheeled vehicles. The gear box, which is driven through a fluid fly wheel, allows the driver to pre-select the next gear to be used and then to make a very rapid change at the right moment. This ability is useful in heavy urban traffic and when driving across uneven country.
Gear Trains on Fixed Axes
The velocity ratio transmitted between two gears is the same as for two pitch circles rolling together. As the circular pitch must be the same for both wheels, the speed ratio is determined by the number of teeth.
i.e. $\displaystyle\frac{n_A}{n_B} = \pm \;\displaystyle\frac{t_B}{t_A}$
- The positive sign is taken for Internal gearing,
- The negative sign is taken for External gearing.
A Simple train is one of three or wheels connected in series.
- The gear ratio is independent of the size of the intermediate wheels which are called Idlers
- For an odd number of externally geared wheels the direction of rotation of the last wheel is the same as the first.
- For an even number of wheels the direction of rotation of the last wheel is opposite to that of the first.
A Compound Chains
- When an intermediate shaft carries two wheels connected in series, the train becomes compound. For instance, if $A$ gears with $B$, which is compounded with $C$, which then drives $D$.
i.e.
Such arrangements can greatly increase or decrease the gear ratio.
- Where the axis of the last wheel is in line with that of the first, the arrangement is known as a reverted train.
Epicyclic Trains.
One or more of the wheels of the gear train are now carried on an arm which itself rotates about the axis of the main wheels.
The central main wheel is called the Sun wheel. The wheels carried by the arm are known as the Planet wheels and the outside internal wheel is called the Annulus. In some epicycilic gear trains the "arm" has a number of "limbs," each carrying a planet wheel. In these cases the "arm" may be referred to as the Spider.
There are a number of ways of analysing epicyclic gear trains. Three are described here:
Tabular Method
- The first line of the table is obtained by fixing the arm and then giving one wheel (e.g. the wheel which is to be fixed later) one revolution.
- Write down the corresponding revolutions of the other wheels from their number of teeth.
- As all the wheels and the arm can be turned about the same axis, an equal number of revolutions may be added (or subtracted) from each wheel.
- Any two lines may be multiplied (or divided) through by the same factor.
- Any two lines may be combined. By these means the given conditions are satisfied and the resulting gear ratio can be determined.
Note. The above may seem confusing and should be read in conjunction with the worked examples; in particular Example (3).
The Relative Velocity Method
If the arm is fixed, then the gear becomes a simple or compound train. Thus the ratio of the speeds of any two wheels can be determined by using the methods of paragraph (2).
i.e.
(If the arm were fixed.)
Where there is an intermediate shaft between $A$ and $B$ carrying, two wheels $C$ and $D$ which are fixed together and arranged so that $A$ drives $C$ and $D$ drives $B$, then the above equation can be modified to:
which can be calculated from the number of teeth etc.
For simple epicyclic operating with a fixed annulus there is nothing to choose between the above two methods. However, if all the wheels are moving, or when dealing with compound epicyclic, the second method gives a quicker and more direct solution. A comparison of the methods can be made from Examples (3) and (4)
The Determination of Torques
There are normally three torques acting on an epicyclic gear. These are:
- $\displaystyle \tau _1,$ the input Torque (in the sense of rotation of the input shaft)
- $\displaystyle \tau _0,$ the output torque (opposite to the sense of rotation of the output shaft)
- $\displaystyle \tau _C\;,$ the Torque on the casing.
Assuming the casing to be fixed, $\eta \;\tau _1\;n_1 = \tau _0\;n_0$
From considerations of the power and for equilibrium ( Taking signs into account )
$\tau _1 + \tau _0 + \tau _C = 0$
- For input and output shafts to have the same direction of rotation
(numerically)
- For input and output to have opposite directions of rotation.
(numerically)
When there is no fixed wheel there will, in general, be three torques to satisfy the following equations. Of the torques one can be found from the horse-power and speed.
$\tau _1 + \tau _2 + \tau _3 = 0$
And, $\tau _1\;n_1 + \tau _2\;n_2 + \tau _3\;n_3 = 0$
Alternatively, the forces acting on each wheel (or compound pair) may be analysed separately since if the gear is running at constant speed, there is no resultant force and no resultant torque on any intermediate wheel.
Example 2 [imperial]
The lead screw of a lathe has a right-hand single thread with a pitch of $\displaystyle\frac{1}{4}$ in. The smallest change wheel has 20 teeth, the largest 120 teeth and the number of teeth on intermediate sizes increases in steps of 5.
Find a gear train suitable for connecting the spindle and the lead screw when:
- a) A right hand screw with 26 threads per inch is to be cut.
- b) A left-hand screw with 35 threads per inch is required.
- a) Since the required screw is right-handed and has 26 threads per inch, it follows that them spindle must make 26 revolutions whilst the saddle moves 1 in. towards the headstock. But the saddle will move 1 in. towards the headstock when the lead screw makes 4 revolutions. Hence the spindle must make 26 revolutions while the lead screw makes 4 revolutions.
Let $n_S = r.p.m.$ of the spindle and $n_1 = r.p.m.$ of lead screw
Then, $\displaystyle\frac{n_S}{n_1} =\displaystyle \frac{26}{4} = \displaystyle\frac{13}{2}$
The only wheel in the set those number of teeth is a factor of 13 is that with 65 teeth
$\frac{n_S}{n_1} = \frac{13}{2} = \frac{65}{5}\times \frac{1}{2} = \frac{65}{25}\times \frac{5}{2} = \frac{65}{25}\times \frac{75}{30}$
The wheels would be arranged as shown in the diagram (a). It can be seen that the spindle and the lead screw revolve in the same direction.
- b)Since the required screw is left handed and has 35 threads per inch, the spindle must make 35 revolutions while the saddle moves 1 in. away from the headstock. The lead screw must therefore make 4 revolutions whilst the spindle makes 35 revolutions in the opposite direction.
$\frac{n_S}{n_1} = \frac{35}{4} = \frac{7}{2}\times \frac{5}{2} = \frac{105}{30}\times \frac{100}{40}$
The wheels would be arranged as shown in diagram (b)
The intermediate or idler wheel $I$ is required in order to make the lead screw rotate in the opposite direction to the spindle. As already noted the intermediate wheel does not affect the speed ratio and so a wheel of any convenient size may be used.
Example 3 [imperial]
In the Epicyclic Gear shown the internal wheels $A$ and $F$ and the compound wheel $C-D$ rotate independently about the axis $O$. The wheels Band $E$ rotate on pins fixed to the arm $L$. The wheels all have the same pitch and the numbers of teeth are : $B$ and $E$ 18, $C$ 28, and $D$ 26.
If $L$ makes 150 r.p.m. clockwise, find the speed of $F$ when :
- a) The Wheel $A$ is fixed.
- b) The Wheel $A$ makes 15 r.p.m. counter-clockwise.
Since the pitch is the same for all the wheels, the diameters are proportional to the number of teeth.
$\therefore \;\;\;\;\;\;\;t_A = t_C + 2\,t_B = 28 + 2\times 18 = 64$
And $t_F = t_C + 2\,t_B = 26 + 2\times 18 = 62$
The required speeds can be calculated from the following Table. The wheels $B$ and $E$ are omitted since they perform only intermediate function.
- a) From line (iii)
$n_F = 9\times \displaystyle\frac{150}{217} = 6.22\;r.p.m.$ Clockwise
- b) From line (v)
$n_F\;= - \displaystyle\frac{1770}{217} = 6.16\;r.p.m.$ Anti-clockwise
- a) $6.22\;r.p.m.$ Clockwise
- b) $6.16\;r.p.m.$ Anti-clockwise
Example 4
In the epicyclic gear train shown in the diagram, wheel $A$ and wheel $E$ (30 teeth) are fixed to a sleeve $Y$ which is free to rotate on spindle $X$. $B$ (24 teeth) and $C$ (22 teeth) are keyed to a shaft which is free to rotate in a bearing on arm $F$. $D$ has 70 teeth. $H$ has 15 teeth and is mounted on a shaft $V$ which rotates at 100 r.p.m. Spindle $X$ makes 300 r.p.m. in the same direction as $V$. All the teeth are of the same pitch.
Find the speed and direction of rotation of $Z$.
Note: $H$ is not part of the epicyclic since it is turning on a fixed shaft not in line with the axis of the arm.
$t_A + t_B = t_C + t_P$
$\therefore \;\;\;\;\;\;\;t_A = 22 + 70 - 24 = 68\;teeth$
In this example it is better to use the Relative Velocity Method which was described earlier. (See equation (2). It can be seen from the diagram that $F$ is the Arm.
The Speed Ratio $Z$ to $E$ = $\displaystyle\frac{n_Z - n_F}{n_E - n_F}$
But if the arm were fixed: The Speed Ratio $Z$ to $E$ = $\displaystyle\frac{t_C}{t_D}\times \displaystyle\frac{t_A}{t_B}$
But $H$ is mounted on shaft $V$ and rotates at 100 r.p.m. It meshes with $E$ which has twice as many teeth and so $E$ rotates at 50 r.p.m. in the opposite direction to $V$.
$\therefore \;\;\;\;\;\;\;\frac{n_Z - 300}{- 50 - 300} = \frac{22}{70}\times \frac{68}{24} = 0.89$
Thus, $\;\;n_Z = 300 - 311.7\;= - 11.7\;r.p.m.$
i.e. 11.7 r.p.m. in the opposite sense to both $V$ and $X$.
- Speed of $Z$ is 11.7 r.p.m. in the opposite sense to both $V$ and $X$.
Example 5 [imperial]
A gear train is shown in the diagram in which the shaft $X$ rotates at 500 r.p.m. Solid with shaft $X$ is the arm $A$ on which the compound bevel wheels $B$ and $D$ can revolve freely together. Wheel $B$ meshes with wheel $C$ and wheel $D$ with wheel $E$ which is also solid with the spur wheel $F$, the latter meshes with $G$ on shaft $Y$ which rotates in the opposite direction to $X$. The bevel wheel $C$ is fixed by keying to the support.
The number of teeth on wheels $C, D, E, F$ and $G$ are 20, 27, 32, 24, and 30 respectively.
Determine the number of teeth required on wheel $B$ to give a speed reduction of 20 to 1 between shafts $X$ and $Y$.
Where bevel gears are turning on axes which are not parallel to that of the Arm, it is difficult to attach any sign to their direction of rotation. For this reason they are omitted from the table below and they are merely made use of to determine the speed of $E$ when $A$ is fixed and $C$ is rotated by 1 rev.
Note that the arm $A$ is integral with $X$ and that $G$ is outside the epicyclic.
$F$ is rotating with $E$ and the speed of $G$ (and $Y$)
$= - \frac{24}{30}\;\left (\frac{20\times 27}{32\,t_B} - 1 \right )= - \frac{n_X}{20}$ (For a speed reduction of 20 to 1.) Taking the value for $\displaystyle n_X$ from the Table as -1,
Hence,
- The number of teeth required on wheel $B$ is $18.$
Example 6 [imperial]
In the epicyclic gear unit shown in the diagram the input shaft is attached to the planet carrier.
$Z$ and the output shaft to the Sun wheel $A$. Sun wheel $A_1$ is fixed to the casing and does not rotate whist the planet wheels $C$ and $C_1$ are compounded and rotate together.
If all teeth are the same pitch and $A$ and $C$ each have 20 teeth, find the teeth on $A_1$ and $C_1$ so that the output shaft runs in the reverse direction and at half the speed of the input shaft.
If the input shaft transmits 14 h.p. at 1500 r.p.m.and the efficiency of the unit is 96% calculate the input and output torques and the fixing torque of the casing.
Tabulate the speeds as follows:
From the question : $\displaystyle\frac{n_Z}{n_A}\;= - \displaystyle\frac{1}{2}$
But from the above table:$- 1\;= - \displaystyle\frac{1}{2}\times\left (\displaystyle\frac{t_{A1}}{t_{C1}} - 1\right )$
$\therefore \;\;\;\;\;\; - 1\;= - \frac{1}{2}\times \left (\frac{t_{A1}}{t_{C1}} - 1\right )$
Thus, $t_{A1} = \displaystyle\frac{3}{2}\;t_{C1}$
But since the pitch is the same for all wheels: $t_{A1} + t_{C1} = t_A + t_C$
Since both $A$ and $C$ have 20 teeth equations (1) and (2)(- from Example (5)) can be solved to give:
$t_{A1} = 24\;teeth$ and $t_{C1} = 16\;teeth$
From the question, $n_Z = 1,500\;r.p.m.$ and so $n_A= - 750\;r.p.m.$
The torques at both input and output are obtained from the horse-powers. (1 h.p. = 550 ft.lb./sec. or 33000 ft.lb./min.)
$\tau _Z = \frac{14\times 33,000}{2\,\pi \times 1500} = 49.1\;lb.\;ft.$
And allowing for the efficiency of the system:
$\tau _A = \frac{0.96\times14\times 33,000}{2\,\pi \times 7,500} = 94.2\;lb.\;ft.$
The fixing torque is the sum of the input and output torques for opposite directions of rotation. ( See equation (4))
Fixing Torque = $143.3\;lb.\;ft.$
- $t_{A1} = 24\;teeth$ and $t_{C1} = 16\;teeth$
- $\tau _Z =49.1\;lb.\;ft.$ and $\tau _A= 94.2\;lb.\;ft.$
- Fixing Torque = $143.3\;lb.\;ft.$
Example 7 [imperial]
The diagram shows a system of gearing. The casing $N$ is fixed. Wheel $H$ is keyed to shaft $B$. Wheels $K$ and $L$ are fixed to each other and are free to rotate on shaft $B$. Wheel $E$ is also free to rotate on shaft $B$. Wheels $F$ and $G$ are fixed to each other and are free to rotate on a pin mounted on wheel $E$. The number of teeth on the various wheels is shown on the diagram and all the teeth are of the same diametral pitch.
If $A$ rotates at 200 r.p.m., whilst $C$ is held stationary, find the speed of $B$.Find also the torque on shaft $C$ to hold it stationary under these conditions if 2 h.p. is being transmitted.
As the casing and shaft $C$ are fixed,it can be seen that both $K$ and $L$ are fixed. Wheel $E$ will turn at a slower rate than wheel $D$ and in the opposite direction.
$n_E\;= - \displaystyle\frac{30}{72}\times 200\;= - \displaystyle\frac{250}{3}\;r.p.m.$ ($E$ is the "arm")
By inspection, $t_H + t_F = t_G + t_F$
$\therefore \;\;\;\;\;t_H = t_G + t_K - t_F = 22\;teeth$
Using equation (3), $\displaystyle\frac{n_H - n_E}{n_L - n_E} = \displaystyle\frac{t_K}{t_G}\times \displaystyle\frac{t_F}{t_H} = \displaystyle\frac{24}{16}\times \displaystyle\frac{18}{22}$ i.e. $\displaystyle\frac{n_H - 250/3}{0 - 250/3} = \displaystyle\frac{27}{22}$
Thus, $n_H = \displaystyle\frac{250}{3}\times \displaystyle\frac{5}{22} = 18.95\;r.p.m.$
As $H$ is keyed to shaft $B$ the above is also the speed of the shaft.
The torque on shaft $B$ is given by: $\tau _B = \displaystyle\frac{2\times33,000}{2\;\pi \times 18.95} = 554\;lb.\;ft.$
The tangential force between the teeth of $F$ and $H$ is given by: $p_1\;\propto \;\displaystyle\frac{\tau _B}{t_H} = 25.2$
Neglecting all losses and since no torque is absorbed by the compound wheel $F-G$ it is possible to take moments about the centre of the wheel. (see diagram (a))
$P_2 = P_1\times \frac{t_F}{t_G} = 28.35$
Similarly by taking moments about the centre of the compound wheel $L_K$ (see diagram (b))
$P_4 = P_2\times \frac{t_K}{t_L} = 32.4$
Then, $\tau _C = P_4\times t_M = 809\;lb.\;ft.$
Check:
The above answer can be checked as follows.
$P_3 = P_2 - P_1 = 3.15$
This acts on $E$ at a radius proportional to $\displaystyle t_H + t_F (= 40)$.
By moments about the axis of $E$ the force between $E$ and $D$ $= P_3\times \displaystyle\frac{40}{t_E} = 3.15\times \displaystyle\frac{40}{72} = 1.75$
$\tau _D = 1.75\times t_D = 52.5\;lb.\;ft.$
Thus, $h.p. at A = 52.5\times 2\,\pi \times\displaystyle\frac{200}{33,000} = 2.0\;h.p.$
Which is of course what was given in the question.
- The speed of $B$ is $18.95\;r.p.m.$
- The torque on shaft $C$ is $809\;lb.\;ft.$