Automatic Transmission

Fluid flywheel or fluid coupling:

The fluid coupling comprises two rotors, each with radial vanes, in a casing filled with light engine oil. The driving member is attached to the crankshaft, and as this rotates the oil contained between the vanes is thrown outwards by centrifugal force and enters the opposing vanes of the driven member, which is connected to the gearbox. The oil passes to the inner portion of the driven rotor and returns to the driving members so that a circulation of oil is produced within the two rotors.

The slip, or difference is speed of the two rotors relative to engine speed, depends upon the engine speed and the load on the driven member. In general, as the engine speed rises from about 500 rev/min to 1000 rev/min the slip decreases from 100 per cent to some 10 per cent. This effect is used to eliminate the action of engagement of a friction clutch, and in addition the fluid flywheel serves to damp torque fluctuations.  

With a further increase in engine speed the slip falls to 1 per cent or 2 per cent, but there is inevitably some loss of energy and resulting in slightly lowered transmission efficiency. Even at the lowest speeds some drag is imposed on the driven rotor, making a simple combination of fluid flywheel and orthodox gearbox unsuitable. So that it is frequently employed with an epicyclic-type gearbox.

 

Fluid friction coupling:

To overcome the limitations of the hydraulic coupling (idling drag and slip) for large trucks applications, a combining shoe and drum centrifugally operated clutch to provide a positive lock-up at higher output speeds with a smaller coreless fluid coupling than would be necessary if the drive was only to be through a fluid coupling. The reduce in size and volume of fluid circulation in the coupling thereby eliminate residual idling drag.

 

 

The torque converter:

The torque converter performs a similar function to the gearbox, increasing torque with a reduction in speed. However, unlike a conventional gearbox the ratio is continuously variable.

In the single-stage converter the driving member or impeller acts as in the fluid coupling and fluid enters the driven member or turbine. There in now, however, a stator or reaction member to redirect the flow from the turbine into the direction of engine rotation before it re-enters the eye of the impeller. The energy remaining in this returning fluid, assisting the impeller flow, provides the torque multiplication.

 

Torque multiplication is at a maximum with the turbine stalled and falls to zero when both members revolve at the same speed. At both these extremes the efficiency is zero, in the first there is no output rotation, in the second no output torque. At intermediate turbine speeds the efficiency rises to some 80-90 per cent depending on design and operation conditions.

The high-turbine-speed inefficiency is a voided by mounting the stator so that it can freewheel in the direction of converter rotation. This occurs as the torque ratio approaches unity and the unit then functions as a fluid coupling, with maximum efficiency of some 95 per cent. An overrunning clutch allows the stator to freewheel (rotate) when the speed of the turbine and impeller reach the coupling point. A lockup torque converter eliminates the 10 percent slip that takes place between the impeller and the turbine at the coupling stage of operation. There are two types of lockup torque converters. The centrifugal lockup clutch (CLC) and piston lockup clutch (PLC), which is the type installed in most automatic transmission. Since the torque converter provides as infinitely variable torque ratio, up to about 2.5:1 for a single stage three element unit, gearbox ratio requirements are simplified. In addition the converter acts as a fluid flywheel, providing an automatic clutch and a torsional damper in the transmission.

Some converters provide torque ratios up to 7.1 by using additional reactor members to redirect the fluid path and impart additional impulses the turbine.

 

Epicyclic gearing:

Epicyclic gearing involves one member of the train both revolving on its own axis and rotation bodily about another axis. In a typical application three planet wheels supported on a carrier are in mesh with an inner sum pinion and an outer annulus or internally toothed ring.

 

Sun, planet system and annulus are the three members of an epicyclic gear train which can form one assembly in an epicyclic gearbox. It all three members are free, no drive is transmitted, providing neutral. If any two members are locked together, usually by a hydraulic operated multiplate clutch, the epicyclic train revolves as a unit, giving a 1:1 ratio. If any one member of the train is held stationary, usually by a hydraulically contracted band around a drum carried by the member, gear ratios will be produced between the other two members.

From this typical train a number of alternative gear ratios are available. In an epicyclic gearbox the range of gear ratios is normally obtained by compounding two or more epicyclic assemblies, the output from one being applied to one member of a second.

From the train discussed, let S be the sun pinion with number of teeth s, A be the annulus with number of teeth a, and C be the planet carrier assembly. Then

 

Number of teeth on planet gear = (a-s)/2

 

Gear ratio of simple epicyclic gearing:

s = number of teeth on sun wheel (S), a = number of teeth on annulus (A),(C) is the plant carrier, NA= annulus rev/min, NC = Planet carrier rev/min, NS = sun rev/min.

 

1-a) Fixing the planet carrier (C)

Condition

S

A

C

Arm fixed

- 1

s/a

0

NA = - (s/a) NS

a NA = - s Ns

Input S: ig = NA/ NS = -a/s [reduction]
 

1-b) Fixing the planet carrier (C):

Condition

S

A

C

Arm fixed

a/s

- 1

0

NS = - (a/s) NA

s NS = - a NA

Input A: ig = NA/NS = -s/a

 

2- Fixing the Sun Gear (S):

Condition

S

A

C

Arm fixed

- 1

s/a

0

Train locked

+ 1

+ 1

+ 1

Sun fixed

0

1 + s/a

1

NA = (1+s/a) NC = [(a+s)/a] NC

NC = [a/(a+s)] NA

(a+s) NC = a NA

Input A: ig = NA/NC = (a+s)/a [reduction]
Input C: ig= NC/NA= a/(a+s)
 

 

3- Fixing the annulus wheel (A):

Condition

S

A

C

Arm fixed

a/s

- 1

0

Train locked

+ 1

+ 1

+ 1

Annulus fixed

1+a/s

0

1

NS = (1+a/s) NC = [(a+s)/s] NC

NC = [s/(a+s)] NS

(a+s) NC = s NS

Input S: ig = NS/NC = (a+s)/s [reduction]
Input C: ig = NC/NS = s/(a+s)
 

 

* The general formula is:

NC = (a NA + s NS) / (a+s) or

(a+s) NC = a NA + s NS

 

In the above equations, put the N of the fixed element (using the brakes) = 0,

then the unite reduction ratio iu = Ninput / Noutput

Using (the clutch) to connect any two elements in the unit will make the unit reduction ratio iu = 1
The gearbox ratio ig = iu1 iu2 i u3 .......

- Find the gearbox ratio in case:

s = number of teeth on sun wheel (S) = 20 teeth  

a = number of teeth  on annulus (A) = 100 teeth

Solution:

Number of plant gear teeth c= (a-s) /2 = (100-20)/2 = 80/2 = 40 tooth

1-a) the carrier (C) is fixed, Input S, ig = - (a/s) = - (100/20) = -  5.0

Output is (A); (NA = NS / -5.0), reduction and reverse.

2- The sun gear (S) is fixed, Input A, ig = (a+s)/a = (100 + 20)/100 = 120/100 = 1.2

Output (C); (NC = NA /1.2), reduction.

3- the annulus (A) is fixed, Input S, ig = (a+s)/s = (100+20)/20 = 120/20 = 6.0

Output (C),(NC = NS /6.0), reduction.

 

 

Compound gearing:

Most automatic gearbox use two or more epicycle units to form a compound gear train.

 

The above figure shows a compound gear train, with a five-position selector, marked PRND21 control the action of two clutches and two brake bands a free wheel, to give three forward gears and a reverse gear.

 

* Excel program to calculate the unit reduction ratio of planetary gear system
 http://www.thecartech.com/KnowYourCar/Transmission/gears/planetary.xls