Advanced Vehicle Technology Episode 1 Part 6 pdf

3.8 Setting gear ratios Matching the engine's performance characteristics to suit a vehicle's operating requirements is pro-vided by choosing a final drive gear reduction and then select

large planet gear absorbs the driving torque reaction

and in the process is made to revolve around the

braked sun gear The overdrive condition is created

by the large planet gears being forced to roll `walk'

about the sun gear, while at the same time revolving

on their own axes As a result, the small planet gears,

also revolving on the same carrier pins as the large

planet gears, drive forward the annular ring gear at a faster speed relative to that of the input

The overall gear ratio step up is achieved by having two stages of meshing gear teeth; one between the large pinion and sun gear and the other between the small pinion and annulus ring gear By using this compound epicyclic gear train, a Fig 3.30 Laycock double epicycle overdrive

relatively large step up gear ratio can be obtained

for a given diameter of annulus ring gear compared

to a single stage epicyclic gear train

Direct drive (Fig 3.30) Direct drive is attained by

releasing the double-sided cone clutch member from

the stationary conical brake and shifting it over so

that it contacts and engages the conical frictional

surface of the annulus ring gear The power flow

from the input shaft and planet carrier now divides

into two paths Ð the small planet gear to annulus

ring gear route and the large planet gear, sun gear

and double-sided clutch member route, again

finish-ing up at the annulus rfinish-ing gear With such a closed

loop power flow arrangement, where the gears

can-not revolve independently to each other, the gears

jam so that the whole gear train combination rotates

as one about the input to output shaft axes It

thereby provides a straight through direct drive It

should be observed that the action of the

unidirec-tional roller clutch is similar to that described for the

single stage epicyclic overdrive

Clutch operating (Fig 3.30) Engagement of

direct drive and overdrive is achieved in a similar

manner to that explained under single stage

epicyc-lic overdrive unit

Direct drive is provided by four powerful springs

holding the double-sided conical clutch member in

frictional contact with the annulus ring gear

Con-versely, overdrive is obtained by a pair of hydraulic

slave pistons which overcome and compress the

clutch thrust springs, pulling the floating conical

clutch member away from the annulus and into

engagement with the stationary conical brake

Hydraulic system (Fig 3.30) Pressure supplied by

the hydraulic plunger type pump draws oil from the

sump and forces it past the non-return ball valve to

both the slave cylinders and to the solenoid valve

and the relief valve

Direct drive engagement When direct drive is

engaged, the solenoid valve opens due to the

sole-noid being de-energized Oil therefore flows not

only to the slave cylinders but also through the

solenoid ball valve to the overdrive lubrication

sys-tem where it then spills and returns to the sump A

relatively low residual pressure will now be

main-tained within the hydraulic system Should the oil

pressure rise due to high engine speed or blockage,

the low pressure ball valve will open and relieve the excess pressure Under these conditions the axial load exerted by the clutch thrust springs will clamp the double-sided floating conical clutch member to the external conical shaped annulus ring gear Overdrive engagement To select overdrive the solenoid is energized This closes the solenoid ball valve, preventing oil escaping via the lubrication system back to the sump Oil pressure will now build up to about 26±30 bar, depending on vehicle application, until sufficient thrust acts on both slave pistons to compress the clutch thrust springs, thereby permitting the double-sided clutch member

to shift over and engage the conical surface of the stationary brake To enable the engagement action

to overdrive to progress smoothly and to limit the maximum hydraulic pressure, a high pressure valve jumper is made to be pushed back and progres-sively open This controls and relieves the pressure rise which would otherwise cause a rough, and possibly sudden, clutch engagement

3.8 Setting gear ratios Matching the engine's performance characteristics

to suit a vehicle's operating requirements is pro-vided by choosing a final drive gear reduction and then selecting a range of gear ratios for maximum performance in terms of the ability to climb gradi-ents, achievement of good acceleration through the gears and ability to reach some predetermined maximum speed on a level road

3.8.1 Setting top gear

To determine the maximum vehicle speed, the engine brake power curve is superimposed onto the power requirement curve which can be plotted from the sum of both the rolling (Rr) and air (Ra) resistance covering the entire vehicle's speed range (Fig 3.31) The total resistance R opposing motion at any speed is given by:

R ˆ Rr‡ Ra

ˆ 10CrW ‡ CDAV2

where Crˆ coefficient of rolling resistance

W ˆ gross vehicle weight (kg)

CDˆ coefficient of aerodynamic resist-ance (drag)

A ˆ projected frontal area of vehicle (m2)

V ˆ speed of vehicle (km/h)

The top gear ratio is chosen so that the

maxi-mum road speed corresponds to the engine speed at

which maximum brake power is obtained (or just

beyond) (Fig 3.32)

Gearing is necessary to ensure that the vehicle

speed is at a maximum when the engine is

develop-ing approximately peak power

Thus Linear wheel speed ˆ Linear road speed

dN

GF ˆ

1000

60 V (m/min)

; Final drive gear ratio GFˆ60 dN100 V

ˆ 0:06 dNV where GFˆ final drive gear ratio

N ˆ engine speed (rev/min)

d ˆ effective wheel diameter (m)

V ˆ road speed at which peak power is

developed (km/h)

Example A vehicle is to have a maximum road

speed of 150 km/h If the engine develops its peak

power at 6000 rev/min and the effective road wheel

diameter is 0.54 m, determine the final drive gear ratio

GFˆ0:06 dNV

ˆ0:06  3:142  0:54  6000

150

ˆ 4:07:1

3.8.2 Setting bottom gear The maximum payload and gradient the vehicle is expected to haul and climb determines the necessary tractive effort, and hence the required overall gear ratio The greatest gradient that is likely to be encountered is decided by the terrain the vehicle is

to operate over This normally means a maximum gradient of 5 to 1 and in the extreme 4 to 1 The minimum tractive effort necessary to propel a vehicle

up the steepest slope may be assumed to be approxi-mately equivalent to the sum of both the rolling and gradient resistances opposing motion (Fig 3.31) The rolling resistance opposing motion may be determined by the formula:

Rr ˆ 10CrW where Rrˆ rolling resistance (N)

Crˆ coefficient of rolling resistance

W ˆ gross vehicle weight (kg) Average values for the coefficient of rolling resistance for different types of vehicles travelling

at very slow speed over various surfaces have been determined and are shown in Table 3.2

Likewise, the gradient resistance (Fig 3.33) opposing motion may be determined by the for-mula:

Rgˆ10WG or 10W sin  where Rgˆ gradient resistance (N)

W ˆ gross vehicle weight (10W kg ˆ WN)

G ˆ gradient (1 in x) ˆ sin  Fig 3.31 Forces opposing vehicle motion over its speed

range Fig 3.32 Relationship of power developed and roadpower required over the vehicle's speed range

Tractive effort ˆ Resisting forces opposing motion

E ˆ R

ˆ Rr‡ Rg (N) where E ˆ tractive effort (N)

R ˆ resisting forces (N)

Once the minimum tractive effort has been

cal-culated, the bottom gear ratio can be derived in the

following way:

Driving torque ˆ Available torque

ER ˆ T GBGFM

; Bottom gear ratio GBˆTGER

FM

where GFˆ final drive gear ratio

GBˆ bottom gear ratio

Mˆ mechanical efficiency

E ˆ tractive effort (N)

T ˆ maximum engine torque (Nm)

R ˆ effective road wheel radius (m) Example A vehicle weighing 1500 kg has a coefficient of rolling resistance of 0.015 The trans-mission has a final drive ratio 4.07:1 and an overall mechanical efficiency of 85%

If the engine develops a maximum torque of

100 Nm (Fig 3.34) and the effective road wheel radius is 0.27 m, determine the gearbox bottom gear ratio

Assume the steepest gradient to be encountered

is a one in four

Rrˆ 10CrW

ˆ 10  0:015  150 ˆ 225N

Rg ˆ10WG

ˆ10  15004 ˆ 3750N

E ˆ Rr‡ Rg

ˆ 3750 ‡ 225 ˆ 3975N

GBˆTGeR

FM

ˆ100  4:07  0:853975  0:27 ˆ 3:1:1 Fig 3.33 Gradient resistance to motion

Fig 3.34 Engine torque to speed characteristics

Table 3.2 Average values of coefficient of rolling

resistance

Coefficient of rolling resistance (C r )

Vehicle type

Concrete Medium hard soil Sand

Note The coefficient of rolling resistance is the ratio of the

rolling resistance to the normal load on the tyre.

i:e: C r ˆ R r

W

3.8.3 Setting intermediate gear ratios

Ratios between top and bottom gears should be

spaced in such a way that they will provide the

tractive effort±speed characteristics as close to the

ideal as possible Intermediate ratios can be best

selected as a first approximation by using a

geo-metric progression This method of obtaining the

gear ratios requires the engine to operate within the

same speed range in each gear, which is normally

selected to provide the best fuel economy

Consider the engine to vehicle speed

character-istics for each gear ratio as shown (Fig 3.35) When

changing gear the engine speed will drop from the

highest NHto the lowest NLwithout any change in

road speed, i.e V1, V2, V3etc

Let G1ˆ 1st overall gear ratio

G2ˆ 2nd overall gear ratio

G3ˆ 3rd overall gear ratio

G4ˆ 4th overall gear ratio

G5ˆ 5th overall gear ratio

where Overall

gear ratio ˆRoad wheel speed (rev/min)Engine speed (rev/min)

Wheel speed when engine is on the high limit NHin

first gear G1ˆNH

G1 (rev/min) Wheel speed when engine is on the low limit NLin

second gear G2ˆNL

G2 (rev/min) These wheel speeds must be equal for true rolling

Hence NGH

1 ˆNGL

2

; G2ˆ G1NL

NH

Also NH

G2 ˆNL

G3

; G3ˆ G2NL

NH and NGH

3 ˆNGL

4

; G4ˆ G3NNL

H

NH

G4 ˆ

NL

G5

; G5ˆ G4NL

NH

The ratioNNL

His known as the minimum to max-imum speed range ratio K for a given engine Now, gear G2ˆ G1NL

NHˆ G1K, since NL

NHˆ k (a constant) gear G3ˆ G2NL

NHˆ G2K ˆ (G1K)K

ˆ G1K2

gear G4ˆ G3NNL

Hˆ G3K ˆ (G1K2)K

ˆ G1K3

gear G5ˆ G4NNL

Hˆ G4K ˆ (G1K3)K

ˆ G1K4: Hence the ratios form a geometric progression

Fig 3.35 Gear ratios selected on geometric progression

The following relationship will also apply for a

five speed gearbox:

G2

G1ˆ

G3

G2ˆ

G4

G3ˆ

G5

G4ˆ

NL

NHˆ K and G5ˆ G1K4 or K4ˆGG5

1

Hence K ˆ GG5

1

 1

or



G5

G1 4

r

In general, if the ratio of the highest gear (GT)

and that of the lowest gear (GB) have been

deter-mined, and the number of speeds (gear ratios) of

the gearbox nG is known, the constant K can be

determined by:

K ˆ GGT

B

  1

nG 1

Bˆ KnG ‡1

; GTˆ GBKnG 1

For commercial vehicles, the gear ratios in

the gearbox are often arranged in geometric

progression For passenger cars, to suit the

chan-ging traffic conditions, the step between the ratios

of the upper two gears is often closer than that

based on geometric progression As a result, this

will affect the selection of the lower gears to some

extent

Example A transmission system for a vehicle

is to have an overall bottom and top gear ratio

of 20:1 and 4.8 respectively If the minimum to maxi-mum speeds at each gear changes are 2100 and

3000 rev/min respectively, determine the following: a) the intermediate overall gear ratios

b) the intermediate gearbox and top gear ratios

K ˆNNL

H

ˆ21003000ˆ 0:7 a) 1st gear ratio G1ˆ 20:0:1 2nd gear ratio G2ˆ G1K ˆ 20  0:7 ˆ 14:0:1 3rd gear ratio G3ˆ G1K2ˆ 20  0:72ˆ 9:8:1 4th gear ratio G4ˆ G1K3ˆ 20  0:73ˆ 6:86:1 5th gear ratio G5ˆ G1K4ˆ 20  0:74ˆ 4:8:1

b) G1ˆ20:04:8 ˆ 4:166:1

G2ˆ14:0 4:8 ˆ 2:916:1

G3ˆ9:84:8ˆ 2:042:1

G4ˆ6:864:8 ˆ 1:429:1

Top gear G5ˆ4:84:8ˆ 1:0:1

4 Hydrokinetic fluid couplings and torque converters

A fluid drive uses hydrokinetic energy as a means

of transferring power from the engine to the

trans-mission in such a way as to automatically match

the vehicle's speed, load and acceleration

require-ment characteristics These drives may be of a

simple two element type which takes up the drive

smoothly without providing increased torque or

they may be of a three or more element unit

which not only conveys the power as required

from the engine to the transmission, but also

multi-plies the output torque in the process

4.1 Hydrokinetic fluid couplings (Figs 4.1 and 4.2)

The hydrokinetic coupling, sometimes referred to

as a fluid flywheel, consists of two saucer-shaped discs, an input impeller (pump) and an output turbine (runner) which are cast with a number of flat radial vanes (blades) for directing the flow path

of the fluid (Fig 4.1)

Owing to the inherent principle of the hydro-kinetic coupling, there must be relative slip between the input and output member cells exposed to each

Fig 4.1 Fluid coupling action

other, and the vortex flow path created by pairs of

adjacent cells will be continuously aligned and

misaligned with different cells

With equal numbers of cells in the two half

members, the relative cell alignment of all the cells

occurs together Consequently, this would cause a

jerky transfer of torque from the input to the output

drive By having differing numbers of cells within

the impeller and turbine, the alignment of each pair

of cells at any one instant will be slightly different

so that the impingement of fluid from one member

to the other will take place in various stages of

circulation, with the result that the coupling torque

transfer will be progressive and relatively smooth

The two half-members are put together so that the fluid can rotate as a vortex Originally it was common practice to insert at the centre of rotation a hollow core or guide ring (sometimes referred to as the torus) within both half-members to assist in establishing fluid circulation at the earliest moment

of relative rotation of the members These couplings had the disadvantage that they produced consider-able drag torque whilst idling, this being due mainly

to the effectiveness of the core guide in circulating fluid at low speeds As coupling development pro-gressed, it was found that turbine drag was reduced

at low speeds by using only a core guide on the impeller member (Fig 4.2) With the latest design

Fig 4.2 Fluid coupling

these cores are eliminated altogether as this also

reduces fluid interference in the higher speed range

and consequently reduces the degree of slip for a

given amount of transmitted torque (Fig 4.6)

4.1.1 Hydrokinetic fluid coupling principle of

operation (Figs 4.1 and 4.3)

When the engine is started, the rotation of the

impeller (pump) causes the working fluid trapped

in its cells to rotate with it Accordingly, the fluid is

subjected to centrifugal force and is pressurized so

that it flows radially outwards

To understand the principle of the hydrokinetic

coupling it is best to consider a small particle of

fluid circulating between one set of impeller and

turbine vanes at various points A, B, Cand D as

shown in Figs 4.1 and 4.3

Initially a particle of fluid at point A, when the

engine is started and the impeller is rotated, will

experience a centrifugal force due to its mass and

radius of rotation, r It will also have acquired some

kinetic energy This particle of fluid will be forced

to move outwards to point B, and in the process

of increasing its radius of rotation from r to R, will

now be subjected to considerably more centrifugal

force and it will also possess a greater amount of

kinetic energy The magnitude of the kinetic energy

at this outermost position forces it to be ejected

from the mouth of the impeller cell, its flow path

making it enter one of the outer turbine cells at

point C In doing so it reacts against one side of the

turbine vanes and so imparts some of its kinetic

energy to the turbine wheel The repetition of fluid

particles being flung across the junction between the

impeller and turbine cells will force the first fluid

particle in the slower moving turbine member

(having reduced centrifugal force) to move inwards

to point D Hence in the process of moving inwards

from R to r, the fluid particle gives up most of its

kinetic energy to the turbine wheel and subsequently

this is converted into propelling effort and motion

The creation and conversion of the kinetic

energy of fluid into driving torque can be visualized

in the following manner: when the vehicle is at rest

the turbine is stationary and there is no centrifugal

force acting on the fluid in its cells However, when

the engine rotates the impeller, the working fluid

in its cells flows radially outwards and enters the

turbine at the outer edges of its cells It therefore

causes a displacement of fluid from the inner edges

of the turbine cells into the inner edges of the

impeller cells, thus a circulation of the fluid will

be established between the two half cell members

The fluid has two motions; firstly it is circulated

by the impeller around its axis and secondly it circulates round the cells in a vortex motion This circulation of fluid only continues as long as there is a difference in the angular speeds of the impeller and turbine, because only then is the cen-trifugal force experienced by the fluid in the faster moving impeller greater than the counter centri-fugal force acting on the fluid in the slower moving turbine member The velocity of the fluid around the couplings' axis of rotation increases while it flows radially outwards in the impeller cells due to the increased distance it has moved from the centre

of rotation Conversely, the fluid velocity decreases when it flows inwards in the turbine cells It there-fore follows that the fluid is given kinetic energy by the impeller and gives up its kinetic energy to the turbine Hence there is a transference of energy from the input impeller to the output turbine, but there is no torque multiplication in the process 4.1.2 Hydrokinetic fluid coupling velocity diagrams (Fig 4.3)

The resultant magnitude of direction of the fluid leaving the impeller vane cells, VR, is dependent upon the exit velocity, VE, this being a measure of the vortex circulation flow rate and the relative linear velocity between the impeller and turbine, VL The working principle of the fluid coupling may be explained for various operating conditions assuming a constant circulation flow rate by means

of velocity vector diagrams (Fig 4.3)

When the vehicle is about to pull away, the engine drives the impeller with the turbine held stationary Because the stalled turbine has no motion, the rela-tive forward (linear) velocity VL between the two members will be large and consequently so will the resultant entry velocity VR The direction of fluid flow from the impeller exit to turbine entrance will make a small angle 1, relative to the forward direc-tion of modirec-tion, which therefore produces consider-able drive thrust to the turbine vanes

As the turbine begins to rotate and catch up to the impeller speed the relative linear speed is reduced This changes the resultant fluid flow direction to 2 and decreases its velocity The net output thrust, and hence torque carrying capacity, will be less, but with the vehicle gaining speed there

is a rapid decline in driving torque requirements

At high turbine speeds, that is, when the output

to input speed ratio is approaching unity, there will

be only a small relative linear velocity and resultant entrance velocity, but the angle 3 will be large This implies that the magnitude of the fluid thrust will be very small and its direction ineffective in

Fig 4.3 Principle of the fluid coupling

Fig 4.4 Relationship of torque capacity efficiency and speed ratio for fluid couplings

Fig 4.5 Relationship of engine speed, torque and slip for a fluid coupling