1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.1

1BA

SIC

PH

YS

ICA

L P

RIN

CIP

LES

1.1U

nit s

yste

m

x

Prefix Abbreviation Decimal mode Power mode

Giga G 1 000 000 000 109

Mega M 1 000 000 106

Kilo k 1 000 103

Hecto h 100 102

Deca da 10 101

Deci d 0.1 10–1

Centi c 0.01 10–2

Milli M 0.001 10–3

Micro µ 0.000 001 10–6

Nano n 0.000 000 001 10–9

Decimal multiples and fractions of SI units

BASIC PRINCIPLES

SI systemIn 1960, an internationally applicable unit system for physical variables came into force: The "Système International d’Uni-tés". It is abbreviated with SI in all langua-ges in the world.

All the details of physical variables, for ex-ample weight or volume data, must be specified in SI units by law in Germany.

The main underlying principle of the SI system is to enable comparisons of diffe-rent technical data.

SI base variablesThe basis for the SI system are base vari-ables with the corresponding base units:

Length in metres [m]

Mass in kilograms [kg]

Time in seconds [s]

Electrical current in Ampère [A]

Temperature in Kelvin [K]

Molality in Mol [mol]

Light intensity in Candela [cd]

Derived SI unitsDerived SI units are formed using the base units. For example, the unit for average speed v is composed of the units of the base variables distance s and time t: If you divide the distance s by the required time t, you receive the average speed v with the unit metres per second.

The fuel consumption b of an engine re-sults as a ratio of the mass of the combus-ted fuel mk and the energy (work) W it ge-nerates in the engine. Consequently, the unit is kilograms per kilowatt hour.

In this way, all the required physical units can be formed as combinations of base variables.

Decimal multiples or fractions of SI unitsIt frequently occurs that the base units are too large or too small for practical calcula-tion. For this reason, it makes sense to ex-press decimal multiples or fractions in the form of mathematical powers.

EXAMPLE

Average speed v:

Fuel consumption b:

Megawatt MW:

1 000 000 Watts = 106 W = 1 MW

Kilometre km:

1000 m = 103 m = 1 km

millisecond ms:

0.001 s = 10–3 s = 1 ms

nanosecond ns:

0.000 000 06 s = 60 ⋅ 10–9 s = 60 ns

⎥⎦

⎤⎢⎣

⎡=

sm

ts

v

⎥⎦

⎤⎢⎣

⎡=

kWhkg

Wm

b K

1.2

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2I

mpo

rtant

phy

sica

l var

iabl

es1.

2.1F

orm

ula

sym

bols

and

uni

ts

MP

12

2col

full

BASIC PRINCIPLES

Formula symbolsSo-called formula symbols are used to simplify and abbreviate mathematical and physical formulae.

The adjacent table contains frequently re-quired variables. Alongside this, the inter-nationally applicable formula symbols and the corresponding variable unit are speci-fied.

UnitsFor comparison, the conversion factors are also specified in common units. In for-mulae and equations, the units of each va-riable are placed in square brackets (e.g. force F [N]).

EXAMPLE

Variable Formula symbols

Units

Length l 1 m = 103 mm = 102 cm

Distance s M

Height h M

Radius r M

Diameter d M

Area A 1 m2 = 10–2 a (are) = 10–4 ha (hectare)

Volume V 1 m3 = 103 dm3 = 103 l

Angle α, β 1° = 0.017 453 rad

Density ρ 1 kg/dm3 = 1 kg/l = 103 kg/m3

Time t s

Frequency f 1 Hz = 1 s–1

Rotational speed n 1 rpm = 1/60 rps

Speed V 1 m/s = 3.6 km/h

Acceleration A m/s2

Fall acceleration G m/s2

Power f 1 N (Newton) = 1 kg ⋅ m/s2

Weight G n

Mass M 1 kg = 103 g, 1 t = 103 kg

Pressure p 1 bar = 105 Pa (Pascal) = 105 N/m2 = 10 N/cm2

Energy, work E, W 1 J = 1 N ⋅ m = 1 W ⋅ s = 1/3.6 ⋅ 10–6 kWh

Torque M N ⋅ m

Power p 1 W (Watt) = 1 J/s = 10–3 kW

Electrical current I A (Ampere)

Electrical voltage U 1 V (Volt)

Electrical resistance r 1 Ω (Ohm)

Quantity of electricity Q 1 C (Coulomb) = 1 A ⋅ sCapacitance c 1 F (Farad) = 1 C/V

Temperature t, θ (T) 0 °C (t, θ) = 273.15 K (T)

Formula symbols for important physical variables

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.3

1.2.

2Def

initi

ons

1.2.

2.1M

ass,

acc

eler

atio

n an

d fo

rce

MP

21

Std

Newton's axiom

a

mF

S

F = m . a

LEGENDa AccelerationF Forcem MassS Centre of gravity

BASIC PRINCIPLES

MassEvery body has a certain mass, measured in kilograms [kg]. The greater the mass of a body, the greater the gravitational force that affects it.

As "prototype" for the mass is the so-called master kilogram, which is kept in Paris. This is a cylinder made of platinum-iridium that defines the mass of 1 kg.

A litre of water at a temperature of 4 °C also has exactly the mass of 1 kg.

AccelerationIf the speed of a moving object changes within any period, this is referred to as ac-celeration.

The unit of acceleration a is "metres per second squared" [m/s2]. It is characteri-sed by a numerical value and direction and is therefore a vectorial (directional) va-riable.

DecelerationA distinction is made between positive and negative acceleration. When a com-mercial vehicle is moving off, for example, the acceleration is positive (the speed in-creases); during braking, it is negative (the speed decreases). Negative acceleration is also referred to as deceleration.

PowerThe force is the cause of the acceleration or deceleration of a a freely moving body.

On a fixed (non-moving) body, a force can lead to a change in shape.

The effect of a force becomes clear, e.g., on striking a billiard ball (acceleration) or the expansion of a cylindrical spring (change in shape). As in the case of acce-leration, force is a vectorial variable.

Newton's lawThe fundamental law of dynamics (New-ton's law) described by Isaac Newton sta-tes:

in order to set a stationary object with a certain mass in an accelerated motion, a force must be exerted on the body. Here, the acceleration is directly proportional to the exerted force, i.e., for example, doub-ling the effective force also doubles the acceleration.

The unit of force is:

1 kg ⋅ m/s2 = 1 Newton [N]

FUNCTION

Calculating the accelerationAcceleration a is the ratio of the change in speed ∆v (in other words: "delta v") and the associated "time window" ∆t:

Calculating the forceForce F is a derived variable from mass and acceleration:

A force has the amount 1 N if it accelera-tes a body with the mass 1 kg from a sta-tionary position to a speed of 1 m/s:

⎥⎦

⎤⎢⎣

⎡=

⋅∆

∆=

2sm

ssm

tv

a

⎥⎦

⎤⎢⎣

⎡=⋅⋅= N

sm

kgamF2

2sm1kg1

N1⋅

=

1.4

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2.

2.2W

eigh

t and

fall

acce

lera

tion

x

Weight

G = m . g

m = 5 kg

LEGENDG Weightg Fall accelerationm Mass

BASIC PRINCIPLES

WeightWeight G is the force that accelerates a body of mass m vertically downwards. It is exerted everywhere on Earth and is fre-quently also referred to as gravitational force (or gravity).

It is directly proportional to the prevailing acceleration due to gravity, i.e. with gro-wing acceleration due to gravity the weight exerted on a body grows to the same degree.

Fall accelerationFall acceleration g is a "natural constant". It depends on the mass of the planet whe-re it occurs. The unit for fall acceleration is m/s2.

The fall acceleration on Earth is also refer-red as "acceleration due to gravity". In our latitudes, the acceleration due to gravity is set at 9.81 m/s2. For approximate techni-cal calculations, the value of 10 m/s2 is sufficient.

FUNCTION

Calculating the weightWeight is calculated as the product of the mass of a body and the acceleration due to gravity exerted on it:

Calculating the fall accelerationThe fall acceleration is calculated as a ra-tio of weight and mass of a body:

EXAMPLE

A body made of cast iron on our planet has the mass m = 5 kg. The weight that the body exerts on its supporting surface is:

G = m ⋅ g = 5 kg ⋅ 9.81 m/s2 = 49.05 N

On the moon, the fall acceleration is only g = 1.26 m/s2. There, a weight of

G = 5 kg ⋅ 1.26 m/s2 = 6.3 N is exerted on an object made of cast iron with the mass m = 5 kg.

A body with the same mass weighs around 8 times less on the moon than on Earth.

⎥⎦

⎤⎢⎣

⎡=

⋅⋅= N

smkg

gmG2

⎥⎦

⎤⎢⎣

⎡==

2sm

kgN

mG

g

1.5

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2.

2.3T

orqu

e

MP

11

12p

ic

Torque

F

r

M = F · r

2

1

Lever principle

=M2

M1 + M2 = 0

2. r–F

=M1 2. rF

M1 = –M2

LEGEND1 Centre of motion2 Line of influenceF ForceM Torquer Lever

BASIC PRINCIPLES

TorqueIf a force is exerted on a body outside its centre of motion, it creates a rotational ef-fect on this body. This depends on what distance or "lever" r the line of influence of the force has from the centre of motion of the body. The rotational effect is referred to as torque M ( Fig.). The unit of torque is "Newtonmetres" [Nm].

Torque can be felt, for example, by pre-venting a small electric motor (e.g. of a fan) from starting up. A similar principle is used to measure the torque of a combus-tion engine ( page 5.71).

Lever principleA rigid body that can be rotated around a fixed point is in equilibrium when the total of all momentum rotating to the left equals the total of all momentum rotating to the right.

For practical application, the lever princip-le means: with low force and a long lever arm, it is possible to apply the same torque as with a high force and a short le-ver arm.

FUNCTION

Calculating the torqueTorque can be calculated as the product of the amount of force F and the distance r from the centre of motion.

M = F ⋅ r [N ⋅ m = Nm]

The lever principle states:

Momentum rotating to the right is recon-ciled as positive; momentum rotating to the left is reconciled as negative.

EXAMPLE

A mechanic uses different extensions for a wrench to generate various torques on a wheel nut.

Case 1:

F1 = 60 N

r1 = 0.4 m

M1 = 60 N ⋅ 0.4 m = 24 Nm

Case 2:

F2 = 30 N

r2 = 0.8 m

M2 = 30 N ⋅ 0.8 m = 24 Nm

Case 3:

F3 = 30 N

r3 = 1.2 m

M3 = 30 N ⋅ 1.2 m = 36 Nm

The mechanic thus requires much greater force to loosen the wheel nut with a short wrench than with a lever extension for the wrench.

∑ ==++ 0MMMM 321

1.6

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2.

2.4W

ork

and

pow

er

MP

11

12p

ic

Work

s

Fm

Lifting work

h

G

m

FLift

LEGENDFLift Lifting powerG Weighth Heightm Masss Distance

BASIC PRINCIPLES

Work (energy)When mechanical work is performed, a force F is used to cover a distance s. As a general principle, the force F is directed towards the distance s. The work W is de-fined as the product of force and distance. Work is also referred to as energy. The unit of work is the Joule.

Work is defined as the released, i.e. avai-lable, work or energy. In physics, distinc-tions are made between a large number of types of energy:

Electrical energy

Thermal energy

Kinetic energy, etc.

PowerThe term "power" s used to refer to perfor-ming work within a certain time.

The power P is the work W gained or con-sumed during a certain time t. The more time required for a certain work, the less power "generated". The unit of power P is the "Watt" (named after James Watt).

FUNCTION

Calculating the workThe following formula is used to calculate work W:

W = F ⋅ s [N ⋅ m = Nm]

The work 1 Joule or 1 Nm is performed when a distance of 1 m is covered with the application of a force of 1 N:

Calculating the powerPower is calculated from the work perfor-med (or energy consumed) divided by the required time:

The unit of power used to be horse power [hp]. To convert hp into the SI unit Kilo-watts [kW]:

1 hp = 736 W = 0.736 kW

1 kW = 1.36 hp

EXAMPLE

A fork-lift truck lifts two wire boxes with different masses m1 and m2 with uniform speed (not accelerates) to the same height h.

g = 9.81 m/s2 ≈ 10 m/s2

m1 = 200 kg

m2 = 500 kg

h = 2 m

The work to be performed or energy re-quired in this case is referred to as lifting work or lifting energy.

W1 = G ⋅ h = m1 ⋅ g ⋅ hW1 = 200 kg ⋅ 10 m/s2 ⋅ 2 m

W1 = 4 000 Nm = 4 000 J

W2 = G2 ⋅ h = m2 ⋅ g ⋅ hW2 = 500 kg ⋅ 10 m/s2 ⋅ 2 m

W2 = 10 000 Nm = 10 000 J

If the lifting operation takes e.g. t = 10 s, the fork-lift truck requires the following po-wers:

P1 = W1/t = 4 000 Nm/10 s = 400 W

P2 = W2/t = 10 000 Nm/10 s = 1 000 W

The power is generated and provided electrically or by a combustion engine.

m1s

mkg1Nm1J1

2⋅== Ws1=

⎥⎦

⎤⎢⎣

⎡=== W

sJ

sNm

tW

P

1.7

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2.

2.5E

ffici

ency

x

Efficiency of energy conversion

Wout

Win

η =

WoutWinη

η2 η3 η4

1 2 3 4

LEGEND1 Diesel fuel2 Engine3 Gearbox4 Axleη EfficiencyWout Eliminated energy (mechanical) Win Applied energy (chemical)

BASIC PRINCIPLES

Energy conversionIn the case of physical operations, diffe-rent types of energy are converted into one another. In the case of the fork-lift truck ( page 1.6), electrical energy (to drive the truck) is converted into lifting work (mechanical lifting of the wire box).

Every energy conversion is subject to loss (dissipation). The physical variable "effi-ciency" is used to quantify the lost energy and characterise the conversion process.

EfficiencyThe efficiency η for energy conversion in machines is defined as a ratio of the elimi-nated and applied energy. It has no unit and is specified either in % or as a deci-mal number.

Energy lossesThe greater the efficiency, the lower the lost energy. The energy losses for com-bustion engines arise above all through friction, heat dissipation and thermal radi-ation ( page 5.72).

FUNCTION

Calculating the efficiencyThe efficiency is the ratio of the eliminated and applied energy:

By multiplying with time t, the efficiency can also be expressed as a ratio of power released and applied:

EXAMPLE

On the MAN D 0836 LF04 engine, the mi-nimum fuel consumption

be = 200 g/kWh.

This means that to generate the energy of 1 kWh the diesel engine requires a fuel vo-lume of m = 0.2 kg.

The specific calorific value of diesel fuel is H = 42 500 kJ/kg. The amount of energy applied to the engine is thus:

The eliminated energy is:

Wout = 1 kWh = 3 600 J

This results in an efficiency for the engine:

Win

Wout=η

tWtW

PP

in

out

in

out

⋅

⋅==η

kJ8 500kgkJ

42 500kg2.0Win == ⋅

%4242.0kJ8 500kJ3 600

WW

in

out ====η

1.8

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2.

2.6R

otat

iona

l spe

ed a

nd a

ngul

ar v

eloc

ity

MP

11

1S

td

Rotational movement

n

n

n

ω

∆ϕ

∆ϕω = ∆t

LEGENDn Rotational speedω Angular velocity

(angular frequency)∆ϕ Rotation angle∆t Time interval

BASIC PRINCIPLES

Rotational speedAn important variable for mass points of bodies that move around a fixed axis of rotation is the rotational speed n. It is measured in "revolutions per second" [s–1] and is thus a direct measure of the speed of the rotational movement. The unit rpm is normally used to specify engine speeds.

Angular velocityIf the rotational movement of a rotating body is examined more closely, it can be seen that a mass point rotating around a fixed axis of rotation runs through a cer-tain angle in a certain time. The ratio of the angle section ∆ϕ and the corresponding time interval ∆t is called the angular velo-city ω (frequently also angular frequency). The angular velocity has the same unit as rotational speed: s–1.

Rotation angle (radian measure)The rotation angle ∆ϕ is usually specified as a multiple of π in the unit radian measu-re (π/2 = 90°). One complete revolution, i.e. 360°, corresponds to the radian measure 2π. The time required for this is the time of a revolution T.

FUNCTION

Calculating the rotational speedDepending on the time unit, the following simple formula is used to calculate rotati-onal speed:

Calculating the angular velocityThe angular velocity ω (in other words: "omega") is the ratio of the rotation angle covered ∆ϕ (in other words: "delta phi") and the associated "time window" ∆t.

If the rotational speed n is specified in rpm, this results in the angular velocity:

EXAMPLE

A combustion engine runs at a rotational speed of n = 3 000 min–1 (rpm). How high is the angular velocity of the rotational mo-vement?

n = 3 000 min–1 = 50 s–1

⎥⎦

⎤⎢⎣

⎡== −1s

s1

sRevolutions

n

⎥⎦

⎤⎢⎣

⎡== −1

Engine minmin1

minRevolutions

n

t∆ϕ∆

=ω 1s−⎦⎤⎣⎡

1sT

n2 −=⋅π=ωπ2

⎦⎤⎣⎡

60n2 Engine⋅π

=ω30

n⋅π= 1s−

⎦⎤⎣⎡

n2t

⋅π=∆

ϕ∆=ω

11 s16.314s50

60n2 Engine⋅π

2 ⋅π=ω

11

s16.31460

min30002 −

−

−

−

=

=

⋅π=ω

=ω

1.9

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.2.

2.7P

ath

spee

d an

d po

wer

in r

otat

iona

l mov

emen

t

MP

11

1S

td

Power on rotational movement

F

r

v

ω

P = M·ω

W = F·2·π · r

LEGENDF Force (towards circuit)P Power on rotational movementr Radius of circuitv Path speedW Energy of rotational movementω Angular velocity

BASIC PRINCIPLES

Path speedIf a mass point rotates at a certain angular velocity ω and fixed distance r (= radius r) around a fixed centre of motion, it has path speed v. This is also referred to as circumferential or peripheral speed. It is calculated as the ratio distance s covered on the on the circuit and the required time t. The unit of path speed is "metres per se-cond" [m/s].

Power on rotational movementThe energy released (or to be applied) on rotational movement is calculated in the same way as the linear movement ( page 1.6) as the product of force F (to-wards the circuit) and the path s covered on the circuit. With the circular movement, the path covered on the circular move-ment is a fraction or multiple of the cir-cumference.

If the time T required for a revolution is di-vided or multiplied by the rotational speed n, the energy indictes the power of the ro-tational movement.

It can be seen that the power on rotational movement rises with increasing torque and increasing rotational speed ( page 5.69).

FUNCTION

Calculating the path speedIn general, the following applies to the path speed v of a ground point on a cir-cuit:

Calculating the rotational powerThe energy ( page 1.6) of the rotational movement is calculated as the product of the force F and the distance s along the circuit (circumference U):

The rotational speed is usually specified in rpm. For the path speed v and power P, this results in kW:

EXAMPLE

The MAN D 2876 LF12 engine generates a torque of M1 = 2 300 Nm at a rotational speed of n = 1 200 min–1 (rpm). How high is the power output P1 at this rotational speed?

A comparison engine generates a torque of

M2 = 2100 Nm at the same rotational speed and thus the power P2:

n1

T;sm

TTU

ts

v =⎥⎦

⎤⎢⎣

⎡===

⋅π r2

nv = ⋅ ⋅π r2

⎥⎦

⎤⎢⎣

⎡=ω⋅= W

sNm

MP

π⋅=⋅π⋅=⋅= 2Mr2FUFW

n2Mnr2FP ⋅π⋅=⋅⋅π⋅=

⎥⎦

⎤⎢⎣

⎡⋅⋅π=

sm

30nr

v9 550

nMP

⋅= kW⎦⎤⎣⎡

min1200n

Nm2300M1

1

=

=−

9550min1200Nm2300

P

9550nM

P

1

1

11

=⋅

=

⋅=

−

kW01.289

kW87.2639550

min1200Nm2100P

9550nM

P

1

2

22

=⋅

=

⋅=

−

1.10

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.3T

orqu

e in

crea

se a

nd p

erfo

rman

ce e

valu

atio

n

x

Torque increase

1000 1200 1400 1600 1800 2000

[Nm]

[kW]

360

330

300

270

240

210 2400

2300

2200

2100

2000

1900

1800

DAFXE 355 C

MAND 2876 LF12

n [min -1]

P

M

∆P

MdA

17 %

MdA

29 %

LEGENDM Torque MdA Torque increasen Rotational speedP Power∆P Power gain

BASIC PRINCIPLES

Torque increaseIn order to be able to achieve high level of average speeds, commercial vehicle engi-nes must have adequate reserves of per-formance, above all when driving uphill and with full loads.

On driving uphill, the rotational speed drops sharply. During the drop in rotatio-nal speed, the torque increases to its ma-ximum value ( page 5.69). The relation-ship between maximum torque and the torque at nominal rotational speed is refer-red to as the torque increase MdA ( Fig.).

Travelling resistanceOn commercial vehicles, the drive charac-teristics with rising travelling resistance are of great significance. An engine with a high level of torque increase provides ade-quate reserves of power, ensuring high average speeds, even uphill.

FUNCTION

Performance evaluationA high value for the torque increase leads to an upward bulge in the performance curve and leads to additional usable po-wer compared to an engine with the same power and lower torque increase.

The greater the torque increase the stee-per the rise in tractive force if the rotational speed drops on reaching an upward incli-ne: The speed decreases more slowly; a constant uphill speed sets in earlier than in the case of engines with low torque incre-ase.

The consequence of this is that higher average speeds are reached: despite the same power. an engine with greater torque increase is more effective.

The torque increase can be calculated using the following simple formula:

EXAMPLE

The course of torque and power of two engines of the 480-hp class from MAN and DAF are compared. The MAN engine (marked in yellow) has a torque increase of 29 %. On the DAF engine, it is only 17 %. The torque increase is calculated by dividing the maximum torque by the torque at nominal rotational speed, multi-plying by 100 and subtracting 100.

Torque increase of the MAN engine:

Torque increase of the DAF engine:

%100100MM

Mnom

ma xdA −⋅= ⎦⎤⎣⎡

%291001001 7502 300

MdA =−⋅=

%171001001 8002 100

MdA =−⋅=

1.11

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.4B

ases

of c

alcu

latio

n fo

r gea

rbox

es1.

4.1G

ear

ratio

and

red

uctio

n ra

tio

x

Gear ratio / reduction ratio

d

d

F

M2 = F · d

n1

z1

n2

n1=2

2

z2

1

2

M1 = F . d2

LEGEND1 Pinion2 Geard Diameter (pinion)F ForceM Torquen1 Rotational speed drive shaftn2 Rotational speed output shaftz1 Number of teeth, pinionz2 Number of teeth, gear

BASIC PRINCIPLES

GearboxAn assembly consisting of one or more pairs of gearwheels as well as a corres-ponding housing with fixed-location mountings is referred to as a gearbox ( page 6.14).

A gearbox with one or more gear stages has the following tasks:

Changing rotational speed (torque)

Changing direction of the rotational movement

Hooking up power take-offs (drive system for other machines)

Pinion and gearThe smaller gear of a gearwheel pair is re-ferred to as a "pinion'; the larger is referred to as a "gear".

The gear ratio or reduction ratio in a gear-box is the relationship between the input rotational speed and the output rotational speed.

Here, the gear ratio means a change in ro-tational speed from "fast" to "slow". A change in rotational speed from "slow" to "fast" is referred to as the reduction ratio.

FUNCTION

Torque in the gearboxA gearbox can convert torque. Here, the propelling force F is transmitted between gear wheels with different diameters. A number of successive gear stages mean that the torque can be increased or redu-ced in multiples.

Gear ratioA gear ratio of i = 2 : 1 means, for examp-le, that the drive shaft has to turn twice for the output shaft to turn once. A gear ratio reduces the rotational speed and increa-ses the torque accordingly. When the ro-tational speed is increased and the torque is reduced, this is referred to as a reduc-tion ratio. A reduction ratio of i = 1 : 5, for example, means that one revolution on the drive shaft leads to five revolutions on the output shaft.

Overall ratioIn the case of multi-stage gearboxes, the overall ratio itotal results from the product of the individual gear ratios.

The gear ratio i of a gearwheel pair with ro-tational speed at the drive and output shaft n1 and n2, the reference diameters d1 and d2 and gear tooth numbers z1 and z2 is calculated as follows:

itotal = i1 ⋅ i2 … ⋅ in (multi-stage gearbox)

The following relationships apply:

n1 ⋅ z1 = n2 ⋅ z2

n1 ⋅ d1 = n2 ⋅ d2

In a gearbox, the driving pinion with refe-rence diameter d with the tangential or pe-ripheral force F produces the torque M1:

The following applies to the driven gear wheel with diameter 2 ⋅ d:

M2 = F ⋅ d [Nm]

The torque has doubled. The gear ratio is i = 2 :1. The rotational speed of the driven gear wheel has halved ( Fig.).

1

2

1

2

2

1

zz

dd

nn

i ===

2d

FM1 ⋅=

1.12

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.4.

2Exa

mpl

e of

a tw

o-st

age

gear

box

x

Two-stage gear ratio in a gearbox

2

1

3

4

F

M A

LEGEND1 Pinion (1st stage)2 Gear wheel (1st stage)3 Pinion (2nd stage)4 Gear wheel (2nd stage)A Axle drive (output end)F power flow in the gearboxM Engine drive (drive end)

BASIC PRINCIPLES

Gearbox efficiencyPower losses in gearboxes occur above all due to friction between the gearbox components. In multi-stage gearboxes, this effect can be seen intensively. This must be borne in mind in the development of a gearbox and the propulsion power set higher accordingly. The power dissipation of gearbox is specified by the gearbox ef-ficiency.

The calculation for a two-stage gearbox is explained in the adjacent example.

EXAMPLE

The driving pinion 1 of the first gear stage of a two-stage gearbox has a reference diameter d1 and 18 teeth. The associated gear 2 has reference diameter d2 = 2 · d1 and 36 teeth. Pinion 1 generates tangen-tial force F1 and generates torque M1:

M1 = F1 ⋅ 0.5 ⋅ d1

The force F1 is transferred by the interme-shed teeth onto gear 2 and creates torque M2 there:

M2 = F1 ⋅ 0.5 ⋅ d2 = F1 ⋅ d1

The ratio of the torques is 2. The torque has doubled:

A the gear has double the number of teeth z2, the rotational speed n2 is reduced to half. The gear ratio i1 is:

In the second gear stage, pinion 3 has re-ference diameter d3 = d1 and 18 teeth. The corresponding gear 4 has reference diameter d4 = 3 ⋅ d1 and 54 teeth.

Torque M3 is generated at pinion 3:

M3 = F3 ⋅ 0.5 ⋅ d3 = M2 = F1 ⋅ d1; d1 = d3

F3 ⋅ 0.5 ⋅ d1 = F1 ⋅ d1

F3 = 2 ⋅ F1

The gear ratio to gear 4 results in the torque M4:

M4 = F3 ⋅ 0.5 ⋅ d4 = F3 ⋅ 0.5 ⋅ (3 ⋅ d1)

M4 = 2 ⋅ F1 ⋅ 1.5 ⋅ d1 = 3 ⋅ F1 ⋅ d1

The gear ratio and ratio of the torques are calculated as follows:

The overall ratio of the gearbox itotal can be calculated as follows:

itotal = i1 ⋅ i2 = 2 ⋅ 3 = 6

2d5.0F

dFMM

11

11

1

2 =⋅⋅

⋅=

2n

n21836

nn

zz

i 12

2

1

1

21 =⇒====

3dF

dF3MM

11

11

3

4 =⋅

⋅⋅=

31854

nn

zz

i4

3

3

42 ====

1.13

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.5P

ress

ure,

vol

ume

and

tem

pera

ture

x

General gas equation

T1 = T2 p1 = p2 V1 = V2

p1

p2

V2

V1

V1, p1 V2, p2 V1, T1 V2, T2 T1, p1 T2, p2

=V1

V2

T1

T2

=p1

p2

T1

T2

=

LEGENDp1 Pressure source state p2 Pressure finish state T1 Temperature source stateT2 Temperature finish stateV1 Volume source stateV2 Volume finish state

BASIC PRINCIPLES

Gas stateThe state of a gas is characterised and described by three physical variables:

Pressure p [bar]

Volume V [l]

Temperature T [K].

They are mutually dependent and change if the state changes (source state 1 and fi-nish state 2).

State change of a gasThe three most important cases of state change of a gas are:

Change in pressure and volume with temperature remaining constant (iso-therm).

Change in temperature and volume with pressure remaining constant (iso-bar).

Change in temperature and pressure with volume remaining constant (iso-chor).

In combustion engines, the changes in pressure, volume and temperature of ga-ses are used to generate energy ( page 5.2).

FUNCTION

General gas equationThe physical relationship between pressu-re p, volume V and temperature T of a gas in two different states (source state 1 and finish state 2) is characterised by the ge-neral gas equation:

If one of the variables is constant, the ge-neral gas equation is reduced.

Gas temperature constant

Gas pressure constant

Gas volume constant

EXAMPLE

The general gas equation can be modified in such a way that the target variable of the corresponding state can be calcula-ted:

A diesel engine has a displacement of Vh = 2.14 l and a compression volume of Vc = 0.133 l. The intake temperature T1 = 60 °C; the final compression tem-perature T2 = 700 °C. The absolute pressure at the start of com-pression is approximately p1 = 1 bar. How high is the final compression pressure p2?

V1 = Vh + Vc = 2.14 l + 0.133 l = 2.273 l

V2 = Vc = 0.133 l

2

22

1

11

TVp

TVp ⋅

=⋅

1

2

2

1

VV

pp

=

2

1

2

1

TT

VV

=

2

1

2

1

TT

pp

=

12

2112 TV

TVpp

⋅

⋅⋅=

bar9.49K15.333l133.0

K15.973l273.2bar1p2 =

⋅

⋅⋅=

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.14

1.6T

herm

odyn

amic

s

MP

21

Std

Energy conversion

Energy in fuel

Energy

EngineMechanically usuable Energy

Heat

Environment

BASIC PRINCIPLES

Thermodynamics (main laws)Thermodynamics is a subarea of physics. It covers all types of energy conversion and attempts to set up laws. The founda-tion is formed by the main laws of thermo-dynamics.

Conservation of energyThe first law of thermodynamics is also re-ferred to as the "law of conservation of en-ergy". Alongside the second law, it descri-bes a natural law:

energy cannot arise or be generated from nothing. Every energy generation is in fact energy conversion.

In the case of the combustion engine, e.g., the energy that is chemically bound in the diesel fuel is converted into mecha-nical energy.

EntropyThe second law of thermodynamics is also referred to as the "law of entropy". The entropy S is a state variable introdu-ced in thermodynamics. Its change is a measure of the "reversibility" of a process. The second law of thermodynamics says that entropy in all energy conversions is naturally greater than zero (S > 0).

Reversible energy conversionThe complete conversion of mechanical energy into thermal or inner energy is pos-sible without restriction. The other way around, however, inner energy, in fuel or

bound as heat, cannot be fully converted into mechanical or electrical energy or re-covered as such. The proportion of the in-ner energy that can be converted is de-scribed by the variable entropy S.

An energy conversion that takes place wi-thout losses is referred to in thermodyna-mics as reversible. In this case, there would be no entropy change (S = 0). Re-versible events are boundary processes. These are idealised processes that serve as the basis of calculation.

In practice, this means that the converted energy cannot be completely recovered.

FUNCTION

Perpetuum mobileThe first law of thermodynamics states that it is impossible to construct a machi-ne that generates work from nothing. This statement is referred to as the prohibition of the "perpetuum mobile type 1" (perpe-tuum mobile is Latin for "something that moves continuously").

A machine that could convert energy in a completely reversible manner would be a perpetuum mobile type 2. The second law of thermodynamics, which describes a natural law, states that a machine of this natures is physically not possible.

Combustion engineIn the case of energy conversion in a com-bustion engine, a part of the energy stored in the fuel is output in the form of heat to the coolant and the environment.

This quantity of energy can no longer be used mechanically. This is referred to as lost energy. An entropy change takes place. Its variable (S > 0) describes the amount of lost energy.

The energy stored in the fuel is converted (1st first), but can only be partially used within of process (2nd law).

1.15

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.7B

asic

prin

cipl

es o

f driv

ing

dyna

mic

s1.

7.1R

ollin

g re

sist

ance

x

Contact material Rolling-resistance coefficient µR

Roller bearing (e.g. ball bearing) 0.001

Wheel on rail (e.g. railway) 0.001 … 0.002

Tram on grooved rails 0.006

Motor vehicle on concrete, asphalt 0.013

Motor vehicle on block pavement 0.015

Commercial vehicle on good asphalt road 0.007 … 0.02

Commercial vehicle on wet asphalt road 0.015 … 0.03

Commercial vehicle on good concrete road 0.008 … 0.02

Commercial vehicle on rough concrete road 0.011 … 0.03

Commercial vehicle on block pavement 0.017 … 0.03

Commercial vehicle on poor road 0.032 … 0.05

Commercial vehicle on compacted mud path 0.150 … 0.94

Commercial vehicle on loose sand 0.150 … 0.30

Rolling-resistance coefficients of various surfaces

BASIC PRINCIPLES

Rolling resistanceRolling resistance arises due to the rolling motion of a vehicle. It counteracts the pro-pelling force of the vehicle and consists of:

Rolling friction between tyres and road surface

Deformation of the tyre (flexing work)

Air friction at the wheel

Friction in the wheel bearing

Rolling-resistance forcePhysically, rolling resistance can be ex-pressed by the rolling-resistance force FR. In practice, the amount of rolling resis-tance on a motor vehicle depends on nu-merous factors:

Driven speed

Vehicle weight

Type and tread of tyres ( page 9.5)

Air pressure in the tyres

Steering geometry and wheel align-ment

Road condition (road surface)

Condition of the brake system

FUNCTION

Calculating the rolling resistanceThe rolling-resistance force FR is the pro-duct of the normal force FN (force the ve-hicle presses onto the road) and the rol-ling-resistance coefficient (or figure) µR ( table):

FR = FN ⋅ µR

On a level stretch of road, the normal force is the equivalent of the weight of the vehicle, which means that the following applies:

FR = µr ⋅ G = µR ⋅ m ⋅ g

Rolling-resistance powerMultiplying with the driven speed means the rolling-resistance force FR can be used to calculate the rolling-resistance power PR. The following applies:

PR = FR ⋅ v = µR ⋅ m ⋅ g ⋅ v

The rolling-resistance power is a form of power dissipation. It is converted into heat and applied, for example, to the wheel be-arings and tyres.

EXAMPLE

The rolling resistance of a commercial ve-hicle truck-trailer unit and of a locomotor are compared.

Both move at a speed of v = 80 km/h and have a mass of 40 t. The weight for the lo-comotor and truck-trailer unit is:

G = m ⋅ g = 40 000 kg ⋅ 10 m/s2 = 400 kN

The rolling-resistance force and power of the locomotor are:

FR1 = µRail ⋅ G = 0.002 ⋅ 400 kN

FR1 = 0.8 kN

PR1 = FR1 ⋅ v = 0.8 kN ⋅ 22.2 m/s

PR1 = 17.8 kW

The rolling-resistance force and power of the truck-trailer unit are:

FR2 = µRoad ⋅ G = 0.011 ⋅ 400 kN

FR2 = 4.4 kN

PR2 = FR2 ⋅ v = 4.4 kN ⋅ 22.2 m/s

PR2 = 97.68 kW

Compared to the locomotor, 5.5 times the rolling-resistance force works against the truck-trailer unit. The rolling-resistance power on the truck-trailer unit is also 5.5 times greater than on the locomotor.

1.16

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.7.

2Aer

odyn

amic

dra

g

x

Object Drag coefficient cw

Thin, level plate(vertical to direction of flow)

1.1

Sphere (depending on size) 0.1 … 0.4

Streamlined body 0.05

Modern passenger car 0.3

Bus 0.6 … 0.7

Truck 0.7 … 1.5

Drag coefficient

BASIC PRINCIPLES

Aerodynamic dragAerodynamic drag arises due to the mo-vement of a vehicle in the atmosphere. It refers to the resistance that the air exerts on the moving object as it moves forward ( page 18.4).

Air resistance forceThe flow resistance force of a vehicle in air is referred to as aerodynamic drag force FL. The aerodynamic drag force depends on the following factors:

Size and shape of the vehicle (frontal area of vehicle, vehicle body sections, geometry of transported loads)

Driven speed (if the speed is doubled, the aerodynamic drag force quadrup-les!)

Air density (with rising temperature, the air density falls; with rising air pres-sure, it increases.)

Wind direction and wind strength

FUNCTION

Calculating the aerodynamic dragThe aerodynamic drag force FL is calcula-ted from the drag coefficient cw, the frontal area A and the speed v of the vehicle as well as the density ρ ("ro") of the air. The calculation formula is:

FL = 0.5 ⋅ cw ⋅ A ⋅ ρ ⋅ v2

The drag coefficient or cw value is determi-ned during test in flow or wind tunnels ( table). It indicates the extent of the aero-dynamic drag of a vehicle due to its sha-pe.

Aerodynamic drag powerIn order to surmount the aerodynamic drag, a vehicle must generate a certain power: the aerodynamic drag power PL results by multiplying the aerodynamic drag force by the driven speed v:

PL = FL ⋅ v = 0.5 ⋅ cw ⋅ A ⋅ ρ ⋅ v3

As the driving speed increases, the aero-dynamic drag force increases to the po-wer of two, the aerodynamic drag power rises to the power of three.

EXAMPLE

Two commercial vehicles with drag coeffi-cients cw1 = 0.8 and cw2 = 1.14 have the same frontal areas A = 8.4 m2 and drive at a speed v of 80 km/h on a level surface. The air density is

ρ = 1.212 kg/m3. How high are each of the aerodynamic drag forces?

The aerodynamic drag force of vehicle 2 is almost 1.5 times the aerodynamic drag force of vehicle 1.

An open platform truck in comparison with a fully panelled semitrailer truck has a poorer cw value. As a result, the fuel con-sumption can be up to 30 % higher, be-cause the greater aerodynamic drag po-wer has to be compensated for by higher engine output.

FL1 = 0.5 · cW1 · A · ρ · v2

FL1 = 0.5 · 0,8 · 8.4 m2 · 1.212 kg/m3 · (22.22)2 m2/s2

FL1 = 2 010.62 N

FL2 = 0.5 · cW2 · A · ρ · v2

FL2 = 0.5 · 1.14 · 8.4 m2 · 1.212 kg/m3 · (22.22)2 m2/s2

FL2 = 2 865.13 N

1.17

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.7.

3Clim

bing

resi

stan

ce

x

Climbing resistance

FST = G . sin αST

αST

G

S

FST

h

l

LEGENDαST Angle of inclination FS Climbing resistance forceG Weighth Altitude of upward inclinel Length of upward incline (horizontal)S Centre of gravity

BASIC PRINCIPLES

Climbing resistanceThe climbing resistance is cause by topo-graphy: upward road inclines exert a counteracting force on the rolling motion of the vehicle that depends on the number of degrees of the upward incline and the total weight of the vehicle. In order to overcome the climbing resistance, the ve-hicle requires greater propelling force. To prevent a drop in speed and to balance out the climbing resistance power PST, higher engine output is necessary.

Climbing resistance forceThe climbing resistance force FST is the product of the sine proportion of the angle of inclination α and the weight being exer-ted on the vehicle G ( Fig.).

The following measures help to overcome the climbing resistance force:

Engaging an appropriate gear in good time on the upward incline

During the upward incline, changing gear as little as possible

Driving with momentum, but keeping the speed within the framework of sta-tutory traffic regulations

FUNCTION

Calculating the climbing resistanceThe climbing resistance force FST is calcu-lated as follows:

FST = G ⋅ sin αST = m ⋅ g ⋅ sin αST

The climbing resistance force can also be expressed by means of the degrees p of the upward incline:

Here, p is expressed in " % of the upward incline':

As a reminder: An upward incline of 5 % means that on a stretch of road of l = 100 metres the increase in altitude h = 5 metre ( Fig.).

Climbing resistance powerThe climbing resistance force provides the climbing resistance power required on the hill:

EXAMPLE

A medium-sized building site truck with load has a total weight of 22 tonnes. It is being driven on a country road with 4 % upward incline. The climbing resistance force FS is then:

The truck drives at a constant speed of v = 70 km/h. The climbing resistance po-wer is calculated as follows:

100p

gm100

pGFST ⋅⋅=⋅=

100⋅ 100⋅lh

tanp =α= %⎦⎤⎣⎡ST

v100

pgmvFP STST ⋅⋅⋅=⋅=

1004

sm

10kg22000F

100p

gm100

pGF

2ST

ST

⋅⋅=

⋅⋅=⋅=

N8800FST =

sm

44.196.3

h/km70v ==

kW171sm

44.19N8800PST =⋅=

1.18

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

1.8C

olle

ctio

n of

form

ulae

xx

FUNCTION

Variable Formula Page

Speed 1.1

Fuel consumption 1.1

Acceleration 1.3

Power 1.3

Weight 1.4

Fall acceleration 1.4

Torque 1.5

Energy / work 1.6

Power 1.6

Efficiency 1.7

Rotational speed 1.8

Angular velocity 1.8

Path speed on rotational movement 1.9

Time of a revolution 1.9

Formulae

⎥⎦

⎤⎢⎣

⎡=

sm

ts

v

⎥⎦

⎤⎢⎣

⎡=

kWhkg

Wm

b K

⎥⎦

⎤⎢⎣

⎡=

⋅∆

∆=

2sm

ssm

tv

a

⎥⎦

⎤⎢⎣

⎡=⋅⋅= N

sm

kgamF2

⎥⎦

⎤⎢⎣

⎡=

⋅⋅= N

smkg

gmG2

⎥⎦

⎤⎢⎣

⎡==

2sm

kgN

mG

g

rFM ⋅= ⎦⎤⎣⎡ NmmN =⋅

⎦⎤⎣⎡ NmmNsFW =⋅⋅=

⎥⎦

⎤⎢⎣

⎡=== W

sJ

sNm

tW

P

Win

Wout=η

⎥⎦

⎤⎢⎣

⎡== −1

Engine minmin1

minRevolutions

n

1sT

n2 −=⋅π=ωπ2

⎦⎤⎣⎡

sm

TTU

ts

v ⎥⎦

⎤⎢⎣

⎡===

⋅π r2

n1

T =

1.19

1

BA

SIC

PH

YS

ICA

L P

RIN

CIP

LE

S

xx

FUNCTION

Variable Formula Page

Work on rotational movement 1.9

Power on rotational movement 1.3

Torque increase of an engine 1.10

Gear ratio (gearbox) 1.11

General gas equation 1.13

Gas pressure and volume at constant gas temperature 1.13

Gas volume and temperature at constant gas pressure 1.13

Gas pressure and temperature at constant gas volume 1.13

Rolling-resistance force 1.15

Rolling-resistance power 1.15

Air resistance force 1.16

Aerodynamic drag power 1.16

Climbing resistance force 1.17

Climbing resistance power 1.17

Formulae

π⋅=⋅π⋅=⋅= 2Mr2FUFW

⎥⎦

⎤⎢⎣

⎡=π ⋅⋅= W

sNm

2 nMP kW9550

nMP

⋅= ⎦⎤⎣⎡

%100100MM

Mnom

ma xdA −⋅= ⎦⎤⎣⎡

1

2

1

2

2

1

zz

dd

nn

i ===

2

22

1

11

TVp

TVp ⋅

=⋅

1

2

2

1

VV

pp

=

2

1

2

1

TT

VV

=

2

1

2

1

TT

pp

=

FR = µR · G = µR · m · g

PR = FR · v = µR · g · v

2WL vAc5.0F ⋅ρ⋅⋅⋅=

3WLL vAc5.0vFP ⋅ρ⋅⋅⋅=⋅=

G sinFST ST⋅= α

v100

pgmvFP STST ⋅⋅⋅=⋅=