Gearboxes

Gear Selection OptionsGear Selection OptionsGear Selection OptionsGear Selection Options

When a drive application requires rotary motion to be transferred with speed change and ortorque change a number of options are available controlled by system layout, economics and power capacity. The drive transmission selection is selected from the options below in order of convenience.

1. Belt Drive (flat, vee , synchronous ) 2. Chain drive 3. Gearbox - geared motor 4. Purchased Stock gears- unlubricated(Plastic/metal- plastic/plastic ) 5. Purchased Stock gears- metal lubricated 6. Engineered gears

It must be emphasised that designing and manufacturing a gear system is relatively, expensive and inconvenient if a proprietory system is available which will do the job..

Types Of Gears Types Of Gears Types Of Gears Types Of Gears

Note: More details and pictures of all gear types below are to be found at the links below table

Type Notes Further Notes

Spur

Majority of gears are spur. Relatively easy to design and make. Parallel shafts. High efficiency (99% per train). No side thrust. Can back drive. Single Ratio up to 1:10. Can be made very accurate with low vibration /noise. Normally steel pinions require lubrication. Plastic gears can be used requiring no lubrication

Internal Spur

Similar performance to normal spur. Results in compact drive geometry. Used in manufacture of epicyclic / planetary gears.

Spur Gears

Helical

Single Helical have similar properties to spur. However drive results in axial thrust. Gears are smoother/quieter for the same size/spec. The gears can run at high speeds up to large diameters. Higher torque/life capabilities for same size as spur.

Double-Helical

Similar benefits to single helical but with no generated side thrust. Higher performance compared to single helical

Crossed-Helical

Shaft at 90o. Difficult to make accurately. Smooth drive.

Helical Gears

Worm

Offset shafts at 90o. Very high ratios possible in single stage. Sliding action.

One gear is normally copper allow (bronze). Low efficiency at higher ratios and low speeds. Lubrication essential for mechanical and thermal reasons. Cannot backdrive at high ratios.

Worm Gears

Bevel Gear

Mainly used for drive transmission through 90o. Only low ratios used (4:1 and

less). Lubrication required. Some vibration on spur type: Helical type smoother. Bevel Gears

Spiroid

Perform a similar function to worm boxes but the gears have characteristics which combine those of the bevel and worm gears. High powers and speed ratios are possible and mechanical efficiencies higher than worm boxes for equivalent ratios.

Harmonic Drive

Performance advantages include high-torque capacity, concentric geometry, lightweight and compact design, zero backlash, high efficiency, high ratios (up to 320:1), and back drivability. Harmonic drive systems suffer however, from high flexibility, resonance vibration. Used in robotics

Harmonic Drives

Brief Comparison Brief Comparison Brief Comparison Brief Comparison

Type Normal

Ratio Range Pitch Line

Velocity (m/s) Efficiency Range

Spur 1:1 to 6:1 25 98-99%

Helical 1:1 to 10:1 50 98-99%

Double Helical 1:1 to 15:1 150 98-99%

Bevel 1:1 to 4:1 20 98-99%

Worm 5:1 to 75:1 30 20-98%

Crossed Helical 1:1 to 6:1 30 70-98%

Additional Notes on Gear Efficiency

Gear Box Notes Gear Box Notes Gear Box Notes Gear Box Notes

Gearboxes Epicyclic Gears Differential Gearboxes

Gear Lubrication Gear Heat Transfer

Gear Boxes IntroductionIntroductionIntroductionIntroduction

Prime movers such as Electric motors, internal combustion engines, steam engines and turbines produce rotary motion at certain speeds and with certain torques at optimum efficiency. The motion produced is rarely equal to the motion required to do the necessary work and gear trains are required to translate the motion economically at maximum efficiency. It may be possible to use a low cost method of translating the motion e.g. a timing belt vee belt or chain transmission system. However these methods are limiting in their scope and are subject to regular maintenance and replacement. The engineered gearbox generally provides the optimum solution. Many companies provide motorised gear units with the electric motor mounted directly onto a gearbox providing the drive conditions (torque and speed ) exactly as required by the user. Once installed to the manufacturers instructions the only maintenance required is regular lubrication. Gearboxes can be engineered to allow gear ratio changes to enable output shaft speed while keeping the input speed and torque at the same value. The primary advantage for using a gearbox for changing speed is to enable the full power to be transmitted at the different speeds. Electric motors and other prime movers are rated for a maximum torque at the optimum speed. If the speed is reduced using electronic controls the resulting developed torque is not proportionally increased. Gearboxes also allow the input shaft and the output shaft to be in different directions.

Simple Drive Train RulesSimple Drive Train RulesSimple Drive Train RulesSimple Drive Train Rules

1) For any pair of meshing gears the angular velocity ratio is given by

ω2 /ω1 = ω z1 /z2 = ω d1 / d2

ω1 = Input speed (rads/s)

ω2 = Output speed (rads/s) z1 Number of teeth on input gear z2 Number of teeth on output gear d1 Pitch Circle Dia of the input gear d2 Pitch Circle Dia of the output gear The sign is - (Reversing) if both gears are external and + (Same direction)if one gear is a ring (internal gear)

2) For a train of gear wheels the overall angular velocity ratio is given by

ω2 /ω1

= ω Product of teeth No's Driving gears / Product of teeth No's Driven gears

= ω Product of pitch diameters (Driving gears) / Product of pitch diameters (Driven gears) The sign is - (Reversing) if there are an odd number of pairs of external gears

Gearbox Design FeaturesGearbox Design FeaturesGearbox Design FeaturesGearbox Design Features

The design of the gearbox includes the following features..

• Input and output shaft relative positions and orientation

• Support of external forces on shafts

• Design and rigidity of casing

• Type, dimensions and strength of gears

• Method of changing speed /direction if required

• Design and strength of gear shafts

• Gearbox bearings

• Gearbox Seals

• Lubrication

• Noise and vibration

• Couplings to shaft

• Fixing /support of gearbox

• Heat dissipation

• Maintenance provisions

Gearbox EGearbox EGearbox EGearbox Examples xamples xamples xamples

A gearbox is loosely defined as an enclosure for housing gears. Examples of gearboxes are numerous and some are listed below:

• Watch mechanism

• Bicycle axle gear (Sturmy Archer-3 speed) Sprocket to wheel axle -(Sturmy Archer-3 speed)

• Power tool gear units - allowing speed reduction, change and reversing

• Automobile synchromesh gearbox -5 speed + reverse - Engine to drive shafts

• Machine tool integral - Electric motor drive to spindle and travelling motions

• Wind turbine gearbox - Turbine to generator

• Steam turbine - speed reduction turbine to generator

• Marine - Gearbox - turbine /diesal prime movers to Prop shaft

• Cranes -Gearbox usind for lifting and travelling motions

These are all specialised applications and the notes on this page relate to gear units manufactured as separate units for mounting in transmission systems. The normal method of fixing an enclosed gearbox in industry is to mount it on a rigid horizontal baseplate designed to absorb vibration. The rotary motion is transferred to the input shaft and from the output shafts via flexible couplings. There are a number of variations as listed below

MountingMountingMountingMounting

• Foot mounted on vertical surfaces

• Foot mounted below horizontal surface

• Flange mounted onto the prime mover

• Shaft mounted with a torque arm to prevent rotation of gearbox

The sketches below show examples of gearboxes mounted in different ways. These are only illustrative sketches and should not be considered as gearbox designs..

When using a gearbox in a non-standard mounting position the lubrication system should be checked for suitability

GearbGearbGearbGearbox Casingox Casingox Casingox Casing

The large gearbox casings are generally castings from cast iron or steel. Cast iron is a rigid material with excellent vibration damping properties. Fabricated steel gearbox are used for small batch quantities. Gearboxes used for the transmissions in vehicles are often made from cast aluminium this is primarily to save weight. The tiny gearbox units are made from a variety of materials including cast zinc alloys. The important criteria in the gearbox casing design are listed below..

• Inclusion of safe lifting points to allow installation

• Support of the shaft bearings and hence the gear loadings;

• The transfer of the developed gear forces to supporting structure or further drive element;

• Containment of lubricant and exclusion of foreign matter;

• Providing a safety and noise barrier;

• Dissipation of the heat generated by gear friction

• Aiding testing, installation, and maintenance by containing all element in one unit;

• Providing convenient access to internals for inspection & maintainance

• Aesthetic benefits

• Enable accessible location of nameplate with all of the gear unit details

Gearbox shaftsGearbox shaftsGearbox shaftsGearbox shafts

A variety of shaft designs are available including the following

• Plain shaft with keyway

• Plain shaft suitable for Friction drive coupling system

• Splined shaft

• Hollow shaft with internal keyway

• Flange

The selection of the drive shaft system is generally based on space considerations, on the design of the prime mover of the driven component and on the loading pattern. A proprietary gearbox is design to best fit in with the existing drive arrangements.

Shaft OrientationShaft OrientationShaft OrientationShaft Orientation

The shafts transfer of motion to and from a gearbox can be supplied in a variety of designs some of which are listed below.

• Inline shafts .....Epi-cyclic. spur, helical, harmonic

• Parallel shafts .... spur, helical

• Shafts at angles but non intersecting....Helical, Worm, Hypoid, Spiroid gears

• Shafts at angles and intersecting... Bevel gears

Worm gears and bevel gears are most commonly supplied with shafts at Right angles.

Epicyclic Gears

IntroductionIntroductionIntroductionIntroduction

An epicyclic gear is a planetary gear arrangement consisting of one or more planet(epicyclic) gears (P) meshed and rotating round a central sun gear (S). The planet gears are also meshed and rotate within an internal ring gear (A). The planet gears are fixed to a planet carrier-crank arm(L) designed to rotate on the same centre as the sun gear. Only one planet-carrier /crank arm is used in a single epicyclic gear train. This complicated arrangement (see below) has a number of modes of operation depending on which members are locked. Epicyclic gears can be based on spur gears, helical gears, or bevel gears. Epicyclic gearboxes are generally purchased as complete units from specialist suppliers.

Design FeaturesDesign FeaturesDesign FeaturesDesign Features

1. Co-axial drive and driven shafts 2. Compact drive 3. Low moment of Inertia 4. Torque transmitted up to 10,000 Nm 5. Drive speed up to 15,000 RPM 6. Efficient 7. Quiet in operation

Epicyclic Gearboxes (Planetary gearboxes)have the following features

Typical Epicyclic Gear ArrangementTypical Epicyclic Gear ArrangementTypical Epicyclic Gear ArrangementTypical Epicyclic Gear Arrangement

Calculation of RatiosCalculation of RatiosCalculation of RatiosCalculation of Ratios

Example 1..

Rotation (CW= +ve) from action....................Nx = number of teeth Action

L A P S

Turn whole gear thro 1 rev Clock Wise 1 1 1 1

Fix arm L and rotate A back CCW 1 rev 0 -1 -NA /NP +NA / NS

Add the two motions above 1 0 1 - NA / NP 1 + NA / NS

Summary: Rotation of the planet shaft (L) 1 rev CW results in the rotation of the shaft (S) 1 + NA / NS revs (CW). Ratio 1 : ( 1 + NA / NS )

Example 2..

Rotation (CW= +ve) from action....................Nx = number of teeth Action

S L P A

Fix L and rotate Sun 1 rev CW 1 0 - NS / NP - NA / NS

Summary : Rotation of the Sun shaft(S) 1 rev CW results in the rotation of the shaft (L) NS / NA revs (CCW). Ratio 1: ( - NS / NA )

Example 3..

Rotation (CW= +ve) from action....................Nx = number of teeth Action

L A B-D S

Turn whole gear through 1 rev CW 1 1 1 1

Fix L and rotate A back CCW 1 rev 0 -1 - NA / NB - ( NA / NB).( N D / N S)

Add the two motions above 1 0 1 - NA / NB 1 - ( NA . ND)/( N B . N S)

Summary: Rotation of the Shaft L 1 rev CW results in the rotation of the shaft (S) 1 - ( NA . ND)/( N B . N S) revs (CW). Ratio 1: [1 - ( NA . ND)/( N B. N S)]

Example 4..

Rotation (CW= +ve) from action....................Nx = number of teeth Action

L A B-D S

Turn whole gear through 1 rev CW 1 1 1 1

Fix L and rotate A back CCW 1 rev 0 -1 - NA / NB + ( NA / NB).( N D / N S)

Add the two motions above 1 0 1 - NA / NB 1 + ( NA . ND)/( N B . N S)

Summary: Rotation of the shaft L 1 rev CW results in the rotation of the shaft (S) 1 + ( NA . ND)/( N B . N S) revs (CW). Ratio 1: [1 + ( NA . ND)/( N B . N S)]

Example 5..

Rotation (CW= +ve) from action....................Nx = number of teeth Action

L A B-D S

Turn whole gear through 1 rev CW 1 1 1 1

Fix L and rotate A back CCW 1 rev 0 -1 + NA / NB + ( NA / NB).( N D / N S)

Add the two motions above 1 0 1 + NA / NB 1 + ( NA . ND)/( N B . N S)

Summary: Rotation of the shaft L - 1 rev CW results in the rotation of the shaft (S) 1 + ( NA . ND)/( N B . N S) revs (CW). Ratio 1: [1 + ( NA . ND)/( N B . N S)]

Example 6..

>

Rotation (CW= +ve) from action-...................... Nx = number of teeth Action

L A B-D S

Turn whole gear through 1 rev CW 1 1 1 1

Fix L and rotate A back CCW 1 rev 0 -1 + NA / NB - ( NA / NB).( N D / N S)

Add the two motions above 1 0 1 + NA / NB 1 - ( NA . ND)/( N B . N S)

Summary : Rotation of the shaft L 1 rev CW results in the rotation of the shaft (S) 1 - ( NA . ND)/( N B . N S) revs (CW). Ratio 1: [1 - ( NA . ND)/( N B . N S)]

EpiEpiEpiEpi----cyclic gear Train Variationscyclic gear Train Variationscyclic gear Train Variationscyclic gear Train Variations

The figure below shows the range of possible epicyclic gear arrangements.. Those in section I & III are classed as simple arrangements because the planet gears mesh with both sun gears. Those in sections II & IV are classed as complex trains because the planet gears partially match with each other and partially mesh with the sun gears.

Calculation of epicyclic Gear RatioCalculation of epicyclic Gear RatioCalculation of epicyclic Gear RatioCalculation of epicyclic Gear Ratio

1) First calculate the ratio of the gears with the planet carrier fixed..

r f = (±) Product of

Driving Gear Teeth /Product of Driven Gear Teeth

Note: when two external gears are in contact there is a sign change (change of direction) when an internal gear meshes with an external gear both gears rotate in the same direction and there is no

change in direction.. 2) Designate and input gear (x) and an output gear (y).

The speed of the input gear relative to the carrier arm = ω x - ω L

The speed of the output gear relative to the carrier arm = ω y - ω L 3) The train value with the carrier fixed =

r f = ( ω y - ω L ) / ( ω x - ω L )

This relationship is used to solve the planetary gear train ratios. Using this method for the examples above

Example 1. 1) r f = (NA /NP ).( - NP /NS )= - NA /NS

2) Select ω A as input speed = 0 and ω S as output speed (unknown) .

3) r f = ( ω S - ω L ) / ( ω A - ω L ) Therefore - r f.ω L = ( ω S - ω L ) Therefore ω S/ω L = 1 - r f

ω S/ω L = 1 + NA /NS

Example 2. 1) r f = (-NS /NP ).( NP /NA )= - NS /NA

2) Select ω S as input speed and ω A as output speed .ω L =0

3) r f = ( ω A - ω L ) / ( ω S - ω L ) Therefore ω A /ω S = r f

ω A/ω S = - NS /NA

Example 3. 1) r f = (NA /NB ).(ND /NS )= ( N A.N D) / (NB .NS )

2) Select ω A as input speed = 0 and ω S as output speed. (solve for ω S/ ω L)

3) r f = ( ω S - ω L ) / ( ω A - ω L ) Therefore 1 - r f = ω S /ω L

ω S/ω L = 1 - ( N A.N D) / (NB .NS )

Example 4. 1) r f = (NA /NB ).( - ND /NS )= - ( N A.N D) / (NB .NS )

2) Select ω A as input speed = 0 and ω S as output speed. (solve for ω S/ ω L)

3) r f = ( ω S - ω L ) / ( ω A - ω L ) Therefore 1 - r f = ω S /ω L

ω S/ω L = 1 + ( N A.N D) / (NB .NS )

Example 5. 1) r f = (- NA /NB ).( ND /NS )= - ( N A.N D) / (NB .NS )

2) Select ω A as input speed = 0 and ω S as output speed. (solve for ω S/ ω L)

3) r f = ( ω S - ω L ) / ( ω A - ω L ) Therefore 1 - r f = ω S /ω L

ω S/ω L = 1 + ( N A.N D) / (NB .NS )

Example 6. 1) r f = (- NA /NB ).(- ND /NS )= ( N A.N D) / (NB .NS )

2) Select ω A as input speed = 0 and ω S as output speed. (solve for ω S/ ω L)

3) r f = ( ω S - ω L ) / ( ω A - ω L ) Therefore 1 - r f = ω S /ω L

ω S/ω L = 1 - ( N A.N D) / (NB .NS )

Epicyclic gear trains

An epicyclic train is often suitable when a large torque/speed ratio is required in a compact envelope. It is made up of a

number of elements which are interconnected to form the train. Each element consists of the three components

illustrated below :

• a central gear ( c) which rotates at angular velocity ωc about the fixed axis O-O of the element, under the

action of the torque Tc applied to the central gear's integral shaft; this central gear may be either an external gear

(also referred to as a sun gear) Fig 1a, or an internal gear, Fig 1b

• an arm ( a) which rotates at angular velocity ωa about the same O-O axis under the action of the torque, Ta - an

axle A rigidly attached to the end of the arm carries

• a planet gear ( p) which rotates freely on the axle A at angular velocity ωp, meshing with the central gear at the

pitch point P - the torque Tp acts on the planet gear itself, not on its axle, A.

The epicyclic gear photographed here without its arms

consists of two elements. The central gear of one element is

an external gear; the central gear of the other element is an

internal gear. The three identical planets of one element are

compounded with ( joined to ) those of the second element.

We shall examine first the angular velocities and torques in a

single three-component element as they relate to the tooth

numbers of central and planet gears, zc and zp respectively.

The kinetic relations for a complete epicyclic train consisting

of two or more elements may then be deduced easily by

combining appropriately the relations for the individual

elements.

All angular velocities, ω, are absolute and constant, and the torques, T, are external to the three-component element; for

convenience all these variables are taken positive in one particular sense, say anticlockwise as here. Friction is presumed

negligible, ie. the system is ideal.

Separate free bodies of each of the three components - including the torques which are applied one to each component - are

illustrated in Figs 2a and 2b for the external and internal central gear arrangements respectively. Also shown are the shaft

centre O and axle A, the radii Rc & Rp of the central and planet pitch cylinders, the radius of the arm Ra.

There are two contacts between the components :

• the planet engages with the central gear at the pitch point P where the action / reaction due to tooth contact is the

tangential force Ft, the radial component being irrelevant;

• the free rotary contact between planet gear and axle A requires a radial force action / reaction; the magnitude of

this force at A must also be Ft as sketched, for equilibrium of the planet.

With velocities taken to be positive leftwards for example, we have for the external central gear :

o geometry from Fig 2a : Ra = Rc + Rp

o velocity of P : vP = vA + vPA so with the given senses : ωcRc = ωaRa - ωpRp

o torques from Fig 2a : Ft = -Tc / Rc = -Tp / Rp = Ta / Ra

and for the internal central gear :

o geometry from Fig 2a : Ra = Rc - Rp

o velocity of P : vP = vA + vPA so with the given senses : ωcRc = ωaRa + ωpRp

o torques from Fig 2a : Ft = -Tc / Rc = Tp / Rp = Ta / Ra

Substituting for Ra from the geometric equations into the respective velocity and torque equations, and noting

that Rc/Rp = zc/zp, leads to the same result for both internal and external central gear arrangements. These are the desired

relations for the three-component element :

( 2a) ( ωc - ωa ) zc + ( ωp - ωa ) zp = 0

( 2b) Tc / zc = Tp / zp = -Ta / ( zc + zp )

. . . . in which zc is taken to be a positive integer for an external central gear, and a negative integer for an internal

central gear.

It is apparent that the element has one degree of kinetic (torque) freedom since only

one of the three torques may be arbitrarily defined, the other two following from the

two equations ( 2b). On the other hand the element possesses two degrees of

kinematic freedom, as any two of the three velocities may be arbitrarily chosen, the

third being dictated by the single equation ( 2a).

From ( 2b) the net external torque on the three-component element as a whole is :

ΣT = Tc + Tp + Ta = Tc { 1 + zp / zc - ( zc + zp )/zc } = 0

which indicates that equilibrium of the element is assured.

Energy is supplied to the element through any component whose torque and velocity senses are identical. From ( 2) the total

external power being fed into the three-component element is :

ΣP = Pc + Pp + Pa = ωcTc + ωpTp + ωaTa = Tc { ωc + ωp zp /zc - ωa ( zc + zp )/zc }

= Tc { ( ωc - ωa ) zc + ( ωp - ωa ) zp } / zc = 0

confirming that energy is conserved in the ideal element.

In practice, a number of identical planets are employed for balance and shaft load

minimisation. Since ( 2) deal only with effects external to the element, this

multiplicity of planets is analytically irrelevant provided Tp is interpreted as being

the total torque on all the planets, which is shared equally between them as suggested

by the sketch here. The reason for the sun- and- planet terminology is obvious; the

arm is often referred to as the spider or planet carrier.

Application of the element relations to a complete train is carried out as shown in the

example which follows. More complex epicyclic trains may be analysed in a similar

manner, but the technique is not of much assistance when the problem is one of gear

train design - the interested designer is referred to the Bibliography.

EXAMPLE

An epicyclic train consists of two three-component elements of the kind examined above. The first element comprises the

external sun gear 1 and planet 2; the second comprises the planet 3 and internal ring gear 4. The planets 2 and 3 are

compounded together on the common arm axles.

Determine the relationships between the kinetic variables external to the train in terms of the tooth numbers z1, z2, z3 & z4.

The train is analysed via equations ( 2) applied to the two elements in turn, together with the appropriate equations which set

out the velocity and torque constraints across the interface between the two elements 1-2-arm and 3-4-arm.

1-2-arm :

( ω1 - ωa ) z1 + ( ω2 - ωa ) z2 = 0 from ( 2a)

T1 / z1 = T2 / z2 = - T a2 / ( z1 + z2 ) from ( 2b)

3-4-arm :

( ω4 - ωa ) ( -z4 ) + ( ω3 - ωa ) z3 = 0 in which z4 is a positive integer as

T4 / ( - z4 ) = T3 / z3 = - Ta3 / ( - z4 + z3 ) central gear is internal

Ta2 and Ta3 are the parts of the total external torque on the arm, Ta, which are applied individually to the two

elements 1-2-arm and 3-4-arm.

Interface :

ω3 = ω2 since the planets 2 & 3 are coupled

T3 = - T2 since the planets 2 & 3 are coupled (action/reaction)

Ta = Ta2 + Ta3 as the arm is common to both elements 1-2-arm and 3-4-arm

Solution :

The basic speed ratio, io, of an epicyclic train is defined as the ratio of input to output speeds when the arm is held

stationary.

Neither input nor output is defined here - indeed this terminology can be confusing with multiple degrees of freedom - so for

example select gear 1 as input, gear 4 as output.

It follows that io = ( ω1/ ω4 )ωa=0.

Solving the three velocity equations and the six torque equations leads to the desired relations :

Velocities : ( ω1 - ωa ) = io ( ω4 -ωa ) where io = - z2 z4 /z1 z3

Torques : T1 = -T4 /io = Ta /( io - 1 )

Evidently this train possesses the same degrees of freedom as an individual element.

Thermal Design of Gearboxes

IntroductionIntroductionIntroductionIntroduction

When designing a gearbox it is necessary to consider the need to remove the heat that is generated inside the gearbox due to the mechanical inefficiency of the gear action. This is not too important for the efficient gears such as spur gears and helical gears. Generally the surface area of the enclosing box for spur and helical gears is more than sufficient to disperse the heat to the local environment by radiation, convection and conduction. In designing worm gearboxes the thermal design of the gearbox is often a factor that significantly affects the system design. A worm gearbox can be compact for a high reduction high power unit while the heat to be removed may be high because the gear is relatively inefficient. A worm gearbox will dissipate more heat as the external temperature rises and the ambient

temperature falls. A reasonable design ambient temperature is about 30o C. a reasonable maximum

gearbox surface temperature is about 80o C. This reflects a temperature difference between the

lubricating oil and the ambient of nearly 50o C assuming a low temperature drop through the gearbox

walls. For the conditions above it is reasonable to assume that a gearbox will dissipate approximately 1kW

per m2 of surface area. With a fan mounted on the worm shaft providing a reasonable relative air

flow this can be increased to about 1.7kW per m2 of surface area. These values assume flat surfaces

and do not include for conduction through the base and shafts and do not include for the benefits from adding fins and ribs to the box external surfaces. I have included some notes below to justify, to some extent these values. These notes relate to very approximate methods of determining the heat dissipated from a gearbox. For serious high risk design projects it is recommended that detailed heat transfer calculations are completed together with tests. For large power compact gearbox with high mechanical losses to be dissipated there may be a need for integral lubrication oil cooling systems .

Basic Heat Transfer EquationBasic Heat Transfer EquationBasic Heat Transfer EquationBasic Heat Transfer Equation

A very simple heat transfer equation combining the effects of radiation and convection is provided below

:

H = C cr A c∆ t

H = Energy dissipated through the housing . (kW)

C cr = Combined (radiation/convection )heat transfer coefficient kW/( m 2. deg.C)

∆ t = Temperature difference between oil and surroundings Ac = Area of case exposed to ambient air temperature. A chart below provides combined heat transfer values based on gearbox exposed area..

This graph is based on information from ..Machine design Theory and Practice .A.D.Deutschman, W.A Michels & C.E. Wilson.. MacMillan Publishing 1975.

For a small gearbox with a surface temperature of 50o above ambient C a heat transfer of about

1,26kW /m2 is achievable.

For a large gearbox with a surface temperature of 50o above ambient a heat transfer of about 0.8kW

/m2 is achievable.

If a fan is provided on the wormshaft and improvement in C cr of about 60% is possible Ref. watlow.com Includes surface heat transfer information. From this source a typical combined heat transfer coefficient for oxidised steel for an ambient air

temperature 21o. C and a body temperature of 93

o C. i.e 72

o C. temperature difference is 1.0kW per

m2.

Ref. Spirax Sarco Includes a chart providing overall heat transfer values from steel tanks walls.

U - at different relative air flows

Side Walls top Temperature surface and ambient

W / (m2.oC ) W / (m

2.oC )

deg.C Static 1m/s 3m/s Static 1m/s 3m/s

30 10.5 18 31.5 14 24 42

50 11.5 20 34,5 15 25,5 45

70 12 20,4 36 16 27,2 48

This table includes for static air conditions and if the air flow adjacent to the surfaces is about 1m/s then the static values are increased by a factor of 1.7 and if the air flow is 3m/s then the static values are increased by a factor of 3.

From this source a heat transfer coefficient of about 1 kW /m2 is achievable from a tank with side

walls and a top surface with a low ambient air movement at a temperature difference of 50o C.. This

Gear Efficiency

IntroductionIntroductionIntroductionIntroduction

Important Note: The equations used are mainly derived from the technical information provided in the SDP/SI technical library see links below. I recommend that for serious work the linked information is more suitable.

The efficiency of a gear system is simple calculated as the

[output shaft power /Input shaft power ].100 %.

The output power is the (input power - the power losses). Power losses in gear systems are associated primarily with tooth friction and lubrication churning losses. Churning losses are relatively independent of the nature of the gears and the gear ratios - they are primarily realated to the peripheral speed of the gears passing through the fluid. Churning losses are difficult to calculate and estimates based on experience are often used in initial gear design. The frictional losses are related to the gear design,the reduction ratio,the pressure angle, gear size, and the coefficient of friction. The notes below relate primarily to estimating /calculating the the part of the efficiency of gear trains which is associated with the tooth friction. A simple table is provided below showing the efficiencies of various gear types. These efficiencies related to tooth friction losses only for single tooth meshes. For drive trains the efficiencies or each mesh in the line is multiplied together ( 2 gears pairs of 90% efficiency result in a gear train efficiency of 81%.

Type Normal

Ratio Range Pitch Line

Velocity (m/s) Efficiency Range

Spur 1:1 to 6:1 25 98-99%

Helical 1:1 to 10:1 50 98-99%

Double Helical 1:1 to 15:1 150 98-99%

Bevel 1:1 to 4:1 20 98-99%

Worm 5:1 to 75:1 30 20-98%

Crossed Helical

1:1 to 6:1 30 70-98%

Symbols Symbols Symbols Symbols

R g = Gear ratio Ro = Outside Diameter of gear (m) ro = Outside Diameter of pinion(m) Rp = Pitch Diameter of gear (m) rp = Pitch Diameter of pinion(m) E = efficiency of gear pair (%) P = Power loss of gear pair as a percentage of input power (%)

α n = normal pressure angle

vs = gear surface sliding velocity (m/s)

α = pressure angle

β = helix angle (deg)

γ = worm lead angle (deg)

γ = worm lead angle (deg)

µ = coefficient of friction

Spur Gears / Helical GearsSpur Gears / Helical GearsSpur Gears / Helical GearsSpur Gears / Helical Gears

Considering a spur gear a good first approximation for average operating conditions is that the power loss at each mesh can be approximated as P% of the potential power transmitted through the mesh. The efficiency is therefore..

Clarification of formula terms.

Gear Type F sign of Rg term

External spur cos α (Rg+1)

Internal Spur α (Rg-1)

Single Helical

cos φn / cos 2 β

tan φn = tan α.cos

β

β = helix angle

(Rg+1)

Worm Gears Worm Gears Worm Gears Worm Gears

The theoretical efficiency of a worm gear is provided on page worm gears and is shown again below as.

Chart of Worm Gear EfficiencyChart of Worm Gear EfficiencyChart of Worm Gear EfficiencyChart of Worm Gear Efficiency

The graph below shows a worm gear efficiencies plotted against the lead angle for different coefficient of friction Notes: This chart agrees with the equation for worm gears having a normal pressure angle of 20 degrees... An approximation for the friction coefficient for worm gears (Bronze -steel) is

µ = 0,04 vs-0,25

vs = sliding velocity (m/s)

Epicyclic GearsEpicyclic GearsEpicyclic GearsEpicyclic Gears

Consider the two meshing gears below which are part of an epicyclic gear train. This is provided as and example of a typical component in an epicyclic gear. The two gears and the arms are rotating as shown .

Now it the arm was stationary the contact point P would have a instantaneous velocity = -ω2

R2. (velocities to the right are positive ). Now if the whole system was rotating as a rigid

assembly with the velocity of the arm the instantaneous velocity -ω3 R2. Combining these two

motions together, the linear velocity of the tooth engagement (gear 2 -> gear 1)is therefore

v12 = - ( ω2 R2 - ω3 R2) = - (ω2 - ω 3)R 2

Now the magnitude of the transmitted tangential force F12 x the tooth engagement velocity v12 is called the potential power and the power loss due to tooth friction is proportional to this power. Generally for spur gears ( and helical gears) it is sufficient to estimate the power loss as 1% of the potential power. For more accurate estimations the equations above can be used.

P = 0,01 F12v12

The potential power is not the actual power but is the but is the power transmitted by the same

gears operating on fixed centres at angular velocities of ( ω2 - ω3 ) for gear 2 and ( ω1 - ω3 ) for

gear 2.

The actual pitch line velocity of the gear mesh is -(ω2) and therefore the ration of the potential

power to the actual power is

Now in cases where the arm is rotating faster than the gear the potential power can be greater than the actual power and the losses proportionally greater.

Example Epicyclic Gear efficiency calculation.Example Epicyclic Gear efficiency calculation.Example Epicyclic Gear efficiency calculation.Example Epicyclic Gear efficiency calculation.

Consider the epicyclic gear chain shown below. The input speed = 250 RPM (ACW) and the input torque = 2.5 Nm

ω2 = (250.2.π)/60 = 26,18rads/s. and ω1 = 0

The power into the gear = Tω2 = 2,5.26,18 = 65 Watts. Calculating forces

The torque on the arm 2 = M2 = 2.5Nm The force (F2) at Radius R2 (= R1 + R3) : = -(M2/R2) : F2 = -2.5 /(0,1m+.025m) = -20 N (forces to right are positive) Now all forces and torques on the link between Arm(2) and gears (3) & (4) are in equilibrium therefore

For the gear mesh between gear 3 and the fixed sun 2 the velocity of tooth engagement is

calculated by the product of the angular velocity of the arm 2 at pitch radius of R2. = ω2R1 =

26,18.0,075 = 1,96m/s The potential power relates to the gear pair operated in isolation - in this case for gear 3 to be in equilibrium (With forces and moments) F2.R3 + F3.R3 = 0 and F2.R3 + F3R3 = 0 therefore F3 = -F2 = 20N The potential power of the mesh between the sun gear 1 and the planet gear 3 = the gear engagement velocity(1,96m/s) x the tangential force (F3= 20N) = 1,96.20 =3,92W Allowing 1% loss on a typical spur gear engagement the power loss at this engagement = 0,39W. For the gear engagement between the gears 4 and 5 the angular velocity of gear five =

The velocity of tooth engagement between gear (4) and gear (5) =

R5 (ω5 - ω2) = 0,068(-5,95- 26,18)=-2.18 m/s

The potential power at this tooth engagement = the gear engagement velocity (2,18 m/s). x the tangential force (F4 = -167N) = 2,18.167 = 364 W Assuming 1% power loss then the loss at this tooth engagement = 3,64 W The power input to the epicyclic gear = 65 W therefore the Gear efficiency = 100(1 - (3,64 + 0,39) /65 ) = 94%

Reference:

http://www.roymech.co.uk/index3.htm