# Kinematic Analysis of Gears

August 8, 2017 | Author: Rob Morien | Category: Gear, Kinematics, Euclidean Geometry, Geometry, Mechanics

#### Description

Kinematic Analysis of Gears

Kinematic Analysis of Gears Gearing is the most convenient and most generally used means of transmitting mechanical energy from one rotating body to another where the parts to be so connected are not too far apart.

Example: The drive mechanism for the paper feed rollers of a fax machine. In this application, an electric motor drives a small gear that drives larger gears to turn the feed rollers. The feed rollers then draw the document into the machine’s scanning device. In addition to transmitting the motion, gears are often used to increase or reduce speed, or change the direction of motion from one shaft to the other.

Types of Gears Spur Gears : The teeth of the spur gear are parallel to the axis of rotation. Spur gears are used to transmit motion between parallel shafts, which covers the majority of applications. Spur gears are used to change the speed and force of a rotating shaft. Pinion:smaller of two gears (typically on the motor) drives a gear on the output shaft. Gear or Wheel:Larger of the two gears.

How much the speed and force change depends on the gear ratio. The gear ratio is the ratio of the number of teeth on the pair of gears that are meshed. The first gear in the pair is on the input shaft. For example, this could be the gear on the shaft of the motor. The second gear in the pair is on the output shaft. This could be the shaft of the wheel. The gear ratio is the ratio number of teeth of the gear on the output axle to the number of teeth of the gear on the

For example, this picture shows an 8 tooth spur gear meshed with a 40 tooth spur gear. If the 8 tooth gear is on the input axle, and the 40 tooth gear is on the output axle, then the gear ratio for this gear pair is 40 to 8. This can be simplified to 5 to 1.

What this means is it takes 5 revolutions of the input gear to make 1 revolution of the output gear. This results in a slowdown of the output gear by a factor of 5. It also increases the force of the output gear by a factor of 5. If the input and output shafts are reversed, then the gear ratio would be 1 to 5. That means the output shaft would rotate 5 times faster than the input shaft, but have 5 times as less force. Spur gears change the direction of rotation. If the input shaft rotates

Rack and pinion A rack is a special case of spur gear where the teeth of the rack are not formed around a circle, but laid flat. Therefore the rack can be perceived as a spur gear with an infinitely large diameter. When a rack mates with a spur gear , translating motion is produced. A rack and pinion is one of the least expensive methods of converting rotary motion to linear motion

Internal or Annular Gears Internal gears have the teeth formed on the inner surface of a circle.

Helical Gears Helical gears are similar to the applications as spur gear. Here the teeth of a helical gear are inclined to the axis of rotation. The angle of inclination is termed as the helix angle. This angle provides a more gradual engagement of the teeth during meshing and produces impact and noise. Because of smother action, helical gears are proffered in high speed applications.

Spur gears are generally not run at peripheral speed of more than 10m/s. Helical gears can be run at speed exceeding 50m/s when accurately machined and balanced.

Herringbone Gear / Double Helical Gears The herringbone Gear appears as two opposite-hand helical gears butted against each other. This complex configuration counterbalances the thrust force of a helical gear.

Bevel Gear Bevel gears have teeth formed on a conical surface and are used to transmit motion between nonparallel shafts. Most of their application involve connecting perpendicular shaft. However bevel gears can be used in applications that require shaft angles that are both higher and smaller than 900.

Miter Gear Miter gear is a special case of bevel gear where the gears are of equal size and the shaft angle is 900.

Worm and Worm Gear A worm and worm gear is used to transmit motion between nonparallel and non-intersecting shafts. The worm has one tooth that is formed in a spiral around a pitch cylinder. This one tooth is also referred to as the thread because it resembles a screw thread. The transmission ratio is a function of the worm pitch and the worm gear pitch diameter. As the worm rotates, its thread pushes

Spur Gear Terminology

Pitch Cylinder: It is an imaginary friction cylinder which by pure rolling, transmit the same motion as that of a pair of gear. Pitch surface: The surface On which teeth are cut to ensure positive drive is called pitch surface

Pitch circle: The intersection of pitch surface with a plane perpendicular to the axis of rotation is called the pitch circle. Pitch point: The contact point of two pitch circle is called the pitch point. Addendum circle: It is the circle which bounds the outer the outer end of the teeth. In other words, it is the diameter of a blank on which teeth are cut. Addendum: The radial distance between the pitch circle and addendum circle is called the addendum. Generally, this distance is kept equal to one module in a 20° full depth involute teeth gear and 0.8 times the module

Dedendum circle: It is a circle which bounds the bottom of the teeth. Dedendum: The radial distance between the pitch circle and the dedendum circle is called the dedendum. The standard value of dedendum is 1.25 times module in 200 full depth teeth gear and that is 0.8 times module in 20° stubbed teeth gear. Total depth of teeth: The sum of addendum and dedendum is called the total depth of the tooth. Clearance: The difference between dedendum and addendum of a mating gear teeth is commonly referred to as clearance.

Tooth thickness: The chord length measured along the pitch circle between the two opposite faces of the same tooth is called tooth thickness. Top land: It is the surface of the top of the tooth. Bottom land: The surface at the bottom of the tooth between the adjacent fillets. Face: Tooth surface between the pitch circle and the top land is known as face. Flank: Tooth surface between the pitch circle and the bottom land is known flank. Circular pitch: The distance measured along the pitch circle from a point on one tooth to the

πd1 πd 2 p= = Z1 Z2

Where d1, d2 = pitch circle diameters of pinion and gear respectively And Z1, Z2 = number of teeth on the pinion and gear respectively Module: The ratio of the pitch circle diameter to the number of teeth is called module and is denoted by m. d m=

1

Z1

Diametral pitch: The ratio of number of teeth in the pitch circle diameter is called the pitch. It is reciprocal of the module. Z1 1 DP = = d1 m

Gear Ratio: It is the ratio of number of teeth on the gear to that on the pinion. Z2 G= Z1

Velocity ratio: It is the ratio of the angular velocity of the driven gear to the angular velocity of the driving ω2 N 2 pinion. d1 Z1 1 VR =

ω1

=

N1

=

d2

=

Z2

=

G

Law of Gearing In a gear drive, the action of tooth profile is to transmit motion at constant angular velocity ratio for which the gears must satisfy the fundamental law of gearing. Accordingly, the common normal at the point of action between two teeth must always pass through a point, called pitch point, in such a way that it divides the line joining the centres of two matting gears in the inverse ratio of angular velocities. Consider two rigid bodies 1 and 2 representing a portion of two gears in mesh rotating about fixed centres O1 and O2. The point A on the rigid body 1 is in contact with a point B on the rigid body 2. Let X - X and Y - Y represent common tangent and

Let ω1 = angular velocity of the rigid body ω2 1 = angular velocity of the rigid body 2

At a given instant, the point A on the rigid body 1 is moving in the direction perpendicular to ω1 O1A with a velocity Va = . O1A . At the same time, the point B body 2 is moving in the ωto direction perpendicular 2 O2B with a velocity Vb = .O2B If the two rigid bodies are always to remain in contact, their relative velocities at any instant along the common normal must be zero, i.e. the componentVaofcosVa Vb βalong common α =and Vb cos normal must beωequal. O A cos α = ω O B cos β 1

1

2

2

O1C O2 D ω1 x O1A x = ω2 x O 2 B x O1A O2 B ω1 x O1C = ω2 x O2 D ω1 O2 D = ω2 O1C

Let P is a point of intersection of line joining the centres of rigid bodies and common normal. Therefore, from similar triangles O1PC and O2DP O 2D O 2P = O1C O1P

Equating the two equations, we get: ω1 O 2P = ω2 O1P

Thus for a constant angular velocity ratio of a gear pair in mesh the normal at the point contact divides the line joining the center of rotation in the inverse ratio of the angular velocities. Thus the dividing point P is called pitch point. This is the fundamental law of gearing which must be satisfied by the profiles adopted for the teeth of gear in mesh. Since all points of contact lie on the common

The gear tooth action is shown in Figure in the next slide in which a line AB, which is normal to the line joining the centers of mating gear, meet at the pitch point P and another line CD, which is tangent to the base circle of gear and pinion, also passes through the pitch point and is normal to teeth in action. The angle between α the lines AB and CD is called the pressure angle ( ) and the normal force that one tooth exerts on the other passes through the pressure line. The pressureαline in a gear pair can be located by rotating line AB through the pressure angle in the direction opposite to the direction of

VELOCITY OF SLIDING The relative velocity of points A and B along the common tangent is known as velocity of sliding. In other words, it represents the sliding velocity of the surface of rigid body 2 relative to the surface of rigid body 1 at the point of contact. Velocity of sin sliding: Vs = V sin β V α b a

Vs = ω2 x O 2 B x sin β - ω1 x O1A x sin α BD AC Vs = ω2 x O 2 B x - ω1 x O1A x O2 B O1A

Vs =ω2 x BD - ω1 x AC Vs =ω2 x ( BP +PD) - ω1 x ( PC - AP) Vs =ω ( 1+ω

2

) x AP +ω

x PD - ω 1x PC

2

ω1 O 2P We know that = ω2 O1P From the similar ∆O1CP and ∆O 2DP O 2P DP = O1P CP Therefore ω2 x PD =ω1 x CP

Thus velocity of sliding: Vs =ω ( 1+ω 2 ) x A P The maximum velocity of sliding occurs at the first or last point of contact. FORMS OF GEAR TEETH When two gears teeth are in mesh, the profile of anyone tooth can be chosen of arbitrary shape and the profile for the other may be determined to satisfy the law of gearing. Such gear teeth are called conjugate teeth. Although gears with conjugate teeth transmit the desire motion, they require special cutter which obviously increase the difficulty in manufacturing. Therefore, conjugate teeth are not in normal use. Usually, the following geometrical curves which satisfy the law of

Involute Profile: An involute is the locus of a point on a straight line which rolls on the circumference of a circle without slippage. In other words, a point on a taut rope, when unwound from a cylinder, would trace an involute curve. To illustrate, consider a cylinder around which a cord abc, which held tightly, is wrapped (as shown in Figure) The point e on the cord represents tracing point. When the cord is unwrapped about the cylinder, the point e will trace out the involute curve d e f. The radius of curvature of the involute is zero at point d and maximum at point f. The radius of curvature at point e is equal to the

The involute profile has the following properties: The shape of the involute profile is dependent only on the dimension of the base circle. If one involute rotates at a uniform rate of motion and is in contact with another, it will transmit a uniform angular motion to the second irrespective of the centre distance between the two corresponding base circles. Angular velocity ratio of involute profile teeth is not sensitive to centre distance of their base circles. When two involutes are in mesh, the angular velocity ratio is inversely proportional to the size of the base circles. The pressure angle of two involutes in mesh is

Cycloidal Profile: A cycloid is the locus of a point on the circumference of a circle which rolls without slipping on a fixed straight line. It has two variants - epicycloid and hypocycloid. An epicycloid is the locus of a point on the circumference of a circle, which rolls without slipping on outside circumference of another circle of finite radius. Similarly, a hypocycloid is a locus of a point on the circumference of circle which rolls without slipping on inside circumference of another circle. In a gear having cycloidal teeth, the face of·

ARC OF CONTACT When two gears start transmitting motion, the initial contact occurs at a point where the flank of the driving gear (pinion) tooth comes in contact with the face tip of the driven gear tooth and the contact ends when the face tip of the pinion tooth comes in contact with the flank of the driven gear tooth. In other words, the contact between the two gears starts when the addendum circle of the driven gear cuts the normal pressure line at point E and ends when the addendum circle of the pinion cuts the normal pressure line at point F. (shown in Figure) The distance between these two points, EF is called path of contact or length of contact. Usually, the path of contact is divided into two

Path of approach. A portion of path of contact from the beginning of engagement to pitch point. i.e., the length EP, is called path of approach Path of recess. The portion of path of contact from pitch point to the end of engagement, i.e., the length PF, is called path of recess. Let rl = pitch circle radius of the pinion r2 = pitch circle radius of the gear

ra1 = addendum circle rarius of the pinion

and = pressure angle, Path of contact = Path of approach + Path of

or Path of approach

EF = EP + PF EP = ED – PD 2

2

EP = O 2 E - O2 D - PD 2

2

2

EP = ra2 - r2 cos α - r2 sin α The maximum possible length of path of approach is: r1 sin α CP = Similarly the path of recess is2 PF = FC – PC 2

PF = O1F - O1C - PC 2 a1

PF = r

2 1

2

- r cos α - r1 sin α

r2 sin of α path of recesss The maximum possible length

Therefore, the path of contact:

EF = ra2 - ra2 cos2 α + ra1 - ra1 cos2 α - ( r1 + r2 ) sin α 2

2

2

2

The maximum length of path of contact occurs when point E lies at point C and point F lies at CD = ( r1 + r2 ) sin α point D. Thus The arc of contact is the distance travelled by a point on either pitch circle of gear or a pinion during the period of contact of a pair of teeth. The contact on pitch circle of gear begins at point G and ends at point H. Similarly, on the pitch circle of pinion, the contact is between K and L. Therefore, the arc of contact is: arc GH = arc KL

But we know from the geometry of the involute curve:

path of contact arc of contact, GH = cos α EF = cos α Since the arc of contact is defined as the length of the pitch circle during the mating of teeth, the number of teeth lying in between the arc of contact GH will be meshing with the teeth on the pinion. Therefore

arc GH Number of teeth in contact = Circular pitch EF = p cos α EF 1 or Number of teeth in contact = x cos α πm

Number of teeth in contact at any instant is called contact ratio. For continuous transmission of power, atleast one pair of teeth should always remain in contact. Generally gears are designed for contact ratio from 1 to 1.6. : Problem: A pair of gears having 200 Involute teeth is required to transmit motion at a velocity ratio of 1:4. If the module of both pinion and gear is 5 mm and centre distance is 250 mm. determine the number of teeth and base circle radius of pinion and gear.

Problem: A pair of spur gear having 20 and 40 teeth are in mesh. The pinion being driving element rotates at 2000 rpm. Find the sliding velocity between the teeth faces (i) at the point of engagement (ii) at the pitch point and (iii) at the point of disengagement. Assume that gear teeth are of 20° involute form. Addendum is 5 mm and module is 5 mm. Find also the angle through which pinion turns while one pair of teeth are in contact.

Problem: The following data refer to a pair of spur gear in mesh having 200 involute profile teeth: Number of teeth on pinion : 24 Number of teeth on pinion : 48 Speed of pinion : 300 rpm Module : 6 mm If the addendum on each gear is such that the path of approach and path of recess are half of their maximum possible values find: (i) The addendum on gear and pinion (ii) The length of arc of contact (iii) The maximum velocity of sliding of gears

INTERFERENCE When a pair of gear transmits power, the normal force is passed through common normal to the two involutes at the point of contact. which is also a tangent to the base circles of mating gear pair. If, by any reason, any of the two surfaces is not involute, the two surfaces would not touch each other tangentially and transmission of power would not be proper. The mating of gear teeth will violate the fundamental law of gearing and this action is called interference. To illustrate. let us consider a pair of gear and pinion in mesh as shown in Figure, in which the path of contact is EF. If, now, the radius of addendum circle of the gear is increased, the point of engagement E shifts along PC towards

The limiting position of point E is the point C. Any further increase in the value of addendum circle radius will shift the point of contact E inside the base circle of the pinion. Since there is no involute profile below the base circle of the pinion, it will form a non-conjugate contact and resulting phenomenon is called interference. In other words, the condition of interference arises when contact between teeth occurs outside the points C and D, i.e. when path of contact EF greater than distance CD. The condition of interference with revised addendum circles marked with dashed line and path of contact E'F' greater than distance CD is shown in Figure. The phenomenon interference can be avoided by employing some preventive

Undercutting: When gear teeth are manufactured by a generating process, a portion of tooth flank which causes interference is cut away by the cutting tools. Thus there will be no conjugate action between teeth. However, by undercutting, the actual ratio of contact decreases which causes more noisy and rough gear action. Secondly, it also reduces the thickness of tooth which ultimately reduces the beam strength of tooth.   Stubbed tooth. When a portion of a tooth near the top is cut away, such a tooth is called stubbed tooth. Such a measure prevents interference but it reduces the contact ratio.

Number of teeth. Interference in a gear pair can be avoided by increasing number of teeth on the gears. This makes the gear larger in diameter and also increases the pitch line velocity. This increased pitch line velocity makes noise gear action and also reduces the power transmission to some extent. Pressure angle. Increasing the pressure angle decreases the base circle diameter of the gear. which means that it increases the involute portion of the tooth profile and hence eliminates the chances of interference. This also demands for smaller number of teeth on gear. However, it will increase the radial force component which may try to dislodge the gear.

Backlash: The backlash ‘B’ is the amount that the width of a tooth space exceeds the thickness of a gear tooth, measured on the pitch circle. i.e., it is the amount that a gear can turn without mating gear turning. Although backlash may seem undesirable, some backlash is necessary to provide for lubrication on the gear teeth.

Force Analysis of a Spur Gear A gear drive is generally specified by power to be transmitted, the speed of the driving shaft and the velocity ratio. The power is transmitted by means of a force exerted by the tooth of driving gear on the mating driven gear. According to the law of gearing, this force ‘Fn’ is always normal to the tooth surface and acts along the pressure angle line. This normal force is designated by two subscripts, for example F12 which means the force exerted by gear 1 against gear

In the above Figure (a) a pair of spur gear in mesh mounted on respective shafts. The driving gear 2 is mounted on shaft 1 and rotates in the clockwise direction. The driven gear 3 is mounted on shaft 4. The free body diagram of the forces acting upon two gears along the pressure line is shown in Figure (b). The driving gear 2 exerts a force F23 on the driven gear 3. Similarly, the driving gear experiences a reaction force F32 .

The normal force Fn (or F23 ), as shown in Figure acting along the pressure line can be resolved into two components-the tangential force Ft and radial force Fr Thus Ft = Fn cos α

Fr = Fn sin α = Ft tan α

α

Where is the pressure angle. The tangential component of force Ft IS mainly responsible for transmitting torque and consequently the power. The radial force Fr is called the separating force, which always acts towards the centre of the gear.

Problem: A layout of gear train having three spur gears is shown in the Figure. Gear A receives 3 kW power at 720 rpm through its shaft and rotates in clockwise direction. Gear B is idler and gear C is driven gear. If number of teeth on gears A, B and C are 20, 50 and 30 respectively. determine the component of gear tooth forces. Module is 5 mm.

GEAR TRAINS A gear train is a mechanism which transmits power or motion from one shaft to another by means of combination of gears. The gear train enables to either step-up or step-down the speed of prime mover and to obtain different range of speeds as per requirement of driven machine. For example, in lathe machine, different materials are cut economically at different cutting speeds. These speeds can be obtained by a suitable gear train. Gear trains are widely used in mechanical clocks, automotive vehicles, turbine generator sets, machine tools etc.

In gear train terminology, a term called train value is quite often used to designate the characteristics of a gear train. It is defined as a ratio of speed of driven gear to that of driving gear. In other words, it is reciprocal of speed ratio (SR). That is

Speed of driven gear T= Speed of driving gear Train value:

Gear trains, depending upon the functional requirement of an application, are classified in the following three categories: Simple gear train Compound gear train

SIMPLE GEAR TRAIN In a simple gear train, a series of gears are mounted on individual shafts to receive and transmit motion (Figure a). In a gear train, the sense of rotation depends upon the direction of rotation of driving gear. For example, in a simple gear train, if the direction of rotation of driving gear is clockwise and it is designated as positive direction, then the direction of mating gear, which is naturally anticlockwise, is termed as negative direction. In other words, all odd number of gears rotate in one direction and even number of gears rotate in other direction.

Figure a shows a simple gear train consisting of four gears each having Z1, Z2, Z3 and Z4 number of teeth and N1, N2, N3 and N4 rpm respectively. Suppose gear 1 is mounted on driving shaft rotating in clockwise direction and gear 4 is mounted on driven shaft, which should rotate in anticlockwise direction.

For gear pan 1-2, the pitch line velocity:

πd1 N1 = πd 2 N 2 Train Value: T1-2

N2 d1 Z1 = = = N1 d2 Z2

Similarly. for pair 2-3 and 3-4, the train values are: N3 Z3 Z2 N4 T2-3 =

N2

=

Z3

and T3-4 =

N3

=

Z4

The train value of complete gear train can be found by multiplying the train values of T = T1-2 × T2-3 × T3-4 individual pair as follows: N2 N3 N4 N4 Z1 Z2 Z3 Z1 = × × = = × × = N1 N2 N3 N1 Z2 Z3 Z4 Z4

N4 Z1 Train Value =T = = N1 Z4 N1 Z4 Speed ratio =SR = = N4 Z1

we observed that the intermediate gears 2 and 3 have no effect on the train value. Therefore, these gears are called idler gears. Idler gears are used for changing the direction of rotation of the driven shaft.

COMPOUND GEAR TRAIN In a compound gear train, a series of gears are connected in such a way that intermediate shaft carries two gears which are fastened together rigidly. Such a gear train is called compound gear train. Figure b shows a compound gear tram constituted of four gears in which gears 2 and 3 are compounded on the intermediate shaft.

For a gear pair 1-2 the train value: T1-2

N2 Z1 = = N1 Z2

Similarly for gear pair 3-4 the train value: T3-4

Z3 N4 = = N3 Z4

Multiplying the above two equations, we get the train value of the compound gear train as: N2 N4 Z1 Z3 T= × = × N1 N3 Z2 Z4

As gears 2 and 3 are compound gears and mounted on the same shaft, their speed will be equal i.e. N2 = N3 . The train value of

N4 Z1 Z 3 T= = × N1 Z2 Z4 In other words, the train value of a compound gear train is the quotient of the product of teeth on the driver gears to that of the product of teeth on the driven gears, of each pair in mesh. Product of number of teeth on driving gears T= Product of number of teeth on driven gears

Reverted Gear Train It is a special type of compound geartrain in which the axes of the driving and driven gear shaft coincide. Figure c shows a riverted gear train in which the driving shaft gear 1 drives gear 2 mounted on the intermediate shaft. Gears 2 compound gears. Gear 3 drives gear 4 which is mounted on the driven shaft coinciding with the axis of the driving shaft. These types of gear trains are most widely used in me clocks and back gear assembly of lathe machine.

3 2

1

4

EPICYCLIC GEAR TRAIN In the types of gear train discussed so far, the axes of the gears remain fixed and there is no relative motion between the axes. In epicyclic gear train, there exists a relative motion between two axes of the gears constituting the train. An epicyclic gear train usually consists of three elements--driving gear, driven gear and an arm which is pivoted about a fixed centre as shown in Figure ‘d’. In this gear train, if the arm A is held fixed, the driving gear 1 and driven gear 2 constitute a simple gear train. However, if gear 2 is held fixed, the arm A can revolve about the centre of gear 2 and gear 1 rolls around the pitch circle circumference of the stationary gear 2. In such a gear train, the driving gear 1 rolls around the driven gear 2 and traces an epicyclic path; hence it is called epicyclic gear

In some gear trains, the fixed gear may be internal gear and pinion which rolls inside of internal gear traces hypocycloid. However, it has become customary to call even these

Tabulation Method In tabulation method, the complex motion of the gear train is splitted into qifferent motions of individual' gear pair and their train value or speed ratio is written in tabular form. Finally these splitted motion segments are added as per their connectivity. A detailed stepwise procedure is seminaries as follows (Figure ‘d’): Step-1: Assume that the arm is locked and all other gears are free to rotate. Step-2: Mark any gear (say gear 2) as reference gear and rotate it through one revolution in the clockwise direction which is designated as positive direction.

Step-3: Calculate the number of revolutions made by all the gears and record them in Table. This can be calculated by known values of number of teeth. Step-4: Multiply each train value by x, assuming that reference gear rotates at x rpm. Write down the corresponding number of revolutions of all the gears in second row of the Table . Step-5: Now it is assumed that arm is unlocked and allowed to rotate in clockwise direction by y rpm. Thus add y to all the elements of second row and write down in the third row. Step-6: Apply the given boundary conditions and find the values of x and y. For example, if

Operation 1. Arm A is locked. Gear 2 is given one turn in clockwise direction. 2. Multiply by ‘x’ 3. Unlock arm and rotate it by y turns. Add ‘y’.

Arm ‘A’ 0

Number of revolution Gear 2 ‘Z2’ +1

0

x

y

x+y

Gear 1 ‘Z1’ Z − 2 Z1

Z2 ×x Z1 Z y- 2 × x Z1 −

x+y=0 And y = 1 Solving Eqs. we get x = -1 and y = 1 Therefore, the revolution of gear 1 for one revolution of arm A and gear 2 being fixed, is.

Z2 N 1 =y × x Z1 Z2 N 1 =1 + Z1

Problem: An epicyclic gear train arrangement is shown in Figure below. Gear E is a fixed gear and gears C and D are compounded and mounted on one shaft. If arm A makes 60 rpm in counterclockwise direction, determine the speed and direction of rotation of gears Band F. The number of teeth on different gears are as given below: Zb = 25, Zc = 15, Zd = 50, Ze = 20 and Zf = 30.

Problem: In an epicyclic gear train as shown in Figure below, the arm A is fixed to the shaft S. The gear B having 80 teeth rotates freely on the shaft S and gear D with 120 teeth is separately driven. If the arm A runs at 100 rpm and gear D at 50 rpm in same direction, find the speed of gear B.