1. Govt. Engineering College Palanpur Name:-Parmar Akash B 130610119031 Salvi Bhargav B 130610119056 1

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3.  Introduction  Gear Tooth Nomenclature  Gear Types  Simple Gear Trains  Compound Gear Trains  Reverted Gear Trains  Epicyclic or Planetary Gear Trains  Examples 3

4. Gear trains are widely used in all kinds of mechanisms and machines, from can openers to aircraft carriers. Whenever a change in the speed or torque of a rotating device is needed, a gear train or one of its cousins, the belt or chain drive mechanism, will usually be used. 4

5. 5 Circular Pitch: The circular pitch is the arc length along the pitch circle circumference measured from a point on one tooth to the same point on the next. The circular pitch defines the tooth size. where d = pitch diameter and N = number of teeth.

6. 6 Base Pitch: The tooth pitch can also be measured along the base circle circumference and then is called the base pitch p b. Diametral Pitch: The diametral pitch p d is: p d = N/d Module: It is the reciprocal of diametral pitch with pitch diameter measured in millimeters. m = d/N In terms of diameters and number of teeth, the Velocity and Torque ratios are: Gear Ratio: The gear ratio m G is always > 1 and can be expressed in terms of either the velocity ratio or torque ratio depending on which is larger than 1. Thus m G expresses the gear train's overall ratio independent of change in direction of rotation or of the direction of power flow through the train when operated either as a speed reducer or a speed increaser.

7. 7 SPUR GEARS: SPUR GEARS are ones in which the teeth are parallel to the axis of the gear. This is the simplest and least expensive form of gear to make. Spur gears can only be meshed if their axes are parallel. HELICAL GEARS: HELICAL GEARS are ones in which the teeth are at a helix angle ψ with respect to the axis of the gear.

8. 8 HERRINGBONE GEARS : HERRINGBONE GEARS are formed by joining two helical gears of identical pitch and diameter but of opposite hand on the same shaft. These two sets of teeth are often cut on the same gear blank. WORMS AND WORM GEARS: If the helix angle is increased sufficiently, the result will be a worm, which has only one tooth wrapped continuously around its circumference a number of times, analogous to a screw thread. This worm can be meshed with a special worm gear (or worm wheel), whose axis is perpendicular to that of the worm. Because the driving worm typically has only one tooth, the ratio of the gear-set is equal to one over the number of teeth on the worm gear. “Perhaps the major advantage of the worm-set is that it can be designed to be impossible to back drive.”

9. 9 RACK AND PINION: Sometimes, the gear of a shaft meshes externally and internally with the gears in a straight line. Such type of gear is called rack and pinion. The straight line gear is called rack and the circular wheel is called pinion. A little consideration will show that with the help of a rack and pinion, we can convert linear motion into rotary motion and vice-versa. BEVEL GEARS: For right-angle drives, crossed helical gears or a worm-set can be used. For any angle between the shafts, including 90°, bevel gears may be the solution. Just as spur gears are based on rolling cylinders, bevel gears are based on rolling cones. The angle between the axes of the cones and the included angles of the cones can be any compatible values as long as the apices of the cones intersect. If they did not intersect, there would be a mismatch of velocity at the interface.

10. 10 SPIRAL BEVEL GEARS: If the teeth are parallel to the axis of the gear, it will be a straight bevel gear. If the teeth are angled with respect to the axis, it will be a spiral bevel gear, analogous to a helical gear. The cone axes and apices must intersect in both cases. The advantages and disadvantages of straight bevel and spiral bevel gears are similar to those of the spur gear and helical gear, respectively, regarding strength, quietness, and cost.

11. A gear train is any collection of two or more meshing gears. A simple gear train is one in which each shaft carries only one gear. The velocity ratio (sometimes called train ratio) of this gear-set is found by expanding equation: Consider a simple gear train with four gears in series. The expression for this simple train's velocity ratio is: m v = (-N1/N2) (-N2/N3) (-N3/N4) = -N1/N4 It may be noted that when the number of intermediate gears are odd, the motion of both the gears is like. But if the number of intermediate gears are even, the motion of the driven or follower will be in the opposite direction of the driver. These intermediate gears are called idle gears. “Simple Gear Trains are limited to a ratio of about 10: 1”. 11

12. 12 When there are more than one gear on a shaft, it is called a compound train of gear. Since the idle gears, in a simple train of gears do not effect the speed ratio of the system. But these gears are useful in bridging over the space between the driver and the driven. To get a train ratio of greater than about 10:1 with spur, helical, or bevel gears (or any combination thereof) it is necessary to compound the train. The speed ratio of compound gear train is obtained as: m v = (-N1/N2)x(-N3/N4)x(-N5/N6) “ The advantage of a compound train over a simple gear train is that a much larger speed reduction from the first shaft to the last shaft can be obtained with small gears.” *Since gears 2 and 3 are mounted on one shaft B, therefore N2 = N3. Similarly gears 4 and 5 are mounted on shaft C, therefore N4 = N5.”

13. 13 When the axes of the first gear (i.e. first driver) and the last gear (i.e. last driven or follower) are co-axial, then the gear train is known as reverted gear train LetT1 = Number of teeth on gear 1, r1 = Pitch circle radius of gear 1, and N1 = Speed of gear 1 in r.p.m. Similarly, T2, T3, T4 = Number of teeth on respective gears, r2, r3, r4 = Pitch circle radii of respective gears, and N2, N3, N4 = Speed of respective gears in r.p.m. Since the distance between the centers of the shafts of gears 1 and 2 as well as gears 3 and 4 is same, therefore r1 + r2 = r3 + r4 m v = N1/N4 = T2 xT4/T1x T3

14. 14 The conventional gear trains described in the previous sections are all one- degree-of-freedom (DOF) devices. Another class of gear train has wide application, the epicyclic or planetary train. This is a two-DOF device. Two inputs are needed to obtain a predictable output. The epicyclic gear trains are useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space. They are used in the back gear of lathe, differential gears of the automobiles, hoists, pulley blocks, wrist watches etc.

15. 15 Velocity Ratios of Epicyclic Gear Trains 1. Tabular method.2. Algebraic method. 1. Tabular method Consider an epicyclic gear train: Let TA = Number of teeth on gear A, and TB = Number of teeth on gear B. First of all, let us suppose that the arm is fixed. Therefore the axes of both the gears are also fixed relative to each other. When the gear A makes one revolution anticlockwise, the gear B will make TA / TB revolutions, clockwise. Assuming the anticlockwise rotation as positive and clockwise as negative, we may say that when gear A makes + 1 revolution, then the gear B will make (– TA / TB) revolutions. Secondly, if the gear A makes + x revolutions, then the gear B will make – x × TA / TB revolutions. Thirdly, each element of an epicyclic train is given + y revolutions. And Finally, the motion of each element of the gear train is added up.

16. 16 Example 1: Find the gear ratio, circular pitch, base pitch, pitch diameters, pitch radii, center distance, addendum, dedendum, whole depth, clearance, and outside diameters of a gear-set with the given parameters. Given: Diametral Pitch P d = 6, 19-tooth pinion is meshed with a 37-tooth gear. Solution: 1. Gear Ratio 2. Circular Pitch 3. Base Pitch 4. Pitch Diameters and Pitch Radii

17. 17 6. Center Distance 7. Addendum and Dedendum 9. Whole Depth 10.Clearance 11. Outside Diameters

18. 18 Example 2: The gearing of a machine tool is shown in Figure. The motor shaft is connected to gear A and rotates at 975 r.p.m. The gear wheels B, C, D and E are fixed to parallel shafts rotating together. The final gear F is fixed on the output shaft. What is the speed of gear F ? The number of teeth on each gear are as given below : No. of teeth on each Gear, T A = 20, T B = 50, T C = 25, T D = 75, T E = 26, T F = 65 Solution: N A = 975 r.p.m. ; N F = ? m v = N A /N B x N C /N D x N E /N F = N A /N F Or N A /N F = T B /T A x T D /T C x T F /T E So, N A /N F = 50 /20 x 75 /25 x 65 /26  N F = N A /(18.75) N F = 975 /(18.75) N F = 52 r.p.m. Ans

19. 19 Example 3: The speed ratio of the reverted gear train, as shown in Figure, is to be 12. The module pitch of gears A and B is 3.125 mm and of gears C and D is 2.5 mm. Calculate the suitable numbers of teeth for the gears. No gear is to have less than 24 teeth. Given : Speed ratio, N A /N D = 12 ;m A = m B = 3.125 mm ; m C = m D = 2.5 mm Solution: Let N A = Speed of gear A, T A = Number of teeth on gear A, r A = Pitch circle radius of gear A, N B, N C, N D = Speed of respective gears, T B, T C, T D = Number of teeth on respective gears, and r B, r C, r D = Pitch circle radii of respective gears. Since the speed ratio between the gears A and B and between the gears C and D are to be same, therefore N A /N B = N C /N D = √12 = 3.464 Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth, therefore T B /T A = T D /T C = √12 = 3.464 We know that the distance between the shafts x = r A + r B = r C + r D = 200mm or m A. T A /2 + m B. T B /2 = m C. T C /2 + m D. T D /2 = 200 After solving we have, T A = 28, T B = 100, T C = 36, T D = 124Ans and Speed ratio, N A /N D = T B /T A x T D /T C = 100/28 x 124/36 = 12.3

20. 20 Example 4: In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B. Given : T A = 36; T B = 45; N ARM = 150 r.p.m (CCW); N A = 0 Solution: At first we consult the table for epicyclic gear train, Since, N ARM = 150 r.p.m (CCW); and N A = 0 So, x + y = 0 or, x = – y = – 150 r.p.m. =>Speed of gear B, N B = y – x. T A / T B and N B = 150 + 150. 36/45 N B = 270 r.p.m (CCW)

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