Gear trains (Chapter 6) Change torque, speed Why we need gears Example: engine of a containership Optimum operating speed of the engine about 400 RPM Optimum operating speed of the propeller about 100 RPM Need reduction gear 1

Connecting the main engine to the propeller through a reduction gear Propeller, Output flange operates at about 100 RPM Gea r Engine

Engine operates at about 400 RPM 2 Types of gears 3 Gear box Stick shift Synchronizers The gear box is in first gear, second gear

4 Gear Nomenclature (6.1) 5 Important definitions Velocity ratio=mV=angular velocity of output gear/ angular velocity of input gear=pitch diameter of input gear/pitch diameter of output gear Torque ratio=mT=torque at output gear/torque at input gear mT=1/mV

Gear ratio=mG=Ngear/Npinion, mG is almost always greater than one 6 Fundamental law of tooth gearing (6.2 and 6.3): velocity ratio must be constant as gears rotate Angular velocity ratio= ratio of distances of P from centers of rotation of input and output gear. 2

3 O2 P 2 O3P T 3 If common normal were fixed then the velocity ratio would be constant. 3

7 If gear tooth profile is that of involute curve then fundamental law of gearing is satisfied Involute curve: path generated by a tracing point on a cord as the cord is unwrapped from base cylinder 8 Generating gear teeth profile Steps:

Select base circles Bring common normal AB Draw involutes CD, EF P 9 Gear action Angular velocity of Gear 3 / angular Velocity of gear 2 = O2P/O3P = constant

10 Fundamental law of gearing: The common normal of the tooth profiles at all points within the mesh must always pass through a fixed point on the line of the centers called pitch point. Then the gearsets velocity ratio will be constant through the mesh and be equal to the ratio of the gear radii. 11 Base circle radius =

Pitch circle radius cos 12 Initial contact: B Path of approach: BP=ua=[(r3+a)2-rb32]1/2-r3sin Path of recess: PC=ur=[(r2+a)2-rb22]1/2-r2sin Final contact: C 13 Standard gears: American Association of

Gear Manufacturers (AGMA) (6.4) Teeth of different gears have same profile as long as the angle of action and pitch is the same. Can use same tools to cut different gears. Faster and cheaper product. Follow standards unless there is a very good reasons not to do so. 14 Template for teeth of standard gears 15 AGMA Specifications

Diametral pitch, pd=1, 1.25, 1.5,,120 Addendum of pinion = addendum gear Observations The larger the pitch, the smaller the gear The larger the angle of action: the larger the difference between the base and pitch circles, the steeper the tooth profile, the smaller the transmitted force. 16 AGMA Standard Gear Specifications Parameter Pressure angle,

Coarse pitch (pd=N/ d<20) 200 or 250 (not common) Fine pitch (pd=N/ d>20) 200 Addendum, a 1/pd 1/pd

Dedendum, b 1.25/pd 1.25/pd Working depth 2.00/pd 2.00/pd Whole depth

2.25/pd 2.2/pd+0.002 Circular tooth thickness 1.571/pd (circular pitch/2) 1.571/pd Fillet radius 0.30/pd

Not standardized Clearance 0.25/pd 0.25/pd+0.002 Minimum width at top land 0.25/pd Not standardized

Circular pitch /pd /pd 17 Min: 0.25/pd 1/pd 1.25/pd

/pd 1.571/pd 0.25/pd d=N/pd 0.3/pd 18 Planetary (or Epicyclic) Gears (10.4) Gears whose centers can move Used to achieve large speed reductions in compact space Can achieve different reduction ratios by

holding different combinations of gears fixed Used in automatic transmissions of cars 19 Planetary gear 20 Components of a planetary gear Planet Carrier Input shaft Sun gear

Ring gear 21 A variant of a planetary gear Carrier 22 Planetary gears in automotive transmission Planetary gears

23 Velocity Analysis Of Planetary Gears (10.6, 10.7) Two degrees of freedom Given the velocities of two gears (e.g. sun and carrier) find velocities of other gears Approach Start from gear whose speed is given Use equation gear = car+ gear/car 24 Velocity analysis of planetary gear

25 T h is p r o g r a m fin d s t h e v e lo c it ie s o f t h e r e m a in in g lin k s g iv e n t h e v e lo c it ie s o f tw o lin k s in a p la n e t a r y g e a r . In p u t: N u m b e r o f te e th o f s u n , p la n e t a n d r in g g e a r s , N 1 , N 3 , N 4 , r e s p e c t iv e ly . N1 3 0 N3

N4 35 100 V e lo c itie s o f t w o lin k s : s p e c ify t h e k n o w n v a lu e s o f t h e in p u ts a n d g u e s s th e v a lu e s o f th e o u tp u ts : 3 1 0 0

4 1 2 0 1 100 2 100 100 12

Giv en 3 0 4 1

1 2 12 26 1 2 12 N1 3 2 12 N3 4 2 N1

12 N4 27 Find 1 2 3 4 12 3.333 1.538

0 1 1.795 28