IM No.: 04, EE 9 2SEM-2022-2023 Gears Gears perform many functions, in this section we look at gears that increase or reduce angular velocity (while simultaneously decreasing or increasing torque, such that energy is conserved). In many ways gears act in rotating systems as do levers in translating systems. A picture of gears is shown below, along with a schematic representation. We will specify gears by their radii. When you buy gears you normally specify the number of teeth and the number of teeth per inch (or centimeter). These two measures are equivalent: the number of teeth and teeth per inch give the circumference, which in turn determines the radius. If the number of teeth are doubled, the radius is doubled. If we consider two gears in equilibrium and in contact with each other, we can two very useful relationships. First we note the geometric relationship that results from the path that the arc lengths along their circumference must be equal as the gears turn. Since the arc lengths (shown with a heavy blue line) must be equal Since the arc lengths (shown with a heavy blue line) must be equal r1θ1 = r2θ2 = arc length If we had defined θ2 in the opposite direction, this expression would have a negative sign (r1θ1 = -r2θ2). We derive a second relationship from a torque balance. Before we can do so we must define a force between the gears termed a "contact force." This force must be equal and opposite across the interface between the two gears, but its direction is arbitrary. We start by drawing free body diagrams with a contact force where the gears meet. The contact force is tangent to both gears and so produces a torque that is equal to the radius times the force. Page 1 of 6 IM No.: 04, EE 9 2SEM-2022-2023 We can do a torque balance on each of the two gears Gear 1 Gear 2 We are not usually interested in fc, so we remove from the equations and we get It is easy to show that the same relationship exists even if the contact forces are defined in the opposite direction (up on gear 1, down on gear 2). Transfer Functions for Systems with Gears The linearized interaction between two gears is depicted in figure below. An input gear with radius π1 and π1 teeth is rotated through angle π1 due to a torque, π1 (π‘). An output gear with radius π2 and π2 teeth responds by rotating through angle π2 (π‘) and delivering a torque, π2 (π‘). Let us now find the relationship between the rotation of Gear 1, π1 (π‘), and Gear 2, π2 (π‘) A gear train A gear train, lever, or timing belt over a pulley is a mechanical device that transmits energy from one part of the system to another in such a way that force, torque, speed, and displacement may be altered. These devices can also be regarded as matching devices used to attain maximum power transfer. Two gears are shown coupled together. The inertia and friction of the gears are neglected in the ideal case considered. The relationships between the torques π1 and π2 , angular displacement π1 andπ2 , and the teeth numbers π1 and π2 of the gear train are derived from the following facts: We conclude that the ratio of the angular displacement of the gears is inversely proportional to the ratio of the number of teeth. The distance traveled along each gear circumference is the same. π1 π1 = π2 π2 and since the ratio of the number of teeth along the circumference is in the same proportion as the ratio of the radii. The number of teeth on the surface of the gears is proportional to the radii π1 , and π2 of the gears; that is, π1 π1 = π2 π2 ππ π1 π2 = π2 π1 Page 2 of 6 IM No.: 04, EE 9 2SEM-2022-2023 Assume the gears are lossless, that is they do not absorb or store energy, the energy into Gear 1 equals the energy out of Gear 2. Since the translational energy of force times displacement becomes the rotational energy of torque times angular displacement, the work done by one gear is equal to that of the other. π1 π1 = π2 π2 If the angular velocities of the two gears π1 and π2 are brought into the picture π1 π2 π1 π2 π1 = = = = π2 π1 π2 π1 π2 gears do have inertia and friction between the coupled gear teeth that often cannot be neglected. An equivalent representation of a gear train with viscous friction, Coulomb friction, and inertia considered as lumped parameters. where π denotes the applied torque, π1 and π2 , are the transmitted torque, πΉπ1 and πΉπ2 are the Coulomb friction coefficients, and π΅1 and π΅2 are the viscous friction coefficients. The torque equation for gear 2 is indicates that it is possible to reflect inertia, friction, compliance, torque, speed, and displacement from one side of a gear train to the other. The following quantities are obtained when reflecting from gear 2 to gear 1: see what happens to mechanical impedances that are driven by gears. Figure (a) shows gears driving a rotational inertia, spring, and viscous damper. For clarity, the gears are shown by an end on view. We want to represent Figure (a) as an equivalent system at π1 without the gears. In other words, can the mechanical impedances be reflected from the output to the input, thereby eliminating the gears? a. Rotational system driven by gears; b. equivalent system at the output after reflection of input torque; c. equivalent system at the input after reflection of impedances From Figure (b), π1 can be reflected to the output by multiplying by π2 = π1 . The result is shown in Figure (b), from which we write the equation of motion as: π2 (π½π 2 + π·π + πΎ)π2 (π ) = π1 (π ) πππ 1 π1 Now convert π2 (π ) into an equivalent π1 (π ), so that Eq. (1) will look as if it were written at the input. Using Figure 2 (a) to obtain π2 (π ) in terms of π1 (π ),, we get π1 π2 (π½π 2 + π·π + πΎ) π1 (π ) = π1 (π ) π2 π1 Simplify [π½π 2 ( π1 2 π1 2 π1 2 ) + π·π ( ) + πΎ ( ) ] π1 (π ) = π1 (π ) π2 π2 π2 The equivalent system at the input and without gears shown in Figure (c). Thus, the load can be thought of as having been reflected from the output to the input. Generalizing the results, we can make the following statement: Rotational mechanical impedances can be reflected through gear trains by multiplying the mechanical impedance by the ratio Page 3 of 6 IM No.: 04, EE 9 2SEM-2022-2023 2 ππ’ππππ ππ π‘πππ‘β ππ ππππ ππ πππ π‘ππππ‘πππ π βπππ‘ ( ) ππ’ππππ ππ π‘πππ‘β ππ ππππ ππ π ππ’πππ π βπππ‘ where the impedance to be reflected is attached to the source shaft and is being reflected to the destination shaft. The next example demonstrates the application of the concept of reflected impedances as we find the transfer function of a rotational mechanical system with gears. Example: Transfer Function—System with Lossless Gears π (π ) PROBLEM: Find the transfer function, 2 ⁄π (π ), for the system of Figure (a) 1 Solution: First reflect the impedances (π½1 and π·1) and torque (π1 ) on the input shaft to the output as shown in Figure (b), where the impedances 2 π π are reflected by ( 2⁄π ) and the torque is reflected by ( 2⁄π ). 1 1 The equation of motion can now be written (π½π π 2 + π·π π + πΎπ )π2 (π ) = π1 (π ) π2 π1 Where in you are going to find the value of equivalent inertia π½π π 2 , equivalent viscous damper π·π π , πππ equivalent spring πΎπ π½π = π½1 ( Solving for the transfer function π2 2 ) + π½2 ; π1 π2 2 π·π = π·1 ( ) + π·2 ; πΎπ = πΎ2 πππ π€ππ π1 π2 (π ) ⁄π (π ) 1 π2 ⁄π π2 (π ) 1 πΊ(π ) = = π1 (π ) π½π π 2 + π·π π + πΎπ πππ π€ππ Example: In order to eliminate gears with large radii, a gear train is used to implement large gear ratios by cascading smaller gear ratios. A schematic diagram of a gear train is shown in Figure below. Next to each rotation, the angular displacement relative to π1 has been calculated., Page 4 of 6 IM No.: 04, EE 9 2SEM-2022-2023 For gear trains, we conclude that the equivalent gear ratio is the product of the individual gear ratios. We now apply this result to solve for the transfer function of a system that does not have lossless gears. Example Transfer Function—Gears with Lossless π (π ) Find the transfer function, 1 ⁄π (π ), for the system 1 SOLUTION: This system, which uses a gear train, does not have lossless gears. All of the gears have inertia, and for some shafts there is viscous friction. To solve the problem, first reflect all of the impedances to the input shaft, π1 . The gear ratio is not the same for all impedances. 2 π π·2 is reflected only through one gear ratio as π·2 ( 1⁄π ) , 2 And π½4 plus π½5 is reflected through two gear ratios as 2 (π½4 +π½5 ) (π3⁄π ) (π1⁄π ) 4 2 The result of reflecting all impedances to π1 (π ) is shown, from which the equation of motion is (π½π π 2 + π·π π + πΎπ )π1 (π ) = π1 (π ) π½π = π½1+ (π½2 + π½3 ) ( π1 2 π1 π3 2 ) + (π½4 +π½5 ) ( ) π2 π2 π4 πππ π€ππ π1 2 π·π = π·1 + π·2 ( ) πππ π€ππ π2 The transfer function is . πΊ(π ) = π1 (π ) 1 = 2 π1 (π ) π½π π + π·π π πππ π€ππ Practice Problem Find the transfer function, πΊ(π ) = π2 (π ) ⁄π(π ) for the rotational mechanical system with gears shown below. Page 5 of 6 IM No.: 04, EE 9 2SEM-2022-2023 VIII. REFERENCES Golnaraghi, F, Kuo, B. (2010) Automatic Control Systems, John Wiley and Sons Inc., M Gopal (2004) Control Systems: Principles and Design, Nise, N. (2011) Control Systems Engineering, Addison-Wesley Publishing Company Jagan, N.C.(2008) Control Systems, BS Publication http://www-control.eng.cam.ac/extras/Virtual_Library/Control_VL.html Introduction to control systems objectives https://webstor.srmist.edu.in/web_assets/srm_mainsite/files/files/EC0303%20Control%20System.pdf Control Systems Lectures - Closed Loop Control https://www.youtube.com/watch?v=O-OqgFE9SD4 Mandal, A. (2006) Introduction to Control Engineering, Modeling, Analysis and Design, New age International Publishers Rotational System with Gears to Transfer Function https://www.youtube.com/watch?v=14aXX9rvGWk Page 6 of 6
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