4. Mechanical Power Transmissions II.pptx

Mechanical Power Transmissions II  Gear Ratios Gears are not just used to transfer power, they also provide an opportunity to adjust the mechanical advantage of a mechanism. As discussed in the introduction to this unit, there are cases where a motor itself...

Mechanical Power Transmissions II  Gear Ratios Gears are not just used to transfer power, they also provide an opportunity to adjust the mechanical advantage of a mechanism. As discussed in the introduction to this unit, there are cases where a motor itself is powerful enough for an application but the motor’s output characteristics are not well suited to the application. A motor that is VERY fast but has only a little bit of torque would not be suitable to lift a heavy load; in these cases it is necessary to use gear ratios to change the outputs to a more appropriate balance of torque and speed. Think of a bicycle: the rider has limited power, and wants to ensure the power gets harnessed as much as possible at all times.  As the mechanical advantage changes, the speed of motion also changes. Power is the rate at which work is done. If the amount of work increases, the speed at which it gets done decreases. In this example, one can see that if the input side of the lever moves 1 meter then the output moves 4 meters. This difference is proportional to the ratio between the lengths of the levers. Thus, output length / input length = 8 / 2 = 4. The interesting thing about this is that it moves these distances at the same time. Let’s say that it takes one second to move the input one meter, and the input is moving at one meter per second. At the same time, the output moves four meters in one second, and is moving at a speed of eight meters per second. The output is moving FASTER than the input, by the ratio  In this example, the same system shown in the previous example, now has a 4 Newton force applied to the input. How much force then results at the output? The first step is to calculate the applied torque on the center of rotation caused by the input force. Using the formulas from Unit 7: Torque = Force x Distance from Center of Rotation = 4 N x 2 meter = 8 N-m The second step is to calculate the resultant force that this torque now has on the output: Force = Torque / Distance = 8 N-m / 8 meter = 1 Newton So looking at the above two examples, if the lever system above has an input Force of 4 Newtons and moves 1 meter, the output will have a force of 1 Newton and moves 4 meters – it moves faster, with less force! One can see how mechanical advantage (in the form of levers) can be used harness a fixed input force to accomplish a desired output. Gears work in the same manner. A spur gear is basically a series of levers; the larger diameter the gear, the longer the levers.  As seen in this example , torque applied to the first gear results in a linear force at the tip of the gear teeth. That same force is applied onto the tip of the tooth of the gear it is mated with, which in turn results in a torque rotating this gear. The diameters of the gears become the lengths of the levers, and the resulting change of torque is equivalent to the ratio of the diameters. Small gears driving large gears result in a torque increase. Large gears driving small gears result in a torque decrease.  In example, if the input 36-tooth gear is rotated 1 tooth (d = 1 tooth width) then the gear is rotating 1/36 th of a revolution (a1 = 360 / 36 = 10 degrees). As it advances it moves the 60-tooth gear 1 tooth also. However, on the 60 tooth gear this is only 1/60th of a revolution (a2 = 360 / 60 = 6 degrees). As the small gear turns a certain amount in a given time the larger gear turns a smaller amount. This means the larger gear is spinning slower than the small gear. This concept works both ways. Small gears driving big gears result in a speed decrease. Large gears driving small gears result in a speed increase. Combining the lessons of examples 8.1 through 8.4, one can see that the ratio between the sizes of two gears meshing is proportional to the resulting torque change and speed change between them. This is known as the Gear Reduction. As discussed previously, the number of teeth a gear has is proportional to its diameter, so instead of using diameters to calculate Gear Reduction, one can just use tooth counts. The Gear Ratio is denoted as (Driving Gear Teeth):(Driven Gear Teeth), so the above pair of gears could be described as 36:60 (or 36 to 60). The Gear Reduction is calculated as Driven Gear Teeth / Driving Gear Teeth So Gear Reduction = Driven Gear Teeth / Driving Gear Teeth = 60 / 36 = 1.67  As discussed above, the Gear Ratio is denoted as (Driving Gear Teeth):(Driven Gear Teeth), so the above pair of gears could be described as 12:60 (or 12 to 60). The Gear Reduction is calculated as Driven Gear Teeth / Driving Gear Teeth So Gear Reduction = Driven Gear Teeth / Driving Gear Teeth = 60 / 12 = 5 Looking at the above example… The stall-torque of the second shaft can be calculated using the following formula: Output Torque = Input Torque x Gear Reduction Output Torque = 1.5 N-m x 5 = 7.5 N-m The free-speed of the second shaft can be calculated using the following formula: Output Speed = Input Speed / Gear Reduction = 100 RPM / 5 = 20 RPM So the secondary shaft spins with a free speed of 20 RPM and the stall torque is 7.5 N- m. The speed decreased, but the torque increased.  This second example can be calculated the same way. Gear Reduction = Driven Gear Teeth / Driving Gear Teeth = 12 / 60 = 0.2 Output Torque = Input Torque x Gear Reduction = 1.5 N-m x 0.2 = 0.3 N-m Output Speed = Input Speed / Gear Reduction = 100 RPM / 0.2 = 500 RPM So the secondary shaft spins with a free speed of 500 RPM and the stall torque is 0.3 N-m. The speed increased, but the torque decreased.  Motion Reversal & Idler Gears One other aspect of gears designers must consider is how they reverse motion: the driven gear spins in the opposite direction as the driving gear. To make the output gear spin in the same direction as the input gear, some designers will use something called an idler gear.  The idler gear has no effect on the overall reduction. In a previous example it was shown that a 12:60 ratio results in a gear reduction of 5. One can similarly calculate the gear reduction on the idler gear set in two stages. Gear Reduction 1 = 36 / 12 = 3 Gear Reduction 2 = 60 / 36 = 1.667 Overall Gear Reduction = Gear Reduction 1 x Gear Reduction 2 = 3 x 1.667 = 5 The only gears that matter to reduction in a system like this are the first gear and the last gear.  Idler gears can be very useful, especially for spanning long distances; the below example shows how they can be used in a competition robot drivetrain. Note that if a designer uses the wrong number of idlers, the drivetrain won’t function quite as well: In the above example, the two wheels are spinning in opposite directions. Robots won’t drive anywhere when built this way.  Compound Gear Reduction In certain situations, a design may require more mechanical advantage than a single gear ratio can provide or is otherwise impractical. For example, if a VEX Robot Design requires a 12:500 gear ratio it is a problem because there is no 500-tooth gear available. In this situation, a designer can use multiple gear reductions in the same mechanism. This is called a compound gear reduction.  n a compound gear system, there are multiple gear pairs. Each pair has its own gear ratio, and a shared axle connects the pairs to each other. The resulting compound gear system still has a driving gear and a driven gear, and still has a gear reduction (now called a “compound gear reduction”). The compound gear ratio is calculated by multiplying the gear reductions of each of the individual gear pairs. For the above example the overall gear reduction is calculated as follows: Compound Gear Reduction = Reduction 1 x Reduction 2 = (60 / 12) x (60 / 12) = (5) x (5) = 25 That means the output shaft is 25 times slower than the input shaft with 25 times as much torque. Compound gear ratios add up quickly!  The example is a gearbox with twelve 12:60 reductions as part of one compound reduction. This produces an overall reduction of 244,140,625, almost a quarter of a billion to 1. This means someone would need to spin the input 244,140,625 times just to get the output to spin once! Fun Fact: spinning the input once per second, it would take approximately 7 years and 9 months before the output spun once.  Other Types of Reductions Gears aren’t the only mechanisms in competition robotics that provide gear reduction. The same principles apply to sprockets & chains and pulleys & belts. Similar to gears, sprocket ratios can be calculated by counting their teeth. Pulleys and belts don’t have teeth, but their ratio can be calculated by comparing their diameters, as shown above. Both of these mechanisms provide more options to designers working with mechanical power transmission. These two options work great in situations where torque needs to be transferred over long distances. Unlike gears, these systems do not reverse the  Applying Gear Ratios to DC Motor Systems Based on the lessons learned before, adjustments to mechanical advantage are important to the design of DC motor systems. DC motors sometimes have current limits they must stay under or other load limits. Designs sometimes require certain speeds, which motors must be geared up or down to achieve. The first step in these types of design problems is to calculate the load the motor must be under to meet the design criteria. This is done using the motor characteristics, the design criteria, and the formulas and lessons used in Unit 7. After this, it is a matter of taking the output requirements and the input limits and calculating the ratio required. For the previous arm configuration, assume the system has one VEX 2-wire Motor 393, and that this motor cannot draw more than 2.5 amps at any time. What gear ratio is needed to lift the 30 Newton object and not exceed this current draw? The first step is to calculate the torque applied by the arm: Torque = Force x Distance = 30 N x 0.5 m = 15 N-m The next step is to calculate the torque load which will cause the motor to exceed 2.5 amps: Torque Load = (Given Motor Current – Free Current) x Stall Torque / (Stall Current – Free Current) Torque Load = (2.5 amps – 0.37 amps) x 1.67 N-m / (4.8 amps – 0.37 amps) Torque Load = (2.13 amps) x 1.67 N-m / (4.43 amps) Torque Load = 0.803 N-m So if the torque load at the output of the gearbox is 15 N-m, and the load on motor can’t be more than 0.803 N-m, what gear reduction is required? Output Torque = Input Torque x Gear Reduction Re arranging: Gear Reduction = Output Torque / Input Torque Gear Reduction = 15 N-m / 0.803 N-m = 18.68  Based on this example the gear reduction required is 18.68. However, as discussed above it is not always possible to achieve specific gear ratios with the gears on hand. Assuming the designer only has 12-tooth, 36-tooth, 60-tooth, and 84-tooth gears, calculate a compound reduction that creates the overall gear ratio required for the above example. There are many different solutions possible; the most important thing is that the compound ratio chosen results in a reduction of more than 18.68 to achieve the design goals shown above. Considerations: Designers should try to achieve the required reduction in as few stages as possible, and should try to get as close to 18.68 as possible without going under. HINT: One can look at the big list of reduction options shown previously in this unit, and see which ones multiplied together as a compound reduction come close to 18.68.  How about 12:60, then 12:60? 60 / 12 = 5 60 / 12 = 5 5 x 5 = 25 25 is greater than 18.68, so it is acceptable – however, 25 is a lot more reduction than required, so the arm will move slower than is necessary to achieve the design goal. Is there another option which is closer to 18.68? How about 12:36, then 12:84, as a simple 2- stage compound gear reduction? 36 / 12 = 3 84 / 12 = 7 3 x 7 = 21 21 is greater than 18.68, so 12:36 then 12:84 is an acceptable choice, this option is much closer to the minimum required reduction than 12:60, 12:60!  Arm Design Designers should create a single motor, gearbox, and arm system that can lift the weight of a single game object - the manipulator from Unit studied before. Designers must choose an arm length appropriate for the game, which fits within the 18” robot size requirement. The gear ratio should be calculated so that the motor can be loaded such that it draws no more than one amp of current. After calculating the necessary ratio, users must design a compound gearbox that achieves this ratio, and then calculate the final speed of the arm. CONCLUSION Mechanical power transmission systems are very important in the design and construction of competition robots. A designer’s ability to vary the gear ratio and the mechanical advantage in a system gives them the versatility necessary to accomplish whatever work needs to be done, with whatever motors they have (at the expense of speed.) At the simplest level, designers just need to determine their inputs and outputs, calculate the difference between them, and set their gear ratio accordingly. This simple method can be applied to any number of motor/gearing systems.  Modeling an Articulating Scoop Testing of the scoop bucket prototype has determined that it is difficult to control the motion of the scoop. The direct drive mechanism from the motor to the lock bar is moving too quickly and the robot operator is having difficulty scooping up the game object. To resolve this design problem, a gear drive is designed and tested using Autodesk Inventor software.  In this project, there are seven videos including the Overview and Summary. To use the videos, click the link and review the content. Pause, rewind, fast forward, and stop features are available as the student reviews the content. The workflow in these videos includes: Modifying the scoop bucket frame. Modeling the driveshaft assembly. Modeling a gear assembly. Animating the assembly. To be able to complete this unit you should have a basic understanding of the Autodesk Inventor user interface, navigation, and know how to work with Assemblies.