Gears and Sprockets for Basic Robotics

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Gears and Sprockets for Basic Robotics Written by George Gillard Published: 24-May-2016

Introduction Gears and Sprockets are powerful tools in robotics. They can be used to make something spin or move faster, or transfer more torque to a particularly highly loaded mechanism. Gears work by interlocking teeth with a neighbouring gear, while sprockets work by using teeth to grip into a chain, which transfers a force to another sprocket somewhere further along a mechanism. Each have their own advantages and disadvantages, but the primary purposes are much the same: to make another shaft spin, often at a different speed or torque. Understanding how the sizes of gears or sprockets relate to the speed and torque of an output shaft are critical to enable a robot builder to construct superb mechanisms, and showing relevant calculations will improve the quality of a design notebook. This guide will explain all of the relevant terms and relationships, with examples and exercises to assist the understanding of the content. The guide begins with explaining the designs of gears and sprockets, and then some very simple physics, providing the background needed to calculate suitable gear ratios to use in the design of a robot. I hope this document is useful to many who read it, and provides some level of inspiration to the design process of your robots. Thank you to those of you who have helped me edit this document, by helping me create this, I hope that we together have helped many others. Sections: 1. Gear and Sprocket Design 2. Speed and Torque 3. Ratios and Reductions 4. Choosing Ratios 5. Exercises and Solutions Page 2 Published: 24-May-2016 George Gillard

1 Gear and Sprocket Design Gears are designed to interlock with each other, which causes them to spin in opposite directions as they transfer mechanical power from one place to another. They do this with teeth around their edges, however there are many different types of gears. Let s have a look at a few types that you can use in the VEX Robotics System. 1.1 Spur Gears These are the classic gears. A simple wheel-shape, with little pointy teeth around the circumference. The edges are parallel with the axis through the centre of the gear, although the edges along the sides of the teeth are rarely actually straight they have a slight curve to them. The purpose of this curve is all about maximising the efficiency of the gear, and reducing any variations in speed or torque as the teeth pass over each other. Spur gears mesh with each other very simply, by just being placed next to each other on the same plane. Figure 1-1.1.1: VEX 60-tooth Gear Figure 1.1.2: VEX 36 and 60-tooth Gears Meshing Together These gears are the simplest to use in a robot design, and by far the most efficient and common. Typically they are used on arms. The VEX Spur Gears can also be used with the VEX Rack Gears to create linear motion. Figure 1.1.3: VEX Rack Gear Figure 1.1.4: VEX Rack and Spur Gear Page 3 Published: 24-May-2016 George Gillard

1.2 Bevel Gears A bevel gear could be interpreted as a hybrid between a spur gear and a cone. Like a spur gear, they are circular shaped from the front view, with teeth around the circumference. Have a look at a side view however, and you notice that the edges of the teeth are not at all parallel with the axis through the centre of the gear, but instead on an angle, such that the entire shape looks somewhat like a cone with the tip cut off. Due to this unique shape, you can interlock two bevel gears and have them mesh, when the gears are not parallel but instead at up to quite steep angles to each other. The VEX bevel gears are designed to operate at 90 to each other. Figure 1.2.1: VEX 24-tooth Front Figure 1.2.2: VEX 24-tooth Side Figure 1.2.3: Two Bevel Gears These gears are rarely used on VEX Robotics Competition robots. They re mostly used on smaller and less critical mechanisms, i.e. object manipulators/intakes. The VEX bevel gears are also used inside the VEX differential (again, very rarely used in competition). Figure 1.2.4: VEX Differential Page 4 Published: 24-May-2016 George Gillard

1.3 Worm Gears Worm gear systems compose of two very different types of gear. First, there is the Worm (also known as a Leadscrew), which looks like a cylinder with a spiralled thread around the outside. Then there is the Worm Gear, which looks very similar to a Spur Gear, except that the teeth are slightly twisted from the central axis of the gear, such that the teeth align with the spiral of the worm. As the worm rotates, the teeth push against the teeth of the worm gear, causing the worm gear to rotate. Worm gear systems allow the creation of a very high gear reduction in a very compact form, and they also resist backdriving (that is, you would not be able to cause the worm to rotate by spinning the worm gear). Figure 1.3.1: VEX Worm Figure 1.3.2: VEX Worm Gear Figure 1.3.3: Worm Gear System Worm gear systems are not commonly used in the VEX Robotics Competition. Most of the time, they are used for winches, due to their convenience of resisting backdriving, however ratchets are also commonly used for this purpose with regular spur gears. Page 5 Published: 24-May-2016 George Gillard

1.4 Sprockets Sprockets are used for many of the same purposes as gears, except that they transfer power through a chain, instead of directly from one to another via teeth like gears do. Whilst gears spin in opposite directions due to the contact between the two, sprockets spin in the same direction as each other. Since the section of chain that meshes with a sprocket is cylindrical, the teeth on a sprocket are curved such that this cylindrical shape can sit between the teeth. Figure 1.4.1: VEX 18-tooth High Strength Sprocket Figure 1.4.2: A simple sprocket and chain assembly Sprockets and chain are most commonly used on drivetrains, to connect multiple wheels together such that they spin together. They are also often used on object manipulators/intakes. It is worth noting that whilst all of the sprockets on the inside of the loop of chain will spin in the same direction, any sprockets on the outside of the loop of chain will spin in the opposite direction. In the example below (Fig 1.4.3), the smaller two sprockets turn in the opposite direction to the larger two sprockets. Figure 1.4.3 A chain and sprocket assembly Page 6 Published: 24-May-2016 George Gillard

2 Speed and Torque 2.1 Speed The term speed is really quite simple it s just the rate of change of displacement over time. For example, you might think of the speed of your car in kilometres per hour, or the speed of your robot in feet per second. If there is a given direction, we refer to the speed as a velocity. For the purposes of gears and sprockets on a robot, we tend to be more interested in the rotational speed. This could be measured in degrees per second, or radians per second, but for VEX we tend to use rotations per minute (RPM). For reference, a VEX 393 Motor geared internally for torque (standard) spins at up to 100 RPM. 2.2 Torque Perhaps the easiest way to think about Torque is by considering it as the rotational equivalent of a force. If you were to push a heavy box across a surface or floor, you would be exerting a force on the box to cause it to accelerate. This force is measured in Newtons (N). Weights are also forces they are the result of the gravitational pull on a n object with a certain mass. To find the force, you ll most likely be wanting this equation: Which is also simply written as: Force (N) = mass (kg) acceleration (m/s 2 ) F = ma The standard acceleration due to gravity is typically accepted as 9.81 m/s 2. Sometimes, this is approximated as 10 m/s 2. Note that the actual acceleration due to gravity varies depending on your location, however these variations are negligible. Example 1: Find the force required to accelerate a 10 kg box at 2 m/s 2. Force = 10 2 = 20 N Example 2: Find the weight force of a 5 kg box. Force = 5 9.81 = 49.05 N Page 7 Published: 24-May-2016 George Gillard

So, now that we know about force, we want to be able to figure out it s rotational equivalent torque. Try holding a heavy book in your hand, with your arm outstretched such that it is level with your shoulder. Now try pulling the book closer in towards your shoulder. Compare what you feel in your shoulder it s easier to hold the book up when it s closer, right? If you re still not convinced, hold your arm straight, level with your shoulder. Using your other arm, pull down on your outstretched arm at different points such as your wrist and elbow, and you should feel a pretty big difference. It s a lot harder to resist when the distance between the force and your shoulder is larger. That is, as the distance increases, the torque increases, too. We also know that the larger the force, the harder it is to resist again, the torque becomes greater. Torque is given by the following equation: Or, using symbols: Torque (N m) = Force (N) distance (m) τ = Fd (Note that the symbol for Torque is the Greek letter Tau, τ). The distance used to find the torque isn t just the distance between the points though, it is the perpendicular distance between the joint (your shoulder for example), and the direction of the force (such as the weight force). From this diagram on the left, it should be pretty clear that for an arm supporting a weight force, the torque will be greatest when the arm is horizontal, since this is when the perpendicular distance is the longest. Figure 2.2.1: Showing the Perpendicular Distance (orange) Page 8 Published: 24-May-2016 George Gillard

Example 1: Find the torque about a horizontal arm when an object with a weight of 20 N is placed 0.5 m away from the joint. Torque = 20 0.5 = 10 N m Example 2: Find the maximum torque about an arm with a length of 1.2 m due to the weight force of a 5 kg box. Force = 5 9.81 = 49.05 N Torque = 49.05 1.2 = 58.86 N m Example 3: Find the torque about an arm which is 1.2 m long, due to the weight force of a 5 kg object placed at the end, when the arm is 30 above the horizontal. Whilst you should always draw a simple diagram alongside your calculations, for this example it is particularly useful to keep things clear. Compare this answer to the answer from Example 2. F Force = 5 9.81 = 49.05 N distance ('d') = 1.2 cos(30) = 1.04 m Torque = 49.05 1.04 = 50.97 N m 30 The torque when the arm is at 30 is less than when the arm is horizontal (maximum torque). d Note that we use cos(30 ) since it is the adjacent length that we wish to find, by rearranging cos(θ) = adjacent hypotenuse Page 9 Published: 24-May-2016 George Gillard

2.3 Relationship between Speed and Torque First of all, we need to cover a bit of terminology how we name the gears that work together (known as a gear train ). The gear or sprocket which is attached to the motor is referred to as the Driving Gear, and the gear which is attached to the output such as a wheel or an arm is known as the Driven Gear. It is also possible to have more gears between the driving and driven gears in a single row or stage, known as Idler Gears. Idler gears have no effect on the speed or torque at all, but could be used when you need some more space between the Driven and Driving Gears, or if you need to reverse the direction. In the following image (figure 2.3.1), the larger gear on the left is the Driving Gear and the smaller gear on the right is the Driven Gear. Figure 2.3.1: Showing the driving and driven gears Using two different sized gears or sprockets results in a change in both speed and torque. As the speed increases, the torque decreases, and vice versa. Speed and torque have an inverse relationship it is impossible to increase both in a gear train. When the driving gear is bigger than the driven gear, the speed increases, and the torque decreases. When the driving gear is smaller than the driven gear, the torque increases and the speed decreases. Now for a slightly more in-depth look at why the speed and torque change depending on the gear sizes. Page 10 Published: 24-May-2016 George Gillard

2.3.1 Effect on Speed When the driving gear is bigger than the driven gear, the speed increases. This follows the equation: ω Driven = N Driving N Driven ω Driving Where N is the number of teeth on a gear, and ω is the rotational speed. An explanation on this is below. For a given rotational speed (known as the angular speed or angular velocity if there is a given direction) on the Driving Gear, we can calculate the tangential speed how fast the teeth are moving in a linear direction at the edge. For this, we use the following equation: tangential speed (m s) = radius (m) angular speed (rad s) Or in symbols, where the angular speed symbol is the Greek letter Omega (ω): v = rω v ω Driving ω Driven Figure 2.3.1.1: 36-tooth Driving Gear and a 60-tooth Driven Gear To use this equation, we need to know the angular speed in radians per second, but typically we use the units of revolutions per second (RPM). To convert RPM to rad/s, we just need to divide it by 60 to get revolutions per second, and then multiply it by 2π to get radians per second. Therefore, we could rewrite our equation for tangential speed to be: tangential speed (m s) = radius (m) (angular speed (RPM) tangential speed Driving = radius Driving angular speed Driving 2π 60 (s) ) tangential speed Driven = radius Driven angular speed Driven Page 11 Published: 24-May-2016 George Gillard

Since the tangential speeds for both driving and driven gears must be the same due to the contact between the teeth which cannot slip past each other. Therefore: tangential speed Driving = tangential speed Driven radius Driving angular speed Driving = radius Driven angular speed Driven Rearranging this gives us the very simple relationship: Or, angular speed Driving angular speed Driven = radius Driven radius Driving angular speed Driven = radius Driving radius Driven angular speed Driving Since the radius is also proportional to the number of teeth (which we tend to know, but we don t tend to know the radius of the gears), we can write: angular speed Driven = No. teeth Driving No. teeth Driven angular speed Driving Or, ω Driven = N Driving N Driven ω Driving Example 1: If a 60-tooth gear is attached to a motor spinning at 100 RPM, and meshes with a 36-tooth gear attached to a wheel, determine the speed of the wheel. Example 2: ω wheel = 60 36 100 = 166.67 RPM If a 12-tooth gear is attached to a motor spinning at 100 RPM, and meshes with an 84-tooth gear attached to an arm, determine the speed of the arm. ω arm = 12 84 100 = 14.29 RPM Page 12 Published: 24-May-2016 George Gillard

2.3.2 Effect on Torque When the driving gear is smaller than the driven gear, the speed increases. This follows the equation: τ Driven = N Driven N Driving τ Driving Where N is the number of teeth on a gear, and τ is the torque. An explanation on this is below. For a given torque on the Driving Gear, we can calculate the force produced at a tooth. For this, we use the following equation that we met earlier: Or, using symbols: Torque (N m) = Force (N) distance (m) τ = Fd The force exerted by a tooth on the driving gear will be equal and opposite to the force applied to a tooth on the driven gear. F τ Driving τ Driven Figure 2.3.2.1: A 12-tooth gear meshing with an 84-tooth gear We can rearrange the above formula to set force as the subject, as follows. Note that the distance in the above formula becomes the radius of the gear. Force Driving = Torque Driving radius Driving Force Driven = Torque Driven radius Driven Page 13 Published: 24-May-2016 George Gillard

Since the forces for both driving and driven gears must be the same, we can express this as: Therefore: Or, rearranged: Force Driving = Force Driven Torque Driving radius Driving = Torque Driven radius Driven Torque Driven = radius Driven radius Driving Torque Driving Since the radius is also proportional to the number of teeth (which we tend to know, but we don t tend to know the radius of the gears), we can write: Or, Torque Driven = No. teeth Driven No. teeth Driving τ Driven = N Driven N Driving τ Driving Torque Driving Example 1: If a 60-tooth gear is attached to a motor with a torque of 1.67 N m, and meshes with a 36-tooth gear attached to a wheel, determine the torque of the wheel. Example 2: τ wheel = 36 60 1.67 = 1.002 N m If a 12-tooth gear is attached to a motor with a torque of 1.67 N m, and meshes with an 84-tooth gear attached to an arm, determine the torque of the arm. τ arm = 84 12 1.67 = 11.69 N m Page 14 Published: 24-May-2016 George Gillard

3 Ratios and Reductions 3.1 Ratios A ratio is simply a comparison between two values, such as x:y (pronounced x to y ). In terms of gears, we use the number of teeth as the quantities, in the order Driving:Driven. Oftentimes, this ratio is simplified down, and it is common in VEX for teams to simplify ratios down such that the lower value is 1. Quite often there is a lot of confusion as to the order of the numbers in the ratio. To overcome this, it is common to see a for speed or for torque suffix, which clarifies whether the purpose of the gear train is for an increase in speed, or an increase in torque. Example 1: Express the gear ratio for a gear train where there is a 36-tooth driving gear and a 60-tooth driven gear. 36:60 or 3:5 or 1:1.667 (for torque) Example 2: When a 24-tooth sprocket attached to a motor powers a wheel with a 12-tooth sprocket attached to it, what is the ratio? 24:12 or 2:1 (for speed) 3.2 Reductions A reduction is much like a ratio, except that it is expressed in a fraction form, and that the order is Driven/Driving (the opposite of the gear ratio). A larger value for a reduction ( > 1/1, such as 4/3) represents a reduction in speed, and an increase in torque, and a smaller value for a reduction (< 1/1, such as 2/3) represents a reduction in torque, and an increase in speed. Like a ratio, for a reduction such as y/x, it would be pronounced y to x. Example 1: Express the gear reduction for a gear train where there is a 36- tooth driving gear and a 60-tooth driven gear. 60/36 or 5/3 or 1.667/1 (for torque) Page 15 Published: 24-May-2016 George Gillard

Example 2: When a 24-tooth sprocket attached to a motor powers a wheel with a 12-tooth sprocket attached to it, what is the reduction? 12/24 or 1/2 (for speed) In the VEX Robotics Competition, generally teams refer to a gear ratio more than a gear reduction. 3.3 Compound Gearing Compound gearing is where the driven gear shares a shaft/axle with another gear, which in turn acts as a driving gear for an additional stage of gearing. 60-tooth Driven Gear 12-tooth Driving Gear 36-tooth Driving Gear 84-tooth Driven Gear Figure 3.3.1: Compound Gearing where the first stage is 12:60, second stage is 36:84 The purpose of compound gearing is to achieve even greater gear ratios or reductions beyond what a simple one-stage gear train could achieve. Due to the extra stage of gears, they do however have a slightly lower efficiency. Page 16 Published: 24-May-2016 George Gillard

To find the gear ratio/reduction of a compound gear train, simply multiply together the ratios/reductions for the two separate stages. For Ratios: For Reductions: (N Driving,1 : N Driven,1 ) (N Driving,2 : N Driven,2 ) (N Driving,1 N Driving,2 ) (N Driven,1 N Driven,2 ) (N Driven,1 / N Driving,1 ) (N Driven,2 / N Driving,2 ) (N Driven,1 N Driven,2 ) / (N Driving,1 N Driving,2 ) The same rules apply with sprockets, as the ratios and reductions work the same way the only difference is how they physically transfer the power. Example 1: Find the gear ratio and reduction for the gear train shown in Figure 3.3.1 on the previous page (gear sizes from back left to bottom front are: 12, 60, 36 and 84-tooth respectively). Gear ratio: First stage is 12:60 = 1:5, second stage is 36:84 = 3:7 Combined is (1 3) : (5 7) = 3:35 or 1:11.67 (for torque) Gear Reduction: First stage is 60/12 = 5/1, second stage is 84/36 = 7/3 Combined is (5 7) / (1 3) = 35/3 or 11.67/1 (for torque) Page 17 Published: 24-May-2016 George Gillard

4 Choosing Ratios This is definitely the most challenging part of working with gears and sprockets from a mathematical view. This process can be made quite simple; however, it is also entirely possible for one to go into a deep analysis to find suitable gears/sprockets with more advanced methods. Generally, simple methods are perfectly reasonable and work well breaking out a spreadsheet to calculate all you could ever dream of is often entirely unnecessary unless you really wish to! Over time with more experience, it is quite normal to be able to predict a suitable ratio before doing any calculations at all. Trial and error is probably the most common method of choosing a ratio, but to potentially save time and have a more optimised finish to the overall design, it is well worth doing some calculations. For this section, we ll work with the classic example of choosing a suitable gear ratio for an arm, lifting a certain weight. 4.1 What do we need to know? Obviously, to calculate what we need in terms of gearing, we need to know the background of the situation. How much weight do we need to lift? How long is the arm? How many motors, and how much torque can each motor provide? Let s pretend we re playing a game in the VEX Robotics Competition where each object has a mass of 1.5 lbs as per the Game Manual, and we ve designed a robot where we want to lift five of these objects, on an arm which is 16 inches long. On the arm, we ll use four VEX 2-wire 393 Motors. 4.2 The Process The very first thing we want to do is convert all of this to metric units as we proceed with our calculations. In order of the process: 1. Find the weight force we need to lift 2. Calculate the torque produced by the weight force about the arm 3. Figure out how much torque we have available 4. Calculate a ratio/reduction Page 18 Published: 24-May-2016 George Gillard

4.2.1 Weight Force Note: There are approximately 2.2 lbs per kilogram. Acceleration due to gravity is approximately 9.81m/s 2. The mass of any object manipulator/intake can be neglected if a suitable amount of elastic support is provided. Total Mass = 1.5 lbs 5 = 3.409 kg 2.2 lbs kg Force = 3.409 9.81 = 33.443 N 4.2.2 Torque due to Weight Note: There are approximately 2.54 cm per inch, which is equal to 0.0254 metres per inch. Arm Length = 16 0.0254 = 0.4064 m Torque MAXIMUM = 33.443 0.4064 = 13.591 N m 4.2.3 Torque from the Motors This is where things start to get complicated. First, let s look at the data provided about a VEX 393 Motor: Output Stage Driving Gear Output Stage Driven Gear Output Speed (RPM) Output Stall Torque (N*m) IME Ticks per Revolution Standard Motor 393 10t 32t 100 1.67 627.2 Gearing High Speed Option 14t 28t 160 1.04 392 (included with Motor 393) Turbo Gear Set (sold separately) 18t 24t 240 0.7 261.333 Figure 4.2.3.1: VEX 393 Motor Data, Source: http://www.vexrobotics.com/motors.html The first thing to note is that there are three different internal gearing options, which result in three different options for the Output Speed and the Output Stall Torque. Perhaps the easiest way to deal with this is to only consider the Standard Gearing, and then we can factor in the differences later on with our final gear ratio. However, the complications hardly stop there! We re given the Output Stall Torque, but what actually is this? To help understand this, it s helpful to refer to a graph which shows the relationship of torque and speed. Page 19 Published: 24-May-2016 George Gillard

Speed (RPM) Power (W) Speed (RPM) Speed vs Torque for a VEX 393 Motor 120 100 80 60 40 20 0 0 0.5 1 1.5 2 Torque (N.m) Figure 4.2.3.2: Speed vs Torque Plot As seen, as the torque provided by the motor increases, the speed at which it spins decreases. The stall torque is the amount of torque the motor can provide at the point where it stalls. This is a bit of a dilemma for us, as we don t want it to spin slow, but we also want to make the best use out of the torque available from the motor. To help look at the ideal point for us, knowing what the power output is will give us some insight. Note: Power (W) = Torque (N m) Angular Velocity (rad s) Speed vs Torque for a VEX 393 Motor 120 100 80 60 40 20 0 0 0.5 1 1.5 2 Torque (N.m) 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 Speed Power Figure 4.2.3.3: Speed vs Torque plot incl. Power We can see from this plot that the power is maximum when the torque is at half of stall torque, and the motor runs at 50 RPM. That s great, right? We ve found the ideal power we wanted to maximise that. Page 20 Published: 24-May-2016 George Gillard

Speed (RPM) Power (W), Current (A) Well actually, no. There s another factor to consider here, and that is the current drawn by the motor. The current in a VEX 393 Motor increases linearly from 0.37 A to 4.8 A [1]. The PTCs inside the VEX 393 Motor are said to trip at about 1.8 A [2], and at maximum power, the current we d be drawing would be at about 2.6 A (see Figure 4.2.3.4). We don t want to pass over this trip current level for very long the PTCs (a protection device inside the motor) will trip at some point, and our motors will stop working until the PTC resets after 5 or so seconds, and even then, the motors will more susceptible to tripping again until they have completely cooled down. Speed vs Torque for a VEX 393 Motor 120 6 100 80 5 4 60 40 20 3 2 1 Speed Power Current 0 0 0.5 1 1.5 2 Torque (N.m) 0 Figure 4.2.3.4: Speed vs Torque plot incl. Power and Current So, we want to optimise our power, and we want the current to be less than 1.8 Amps. From the plot, we can determine that this ideal torque comes out as just over 0.5 N m, when the speed is approximately 70 RPM. So there we have it, a figure for the torque per motor 0.5 N m. You may choose to push the current a little more towards the limits, so long as you don t need to drive the motors at this current for too long, or you may choose to play it safer. From experience, assuming a good standard of build quality and minimal friction, you can push a 393 motor on standard internal gearing up to 0.7-0.8 N m, although certainly not for very long. Ideally, you need to incorporate the efficiency of your mechanical system into your calculations some physical tests may be useful to determine this. 1 VEX Robotics. (n.d.). Motors. Retrieved from http://www.vexrobotics.com/motors.html 2 jpearman. (2016, Jan 4). Motor torque-speed curves REV2 [Online forum comment]. Retrieved from http://www.vexforum.com/index.php/8089-motor-torque-speed-curves-rev2 Also useful: AURA. (2015, Jan 13). Motor Torque, Speed and Current [Video file]. Retrieved from https://youtu.be/stdonyfi2c4 Page 21 Published: 24-May-2016 George Gillard

4.2.4 Calculating a Ratio/Reduction At this point, we know the maximum amount of torque we will need (13.591 N m as per section 4.2.2), and we also know how much torque we can expect from each motor safely (0.5 N m as per section 4.2.3) which we have four of as per the design discussed. We need to use a gear reduction to magnify the torque from the motors up such that it meets the demand of the torque due to the weight force. Total Available Torque = 4 0.5 = 2.0 N m Torque Required = 13.591 N m Dividing the torque required by the torque available will result in a number which corresponds to the gear reduction that we need. Reduction = Torque Required = 13.591 = 6.8 1 Torque Available 2.0 Therefore, our reduction is 6.8/1 (minimum), meaning that we need of a gear ratio of at least 1:6.8. When selecting gears, you want to aim to maximise efficiencies having extra stages of compound gearing adds some more losses due to friction, so ideally you want to use a simple single stage reduction. In the VEX Robotics system, there are 12 and 84-tooth gears. We know this results in a 12:84 gear ratio, which can be simplified to 1:7. This is very close, and as close as we can get with a single stage reduction. 4.2.5 Extra Notes When considering the effects of using other internal gearing options for the 393 Motors, with relation to the standard gearing, the other options are the equivalent of 1.6:1 ( speed ), and 2.4:1 ( turbo ). Calculating a ratio using the standard internal gearing of 1:1, and then applying these two other ratios, will lead to more options. In the VEX Robotics system, the largest single stage gear ratio you can achieve is 1:7 as of the present time. For any ratios larger than this, you will need compound gearing. If the ratio calculated is for a load which you ll only rarely lift, it may be worthwhile picking a ratio that pushes the limits on the current a bit more such that the final ratio is faster, with less torque. Whilst these maximum load cases may struggle, the benefits of a faster lift for the majority of cases may be more beneficial in a competition environment when speed is critical. Going through all of the processes outlined in this section are crucial to a good design process, and showing these calculations would improve the quality of a design notebook significantly. Page 22 Published: 24-May-2016 George Gillard

5 Exercises and Solutions In case you want to check your understanding on any topics, have a go at some of the following exercises. Solutions are provided in Section 5.5. 5.1 Gear and Sprocket Design 1. Which direction does gear on the far right turn if the gear on the far left turns clockwise? 2. If the far left sprocket spins clockwise, which direction do the sprockets labelled A and B spin? A B 3. Would you expect a spur gear meshing with a worm gear to work well? 4. True or false: Rack gears, together with spur gears, can be used to convert between linear and rotational motion. 5. In the differential, would we expect the two opposite bevel gears to spin in the same or opposite direction, in relation to each other? 5.2 Speed and Torque 1. Find the speed of a driven 36-tooth gear, when a 60-tooth driving gear spins at 100 RPM. 2. Find the speed of an 84-tooth driven gear, when a 36-tooth driving gear spins at 100 RPM and there is a 12-tooth idler gear in the middle. 3. Find the weight force due to a 2.5 kg object. 4. Find the torque due to the weight force calculated in (3), when the object is placed on a horizontal arm 0.6 m from the joint. 5. Find the torque about an 84-tooth driven gear, when a 12-tooth driving gear provides 2.1 N m of torque. Page 23 Published: 24-May-2016 George Gillard

5.3 Ratios and Reductions 1. If a 36-tooth driving gear meshes with an 84-tooth driven gear, what are the ratio and reductions? 2. If a 12-tooth driving gear meshes with a 36-tooth gear, which shares a shaft with another 12-tooth gear which finally powers a 60-tooth driven gear, what are the ratio and reductions? 3. For a ratio of 2:1, does this suggest the purpose of the gear train is for an increase in speed, or torque? 4. Using a selection of gears (12, 36, 60 and 84-tooth) and sprockets (6, 12, 18, 24 and 30-tooth), how could you create a ratio of 1:14? 5. Using a selection of gear (12, 36, 60 and 84-tooth) and various motor configurations (100, 160, and 240 RPM), determine a suitable arrangement which results in an output speed of 266.67 RPM. 5.4 Choosing Ratios 1. For an arm which has a torque applied to it of up to 10 N m, and an available torque of 2.0 N m, what would be the necessary gear reduction? 2. For a 0.3 m arm with four motors each with 0.5 N m of torque and a gear ratio of 3:35 (for torque), how much mass could this system lift? 3. If you had two motors with a torque of 0.5 N m each, and you need to lift a 2 kg mass at the end of a 0.4 m arm, determine a suitable gear ratio. 4. If an arm has an applied torque of 16.5 N m due to a weight force, and you can use a single stage gear train consisting of 12, 36, 60 or 84-tooth gears, with access to motors with an available torque of between 0.5 and 0.6 N m each, what is the minimum number of motors you would need? 5. For a 16 inch horizontal arm supporting a 3 lb mass at the end, how many motors, and what gear ratio would you need, to lift the arm with a speed of 12 inches per second vertically at this instant? Use motors which spin at 70 RPM with 0.5 N m of torque each. Page 24 Published: 24-May-2016 George Gillard

5.5 Solutions 5.5.1 Gear and Sprocket Design 1. Anticlockwise. Adjacent gears spin in opposite directions to each other. 2. A spins anticlockwise, B spins clockwise. 3. No. The edges of a spur gear are parallel to the central axis of the gear, but for the worm gear they are slightly twisted. Therefore, the teeth wouldn t align very well. 4. True. Rack gears are designed for linear motion; spur gears are designed for rotational motion. The two are designed to work together. 5. Opposite directions. 5.5.2 Speed and Torque 1. 60 36 100 = 166.67 RPM 2. Idler gear has no effect. 36 84 3. Force = 2.5 9.81 = 24.525 N 4. Torque = 24.525 0.6 = 14.715 N m 5. Torque = 84 2.1 = 14.7 N m 12 100 = 42.86 RPM 5.5.3 Ratios and Reductions 1. Ratio is 36:84 or 3:7 or 1:2.33 (for torque). Reduction is 84/36 = 7/3 or 2.33/1. 2. First stage is 12:36 or 1:3. Second stage is 12:60 or 1:5. Final ratio is 1:15, or a reduction of 15/1. 3. Increase in speed. 4. Compound setup, with 1:7 (gears, 12:84), and 1:2 (sprockets, 6:12 or 12:24). 5. 60:36 or 5:3, with motors configured for 160 RPM. 5.5.4 Choosing Ratios 10 1. = 5/1 2 2. Torque about arm = 4 0.5 35 3 = 23.33 N m Force = 23.33 = 38.89 kg 0.6 Mass = 38.89 = 3.96 kg 9.81 3. Reduction = 2 9.81 0.4 = 7.85 1 Ratio = 1: 7.85 2 0.5 4. Min motors will be with a maximum gear reduction (84/12 = 7/1) and max torque per motor (0.6 N m). 16.5 No. Motors = 7 0.6 4 motors 5. Speed = 12 0.0254 = 0.75 rad s, 0.75 60 = 7.16 RPM 16 0.0254 2π 70 Reduction = 7.16 = 9.77 1 Torque needed = Force distance = (3 2.2 9.81) (16 0.0254) = 26.31 N m 23.31 No. Motors = = 5.38 6 motors (rounding up) 9.77 0.5 Page 25 Published: 24-May-2016 George Gillard