1 00:00:04,620 --> 00:00:09,950 Servomotors are widely used electro-mechanical components. For example, they are used in 2 00:00:09,950 --> 00:00:16,740 steering systems for scale model helicopters and other radio controlled cars, robots, and 3 00:00:16,740 --> 00:00:19,790 aircraft, including military "drones". 4 00:00:19,790 --> 00:00:25,919 A servomotor includes an electric motor, circuitry to control its speed and direction, and gearing 5 00:00:25,919 --> 00:00:31,759 to attain the high torques needed to apply moderately large forces over relatively short 6 00:00:31,759 --> 00:00:38,759 linear displacements. The gearing inside this servo is a compound gear train with four stages. 7 00:00:39,010 --> 00:00:44,499 In this video, we will understand how linear algebra can help us to predict the losses 8 00:00:44,499 --> 00:00:49,929 and understand the design trade-offs when converting high speeds and low torques into 9 00:00:49,929 --> 00:00:54,670 low speeds and high torques using a gear train. 10 00:00:54,670 --> 00:01:00,129 This video is part of the Linearity Video Series. Many complex systems are modeled or 11 00:01:00,129 --> 00:01:04,110 approximated linearly because of the mathematical advantages. 12 00:01:04,110 --> 00:01:10,590 Hi, my name is Dan Frey, and I am a professor of Mechanical Engineering and Engineering 13 00:01:10,590 --> 00:01:12,710 Systems at MIT. 14 00:01:12,710 --> 00:01:17,570 Before watching this video, you should be able to identify and describe the forces and 15 00:01:17,570 --> 00:01:20,760 torques that act on a system. 16 00:01:20,760 --> 00:01:25,200 After watching this video and some practice, you will be able to: 17 00:01:25,200 --> 00:01:31,110 Model the forces and torques in a gear train as a system of linear equations. 18 00:01:31,110 --> 00:01:36,950 Combine these linear equations into a matrix equation. 19 00:01:36,950 --> 00:01:43,950 First, let's get a quick overview of the design. Here's a servomotor with the case opened. 20 00:01:46,229 --> 00:01:52,960 Inside the case is a DC motor with a small pinion gear on it. The pinion has 10 teeth 21 00:01:52,960 --> 00:01:59,960 and is mated to a gear with 72 teeth. So the first pair provides a speed reduction of 7.2:1. 22 00:02:01,800 --> 00:02:08,769 This large, 72 tooth gear is part of a single molding with a 10 tooth gear. 23 00:02:08,769 --> 00:02:15,120 The 10 tooth gear on top of the 72 tooth gear is mated to a plastic gear with 48 teeth as 24 00:02:15,120 --> 00:02:22,120 you see here. So that stage provides a speed reduction of 4.8:1. The 48 tooth gear is molded 25 00:02:23,760 --> 00:02:25,849 to another 10 tooth gear. 26 00:02:25,849 --> 00:02:32,849 The 11-tooth gear is mated to a black 36-tooth gear. This 36-tooth gear is molded to a 16-tooth 27 00:02:33,280 --> 00:02:40,280 gear. The 16-tooth gear is mated to this black 42 tooth gear, which is splined onto the output 28 00:02:43,269 --> 00:02:45,090 shaft. 29 00:02:45,090 --> 00:02:52,090 The overall sequence of gear pairs is: 10 teeth mated to 72 teeth, 10 mated to 48, 11 30 00:02:53,079 --> 00:03:00,010 mated to 36 and finally 16 mated to 42 teeth on the splined output shaft. 31 00:03:00,010 --> 00:03:06,459 The overall gear ratio is the product of all the gear ratios of the gear pairs in the train. 32 00:03:06,459 --> 00:03:11,769 Therefore about 326 to 1 as shown by this formula. 33 00:03:11,769 --> 00:03:16,879 Based on the motor's output and some measurement and calculation, we would expect an output 34 00:03:16,879 --> 00:03:22,099 torque of 6.8 Newton meters in an idealized gear train. 35 00:03:22,099 --> 00:03:27,180 But there are frictional losses at every stage of the power transmission process, so we guess 36 00:03:27,180 --> 00:03:33,049 the output shaft will provide substantially less torque than that. So let's start by measuring 37 00:03:33,049 --> 00:03:38,150 the maximum force at the output shaft, and convert that to an output torque. 38 00:03:38,150 --> 00:03:42,659 We attach weights to a servo horn that has been connected with the output shaft on the 39 00:03:42,659 --> 00:03:48,168 servomotor. An electrical power source (in this case, a battery) is connected through 40 00:03:48,168 --> 00:03:50,340 a radio controller. 41 00:03:50,340 --> 00:03:57,340 Here you see the servomotor lifting 1 kg. Here it is lifting about 2kg. And here it 42 00:03:58,150 --> 00:04:05,150 lifts even 3kg! This lightweight (100 gram) servomotor produces large amounts of torque, 43 00:04:07,730 --> 00:04:12,069 which can apparently lift more than 30 times its own weight. 44 00:04:12,069 --> 00:04:18,180 In this set-up, the servomotor will lift a 4.42kg weight starting from a 90 degree angle. 45 00:04:18,180 --> 00:04:22,979 As the motor lifts the weight, the readout on the scale should change. 46 00:04:22,979 --> 00:04:29,949 Here you see the initial weight, with slack in the string, is 9.75 lbs. This is about 47 00:04:29,949 --> 00:04:36,949 4.42kg. With the motor pulling at maximum capacity, the scale reads 3.70 lbs, which 48 00:04:38,530 --> 00:04:42,190 is 1.68 kg. 49 00:04:42,190 --> 00:04:47,340 A calculation tells us that the maximum torque available at the output shaft according to 50 00:04:47,340 --> 00:04:54,340 our tests is 1.47 Newton meters. 51 00:04:55,979 --> 00:05:01,310 We can use linear algebra to work out a better estimate. The force transmission and frictional 52 00:05:01,310 --> 00:05:08,220 losses in each step of the train can be modeled by a set of linear equations, (summing forces 53 00:05:08,220 --> 00:05:13,520 and summing torques on the rigid body). For the three rigid bodies that are comprised 54 00:05:13,520 --> 00:05:20,169 of two gears molded together, we will need three equations (sums of separate x and y 55 00:05:20,169 --> 00:05:23,840 forces and sums of torques). 56 00:05:23,840 --> 00:05:28,860 The pinion gear and last gear in the train are simpler, because there is only one gear 57 00:05:28,860 --> 00:05:34,080 in the body, so we can combine each of those into a single equation. 58 00:05:34,080 --> 00:05:39,909 The equations modeling the entire train can be assembled by linking together the five 59 00:05:39,909 --> 00:05:46,909 sets of equations with four additional equations to link each mating pair, so the overall system 60 00:05:47,199 --> 00:05:54,199 will have 15 equations--- 3 sets of 3 each (9 total) plus 1 each for the input and output 61 00:05:54,389 --> 00:06:01,229 gears plus 4 equations that model the connections between the mating gears. The solution to 62 00:06:01,229 --> 00:06:06,919 that set of equations helps us to estimate performance of the machine and also gives 63 00:06:06,919 --> 00:06:10,720 us insight into its design. 64 00:06:10,720 --> 00:06:16,349 Let's make the model of just one gear body now, say the 10 tooth and 48 tooth molded 65 00:06:16,349 --> 00:06:23,039 gear seen here. One way to model the system is to posit that there are three unknown forces 66 00:06:23,039 --> 00:06:28,110 acting on this single body of two gears molded together. 67 00:06:28,110 --> 00:06:35,110 We can name these forces F-TL2, F-TS2 , F-ShaftX2, and F-ShaftY2. The force F-TL2 is a force 68 00:06:42,210 --> 00:06:48,949 tangent to the gear pitch circle on the larger of the two gears. The T is for Tangent, the 69 00:06:48,949 --> 00:06:55,949 L is for Large, and the 2 indicates that this is the second compound gear in the train. 70 00:06:57,080 --> 00:07:03,969 The force F-TS2 is a force tangent to the gear pitch circle on the smaller of the two 71 00:07:03,969 --> 00:07:09,289 gears. Again the T means tangent, and the S here means small. 72 00:07:09,289 --> 00:07:16,270 Recall that these tangential forces really come from forces normal to the gear teeth. 73 00:07:16,270 --> 00:07:21,000 These gears are designed so that all of the gears in this servomotor have a pressure angle 74 00:07:21,000 --> 00:07:28,000 of 20 degrees. So there is a X component to this force, the separation force between the 75 00:07:28,169 --> 00:07:35,169 mating gears, which has magnitude given by FTL2 and FTS2 times tan 0.35, where 0.35 is 76 00:07:40,849 --> 00:07:44,990 20 degrees expressed in radians. 77 00:07:44,990 --> 00:07:51,990 The forces F-ShaftX2 and F-ShaftY2 are the X and Y components of the normal force to 78 00:07:53,060 --> 00:07:58,060 the shaft, which is applied to the gear it supports. 79 00:07:58,060 --> 00:08:03,539 There are also frictional forces associated with the normal force supporting the shaft, 80 00:08:03,539 --> 00:08:10,539 but they are not separate unknowns, they link the X and Y components normal forces by the 81 00:08:18,840 --> 00:08:22,340 friction coefficient, mu2. 82 00:08:22,340 --> 00:08:27,650 This completes the description of the unknown forces in our model. Those four forces appear 83 00:08:27,650 --> 00:08:32,909 in three different linear equations as represented by this matrix: 84 00:08:32,909 --> 00:08:38,260 Take a moment to verify that these equations balance the forces and torques. 85 00:08:38,260 --> 00:08:45,260 The first row in the matrix represents a sum of the forces on the gear in the X direction. 86 00:08:52,279 --> 00:08:57,760 The second row represents a sum of the forces on the gear in the Y direction. The third 87 00:08:57,760 --> 00:09:03,610 row represents a sum of the torques or moments on the gear in the direction parallel to the 88 00:09:03,610 --> 00:09:05,339 gear shaft. 89 00:09:05,339 --> 00:09:10,680 Now let's work on the model of the next gear in the train -- the one with 16 teeth and 90 00:09:10,680 --> 00:09:17,680 36 teeth molded together. Again, we model the system by positing that there are four 91 00:09:17,800 --> 00:09:24,230 unknown forces acting on this single body having two gears molded together and we can 92 00:09:24,230 --> 00:09:31,230 name these forces FTL3, FTS3 , FShaftX3, and FShaftY3. And we obtain this matrix. 93 00:09:38,329 --> 00:09:44,120 How can we link these two matrices together to get a model of the second and third gears 94 00:09:44,120 --> 00:09:50,589 combined? We see that the smaller gear on body 2 is in contact with the larger gear 95 00:09:50,589 --> 00:09:57,350 on body 3. According to Newton's third law, the reaction force on body 2 should be equal 96 00:09:57,350 --> 00:10:00,630 and opposite to that on body 3. 97 00:10:00,630 --> 00:10:07,630 In fact, the situation is more complex. As the gears enter mating they slide into engagement. 98 00:10:10,279 --> 00:10:15,449 As the gears depart contact, they slide back out. There are losses at the interface that 99 00:10:15,449 --> 00:10:21,290 are complex to model, and we will represent them simply using the efficiency in the average 100 00:10:21,290 --> 00:10:23,430 transmitted force. 101 00:10:23,430 --> 00:10:29,040 Now we can join the model of the second gear and the third gear into a single system of 102 00:10:29,040 --> 00:10:33,800 linear equations represented by this matrix: 103 00:10:33,800 --> 00:10:38,529 The row in the middle is like "glue" holding the two models together. The force on the 104 00:10:38,529 --> 00:10:44,820 large gear on body 3 is equal to the force on the large gear on body 2 except a penalty 105 00:10:44,820 --> 00:10:51,790 is applied for inefficiencies. The principal mechanism for loss of power and torque is 106 00:10:51,790 --> 00:10:58,120 sliding friction -- in this case, shearing the lubricant on the gear faces. 107 00:10:58,120 --> 00:11:03,839 Before going on to assemble more of the model, it is worth inspecting our work so far for 108 00:11:03,839 --> 00:11:10,839 patterns. Note that the top right and bottom left of our matrices are filled with 3x4 matrices 109 00:11:11,839 --> 00:11:18,760 of zeros. That is a clue that our matrix is shaping up to be banded in structure. Although 110 00:11:18,760 --> 00:11:24,690 every variable affects every other variable, the influences propagate locally (in some 111 00:11:24,690 --> 00:11:31,149 sense). For example, the frictional losses on shaft 2 do influence the frictional losses 112 00:11:31,149 --> 00:11:38,149 on shaft 3, but only indirectly through their effect on the force between the two bodies. 113 00:11:39,820 --> 00:11:44,730 Because we arranged the variables in our vectors in a way that respects this structure, our 114 00:11:44,730 --> 00:11:50,300 matrix has a band in the middle from the top left corner to the bottom right. We will try 115 00:11:50,300 --> 00:11:54,980 to keep this up as we continue building the model. It will clarify interpretation of the 116 00:11:54,980 --> 00:12:01,980 model and also aid in efficiency and stability of the computations needed to solve it. 117 00:12:02,699 --> 00:12:07,779 Now it's your turn, take a moment to write a matrix that models the force and torque 118 00:12:07,779 --> 00:12:14,779 balance on the first rigid body gear in the gear train. 119 00:12:20,680 --> 00:12:26,740 This matrix is completely analogous to the matrices we obtained in for the 2nd and 3rd 120 00:12:26,740 --> 00:12:27,320 gear bodies. 121 00:12:27,320 --> 00:12:32,560 When we add the pinion gear and fourth gear into the model, we need to take additional 122 00:12:32,560 --> 00:12:38,610 care. These will have slightly different structure. The gear connected to the pinion on the motor 123 00:12:38,610 --> 00:12:45,389 has an applied torque due to the motor. The reaction at the output shaft is an externally 124 00:12:45,389 --> 00:12:52,389 applied torque, due entirely to the large fourth gear. It is possible to combine these 125 00:12:52,970 --> 00:12:55,560 into a single equation for each gear. 126 00:12:55,560 --> 00:13:00,350 Pause the video and try to write out the full system of linear equations for the overall 127 00:13:00,350 --> 00:13:07,350 4 gear train system. 128 00:13:09,420 --> 00:13:14,170 Placing these into the model as the first and last rows, the overall system of linear 129 00:13:14,170 --> 00:13:15,990 equations is: 130 00:13:15,990 --> 00:13:22,440 Note that the matrix has 15 rows and 15 columns. This should be expected as the system should 131 00:13:22,440 --> 00:13:28,940 be neither overdetermined nor underdetermined. Given a particular torque input at the motor, 132 00:13:28,940 --> 00:13:34,350 there should be a single value of output torque consistent with our gear train model and its 133 00:13:34,350 --> 00:13:37,709 deterministically defined parameters. 134 00:13:37,709 --> 00:13:44,709 By solving this system, we find unique values for all the unknowns. When we assembled the 135 00:13:44,880 --> 00:13:49,920 equations, we guessed the direction of each force in the free body diagram. So, it's useful 136 00:13:49,920 --> 00:13:55,800 to inspect the solution for negative values. For the parameter values we chose, the reaction 137 00:13:55,800 --> 00:14:02,800 force at the shaft of the third body in the x direction turns out to be negative. 138 00:14:03,069 --> 00:14:07,350 We should ask ourselves, did we guess wrong about the direction of the force, or is there 139 00:14:07,350 --> 00:14:14,350 something wrong with our matrix? Looking at the forces computed it seems that the friction 140 00:14:18,940 --> 00:14:25,569 in the x direction due to support of the large y reaction on this gear was more dominant 141 00:14:25,569 --> 00:14:27,920 that we expected. 142 00:14:27,920 --> 00:14:32,509 It's sensible to run some sanity checks in a case like this. If the friction on that 143 00:14:32,509 --> 00:14:38,410 gear were very low, would the sign be positive as expected? We ran that scenario, and the 144 00:14:38,410 --> 00:14:44,360 sign did become positive. So our guess about the direction of the reaction force was wrong. 145 00:14:44,360 --> 00:14:50,519 Our assumptions for the magnitude of the friction were too low. But it seems the model is behaving 146 00:14:50,519 --> 00:14:56,569 in reasonable ways. 147 00:14:56,569 --> 00:15:01,360 Now that we have a reasonable level of confidence in the model, we can begin to use it to explore 148 00:15:01,360 --> 00:15:06,420 the design decisions that the engineers made. If we put in reasonable values for the gear 149 00:15:06,420 --> 00:15:12,690 mating efficiency, such as 96%, and reasonable values for the friction coefficients at the 150 00:15:12,690 --> 00:15:19,569 bushings, such as 0.3, we find an overall efficiency of the gear train is about 52%. 151 00:15:19,569 --> 00:15:24,209 This is in reasonable agreement with our simple measurements. We found that the maximum torque 152 00:15:24,209 --> 00:15:30,230 available at the output shaft is 1.47 Newton meters, which would imply an efficiency closer 153 00:15:30,230 --> 00:15:32,110 to 25%. 154 00:15:32,110 --> 00:15:37,930 But our measurement used an overhanging load, which caused the shaft to bend. Our simple 155 00:15:37,930 --> 00:15:44,019 model makes a large and optimistic assumption of loading the servomotor with torques only. 156 00:15:44,019 --> 00:15:49,149 It's not surprising that the answers differ substantially. It suggests installing the 157 00:15:49,149 --> 00:15:55,529 servo so that bending loads are supported elsewhere, not in the servo itself. 158 00:15:55,529 --> 00:16:01,470 The gears in automotive transmissions with similar ratios are much more efficient, perhaps 159 00:16:01,470 --> 00:16:08,060 90% to 95% efficient. But for a compact and inexpensive gear train, this design performs 160 00:16:08,060 --> 00:16:09,810 well, especially since the plastic the gears are molded from adds a great deal of rolling 161 00:16:09,810 --> 00:16:09,920 friction. 162 00:16:09,920 --> 00:16:15,360 To summarize, we built an engineering model of a servomotor gear train using systems of 163 00:16:15,360 --> 00:16:21,110 linear equations. The matrix representations helped us to explore the interactions among 164 00:16:21,110 --> 00:16:25,690 variables in the system like separation forces between gears and friction at the shafts. 165 00:16:25,690 --> 00:16:30,600 Since the 15 by 15 system of equations is so fast and easy to solve on a modern computer, 166 00:16:30,600 --> 00:16:34,240 we could run a large number of "what if" scenarios. 167 00:16:34,240 --> 00:16:39,040 We hope this video helped you see how linear algebra can be used to make and understand 168 00:16:39,040 --> 00:16:44,430 engineering design decisions.