1 00:00:00,540 --> 00:00:01,330 ‫Welcome back. 2 00:00:02,280 --> 00:00:09,390 ‫In this video, I would like to briefly discuss one matter from the previous course in this core series, 3 00:00:09,690 --> 00:00:11,440 ‫the lateral control of the vehicle. 4 00:00:12,240 --> 00:00:18,690 ‫Let's look again and how we formed the lateral control equations of motion. 5 00:00:19,500 --> 00:00:20,970 ‫They looked like this. 6 00:00:21,210 --> 00:00:22,170 ‫Here they were. 7 00:00:23,220 --> 00:00:31,890 ‫This was your net force in the lateral direction, and since we see lateraled, and that means that 8 00:00:31,890 --> 00:00:39,390 ‫we were in the body frame and this was our net moment in the body frame. 9 00:00:39,900 --> 00:00:46,770 ‫And so we equated the net force to this and the net moment to this. 10 00:00:47,610 --> 00:00:55,170 ‫These are dynamics, equations into the you add up all the forces in the lateral direction and you equate 11 00:00:55,170 --> 00:01:00,390 ‫them to the mass times acceleration in the lateral direction. 12 00:01:00,870 --> 00:01:10,080 ‫And you do the same with the moments and you equate the net moment to mass movement of inertia times. 13 00:01:10,080 --> 00:01:21,870 ‫The angular acceleration, the mass remains constant in time, so mass dot equals zero kilograms per 14 00:01:21,870 --> 00:01:22,480 ‫second. 15 00:01:23,220 --> 00:01:26,910 ‫In other words, the vehicle's mass does not change in time. 16 00:01:27,240 --> 00:01:30,750 ‫And the same thing is true for the mass movement of inertia. 17 00:01:31,230 --> 00:01:38,400 ‫It is constant in time as well, because just like in the Jones case, we wrote down these equations 18 00:01:38,970 --> 00:01:40,290 ‫in the body frame. 19 00:01:40,950 --> 00:01:48,050 ‫And that's why this mass movement of inertia will not change as we rotate our vehicle. 20 00:01:48,840 --> 00:01:57,180 ‫Now, in the oven case, we made a distinction between rotating about the inertial Z axis for which 21 00:01:57,180 --> 00:02:06,180 ‫we use the angular velocity side dot radians per second, and about the body frame Z axis for which 22 00:02:06,180 --> 00:02:09,990 ‫we used our radians per second. 23 00:02:10,740 --> 00:02:18,930 ‫We had to make this distinction because the directions of the big Z and the small Z, the inertial and 24 00:02:18,930 --> 00:02:25,830 ‫the body frames the axis, their directions were not guaranteed to be in the same direction. 25 00:02:26,400 --> 00:02:30,390 ‫That's because we worked in 3D in the vehicle case. 26 00:02:30,540 --> 00:02:32,520 ‫We only worked in 2D. 27 00:02:33,560 --> 00:02:43,430 ‫And the direction of all the stations in our 2D world was either out of the screen like this or into 28 00:02:43,430 --> 00:02:51,710 ‫the screen like this, according to the right hand rule, and that is true for rotating about the inertial 29 00:02:51,710 --> 00:02:54,430 ‫ze axis and the body frames the axis. 30 00:02:54,890 --> 00:03:04,760 ‫And therefore we forgot about rotating about the inertial frame Z axis and we used BPCI that radians 31 00:03:04,760 --> 00:03:11,760 ‫per second to describe the angular velocity about the body frame, the Z axis. 32 00:03:12,380 --> 00:03:19,220 ‫There was no need to describe angular velocity about the inertial frame, the Z axis. 33 00:03:19,700 --> 00:03:21,250 ‫We did not have to do it. 34 00:03:21,650 --> 00:03:27,110 ‫We only had to describe our rotation about the body frames, the axis. 35 00:03:27,920 --> 00:03:37,040 ‫But OK, if you remember then we wrote down our lateral acceleration like this, and then I explained 36 00:03:37,040 --> 00:03:40,460 ‫to you that we had two terms here. 37 00:03:41,450 --> 00:03:46,790 ‫The first term showed how the lateral velocity changed with time. 38 00:03:47,420 --> 00:03:54,650 ‫But then you also had this centripetal acceleration that you had to take into account. 39 00:03:55,610 --> 00:04:03,530 ‫I said that this y double dot is the term that describes the lateral acceleration, assuming that the 40 00:04:03,530 --> 00:04:10,040 ‫body frame was not rotating with respect to the inertia frame, the body frame that we attach to the 41 00:04:10,040 --> 00:04:10,730 ‫vehicle. 42 00:04:11,360 --> 00:04:18,920 ‫And then the second term was there, which was the centripetal acceleration which made the object and 43 00:04:18,920 --> 00:04:20,880 ‫therefore the body frame rotate. 44 00:04:21,920 --> 00:04:27,240 ‫Now, that was one way of showing it a more simplified way. 45 00:04:27,530 --> 00:04:36,050 ‫However, now that you know about rotating frames and you know that if you work in the body frame, 46 00:04:36,230 --> 00:04:37,760 ‫you have to consider that. 47 00:04:38,330 --> 00:04:47,870 ‫Now, I can give you a clearer and a more formal explanation as to why this acceleration term has these 48 00:04:47,870 --> 00:04:48,940 ‫two terms here. 49 00:04:49,220 --> 00:04:50,270 ‫So let's look at it. 50 00:04:51,140 --> 00:04:56,730 ‫In our vehicle case, we only considered the forces in the lateral direction. 51 00:04:56,990 --> 00:05:00,680 ‫So these are your two forces here in the lateral direction. 52 00:05:01,430 --> 00:05:07,130 ‫Let's for a moment consider both the longitudinal and lateral direction. 53 00:05:07,610 --> 00:05:14,930 ‫And remember, we still work in the body frame and not in the inertia frame and the body frame is attached 54 00:05:14,930 --> 00:05:15,730 ‫to the vehicle. 55 00:05:16,250 --> 00:05:19,190 ‫This is here in purple Y and X. 56 00:05:20,240 --> 00:05:23,630 ‫So our equations of motion are the following. 57 00:05:24,620 --> 00:05:35,270 ‫The net force in the body frame equals mass times, the dot vector in the body frame and also the net 58 00:05:35,270 --> 00:05:46,850 ‫moment in the body frame equals your inertia tensor in the body frame times the omega dot vector in 59 00:05:46,850 --> 00:05:51,740 ‫the body frame and that was your angular acceleration vector. 60 00:05:52,490 --> 00:06:00,380 ‫Since we are in 2D, there are no forces in the Z direction, nor there are velocities in the Z direction. 61 00:06:01,470 --> 00:06:09,390 ‫That is why the Z component of the force vector is zero Newtons and the Z component of the velocity 62 00:06:09,390 --> 00:06:12,300 ‫vector is zero meters per second. 63 00:06:13,140 --> 00:06:18,890 ‫In addition, the only rotation that occurs happens about the Z axis. 64 00:06:19,920 --> 00:06:28,410 ‫That is why our angular velocity vector has only one non-zero element, and that element is about the 65 00:06:28,410 --> 00:06:29,810 ‫body from the axis. 66 00:06:30,120 --> 00:06:32,930 ‫So that would be our side, that radians per second. 67 00:06:33,630 --> 00:06:44,160 ‫And the inertia tensor becomes mass moment of inertia about the Z axis, about the body from Z axis. 68 00:06:45,300 --> 00:06:50,260 ‫And we also assume that the vehicle is mass symmetric about the body from Z axis. 69 00:06:51,090 --> 00:06:54,500 ‫So our equations of motion in 2D are the following. 70 00:06:55,470 --> 00:07:00,420 ‫You have the two force components here in the X and Y direction. 71 00:07:00,660 --> 00:07:01,730 ‫I and J. 72 00:07:02,070 --> 00:07:06,390 ‫And remember, we're in the body frame equals mass times. 73 00:07:07,200 --> 00:07:17,670 ‫You have X double dot times I which was the time change of the longitudinal velocity plus then you have 74 00:07:17,670 --> 00:07:29,010 ‫the longitudinal velocity times I dot and then plus the time derivative of the lateral velocity times 75 00:07:29,340 --> 00:07:36,420 ‫J plus the lateral velocity itself times J dot. 76 00:07:37,650 --> 00:07:45,300 ‫You know that your eye dot and J dot can be calculated like this with the cross product where you take 77 00:07:45,300 --> 00:07:53,010 ‫the angular velocity vector and cross it with I and J unit vector respectively. 78 00:07:53,010 --> 00:08:04,740 ‫For IDOT and Jadot you can rewrite this entire thing like this where this part here would be your idot 79 00:08:05,250 --> 00:08:09,930 ‫and this part here would be your J Dot. 80 00:08:10,980 --> 00:08:17,250 ‫You can move this X dot here and you can move this Y dot here. 81 00:08:17,400 --> 00:08:24,750 ‫And so you can rewrite this equation like this, you can factor out your angle of velocity vector and 82 00:08:24,750 --> 00:08:33,900 ‫you can just put it here and then you will cross it with this vector here x dot I plus Y that J. 83 00:08:34,740 --> 00:08:41,850 ‫Instead of using unit vectors you can just rewrite this vector like this, which is the same thing equals 84 00:08:41,850 --> 00:08:43,320 ‫mass times. 85 00:08:44,280 --> 00:08:56,370 ‫Then this part you can rewrite like this plus and now this part here you can rewrite it like this here. 86 00:08:56,370 --> 00:09:03,930 ‫Our angular velocity vector still has three elements, but then the rotation about the body frame X 87 00:09:03,930 --> 00:09:06,330 ‫and Y axis didn't exist. 88 00:09:06,630 --> 00:09:17,610 ‫So we only had our side dot and then we crossed it with another vector, which was this one. 89 00:09:17,610 --> 00:09:19,530 ‫And we can rewrite it like this. 90 00:09:19,800 --> 00:09:27,030 ‫We can have X dot here, we can have Y dot here and then we can have zero here. 91 00:09:27,150 --> 00:09:39,810 ‫So it would be like X the I plus Y that J plus zero K like this, but since zero times K equals zero, 92 00:09:39,810 --> 00:09:41,160 ‫then we don't have to write it. 93 00:09:42,000 --> 00:09:46,880 ‫And so I close the parentheses and now we have to find this cross product. 94 00:09:47,820 --> 00:09:53,940 ‫So what I'm going to do, I'm going to take this class product and I'm going to calculate it here. 95 00:09:54,420 --> 00:09:57,000 ‫And the way you calculate it is like this. 96 00:09:57,010 --> 00:10:09,180 ‫So you had your I j and then K and then here I would have zero zero and then bpci dot and then here 97 00:10:09,180 --> 00:10:14,970 ‫I would have x dot, y dot and zero. 98 00:10:15,270 --> 00:10:23,970 ‫And let's see what we're going to get if we take our AI unit vector, then here you would have this 99 00:10:24,180 --> 00:10:27,030 ‫part from which you had to take the determinant. 100 00:10:27,180 --> 00:10:31,980 ‫So it would be zero times zero minus sign the times y dot. 101 00:10:32,100 --> 00:10:44,730 ‫So you would end up with this term here like this and then you would have your minus J and then now 102 00:10:44,730 --> 00:10:55,290 ‫you would consider these elements and these elements and it would be zero times zero minus sign the 103 00:10:55,440 --> 00:10:56,760 ‫times x dot. 104 00:10:57,910 --> 00:11:05,290 ‫Like this, and since you have two minuses here, it will become a plus and then you will have plus 105 00:11:06,070 --> 00:11:12,230 ‫K, but then you can already see your two by two determined will be zero. 106 00:11:12,250 --> 00:11:16,910 ‫So it would be zero times Y, not minus zero times X dot. 107 00:11:17,320 --> 00:11:18,740 ‫So this will be zero. 108 00:11:19,180 --> 00:11:21,310 ‫So you can just say that this term is zero. 109 00:11:22,500 --> 00:11:33,870 ‫That means this entire thing now equals mass times, this vector plus and now this thing here can be 110 00:11:33,870 --> 00:11:36,400 ‫rewritten in this form here. 111 00:11:37,470 --> 00:11:46,560 ‫Now, we said that in our vehicle case when we did our lateral control and we were in a straight road 112 00:11:46,560 --> 00:11:54,630 ‫and we only wanted to change our lanes, then we made an assumption that our longitudinal velocity was 113 00:11:54,630 --> 00:11:55,210 ‫constant. 114 00:11:55,470 --> 00:12:05,370 ‫Therefore, the time derivative of our longitudinal velocity was zero meters per second squared. 115 00:12:06,300 --> 00:12:13,160 ‫Now, if you really wanted to be picky, then you should consider this term here. 116 00:12:13,170 --> 00:12:14,060 ‫Of course. 117 00:12:14,940 --> 00:12:16,500 ‫However, we neglected it. 118 00:12:16,680 --> 00:12:17,210 ‫Why? 119 00:12:17,790 --> 00:12:25,740 ‫Well, if you look at our situation, then in our case we had a car on a straight road attempting to 120 00:12:25,740 --> 00:12:27,890 ‫only change lanes. 121 00:12:28,620 --> 00:12:31,440 ‫So that was our car here on a straight road. 122 00:12:32,440 --> 00:12:42,670 ‫We wanted to perform this maneuver in I think it was seven seconds, our longitudinal velocity was, 123 00:12:42,670 --> 00:12:46,050 ‫I think around 20 meters per second. 124 00:12:46,960 --> 00:12:55,510 ‫And now if you think about the lateral velocity, then you can imagine that in this case, the lateral 125 00:12:55,510 --> 00:13:01,620 ‫velocity would be extremely small compared to the longitudinal velocity. 126 00:13:02,140 --> 00:13:03,940 ‫So you could really say that. 127 00:13:04,150 --> 00:13:09,230 ‫Why that is a lot smaller than X dot. 128 00:13:10,060 --> 00:13:12,220 ‫So this term would be very small. 129 00:13:12,640 --> 00:13:22,030 ‫And also our maneuver was very gradual because we only wanted to go from one lane to another lane on 130 00:13:22,030 --> 00:13:27,970 ‫a straight road and all that had to happen, let's say, in seven seconds. 131 00:13:28,490 --> 00:13:33,720 ‫And that means that we didn't have to do any sudden turns. 132 00:13:34,330 --> 00:13:41,140 ‫So we could also say that our side that was small, our angular velocity was small. 133 00:13:41,990 --> 00:13:49,900 ‫And so in order to simplify our lives, we would like to get rid of those small, annoying terms that 134 00:13:49,900 --> 00:13:55,120 ‫contribute very little to precision, but make our equations more complicated. 135 00:13:55,790 --> 00:14:04,530 ‫And so we assume that this term here is also zero meters per second squared. 136 00:14:05,440 --> 00:14:16,000 ‫That means that our net force in the longitudinal direction was zero Newtons and we are only left with 137 00:14:16,000 --> 00:14:21,910 ‫this second row where you only had the lateral forces and the lateral acceleration. 138 00:14:22,840 --> 00:14:28,200 ‫So in other words, and I'm going to write it here just to squeeze the final piece of information here. 139 00:14:28,780 --> 00:14:41,440 ‫So if I force the net force in the lateral direction equals mass times and then you have your Y double 140 00:14:41,440 --> 00:14:50,050 ‫dot, which goes here and then plus and then you have your side up times X dot. 141 00:14:50,440 --> 00:14:59,560 ‫So it would be here and you see that this is your lateral acceleration. 142 00:15:00,580 --> 00:15:05,290 ‫So you have the same result that you had in the previous course. 143 00:15:06,040 --> 00:15:13,100 ‫Only now you can truly, mathematically justify it using the rotational frames since you're working 144 00:15:13,100 --> 00:15:19,780 ‫in the body frame and your body frame rotates with respect to the inertia frame, you have two terms 145 00:15:19,990 --> 00:15:21,850 ‫in the lateral acceleration. 146 00:15:22,660 --> 00:15:30,190 ‫So y double that is the lateral acceleration, assuming that the body frame is not rotating. 147 00:15:30,580 --> 00:15:38,860 ‫And then the second term X that time side, that corrects for the fact that the body frame is rotating 148 00:15:38,860 --> 00:15:45,100 ‫with respect to the inertia frame and that correction is the centripetal acceleration. 149 00:15:45,940 --> 00:15:52,840 ‫If you had worked in the inertia frame, then you would not have had this correction factor. 150 00:15:53,290 --> 00:15:59,950 ‫But then you must have considered forces and accelerations in the X and Y direction. 151 00:16:00,790 --> 00:16:05,980 ‫Your forces in the X direction would be like this. 152 00:16:06,400 --> 00:16:12,580 ‫That would be your net force in the direction and then in the Y direction. 153 00:16:13,120 --> 00:16:15,340 ‫It would be like this. 154 00:16:15,790 --> 00:16:26,430 ‫And then your accelerations would have been also in the X direction and also in the Y direction. 155 00:16:27,040 --> 00:16:34,810 ‫But since we worked in the body frame, we could only consider our forces in one direction. 156 00:16:35,110 --> 00:16:42,520 ‫But we had this second term because we had to take into account that the body frame was rotating with 157 00:16:42,520 --> 00:16:43,960 ‫respect to the inertia frame. 158 00:16:45,160 --> 00:16:49,010 ‫Finally, let's consider our moment equation. 159 00:16:49,660 --> 00:16:55,360 ‫So this was our net moment here and let's write it down like this. 160 00:16:55,720 --> 00:17:06,010 ‫So the net moment that we had in the body frame equals and now you had your math moment of inertia about 161 00:17:06,010 --> 00:17:11,940 ‫the body frame, the Z axis, and we followed the same derivation. 162 00:17:12,130 --> 00:17:20,410 ‫So you had your side double dot and now you have your K unit vector and it's only a K unit vector because 163 00:17:20,410 --> 00:17:24,250 ‫you're only rotating about the Z axis, right? 164 00:17:24,880 --> 00:17:31,900 ‫You're only rotating about the body from Z axis and therefore you only have this K here, but then. 165 00:17:32,180 --> 00:17:44,000 ‫We should also right here upside down at times like that, so you already know that K dot equals the 166 00:17:44,000 --> 00:17:48,020 ‫angular velocity vector cross K. 167 00:17:49,110 --> 00:18:01,020 ‫That means that this entire thing, you can write it like this double dot, K plus side at times and 168 00:18:01,020 --> 00:18:09,210 ‫then the angular velocity across K and then you can take this side dot and you can put it here and that 169 00:18:09,210 --> 00:18:10,630 ‫would be your K here. 170 00:18:11,730 --> 00:18:16,040 ‫Now if you want to find this, then it would look like this. 171 00:18:16,530 --> 00:18:26,190 ‫So your angular velocity would only have this zero, the zero and then you would have side up here. 172 00:18:27,080 --> 00:18:35,660 ‫And then you would cross it with zero zero and upside that which is the same vector. 173 00:18:36,720 --> 00:18:42,330 ‫Now, of course, you can methodically compute this cross product, however, if you look at it, it 174 00:18:42,330 --> 00:18:50,280 ‫is the same vector and even if it wasn't, but if it was parallel, if these two vectors, if they were 175 00:18:50,280 --> 00:18:53,810 ‫parallel, then what did we say about cross products? 176 00:18:54,720 --> 00:19:00,570 ‫You need to have a perpendicular component of one vector with respect to the other one. 177 00:19:01,380 --> 00:19:04,310 ‫Then you have a non-zero cross product. 178 00:19:05,010 --> 00:19:10,040 ‫If both vectors are parallel to each other, then they're cross product is zero. 179 00:19:10,770 --> 00:19:18,540 ‫And so you can see that not only these two vectors are the same vectors, but then they are also parallel 180 00:19:18,690 --> 00:19:19,390 ‫to each other. 181 00:19:20,100 --> 00:19:25,650 ‫Therefore, this entire thing is zero radians per second squared. 182 00:19:26,550 --> 00:19:29,480 ‫So this second term will disappear. 183 00:19:30,600 --> 00:19:40,350 ‫And so you're left with a net moment in the body frame equals the mass moment of inertia about the body 184 00:19:40,350 --> 00:19:46,950 ‫from Z axis times, BPCI double dot times, the unit vector K. 185 00:19:47,580 --> 00:19:48,500 ‫And there you go. 186 00:19:48,960 --> 00:19:55,110 ‫Now you see how all these equations for the vehicle's lateral control were developed. 187 00:19:55,470 --> 00:19:58,960 ‫In the next video, we will go back to our other e-mail. 188 00:19:59,460 --> 00:20:00,440 ‫Thank you very much. 189 00:20:00,450 --> 00:20:02,010 ‫And see you next on.