1 00:00:00,750 --> 00:00:01,620 ‫Welcome back. 2 00:00:02,190 --> 00:00:09,060 ‫In this video, I will give you an example that should really make clear what we need those rotation 3 00:00:09,060 --> 00:00:10,010 ‫mattresses for. 4 00:00:10,650 --> 00:00:19,680 ‫Let's take our example in which we rotated pie over to radiance about each axis and let's use the R, 5 00:00:19,680 --> 00:00:21,330 ‫X, Y, Z convention. 6 00:00:21,870 --> 00:00:29,670 ‫So you have your inertial frame here and you also have your body frame here that is perfectly aligned 7 00:00:29,670 --> 00:00:30,840 ‫with the inertia frame. 8 00:00:31,380 --> 00:00:38,040 ‫Let's imagine that we have our drone here that has this body frame attached to it. 9 00:00:38,580 --> 00:00:48,420 ‫Remember, when dealing with orientation, then it does not matter how the origins of the frames are 10 00:00:48,420 --> 00:00:50,340 ‫positioned with respect to each other. 11 00:00:50,910 --> 00:00:58,430 ‫What matters is how the axis of the frames are oriented with respect to each other. 12 00:00:59,010 --> 00:01:01,280 ‫So we align their origins. 13 00:01:01,830 --> 00:01:08,220 ‫And so here you can see that the inertial and body frames are perfectly aligned with each other. 14 00:01:08,730 --> 00:01:21,120 ‫And let's say that the drones angular velocity vector that has three components P, Q and R Radians 15 00:01:21,120 --> 00:01:22,020 ‫per second. 16 00:01:22,650 --> 00:01:27,400 ‫These are the three components of the angular velocity vector. 17 00:01:28,020 --> 00:01:33,720 ‫Let's say that each component has a value of one radians per second. 18 00:01:34,140 --> 00:01:42,990 ‫So we can say that Omega with respect to the body frame, which is our angular velocity vector, which 19 00:01:42,990 --> 00:01:51,290 ‫has these components, has these values for each component and of course it's in radians per second. 20 00:01:51,840 --> 00:02:01,050 ‫Remember P q r r the angular velocities in the body frame, you measured them in the body frame. 21 00:02:01,650 --> 00:02:03,720 ‫And so I'm going to switch to green now. 22 00:02:04,350 --> 00:02:13,470 ‫And if P equals one radians per second, well then that's your pride and your one would be here then 23 00:02:13,470 --> 00:02:14,220 ‫that would be. 24 00:02:14,220 --> 00:02:21,830 ‫Q again your one would be here and that would be R again, your one would be here. 25 00:02:22,350 --> 00:02:29,940 ‫And so we can try to draw our resultant vector, which would be this one, the purple one. 26 00:02:30,510 --> 00:02:31,470 ‫That's the result. 27 00:02:31,470 --> 00:02:37,950 ‫And vector in three D has three components, one in the body frame X dimension one in the body from 28 00:02:37,950 --> 00:02:41,520 ‫Y dimension and one in the body frame Z dimension. 29 00:02:42,090 --> 00:02:46,850 ‫And this purple vector will that's your Omega. 30 00:02:46,860 --> 00:02:54,240 ‫With respect to the body frame, it's this one here in the coordinates of this resulting vector in the 31 00:02:54,240 --> 00:02:58,680 ‫body frame is one one and one. 32 00:02:59,350 --> 00:03:00,930 ‫Now a question. 33 00:03:01,560 --> 00:03:08,730 ‫What would the coordinates of this vector be in the inertial axis? 34 00:03:09,240 --> 00:03:15,600 ‫Well, in this case, this is your inertial frame and this is your body frame. 35 00:03:16,230 --> 00:03:20,540 ‫You can see that they're equal to each other in terms of orientation. 36 00:03:21,330 --> 00:03:30,570 ‫Therefore, if you are, which are measured in the body frame, are one one one, which are also the 37 00:03:30,570 --> 00:03:38,520 ‫coordinates of this result vector, then the coordinates of this resultant vector in the inertia frame 38 00:03:39,150 --> 00:03:41,400 ‫is also one one one. 39 00:03:41,940 --> 00:03:43,170 ‫So that's pretty easy. 40 00:03:43,740 --> 00:03:56,160 ‫But now let's rotate about the inertial x axis by plus PI over two radians. 41 00:03:56,820 --> 00:04:00,360 ‫So now our body frame is like this. 42 00:04:00,690 --> 00:04:04,920 ‫However, the P Q are angular velocity components. 43 00:04:05,460 --> 00:04:15,840 ‫They are still one radians per second since p q are variables are measured in the body frame, then 44 00:04:15,840 --> 00:04:24,270 ‫this would be your P in the body frame X direction, this would be your Q in the body frame Y direction 45 00:04:24,870 --> 00:04:28,260 ‫and this would be your R in the body from Z direction. 46 00:04:28,800 --> 00:04:38,250 ‫And so that means that this would be your resultant vector, this would be your omega be the coordinates 47 00:04:38,250 --> 00:04:49,020 ‫of Omega B are still one, one and one because this Omega B is your angular velocity vector in the body 48 00:04:49,020 --> 00:04:49,470 ‫frame. 49 00:04:50,070 --> 00:04:53,640 ‫And so you measure it with respect to the body frame. 50 00:04:54,180 --> 00:05:00,240 ‫By the way, the angular velocity vector is usually called small. 51 00:05:00,340 --> 00:05:10,630 ‫Omega, but because it looks like W. at some point in the course, I accidentally call it W. but I'm 52 00:05:10,630 --> 00:05:12,620 ‫still talking about the same thing. 53 00:05:13,150 --> 00:05:20,640 ‫So either I call it Omega or W. is still the same angular velocity vector in the body frame. 54 00:05:21,130 --> 00:05:25,350 ‫I'm just telling you that so that you would not get confused. 55 00:05:25,810 --> 00:05:35,660 ‫But OK, so now what would the coordinates of this omega vector be in the inertial frame? 56 00:05:36,220 --> 00:05:38,880 ‫Of course, it's easy to see what it is. 57 00:05:39,490 --> 00:05:49,690 ‫You can clearly see that in the inertia frame it's X equals one because your piece here, then Y equals 58 00:05:49,870 --> 00:05:53,500 ‫minus one because your R is here. 59 00:05:54,070 --> 00:06:00,550 ‫But now in the inertial frame, R is in the negative Y direction. 60 00:06:01,150 --> 00:06:06,490 ‫This is your positive Y here and your R is in the negative both direction. 61 00:06:06,940 --> 00:06:09,070 ‫And your Z of course is one. 62 00:06:09,700 --> 00:06:13,900 ‫But let's use rotation matrices to find that answer. 63 00:06:14,350 --> 00:06:26,350 ‫Since we rotated plus pi over to radians about the inertial x axis, we take our R X matrix which looks 64 00:06:26,350 --> 00:06:27,090 ‫like this. 65 00:06:27,700 --> 00:06:38,530 ‫And by the way, C means cosine and X means sine and then we plug in plus power to radians into these 66 00:06:38,530 --> 00:06:42,190 ‫angles here, into our FY angles. 67 00:06:42,670 --> 00:06:53,480 ‫If we do that then cosine pi over two is zero sine power, two is one, the minus sign power two is 68 00:06:53,500 --> 00:06:57,510 ‫minus one and then cosine power two is zero. 69 00:06:58,030 --> 00:07:09,370 ‫And so our accordionists in the inertia frame equals r x times our coordinates in the body frame then 70 00:07:09,370 --> 00:07:20,500 ‫our r x matrix for power to radians times the actual values of P are and of course P Q are. 71 00:07:20,800 --> 00:07:27,280 ‫They are always measured in the body frame and if you multiply this matrix by this vector, then you 72 00:07:27,280 --> 00:07:31,180 ‫will get one minus one and one. 73 00:07:31,840 --> 00:07:41,980 ‫You see through this rotation matrix we found the coordinates of this omega B vector in the inertia 74 00:07:41,980 --> 00:07:51,670 ‫frame and just like you expected, you see X equals one, just like here, Y equals minus one just like 75 00:07:51,670 --> 00:07:55,930 ‫here, and Z equals one just like here. 76 00:07:56,470 --> 00:07:57,700 ‫But let's go further. 77 00:07:58,270 --> 00:08:04,420 ‫Let's rotate pi over to Radians about the inertial y axis. 78 00:08:04,730 --> 00:08:11,970 ‫So this rotation would also be pi over to radians like this. 79 00:08:12,670 --> 00:08:21,160 ‫Now our body frame is like this and again P Q are equal one radians per second. 80 00:08:21,700 --> 00:08:31,720 ‫Now this one would be P, this one would be Q and this one would be R because P, Q are measured in 81 00:08:31,720 --> 00:08:32,590 ‫the body frame. 82 00:08:33,040 --> 00:08:38,800 ‫And so that means that the resulting vector would be something like this. 83 00:08:39,250 --> 00:08:48,610 ‫That would be your Omega B again, it's easy to see what kind of coordinates this vector has in the 84 00:08:48,610 --> 00:08:49,600 ‫inertia frame. 85 00:08:50,110 --> 00:08:53,620 ‫In the body frame it's one one one. 86 00:08:54,250 --> 00:09:00,370 ‫But in the inertia frame it's one minus one and minus one. 87 00:09:01,000 --> 00:09:08,890 ‫You can see that this vector has a positive component in the X direction, a negative component in the 88 00:09:08,890 --> 00:09:17,320 ‫Y direction, and then a negative component in the Z direction because Q points in the positive direction 89 00:09:17,770 --> 00:09:26,260 ‫then are points in the negative Y direction and then P points in the negative Z direction when we talk 90 00:09:26,260 --> 00:09:27,700 ‫about the inertia frame. 91 00:09:28,180 --> 00:09:32,350 ‫But again, let's use the rotation matrices. 92 00:09:32,800 --> 00:09:43,840 ‫In this case, the inertia frame coordinates equal, the matrix are Y times the Matrix R X and then 93 00:09:43,840 --> 00:09:45,820 ‫coordinates in the body frame. 94 00:09:46,390 --> 00:09:50,530 ‫So it's our Y times, our X times this vector here. 95 00:09:50,920 --> 00:09:55,090 ‫So R Y equals this one. 96 00:09:55,690 --> 00:09:59,920 ‫But since Seeta equals Pi over. 97 00:10:00,200 --> 00:10:09,080 ‫To radiance, then this matrix will be like this cosine pi over two will be zero, then zero. 98 00:10:09,100 --> 00:10:12,940 ‫Here, here, this one will be minus one. 99 00:10:13,570 --> 00:10:16,920 ‫Then you will have zero one zero. 100 00:10:17,530 --> 00:10:22,060 ‫And then here you will have one zero and then zero. 101 00:10:22,600 --> 00:10:25,620 ‫And now you have to compute this thing here. 102 00:10:26,080 --> 00:10:30,150 ‫So are Y times are X equals. 103 00:10:30,640 --> 00:10:33,730 ‫This is our Y, this is our X. 104 00:10:34,120 --> 00:10:38,170 ‫So again, this is our Y and our X. 105 00:10:38,450 --> 00:10:46,000 ‫And so if you multiply them together and I'm just going to write it here, OK, this will be their product, 106 00:10:46,480 --> 00:10:48,940 ‫that's your R Y times, our X. 107 00:10:49,240 --> 00:10:58,540 ‫And so now you write it here again, that's your r y times are X times this vector. 108 00:10:59,140 --> 00:11:08,700 ‫And if you perform the matrix vector multiplication then you will get here one minus one and minus one. 109 00:11:09,160 --> 00:11:17,170 ‫And as you can see, you got what you expected, your omega vector that has the coordinates of one one 110 00:11:17,170 --> 00:11:20,890 ‫one in the body frame in the inertia frame. 111 00:11:20,890 --> 00:11:26,000 ‫However, they are one minus one and minus one. 112 00:11:26,530 --> 00:11:38,440 ‫And finally, let's rotate about the inertial Z axis by plus pi over to radians here. 113 00:11:38,470 --> 00:11:47,410 ‫I would ask you to do it yourself after you have rotated pi over to Radians about the inertial Z axis 114 00:11:48,010 --> 00:11:52,710 ‫peak, you are will still be one one and one radians per second. 115 00:11:53,230 --> 00:12:03,900 ‫Remember p q r r angular velocities in the body frame X, Y and Z respectively. 116 00:12:04,390 --> 00:12:14,860 ‫So your task is to use rotation matrices to find the coordinates of this omega B resultant vector in 117 00:12:14,860 --> 00:12:23,620 ‫the initial frame after you had rotated pi over to radians about the initial Z axis, the zero rotation 118 00:12:23,620 --> 00:12:25,590 ‫matrix r z. 119 00:12:26,230 --> 00:12:27,460 ‫It looks like this. 120 00:12:28,020 --> 00:12:34,210 ‫So try to use the same procedure and we will look at the solutions in the next video. 121 00:12:34,780 --> 00:12:35,710 ‫Thank you very much.